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Topology
Stanislav Jabuka
Notes for Math 440/640
University of Nevada, Reno
Contents
Chapter 1. Continuity and convergence in Euclidean spaces
1.1. Continuity in Euclidean space
1.2. Open and closed subsets of Euclidean space
1.3. Continuity and convergence in terms of open and closed sets
1.4. Some properties of open and closed sets in Euclidean space
1.5. Exercises
5
5
8
9
12
13
Chapter 2. Topological spaces
2.1. Definition of a topological space
2.2. Examples of topological spaces
2.3. Properties of open and closed sets
2.4. Bases and subbases of a topology
2.5. Exercises
15
15
16
25
28
33
Chapter 3. Continuous functions and convergent sequences
3.1. Continuous functions
3.2. Convergent sequences
3.3. Uniform convergence of functions
3.4. Space filling curves
3.5. Exercises
37
37
45
50
51
56
Chapter 4. Separation axioms
4.1. Degrees of separation
4.2. Examples
4.3. Topological invariants
4.4. A first application of topological invariants
4.5. The Urysohn Lemma
4.6. Exercises
61
61
63
65
67
70
76
Chapter 5. Product spaces
5.1. Finite products
5.2. Infinite products
5.3. Exercises
79
79
85
88
Chapter 6. Compactness
6.1. Definition and first examples
91
91
3
4
CONTENTS
6.2.
6.3.
6.4.
6.5.
6.6.
Compactness for metric spaces
Properties of compact spaces
The one point compactification
Flavors of compactness
Exercises
94
98
101
104
105
Chapter 7. Connectedness and Path-Connectedness
7.1. Connected topological spaces
7.2. Path-connectedness
7.3. Local connectivity
7.4. Exercises
109
109
114
119
121
Chapter 8. Quotient spaces
8.1. Quotient spaces: Definitions, properties and first examples
8.2. Surfaces as quotient spaces of polygons
8.3. Lens spaces
8.4. Seifert fibered spaces
8.5. Real and complex projective spaces
8.6. Cylinders, cones and suspensions
8.7. Exercises
123
123
131
134
136
136
136
136
Chapter 9. Foundational theorems
9.1. The Tietze extension theorem
9.2. Metrization theorems
9.3. The Borsuk-Ulam theorem
9.4. The Tychonoff compactness theorem
9.5. Exercises
139
139
142
142
142
142
Chapter 10. Spaces of functions
143
CHAPTER 1
T
Continuity and convergence in Euclidean spaces
his chapter is motivational in nature and provides impetus for the definitions
of a topological space X and a continuous function f : X → Y between
two topological spaces X and Y , given in Chapters 2 and 3 respectively.
These definitions are straightforward but upon first encounter seem ad hoc
and arbitrary. To convey a sense of where they are coming from, in Section 1.1 we
first examine the familiar setting of continuity of functions f : Rn → Rm on Euclidean
spaces in terms of its customary definition in the δ, ε language from analysis. In Section
1.2 we introduce two families of special subsets of Rn , the open and closed subsets. The
discussion of this chapter culminates in Section 1.3 where we recast continuity in terms
of open and closed subsets of Rn . The main results of this chapter are Theorems 1.3.1
and 1.3.2. The remaining Section 1.4 scrutinizes some properties of open and closed
subsets of Rn .
1.1. Continuity in Euclidean space
Let R denote the set of real numbers. We define intervals in R as any of the subsets
of R of the form
[a, b] = {x ∈ R | a ≤ x ≤ b},
[a, bi = {x ∈ R | a ≤ x < b},
ha, b] = {x ∈ R | a < x ≤ b},
ha, bi = {x ∈ R | a < x < b}.
The first of these are referred to as segments or closed intervals, the last in turn are
called open intervals.
The n-dimensional Euclidean space Rn is defined as the set of ordered n-tuples of
real numbers:
Rn = {(x1 , ..., xn ) | xi ∈ R },
where, as usual, the symbol R denotes the set of real numbers. We will adopt the
convention of denoting the n-tuple (x1 , ..., xn ) simply by x, the n-tuple (y1 , ..., yn ) by
y, etc. Two elements x = (x1 , ..., xn ) and y = (y1 , ..., yn ) from Rn can be added to each
other and each can be multiplied by a real number λ ∈ R in the familiar ways:
x + y = (x1 + y1 , ..., xn + yn )
and
λ · x = (λ · x1 , ..., λ · xn ),
endowing Rn with the structure of a real n-dimensional vector space.
5
6
1. CONTINUITY AND CONVERGENCE IN EUCLIDEAN SPACES
Definition 1.1.1. For a real number p ≥ 1 and for p = ∞, we define the functions
dp : Rn × Rn → [0, ∞i as

1
 (|x1 − y1 |p + ... + |xn − yn |p ) p
; if p 6= ∞
(1.1)
dp (x, y) =

max {|x1 − y1 |, ..., |xn − yn |}
; if p = ∞.
When p = 2 we call the resulting function d2 the Euclidean distance function. We will
refer to the functions dp as metrics on Rn , something we will justify in Exercise 2.5.6.
For example
d1 (x, y) = |x1 −y1 |+· · ·+|xn −yn |
and
3
4
d 3 (x, y) = |x1 − y1 | + · · · + |xn − yn |
4
3
4
43
,
while d2 takes on the familiar form
p
d2 (x, y) = (x1 − y1 )2 + · · · + (xn − yn )2 .
Definition 1.1.2. Let r > 0 be a real number and let x ∈ Rn be a point in
Euclidean space. We define the open (Euclidean) ball Bx (r) of radius r and with center
x as the subset of Rn given by
Bx (r) = {y ∈ Rn | d2 (x, y) < r}.
Note that this definition of Bx (r) used the Euclidean distance function d2 . We
could equally well have used any of the distance functions dp (including the case of
p = ∞) in this definition. Thus, we’ll let Bxp (r) denote the open ball with center x and
radius r but taken with respect to the metric dp :
Bxp (r) = {y ∈ Rn | dp (x, y) < r}.
p
Figure 1 illustrates the shapes of B(0,0)
(1) ⊂ R2 for the choices of p = 1, 2, ∞.
1
B(0,0)
(1)
2
B(0,0)
(1) = B(0,0) (1)
∞
B(0,0)
(1)
p
Figure 1. Examples of the open balls B(0,0)
(1) ⊂ R2 for p = 1, 2, ∞.
The dotted lines indicate that the boundaries of these shapes are not
part of the balls.
1.1. CONTINUITY IN EUCLIDEAN SPACE
7
Definition 1.1.3. A function f : Rn → Rm is continuous at the point x ∈ Rn if
for every ε > 0 there exists a δ > 0 such that d2 (x, y) < δ implies d2 (f (x), f (y)) < ε.
Said differently, f is continuous at x ∈ Rn if for every ε > 0 there exists a δ > 0 such
that
f (Bx (δ)) ⊆ Bf (x) (ε).
The function f is said to be continuous if it is continuous at every point x ∈ Rn . See
Figure 2 for an illustration.
f
Bf (x) (ε)
f (x)
x
Bx (δ)
Rn
Rm
Figure 2. The function f is continuous at x ∈ Rn if for every ε > 0
there is a δ > 0 such that the ball of radius δ centered at x (shaded disk
on the left) maps into the ball of radius ε > 0 with center f (x) (shaded
disk on the right). The amoeba-like shape inside of Bf (x) (ε) indicates
the image of Bx (δ) under f .
Continuity of functions is closely tied to convergence of sequences, a connection we
briefly review next. Recall that a sequence x in Rn is merely a function x : N → R.
We denoted x(k) by xk and we write {xk }∞
k=1 or simply {xk }k to denote the entire
sequence (rather than writing x, the name of the function, so as to not confuse it with
our notation x for points in Rn ). To indicate that the sequence {xk }k belongs to Rn ,
we write {xk }k ⊂ Rn .
Definition 1.1.4. Let {xk }k be a sequence in Rn .
(a) We shall say that {xk }k converges to x ∈ Rn if for every ε > 0 there exists a
positive integer k0 such that for all k ≥ k0 the relation xk ∈ Bx (ε) holds. In
this case we write limk→∞ xk = x or simply lim xk = x.
(b) The sequence {xk }k is called a Cauchy sequence if for every ε > 0 there is an
integer k0 such that whenever k, m ≥ k0 then d2 (xk , xm ) < ε.
8
1. CONTINUITY AND CONVERGENCE IN EUCLIDEAN SPACES
The following two theorems are standard results discussed in any introductory
analysis course, their proofs are omitted.
Theorem 1.1.5. Let f : Rn → Rm be a function and x ∈ Rn be any point. Then f
is continuous at x if and only if for every sequence {xk }k ⊂ Rn that converges to x in
Rn , the sequence {yk }k with yk = f (xk ), converges to y = f (x) in Rm . Said differently,
f is continuous at x if and only if it commutes with the limit symbol lim when applied
to any sequence {xk }k with limit x:
f lim xk = lim f (xk ).
k→∞
k→∞
n
Theorem 1.1.6. A sequence {xk }k ⊂ R is convergent if and only if it is Cauchy.
1.2. Open and closed subsets of Euclidean space
Given a set X and a subset A ⊂ X, we define the complement of A in X as the set
X − A = {x ∈ X | x ∈
/ A}.
Definition 1.2.1. A subset A ⊆ Rn is called closed if for every convergent sequence
{ai }i ⊂ A the limit limi→∞ ai also lies in A. A subset B ⊆ Rn is called open if Rn − B
is closed.
Example 1.2.2. When n = 1, all closed intervals [a, b] are closed sets while open
intervals ha, bi are open sets in R. The sets R and ∅ are both closed and open while
the intervals [a, bi are neither closed nor open. Any finite subset of R is closed.
Example 1.2.3. When n = 2 examples of closed sets are the closed rectangle
[a, b] × [c, d], the closed circles {(x, y) ∈ R2 | x2 + y 2 ≤ r}, the upper half-plane {(x, y) ∈
R2 | y ≥ 0}, the graph Γf = {(x, f (x)) | x ∈ R} of a continous function f : R → R. The
main examples of open sets are the open balls Bx (r).
The following proposition gives a characterization of open sets independent of the
definition of closed sets.
Proposition 1.2.4. A subset U ⊆ Rn is open if and only if for every x ∈ U there
exists a ε > 0 such that Bx (ε) ⊆ U .
Proof. =⇒ Let U be an open subset of Rn and let x ∈ U be an arbitrary
element. Suppose there were no ε > 0 for which the inclusion Bx (ε) ⊆ U were true.
Then, for any positive integer i, the ball Bx ( 1i ) would have to intersect the complement
V = Rn − U of U . Pick an arbitrary element bi ∈ V ∩ Bx ( 1i ), one for each i ∈ N, thus
creating a sequence {bi }i ⊂ V . The property d2 (x, bi ) < 1i implies that bi converges to
x. However, since U is open, V must be closed and so x = lim bi must lie in V . But
clearly x ∈
/ V , creating a contradiction. Therefore, an ε with the property Bx (ε) ⊆ U
must exist.
⇐= To prove that U is open we need to prove that V = Rn −U is closed. Supposed
that this fails. Then there must exists a convergent sequence {bi }i ⊂ V whose limit
1.3. CONTINUITY AND CONVERGENCE IN TERMS OF OPEN AND CLOSED SETS
9
x = lim bi lies in U . Pick an ε > 0 so that Bx (ε) ⊆ U . Since none of the bi lie in U , we
see that d2 (bi , x) ≥ ε. This however is impossible for a convergent sequence, creating
a contradiction. We are then forced to conclude that V is a closed subset of Rn and
hence that U is open.
Example 1.2.5. The open balls Bx (r) are indeed open set. This follows easily
from the preceding proposition. For if y ∈ Bx (r), let ρ = min{d2 (x, y), r − d2 (x, y)}. If
y 6= x then ρ is positive and By (ρ) ⊂ Bx (r). If y = x we may take ρ = r and the same
conclusion holds.
1.3. Continuity and convergence in terms of open and closed sets
The purpose of this section is to prove that the continuity of a function f : Rn → Rm
either at a point or globally, can be expressed entirely in terms of open or closed sets.
Recall that the preimage f −1 (V ) of a function f : X → Y between two sets X and
Y , and of a subset V ⊂ Y , is defined as
f −1 (V ) = {x ∈ X | f (x) ∈ V }.
In particular, f −1 (V ) is a subset of X. We remark that the notation f −1 used above
does not suggest that f has an inverse function. For example, if f : R → R is the
function f (x) = x2 , then f −1 ([0, 1]) = [−1, 1].
Theorem 1.3.1. Let f : Rn → Rm be a function.
(a) The function f is continous at x ∈ Rn if and only if for every open set V ⊆ Rm
containing f (x), there exists an open set U ⊆ Rn containing x with the property
that f (U ) ⊂ V .
(b) The function f is continuous if and only if for every open set V ⊆ Rm the
preimage U = f −1 (V ) of V under f , is an open subset of Rn .
Proof. (a) =⇒ Assume that f is continuous at x ∈ Rn and let V ⊂ Rm be
any open set containing f (x). Since V is open, by Proposition 1.2.4, there exists an
ε > 0 such that Bf (x) (ε) is contained in V . But continuity of f at x then implies the
existence of a δ > 0 such that f (Bx (δ)) ⊂ Bf (x) (ε). Since Bx (δ) is an open set (see
example 1.2.5) and since Bf (x) (ε) is contained in V , by taking U = Bx (δ), we see that
f (U ) ⊂ V , as claimed. See Figure 3 for an illustration.
⇐= Assume that f has the property that for every open subset V ⊂ Rm that
contains f (x) there is an open subset U ⊂ Rn containing x such that f (U ) ⊂ V .
We’d like to show that then f is continuous at x. Pick an arbitrary ε > 0 and let
V = Bf (x) (ε). Note that V is an open set containing f (x) and so we are guaranteed
the existence of an open set U containing x such that f (U ) ⊂ Bf (x) (ε). But since U is
open, there must exist a δ > 0 such that Bx (δ) ⊂ U (by Proposition 1.2.4). Thus we
found a δ such that f (Bx (δ)) ⊂ Bf (x) (ε) and so f is continuous at x, see Figure 4.
(b) =⇒ Suppose f is continuous and let V ⊂ Rm be any open set. We’d like
to show that U = f −1 (V ) is also an open set. Let x ∈ U be any point. Since f is
10
1. CONTINUITY AND CONVERGENCE IN EUCLIDEAN SPACES
f
V
Bf (x) (ε)
Bx (δ)
x
Rn
Rm
Figure 3. This picture illustrates the portion (a) =⇒ of the proof of
Theorem 1.3.1. V is an arbitrary open set containing f (x) and ε > 0
is chosen so that Bf (x) (ε) (shaded disk on the right) is contained in V .
By continuity of f at x, there is a δ > 0 so that Bx (δ) (shaded disks
on the left) maps into B(f (x) (ε) under f . The image of Bx (δ) under f is
indicated by the smaller amoeba-like shape inside of Bf (x) (ε).
continuous at x, part (a) of the theorem shows that there exists an open set Ux with
f (Ux ) ⊂ V . In particular, Ux ⊂ U . But since Ux is open and x ∈ Ux , there must exist
a δ > 0 so that Bx (δ) ⊂ Ux and so Bx (δ) ⊂ U . According to proposition 1.2.4, this
shows that U is open.
⇐= Suppose that f has the property that f −1 (V ) is an open set whenever V is an
open set. We’d then like to show that f must be continuous. By definition, this means
that we must show that f is continuous at each point x ∈ Rn . Using part (a) of the
theorem, demonstrating the continuity of f at x is equivalent to showing that for each
open V ⊂ Rm containing f (x), there exists an open set U ⊂ Rn containing x and such
that f (U ) ⊂ V . But our working assumption allows us to simply take U = f −1 (V ),
thus completing the proof of the theorem.
The importance of the preceding theorem is that it provides a definition of continuity that only relies on the notion of open sets. This observation will serve as the basis
for the definition of a topological space (in Chapter 2) and the generalization of continuity to such settings. The next theorem testifies that one can also define continuity
in terms of closed sets only.
Theorem 1.3.2. A function f : Rn → Rm is continuous if and only if f −1 (B) is a
closed subset of Rn whenever B is a closed subset of Rm .
1.3. CONTINUITY AND CONVERGENCE IN TERMS OF OPEN AND CLOSED SETS
11
f
V = Bf (x) (ε)
U
Bx (δ)
x
Rn
Rm
Figure 4. This picture illustrates the portion (a) ⇐= of the proof of
Theorem 1.3.1. Given any ε > 0 we set V = Bf (x) (ε) rendering it an open
set containing f (x). By assumption, there is an open set U (amoeba-like
set on the left) containing x and such that f (U ) ⊂ V . But since U is
open, there is a δ > 0 so that Bx (δ) (shaded disk on the left) is contained
in U . The images of U and Bx (δ) are indicated inside of V .
Proof. =⇒ Suppose that f is continuous and let V ⊂ Rm be a closed set. We’d
like to show that A = f −1 (B) is a closed subset of Rn . To see this, note that
f −1 (Rm − B) = Rn − f −1 (B) = Rn − A.
Since Rm − B is open and since according to Theorem 1.3.1 f −1 (Rm − B) then also
must be open, the above equality of sets shows that Rn − A is open. By definition, this
means that A is closed.
⇐= Suppose that f −1 (B) is closed whenever B is closed. Let V ⊂ Rm be any
open set, let U = f −1 (V ) and set B = Rm − V (note that B is closed). Then, on one
hand, A = f −1 (B) is closed while on the other,
f −1 (B) = f −1 (Rm − V ) = Rn − f −1 (V ) = Rn − U.
Thus U must be open and so, according to Theorem 1.3.1, f must be continuous.
We next turn to sequences where, as with continuity, convergence can be expressed
solely in terms of open sets.
Theorem 1.3.3. A sequence {xk }k ⊂ Rn converges to x ∈ Rn if and only if for
every open set U ⊂ Rn containing x, there exists a natural number k0 such that for all
k ≥ k0 we obtain xk ∈ U .
12
1. CONTINUITY AND CONVERGENCE IN EUCLIDEAN SPACES
Proof. =⇒ Suppose that lim xk = x and that U is an open set containing x.
Then there exists an ε > 0 such that Bx (ε) ⊂ U . Find a natural number k0 such that
for all k ≥ k0 the points xk lie in Bx (ε). Clearly then xk ∈ U also for all k ≥ k0 .
⇐= Suppose now that xk is a sequence in Rn with the property that for every
open set U containing x, there is a natural number k0 such that k ≥ k0 implies that
xk ∈ U . Given any ε > 0, we need to demonstrate that there is a k0 so that k ≥ k0
implies that xk ∈ Bx (ε). Of course, choosing U = Bx (ε) finishes the proof.
1.4. Some properties of open and closed sets in Euclidean space
In the proof of Theorem 1.4.1 below, we shall use DeMorgan’s laws (1.2) below.
To state them, let X be any set and let Ui ⊆ Rn be a family of subsets of X with i
running through some indexing set I. Then the following, easy to prove equalities of
sets hold:
X − (∩i∈I Ui ) = ∪i∈I (X − Ui )
(1.2)
X − (∪i∈I Ui ) = ∩i∈I (X − Ui )
Said differently, these relations imply that
“The complement of the intersection is the union of the complements.”
“The complement of the union is the intersection of the complements.”
With these preliminaries in place, we are ready to turn to the main result of this
section.
Theorem 1.4.1. The following are properties of open subsets of Euclidean space:
(a) The sets Rn and ∅ are both open and closed sets.
(b) The union of any number of open sets is an open set.
(c) The intersection of any finite number of open sets is an open set.
Proof. (a) Follows directly from the definitions of open and closed subsets of Rn .
(b) Let I be an arbitrary indexing set and for each i ∈ I, let Ui ⊆ Rn be an open
set. Let U = ∪i∈I Ui . We need to show that U is an open set. Let x ∈ U . Then x ∈ Ui
for some i ∈ I. By Proposition 1.2.4, there must exist a ε > 0 such that Bx (ε) ⊆ Ui .
But then Bx (ε) ⊂ U since Ui ⊆ U . This shows that for every x ∈ U there is a ε > 0
such that Bx (ε) ⊆ U . According to Proposition 1.2.4 this means that U is an open set.
(c) Let U1 , ..., Um be a finite family of open sets and let V = ∩m
j=1 Uj . To see that
V is open, pick an arbitrary x ∈ V . Then x ∈ Ui for every j ∈ {1, 2, ..., m} and so there
exist numbers εj > 0 with the property that Bx (εj ) ⊆ Uj . Let ε = min{ε1 , ..., εm } and
note that ε > 0. Clearly then Bx (ε) ⊆ Uj for every j ∈ {1, ..., m} so that Bx (ε) ⊆ V .
This shows that V is an open set.
Corollary 1.4.2. The intersection of any number of closed subsets of Rn is a
closed set. The union of any finite number of closed subsets of Rn is a closed set.
1.5. EXERCISES
13
Proof. The corollary is a direct consequence of Theorem 1.4.1 and DeMorgan’s
laws (1.2). Namely, given a family Vi ⊂ Rn of closed sets, indexed by an indexing set
I, set Ui = Rn − Vi . Clearly each Ui is then open and so by part (b) of Theorem 1.4.1,
so is U = ∪i∈I Ui . But then V = Rn − U is closed, whereas by DeMorgan’s laws, V
equals
V = Rn − U = Rn − (∪i∈I Ui ) = ∩i∈I (Rn − Ui ) = ∩i∈I (Rn − (Rn − Vi )) = ∩i∈I Vi .
The case of finite unions of closed sets follows similarly and is left as an easy exercise.
There are many examples of infinite families of open sets whose intersection is not
open, and infinite families of closed sets whose union is not closed.
Example 1.4.3. For each i ∈ N let Ui = − 1i , 1i . Then each set Ui is an open
subset of R, but the intersection ∩∞
i=1 Ui = {0} is not open.
Example 1.4.4. For i ∈ N let Vi = 1i , 3 − 1i . Each Vi is a closed subset of R but
their union ∪∞
i=1 Vi = h0, 3i is not closed.
1.5. Exercises
1.5.1. Determine if the following subsets of R are open, closed or neither:
(a) [0, 1i ∪ h1, 2].
(b) Q.
(c) ∪n∈Z [2n, 2n + 1].
(d) The Cantor set C. (The Cantor set C is obtained from [0, 1] by dividing it into 3
segments of equal length, and removing the middle open interval h 13 , 23 i. In the
remaining union of two segments [0, 13 ] ∪ 32 , 1], each is divided into 3 segments of
equal length, and from each the middle open interval is again removed so as to
obtain[0, 19 ] ∪ [ 29 , 31 ] ∪ [ 23 , 79 ] ∪ [ 79 , 1]. The Cantor set is obtained by continuing this
process ad infinitum. Thus at each stage, one divides each remaining segment
into 3 segments of equal length, and removes the open middle interval. See
Figure 5 for the first several iteration of this construction.)
C1 = [0, 1]
C2 = [0, 31 ] ∪ [ 23 , 1]
C3 = [0, 91 ] ∪ [ 29 , 13 ] ∪ [ 32 , 79 ] ∪ [ 98 , 1]
1
2 1
C4 = [0, 27
] ∪ [ 27
, 9 ] ∪ · · · ∪ [ 89 , 25
] ∪ [ 26
, 1]
27
27
1
2 1
C5 = [0, 81
] ∪ [ 81
, 27 ] ∪ · · · ∪ [ 26
, 79 ] ∪ [ 80
, 1]
27 81
81
Figure 5. The first five iterations Cn in the construction of the Cantor set.
1.5.2. Proof the claims made in
14
1. CONTINUITY AND CONVERGENCE IN EUCLIDEAN SPACES
(a) Example 1.2.2,
(b) Example 1.2.3.
1.5.3. Let A be a subset of R that is both open and closed. Show that A is either
the empty set or else A = R. Is the same claim true for subsets of Rn ?
1.5.4. For any p ∈ [1, ∞i∪{∞} once can define the p-open subsets of Rn as the sets
U ⊂ Rn such that for every x ∈ U there exists an ε > 0 with Bxp (ε) ⊂ U . Show that
the 1-open subsets and the ∞-open subsets of Rn are the same as the open subsets of
Rn . More generally, show that the set of p-open subsets is independent of the choice
of p ∈ [1, ∞i ∪ {∞}.
1.5.5. A point x ∈ Rn is said to be an accumulation point of the subset A ⊂ Rn if
every open set U ⊂ Rn that contains x, intersects A nontrivially. Show that A ⊂ Rn
is a closed set if and only if it contains all of its accumulation points.
CHAPTER 2
H
Topological spaces
ere we formally introduce the notion of a topological space X as a nonempty
set X equipped with a topology T . A topology on X is merely a collection
of subsets of X, subject to axioms motivated by the results of Theorem 1.4.1
for the case of X = Rn . We give many examples of topological spaces in
Section 2.2, examples which we rely on in later chapters to examine various properties of topological spaces. In Section 2.3 we introduce the important notions of the
interior, closure and boundary of a subset of a topological space, and verify some of
their properties. The final section of this chapter introduces bases and subbases of a
topology, and defines notions such as first and second countability and separability of a
topological space.
2.1. Definition of a topological space
Definition 2.1.1. A topology T on a set X is a collection of subsets of X subject
to the following three rules, called the Axioms of a topology:
1. The empty set ∅ and all of X belong to T .
2. Let I be an arbitrary indexing set and for each i ∈ I, pick a set Ui ∈ T . Then
we require that ∪i∈I Ui also belong to T . Said differently, T must be closed
under taking arbitrary unions.
3. For elements U1 , ..., Un ∈ T , the set U1 ∩ ... ∩ Un must also belong to T . Said
differently, T must be closed under taking finite intersections.
A topological space is a pair (X, T ) consisting of a set X and a topology T on X. A
subset U of X is called an open set if U ∈ T and a subset V ⊂ X is called closed if
X − V is open. A neighborhood of a point x ∈ X is any open set U ⊂ X that contains
x.
This definition is central to the remainder of the book and so, before moving on
to consider examples, we first pause to elucidate its various aspects. The choice of the
three axioms of a topology should not be too surprising given the results from Chapter
1. Specifically, they are modeled on the three properties of open subsets of Euclidean
space proved in Theorem 1.4.1. Keeping in mind that open subsets of Rn were used
to recast the definition of continuity (Theorem 1.3.1), the attentive reader will have
little difficulty in guessing what the definition of a continuous function between two
topological spaces should be (for an answer, see Definition 3.1.1).
15
16
2. TOPOLOGICAL SPACES
Given a topological space (X, T ) we shall often simply write X when the topology
T is understood from context, and refer to X as a topological space. On the other
hand, when several topological spaces are involved in a discussion, we may label the
topology T by TX to indicate that it belongs with X. For example, we shall write
(X, TX ) or (Y, TY ) to label topological spaces.
The notion of open and closed subset of a topological space X shall be crucial to
all subsequent chapters. Whether or not a given subset A ⊂ X is open or not, depends
on the choice of a topology T on X. As we shall see in the examples below, a set X
admits many different topologies and a subset A ⊂ X may be open with respect to
some but not with respect to other topologies. The most common misconception about
open and closed sets among novices, is the notion that they form a dichotomy.
Remark 2.1.2. In a topological space X, the notions of “open subset”and “closed
subset”do not form a dichotomy. That is, the failure of a subset A ⊂ X to be open
does not typically imply that A is closed. Conversely, the failure of A to be closed does
not typically render it open. Typically, a topological space has numerous subsets that
are neither open nor closed, but can also have subsets that are both open and closed.
The empty set and all of X are examples of the latter.
Definition 2.1.3. Given a set X and two topologies T1 and T2 on X, we say that
T1 is finer than T2 or, equivalently, that T2 is coarser than T1 , if T2 ⊂ T1 . We shall
write T2 ≤ T1 to denote this relation between the two topologies. As usual, we shall
write T2 < T1 to indicate that T2 ≤ T1 and T2 6= T1 .
The relation “≤” gives the set of all topologies on X a partial ordering in that
T1 ≤ T2 and T2 ≤ T1 imply T1 = T2 . Likewise, T1 ≤ T2 and T2 ≤ T3 imply T1 ≤ T3 .
However, given two topologies T1 and T2 on X, neither of T1 ≤ T2 or T2 ≤ T1 has to
necessarily hold.
Before exploring properties of open and closed subsets of a topological space X,
we turn to examine several examples of topological spaces. We recommend that the
reader not skip this next section, it will be used as a testing ground for many of the
concepts touched upon in later chapters.
2.2. Examples of topological spaces
This section is devoted to exploring some of the many examples of topological
spaces. Example 2.2.4 below shows that every set X can be equipped with a topology,
exhibiting that topological spaces are indeed very common animals in the jungle of
mathematics. Examples 2.2.1 – 2.2.3 discussing the Euclidean, the subspace and the
metric topology, are particularly relevant as they make frequent appearances throughout the text.
Example 2.2.1. The Euclidean topology Work from Chapter 1 shows that the
Euclidean space Rn becomes a topological space when equipped with the topology TEu ,
henceforth referred to as the Euclidean topology, defined as
TEu = {U ⊂ Rn | ∀x ∈ U ∃r > 0 such that Bx (r) ⊂ U }.
2.2. EXAMPLES OF TOPOLOGICAL SPACES
17
Recall that Bx (r) denotes the open ball {y ∈ Rn | d2 (x, y) < r} from Definition 1.1.1.
With this definition of TEu , the verification of the three axioms of topology is provided
courtesy of Theorem 1.4.1.
Example 2.2.2. The relative or subspace topology. Let (X, T ) be a given
topological space and let A be a subset of X. Then A automatically inherits the
structure of a topological space from X by equipping it with the relative topology or
subspace topology TA defined as:
TA = {U ∩ A | U ∈ T }
We call the topological space (A, TA ) a subspace of X. Saying that “A is a subspace of
X”means that we have given the subset A of X the subspace topology.
It is quite straightforward to verify that TA satisfies the axioms of a topology on A:
1. Since ∅, X ∈ T then ∅ ∩ A = ∅ and X ∩ A = A belong to TA .
2. Let Ui ∈ TA , i ∈ I be given and set U = ∪i∈I Ui . For each Ui there exists a
Vi ∈ T such that Ui = Vi ∩ A. But then U = V ∩ A where V = ∪i∈I Vi ∈ T
showing that U ∈ TA .
3. Let U1 , ..., Un ∈ TA and find sets V1 , ..., Vn ∈ T such that Ui = Vi ∩ A. Then the
set U = U1 ∩ ... ∩ Un equals V ∩ A where V = V1 ∩ ... ∩ Vn ∈ T and is therefore
contained in TA .
The subspace topology gives us immediately a myriad of examples of topological
spaces by applying it to various subsets of (Rn , TEu ) from the previous example. For
instance, each of
the n-sphere
the graph of f : Rn → Rm
the 2-dimensional torus
S n = {x ∈ Rn+1 | d2 (x, 0) = 1} ⊂ Rn+1 ,
Γf = {(x, f (x)) ∈ Rn × Rm | x ∈ Rn } ⊂ Rn+m ,
T 2 = {x ∈ R4 | x21 + x22 = 1 and x23 + x24 = 1} ⊂ R4 ,
becomes a topological space with the relative Euclidean topology. In the definition of
T 2 , the symbol x was used to denote the ordered quadruple (x1 , x2 , x3 , x4 ).
Example 2.2.3. Metric spaces. A metric space is a pair (X, d) consisting of a
non-empty set X and a function d : X × X → [0, ∞i, referred to as the metric on X,
that is subject to the next three axioms of a metric:
1. d(x, y) = 0 if and only if x = y.
2. Symmetry: d(x, y) = d(y, x) for all x, y ∈ X.
3. Triangle inequality: d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X.
When the metric d is understood from context, we will call X itself a metric space. In
analogy to the case of the Euclidean metric d2 on Rn , here too we can define what we
shall again call the open ball with center x ∈ U and radius r > 0 as
Bx (r) = {y ∈ X | d(x, y) < r}.
Every metric space (X, d) comes equipped with a natural choice of topology Td , called
the metric topology, defined by
Td = {U ⊂ X | ∀p ∈ U ∃r > 0 such that Bp (r) ⊂ U }.
18
2. TOPOLOGICAL SPACES
The reader will have noticed that this definition agrees with the definition of the Euclidean topology from Example 2.2.1. Indeed, the premier examples of metric spaces
are (Rn , dp ) with p ∈ [1, ∞i ∪ {∞} (with dp as given in Definition 1.1.1). The verification of the axioms of a metric for dp is deferred to Exercise 2.5.6. The fact that Td is
indeed a topology on X follows exactly as in proof of Theorem 1.4.1 for the Euclidean
space. Indeed, the proof of the latter never uses the fact that it deals with (Rn , d2 )
explicitly.
We shall encounter more about metric spaces in later chapters and so for now
we limit ourselves to only one additional example. Consider the segment [a, b] ⊂ R
and let X = C 0 ([a, b], R) be the set of continuous functions f : [a, b] → R. Define
d : X × X → [0, ∞i as
Z b
d(f, g) =
|f (t) − g(t)| dt.
a
Then (X, d) is a metric space and thus becomes a topological space. The second
and third axiom of a metric are readily verified (and their verification does not use
continuity of the functions involved):
Z b
Z b
d(f, g) =
|f (t) − g(t)| dt =
|g(t) − f (t)| dt = d(g, f )
a
a
b
Z
|f (t) − h(t)| dt
d(f, h) =
a
b
Z
|(f (t) − g(t)) + (g(t) − h(t))| dt
=
a
Z
b
≤
|f (t) − g(t)| + |g(t) − h(t)| dt
Z b
Z b
=
|f (t) − g(t)| dt +
|g(t) − h(t)| dt
a
a
a
= d(f, g) + d(g, h)
The demonstration of axiom 1 of a metric is left for Exercise 2.5.1.
Example 2.2.4. The discrete and indiscrete topologies. Every set X always
admits topologies. Namely, every set can be made into a topological space by choosing
either the indiscrete topology Tindis (also referred to as the trivial topology) or the
discrete topology Tdis defined as
Tindis = {∅, X},
Tdis = {A | A ⊂ X}.
Thus Tindis only contains the empty set and all of X and is therefore the smallest
possible topology on X (according to the first axiom of a topology) while Tdis equals the
entire power set of X and is consequently the largest topology on X. In the notation of
2.2. EXAMPLES OF TOPOLOGICAL SPACES
19
Definition 2.1.3, any topology T on X satisfies the double inequality Tindis ≤ T ≤ Tdis .
The axioms of a topology are trivially true for Tindis and Tdis .
These two extreme topologies on X do not lead to interesting topological properties.
For example, as we shall see in Chapter 3, every function on (X, Tdis ) is continuous.
Fertile ground for topological exploration lies with those topologies that live between
these two extremes.
Example 2.2.5. Topologies on finite sets. On finite sets, topologies are by
necessity also finite and can be listed by simply listing their elements. For instance, if
X = {1, 2, 3, 4, 5}, then each of the following is a topology on X (Exercise 2.5.2):
(a) T1 = {∅, X, {1, 2}, {3, 4, 5}}
(b) T2 = {∅, X, {1}, {2}, {1, 2}}
(c) T3 = {∅, X, {1, 2, 3}, {2, 3, 4}, {2, 3}, {1, 2, 3, 4}}
(d) T4 = {∅, X, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}
Example 2.2.6. Included point topology. Let X be any non-empty set and let
p ∈ X be an arbitrary point. We define the included point topology on X as
Tp = {U ⊂ X | U is the empty set, or p ∈ U }.
To see that (X, Tp ) is a topological space, we need to verify the axioms of topology for
Tp from Definition 2.1.1.
1. Clearly ∅ ∈ Tp be definition. Also, X ∈ Tp since p ∈ X.
2. Let Ui ∈ Tp with i ∈ I where I is any indexing set and let U = ∪i∈I Ui . If each
Ui is the empty set then so is U and is therefore contained in Tp . If at least one
set Ui is not empty, then p ∈ Ui and thus p ∈ U showing again that U ∈ Tp . So
in either case U must belong to Tp .
3. Let U1 , ..., Un ∈ Tp and let V = ∩ni=1 Ui . If even one of U1 , ..., Un is empty then
V is empty as well and thus a member of Tp . If none of U1 , ..., Un is empty then
they each must contain p and therefore V must contain p as well. So in this
case V is also in Tp .
Example 2.2.7. The excluded point topology. Let X again be any non-empty
set and, as in the previous example, pick an arbitrary point p ∈ X. The excluded point
topology T p on X is then defined to be
T p = {U ⊂ X | U equals X, or p ∈ (X − U )}.
Let’s verify the axioms of a topology:
1. X belongs to T p be definition and ∅ belongs to T p since p ∈
/ ∅.
2. Let Ui ∈ T p , i ∈ I and set U = ∪i∈I Ui . If at least one Ui equals X then U = X
and so U ∈ T p . If none of the Ui equals X then no Ui can contain p and so p
cannot be contained in U either. Thus, in this case too, we get U ∈ T p .
3. Take U1 , ..., Un ∈ T p and let V = U1 ∩ ... ∩ Un . If all of the set Ui happen to
equal X then so does V and is thus automatically contained in T p . Conversely,
if there is at least one Ui not equal to X then that particular Ui cannot contain
20
2. TOPOLOGICAL SPACES
p and consequently neither can V . Thus, in this case too, V is again an element
of T p .
See Exercise 2.5.3 for a generalization of the included/excluded point topology.
Definition 2.2.8. Let X be any non-empty set. A partition P on X is a collection
of subsets of X such that
1. If A, B ∈ P are two distinct elements of P then A ∩ B = ∅.
2. The elements of P cover all of X: ∪A∈P A = X.
Thus a partition P is a way of dividing all of X into mutually disjoint set. The partition
P = {X} consisting of only X shall be referred to as the trivial partition.
Examples of partitions abound. For instance, if we take X = R, then
P1 = {[a, a + 1i | a ∈ Z},
P2 = {Q, R − Q},
(2.1)
P3 = {x + Z | x ∈ hπ, π + 1]},
are all examples of partitions.
Example 2.2.9. Partition topology Let X be a non-empty set and let P =
{Ui | i ∈ I} be a partition on X. We then define the partition topology TP as
TP = {∪j∈J Uj | J ⊂ I}.
Thus elements of TP are obtained by taking unions of set from P. To see that this is
a topology we check the three axioms of a topology.
1. Choosing J = ∅ and J = I renders the set ∪j∈J Uj equal to the empty set and
all of X respectively.
2. Let Jk , k ∈ K be a family of subsets of I giving rise to the sets Vk = ∪j∈Jk Uj
from TP and let V = ∪k∈K Vk . Rewriting this definition of V we see that
V = ∪j∈L Uj
with
L = ∪k∈K Jk ⊂ I
Thus, of course, V belongs to TP .
3. This case proceeds in complete analogy with the previous point. Let V1 =
∪j∈J1 Uj , ..., Vn = ∪j∈Jn Uj and set W = ∩nk=1 Vk . But then
W = ∩j∈L Uj
with
L = ∩nk=1 Jk ⊂ I
we see again that W lies in P.
For instance, let us pick the partion P1 from (2.1) on X = R. Examples of open
sets in the associated partition topology TP1 are intervals of the form [a, bi as well as
[a, ∞i and h∞, bi with a, b ∈ Z. However h0, 1i is not an open set in this topology
(verify this!).
Example 2.2.10. Finite complement topology. On a non-empty set X we
define the finite complement topology Tf c as
Tf c = {U ⊂ X | U = ∅
or
X − U is a finite set }.
2.2. EXAMPLES OF TOPOLOGICAL SPACES
21
If X is itself a finite set then the finite complement topology agrees with the discrete
topology (from Example 2.2.4) but if X is infinite, Tf c and Tdis are rather different
topologies. We leave the verification of the axioms of topology for Tf c for Exercise
2.5.8.
Example 2.2.11. The countable complement topology. The countable complement topology Tcc on a set X is define as
Tcc = {U ⊂ X | U = ∅
or
X − U is a countable set }.
Note that if X is itself countable then Tcc agrees with the discrete topology Tdis on X.
However, for example on X = R, the two topologies are different. The axioms of a
topology for this example are addressed in Exercise 2.5.8.
Example 2.2.12. The Fort topology. This example is obtained by combining
the excluded point topology with the finite complement topology from above. Assume
that X is an infinite set and let p ∈ X be an arbitrary point. We then define the Fort
topology TF ,p as
TF ,p = {U ⊂ X | Either X − U is finite or p ∈
/ U }.
We turn to the verification of the axioms of a topology.
1. Since p ∈
/ ∅ and since X − X is a finite set, we find that ∅, X ∈ TF ,p .
2. Let Ui ∈ TF ,p , i ∈ I be a family of sets in this topology and set U = ∪i∈I Ui .
To see that U too must belong to TF ,p we must consider two cases separately.
Firstly, suppose that none of the sets Ui contains p. In this case U doesn’t
contain p either and so U ∈ TF ,p . In the second case, suppose that at least one
Ui contains p (and therefore U must also contain p). This particular Ui then
has to have finite complement X − Ui . Since X − U ⊂ X − Ui (this follows
for example from DeMorgan’s laws (1.2)) we see that X − U is also finite and
therefore that U ∈ TF ,p .
3. Left as an exercise (Exercise 2.5.9).
For instance, choosing X = R and p = 0, the closed sets of the Fort topology TF ,0
are all finite subsets of R and all subsets of R that contain 0.
Example 2.2.13. The order topology In this example we suppose that the set
X is equipped with an ordering “≤”, a notion which we briefly review before defining
the associated order topology T≤ .
Recall than a relation r on X is simply a subset of X × X. It is customary to write
xry if (x, y) ∈ r. A typical choice for the name of a relation is a “relation symbol”,
such as ∼, ≡, ≤, ≺ etc. For example, if we called a relation ∼ then we would write
x ∼ y to indicate that (x, y) belongs to this relation.
A (total) ordering “≤”on X is a relation on X subject to the conditions
1. “≤” is reflexive: x ≤ x for any x ∈ X.
2. “≤” is antisymmetric: If x ≤ y and y ≤ x then x = y.
3. “≤” is transitive: If x ≤ y and y ≤ z then x ≤ z.
22
2. TOPOLOGICAL SPACES
4. “≤” satisfies the trichotomy law: For all x, y ∈ X, either x = y or x < y or
y < x.
We shall write x < y to mean that x ≤ y but x 6= y.
Given an ordering “≤”on a set X with at least two elements, let us consider the
following special subset of X
Lx = {y ∈ X | y < x}
and
Rx = {z ∈ X | x < z}.
In terms of these we define the order topology T≤ associated to the ordering “≤”as
T≤ = {U ⊂ X | U is obtained by taking arbitrary unions of
finite intersections of the sets Lx and Ry with x, y ∈ X}
Examples of elements from T≤ are Lx ∩ Ry which we shall denote by hx, yi and refer
to as the open intervals of the order topology. Note that hx, yi is an empty set unless
there exists an element z ∈ X with x < z < y.
Since the set T≤ is closed under finite intersection and arbitrary unions (by definition
of T≤ ), axioms (2) and (3) of a topology (definition 2.1.1) are trivially true. To see
that the empty set and X belong to T≤ we proceed as follows. Given two distinct
elements x, y (recall that we assumed that X has at least two elements), we have either
x < y or y < x (according to trichotomy axiom above). Suppose that y < x, then
hx, yi = ∅ showing that ∅ ∈ T≤ . On the other hand, for these same x, y ∈ X we obtain
X = Lx ∪ Ry (Exercise 2.5.11) and so X ∈ T≤ .
As an illustration of the order topology, we consider the lexicographic ordering on
Rn . Let ≤ denote the standard ordering of the real numbers R and extend it to an
ordering on Rn by the following rule: (x1 , ..., xn ) < (y1 , ..., yn ) if
x1 < y1
x1 = y1 and x2 < y2
x1 = y1 , x2 = y2 and x3 < y3
..
.
or,
or,
or,
x1 = y1 , x2 = y2 , ..., xn−2 = yn−2 and xn−1 < yn−1
x1 = y1 , x2 = y2 , ..., xn−2 = yn−2 , xn−1 = yn−1 and xn < yn .
or,
The thus obtained ordering is called the lexicographic ordering on Rn . When n = 1,
this topology equals the Euclidean topology on R (from Example 2.2.1). However,
when n ≥ 2 the resulting order topology on Rn is quite different from its Euclidean
counterpart. For example, when n = 2, consider the interval h(0, 0), (1, 0)i ⊆ R2 given
by
h(0, 0), (1, 0)i = {(x, y) ∈ R2 | 0 < x < 1} ∪ {(0, y) ∈ R2 | y > 0} ∪ {(1, y) ∈ R2 | y < 1}.
Figure 1 shows this interval drawn in the Euclidean plane. Note that the interval
h(0, 0), (1, 0)i does not belong to the Euclidean topology on R2 .
2.2. EXAMPLES OF TOPOLOGICAL SPACES
23
1
Figure 1. The shaded region in R2 represents the interval h(0, 0), (1, 0)i
in the order topology on R2 associated to the lexicographic ordering. The
dotted lines are not part of the region while the full lines are. The region
extends infinitely vertically in both directions.
Example 2.2.14. The lower and upper limit topologies on R. The lower-limit
topology Tll on R is defined as
Tll = {U ⊂ X | U is obtained by taking unions of finite intersections of
sets [a, bi with a, b ∈ R}.
The definition of Tll shows that it is closed under finite intersections and arbitrary
unions while the empty set and R belong to Tll because, for example, ∅ = [0, 1i ∩ [3, 5i
and R = ∪a∈Z [a, a + 1i. Thus, Tll is indeed a topology.
The upper limit topology Tul on R is defined analogously by replacing the sets [a, bi
in the definition of Tll by the sets ha, b].
Notice that the open intervals ha, bi belong both to Tll and to Tul since, for example,
∞ [
1
ha, bi =
a + , b ∈ Tll .
n
n1
The starting value of n in the above union is chosen large enough so that a + 1/n < b.
¿From this observation it is not hard to show that TEu ⊂ Tll and TEu ⊂ Tul . However,
the set [0, 1i belong to Tll but not to TEu showing that TEu 6= Tll . A similar observation
applies to the upper limit topology.
Example 2.2.15. The topologist’s sine curve. This example and the two subsequent ones, do not define new types of topologies, but rather look at subspaces of
(R2 , TEu ) (see Examples 2.2.1 and 2.2.2) with certain special properties that shall be
explored in later chapters.
24
2. TOPOLOGICAL SPACES
We define the topologist’s sine curve to be the subspace X ⊂ R2 given by
X = {(x, sin(1/x)) | x ∈ h0, 1]} ∪ ({0} × [0, 1]).
Thus, the topologist’s sine curve is the union of the graph of sin(1/x) over h0, 1] union
the closed segment [−1, 1] on the y-axis. This space is illustrated in Figure 2a.
1
1
−1
(a)
...
..
.
(b)
(c)
Figure 2. (a) The topologist’s sine curve. (b) The infinite broom. (c)
The Hawaiian earrings.
Example 2.2.16. The infinite broom. Let In ⊆ R2 be the closed straight-line
segment joining the origin (0, 0) to the point (1, 1/n) for n ∈ N. The infinite broom is
the subspace X of (Rn , TEu ) defined by (see Figure 2b)
X = (∪∞
n=1 In ) ∪ ([0, 1] × {0}).
2.3. PROPERTIES OF OPEN AND CLOSED SETS
25
Example 2.2.17. Hawaiian earrings. For n ∈ N let Cn ⊂ R2 be the circle with
center (1/n, 0) and with radius rn = n1 . Note that all off these circles pass through the
origin (0, 0).The Hawaiian earrings is the subspace X of (R2 , TEu ) given as the union
of these circles (see Figure 2c):
∞
2
1 2
1 2
2
X = ∪∞
n=1 Cn = ∪n=1 {(x, y) ∈ R | (x − n ) + y = ( n ) }.
2.3. Properties of open and closed sets
This section discusses some general properties of open and closed sets of a topological space (X, T ). Recall that a subset U ⊂ X is called open if U ∈ T , a subset A ⊂ X
is called closed if X − A ∈ T . While typically a random subset Y ⊂ X is neither open
nor closed, we shall see presently it can be “approximated”by both kinds of sets.
Definition 2.3.1. Let A ⊂ X be an arbitrary subset of X. Define the interior Å
(or Int(A)) and the closure Ā (or Cl(A)) of A as
Å = union of all open sets contained in A,
Ā = intersection of all closed sets containing A.
The boundary or frontier ∂A of A is defined as ∂A = Ā − Å.
Lemma 2.3.2. Let A be a subset of the topological space X. Then
(a) The interior Å of A is an open set and it is the largest open set contained in
A. The equality A = Å holds if and only if A is open.
(b) The closure Ā of A is a closed set and it is the smallest closed set containing
A. The equality Ā = A holds if and only if A is closed.
(c) A point x ∈ X belongs to Ā if and only if every neighborhood of x intersects A.
Proof. These claims follow readily from the definition of interior and closure of a
set.
(a) Since Å is obtained as a union of open sets (those contained in A) it is an
open set. If U is any other open set containing A, then U is one of the sets from the
union which defines Å and is thus contained in Å, showing that the interior of A is the
largest open set contained in A. If Å = A then clearly A must be open since Å is open.
Conversely, if A is open then it is clearly the largest open set containing A and thus
by necessity equal to Å.
(b) By definition, Ā is an intersection of closed set and must therefore be closed.
If B is any closed set containing A, then B occurs in the intersection of sets defining
Ā and hence Ā ⊂ B. This shows that Ā is the smallest closed set containing A. If
Ā = A then A is closed, since Ā is. If A is closed then A itself is the smallest closed
set containing A and is thereby equal to Ā.
(c) Suppose firstly that x ∈ Ā and let U be a neighborhood of x. If we had
A ∩ U = ∅ then Ā − U would be a closed set containing A and would properly be
contained in Ā, an impossibility according to part (b) of the lemma. Thus we must
have A ∩ U 6= ∅.
26
2. TOPOLOGICAL SPACES
Conversely, suppose that x ∈ X is a point such that U ∩ A 6= ∅ for every neighborhood U of x. If we had x ∈
/ Ā, we could take U = X − Ā which would give an
immediate contradiction. Thus we must have x ∈ Ā.
Definition 2.3.1 shows that the inclusions
Å ⊂ A ⊂ Ā,
hold for every subset A of X while Lemma 2.3.2 shows that Å is always an open set
and Ā is always a closed set. It is in this sense that A can be approximated by both
an open set (its interior) and a closed set (its closure). How good an approximation of
A is given by the sets Å and Ā depends heavily on the topology on X.
Example 2.3.3. Consider the set R and its subset A = h0, 1i. Find the interior
Å, the closure Ā and the boundary ∂A of A with respect to the following choices of
topologies on R:
1. The Euclidean topology (Example 2.2.1). With this toplogy A itself is open
and so, by Lemma 2.3.2, the interior of A equals A itself. The closure of A is
a closed set cointaining A. We guess that Ā = [0, 1]. Indeed, [0, 1] is closed
and contains A and the only smaller subsets cotaining A are [0, 1i, h0, 1] and A
itself, neither of which is closed. Thus
Å = A = h0, 1i,
Ā = [0, 1],
∂A = {0, 1},
a result which confirms our Euclidean intuition.
2. The included point topology with p = 0 (Example 2.2.6). Since p ∈
/ A we see
that A is in fact closed so that Ā = A. On the other hand, the only open set
from Tp that is contained in A is the empty set, thus Å = ∅. Therefore,
Å = ∅,
Ā = A = h0, 1i,
∂A = A = h0, 1i.
3. The included point topology with p = 1/2 (Example 2.2.6). Since p ∈ A we see
that Å = A. However, since p ∈ A, the only closed set containing A is all of R
showing that Ā = R. We arrive at
Å = A = h0, 1i,
Ā = R,
∂A = h−∞, 0] ∪ [1, ∞i.
4. The finite complement topology (Example 2.2.10). In this topology, closed set
are finite subsets of R and all of R. This makes is clear that Ā = R. No subset
of A has finite complement showing that Å = ∅. In summary
Å = ∅,
Ā = R,
∂A = R.
The next lemma provides an alternative definition of the boundary ∂A.
Lemma 2.3.4. Let X be a topological space and let A be a subset of X. Then
(a) X − A = X − Å.
(b) Int(X − A) = X − Ā.
(c) ∂A = Ā ∩ X − A.
2.3. PROPERTIES OF OPEN AND CLOSED SETS
27
Proof. (a) By definition, the closure of X − A is the intersection of all closed sets
containing X − A:
X − A = ∩B∈B B
with
B = {B ⊂ X | B is closed and X − A ⊂ B}.
Let C be the collection of subset of X gotten from B by taking complements of elements
from B:
C = {C ⊂ X | X − C ∈ B}.
The key observation now is that C ∈ C if and only if C is open (since X − C must be
closed) and C is contained in A (since X − C contains X − A). Therefore, by definition
Å = ∪C∈C C.
An applicatioin of DeMorgan’s laws 1.2 yields the desired result:
X − A = ∩B∈B B = ∩C∈C (X − C) = X − ∪C∈C C = X − Å.
(b) This part follows in much the same way as part (a). Namely, let now B be the
family of closed set containing A and C be the family of open sets contained in X − A.
As before, there is a bijective correspondence B → C by sending B to X − B. Using
again DeMorgan’s law finishes the proof:
Int(X − A) = ∪C∈C C = ∪B∈B (X − B) = X − ∩B∈B B = X − Ā.
(c) This follows easily from part (a) of the present theorem:
∂A = Ā − Å = Ā ∩ (X − Å) = Ā ∩ X − A.
Definition 2.3.5. Let X be a topological space. A subset A of X is called dense
in X if Ā = X. The space X is called separable if there exists a countable dense subset
A of X.
As many topological spaces X are uncountable as sets, the notion of separability
provides a measure of how big X is as a topological space. A dense subset A ⊂ X
has the property that it intersects every open set of X (Corollary 2.3.7). Thus, if one
can find a countable dense subset of X (i.e. if X is separable), one should think of
X as having “relatively few” open sets and accordingly as being a “relatively small”
topological space.
Example 2.3.6. The subset A = h0, 1i of R is dense with respect to either the
particular point topology Tp with p = 1/2 or with respect to the finite complement
topology Tf c (Example 2.3.3).
Corollary 2.3.7. A subset A of the topological space (X, T ) is dense if and only
if A ∩ U 6= ∅ for all U ∈ T other than U = ∅.
Proof. This is a direct consequence of part (c) of Lemma 2.3.2.
28
2. TOPOLOGICAL SPACES
Example 2.3.8. The set Q of rational number is dense in (R, TEu ) since it intersects
every open interval ha, bi and every open set is a union of open intervals. More generally,
by the same principle, Qn is dense in (Rn , TEu ). Consequently, since Qn is a countable
set for each n ≥ 0, (Rn , TEu ) is a separable topological space for all n ≥ 0.
Example 2.3.9. Consider the discrete topology Tdis on R (Example 2.2.4). In this
topology every subset A ⊂ R is open and closed so that Ā = A for every A ⊂ R.
Therefore the only dense subset of (R, Tdis ) is R itself. We infer that (R, Tdis ) is not
separable.
2.4. Bases and subbases of a topology
The reader familiar with linear algebra will recall that a basis for a real finite
dimensional vector space V is a set of vectors {e1 , ..., en } ⊂ V such that every vector
v ∈ V can be written uniquely as a linear combination v = λ1 e1 + .. + λn en with
λi ∈ R. Thus, while V is typically an infinite set, we can capture the totality of
its vectors with the finite set {e1 , ..., en } by relying on the vector space operations of
“vector addition”and “scalar multiplication”.
As a topology T on a set X is a collection of subsets of X, it comes equipped with
two operations among its elements, namely those of taking unions and taking intersections. It is thus conceivable, in analogy with the vector space basis, that there are
subsets of T which “generate”all of T by means of taking unions and/or intersections
of its elements. This is indeed that case and we shall consider both subsets B ⊂ T
which “generate”all of T by means of only taking unions of elements from B, and
subsets S ⊂ T which will generate T by relying on both unions and intersection. The
first of these cases will lead the notion of a basis for (X, T ), the closest analogy to a
vector space basis. The second will lead to the notion of a subbasis, one that is without
analogue in the world of vector spaces.
Definition 2.4.1. Let (X, T ) be a topological space and let B and S be subsets
of T such that
(a) Every set U ∈ T is a union of sets from B.
(b) Every set U ∈ T is a union of finite intersections of sets from S.
Then B is called a basis for the topology T while S is called a subbasis for the topology
T . We also say that T is generated by B (by mean of taking unions of elements from
B) or that T is generated by S (by means of taking unions of finite intersections of
elements from S.
If B and S are given by B = {Ui ∈ T | i ∈ I} and S = {Vj ∈ T | j ∈ J } with I
and J being two indexing sets, then properties (a) and (b) from Definition 2.4.1 mean
that every set U ∈ T can we written as
(a) U = ∪i∈IU Ui , for some subset IU of I.
(b) U = ∪`∈LU (∩j∈J` Vj ), for some family of indices LU and for finite families of
subsets J` of J with ` ∈ LU .
2.4. BASES AND SUBBASES OF A TOPOLOGY
29
The indexing set LU from (b) above, is of course allowed to be infinite. While we will
typically start out with a topological space (X, T ) and then seek bases and subbases
for T , it is meaningful to ask about going the other way. That is, given a set X and
two collections B, S of subsets of X, under what conditions are B and S a basis and a
subbasis for some topology T . The next two lemmas address this question.
Lemma 2.4.2. A collection B of subsets of a set X is a basis for some topology T
if and only if B has the properties
(a) X = ∪B∈B B.
(b) For every B1 , B2 ∈ B and every point x ∈ B1 ∩ B2 , there exists an element
B3 ∈ B such that x ∈ B3 ⊂ B1 ∩ B2 .
If these two conditions are met, the topology T generated by B consists of all possible
unions of elements from B:
T = {U ⊂ X | U is a union of sets from B}.
Proof. One implication of the lemma is immediate. Namely, if B is a basis for the
topology T then, since X ∈ T , it must be that X = ∪B∈B B. Likewise, if B1 , B2 ∈ B,
then B1 ∩ B2 is an open set and therefore a union of elements from B.
Let’s turn to the other implication. We assume that B is a collection of subsets of
X subject to conditions (a) and (b) from the lemma and let T be the set
T = {U ⊂ X | U is a union of elements from B}.
It is then clear that X ∈ T (by condition (a)) and ∅ ∈ T (the empty set is the empty
union of any sets). By definition of T , it is obviously closed under unions since unions of
unions are again just unions. To see that T is closed under finite intersection it suffices
to show that it is closed under twofold intersections. Thus, let U1 , U2 ∈ T , we need to
show that U1 ∩ U2 is a union of elements from B. Let x ∈ U1 ∩ U2 be any point. By
definition of T , there must exist elements B1 , B2 ∈ B with x ∈ Bi ⊂ Ui , i = 1, 2. But
then by property (b), there is an element Bx ∈ B such that x ∈ Bx ⊂ B1 ∩B2 ⊂ U1 ∩U2 .
But then clearly
[
U1 ∩ U2 =
Bx ,
x∈U1 ∩U2
and so U1 ∩ U2 ∈ T . This shows that T is indeed a topology on X. Moreover, the
definition of T shows that B is a basis for T , as claimed.
Lemma 2.4.3. A collection S of subsets of a set X is a subbasis for a topology T
on X, if and only if X = ∪S∈S S. In this case, the topology T is obtained as
T = {U ⊂ X | U is a union of finite intersections of sets from S}.
Proof. If S is a subbasis for a topology on X, then the condition X = ∪S∈S S is
clearly satisfied.
Conversely, suppose that X = ∪S∈S S and let T be defined as in the statement of
the lemma. Then ∅ ∈ T and X ∈ T by the stated condition on S. The fact that T is
closed under finite intersections and arbitrary unions, follows from the definition of T ,
30
2. TOPOLOGICAL SPACES
showing that T is a topology. Likewise, the fact S is a subbasis of T also follows from
the definition.
Lemmas 2.4.2 and 2.4.3 give us ways to define topologies on a set X by either
picking a basis or a subbasis first and letting them generate the topology. Here are a
few examples.
Example 2.4.4. The sets
B1 = {ha, bi | a, b ∈ R, a < b},
S1 = {h−∞, bi, ha, ∞i | a, b ∈ R},
are a basis and subbasis for (R, TEu ) (recall that TEu denotes the Euclidean topology
on R). Similarly, by relying on the density of the rational numbers Q in R, one can
show that
B2 = {ha, bi | a, b ∈ Q, a < b},
S2 = {h−∞, bi, ha, ∞i | a, b ∈ Q},
are also a basis and a subbasis for (R, TEu ). The key difference between the two
examples is that the sets B2 and S2 are countable sets while both of B1 and S1 are
uncountable.
Example 2.4.5. Consider the included point topology Tp on R (Example 2.2.6).
The sets {p} and {x, p}, for any x ∈ R, are open sets. However, neither {p} nor {x, p}
can be obtained as a union of other nonempty open sets and must therefore be part of
every basis B. Accordingly, every basis B for (R, Tp ) has uncountably many elements.
The smallest possible basis for (R, Tp ) is
B = {{p}, {x, p} | x ∈ R − {p}}.
Example 2.4.6. Let (X, ≤) be an ordered set and consider the order topology T≤
on X (Example 2.2.13). Then a basis and a subbasis for (X, T≤ ) are given by
B = {ha, bi | a, b ∈ X, a < b}
and
S = {La , Rb | a, b ∈ X}.
In the case of X = R and with ≤ being the usual ordering of real numbers, these two
sets agree with B1 and S1 from Example 2.4.4.
Example 2.4.7. Consider the set R and let S be the collection of subsets of R
given by
S = {x + Q+ , y + Q− | x, y ∈ R},
where Q+ and Q− are the sets of the positive and of the negative rational numbers
respectively. According to Lemma 2.4.3, S is a subbasis for a topology TS on R. An
example of an open set in this topology is h−1, 1i ∩ Q.
Just as vector spaces are divided into finite dimensional and infinite dimensional
examples according to whether or not they possess a finite basis or not, so too topological spaces can be group into two distinct categories. The first attempt to define the
2.4. BASES AND SUBBASES OF A TOPOLOGY
31
analogue of a finite dimensional vector space for topological spaces, might be to demand the existence of finite basis for the topology. However, since a topology generated
by a finite basis is by necessity finite, this definition would exclude most interesting
examples (for instance, no Euclidean space (Rn , TEu ) with n ≥ 1 admits a finite basis).
Instead, we will divide topological spaces into two categories according to whether or
not they possess a countable basis or not.
Definition 2.4.8. A topological space (X, T ) is called second countable if it possesses a countable basis B.
We should think of a second countable topological space as being “small” and of
one that isn’t second countable, as being “large”. Another measure of “size” for a
topological space we encountered previously was that of being separable (Definition
2.3.5). The next lemma explicates the relation between the two.
Lemma 2.4.9. A second countable topological space is separable. The converse is
false in general (Example 2.4.10).
Proof. Let B = {Ui | i ∈ N} be a countable basis for the second countable topological space (X, T ). Without loss of generality we can assume that Ui 6= ∅ for any
i ∈ N. Let ai ∈ Ui be any element and let A = {ai | i ∈ N} ⊂ X. Then A is a countable
set and we claim that Ā = X. For if U ⊂ X is any nonempty open set then there exists
some j ∈ N with Uj ⊂ U . But then A ∩ U is nonempty (as it contains aj ) showing that
A is dense according to Corollary 2.3.7.
Example 2.4.4 shows that the Euclidean space (R, TEu ) is second countable while
Example 2.4.5 shows that (R, Tp ) is not second countable.
Example 2.4.10. We just saw that the set of real numbers R with the included
point topology Tp isn’t a second countable space. On the other hand, let A = {p} ⊂ R
and notice that Ā = R (since closed sets in this topology are sets either not containing
p or else all of R). Thus (R, Tp ) is separable.
We finish this section by considering a local version of the notion of second countability.
Definition 2.4.11. Let (X, T ) be a topological space and let x ∈ X be a point in
X. A neighborhood basis around x is a collection Bx of neighborhoods of x such that
for every neighborhood U of x there is an element V ∈ Bx with x ∈ V ⊂ U . We say
that (X, T ) is first countable if every point x ∈ X possesses a countable neighborhood
basis.
It should be clear that a second countable space is automatically first countable.
The converse is false as the next example demonstrates.
32
2. TOPOLOGICAL SPACES
Example 2.4.12. Consider the space (R, Tp ) where Tp is the included point topology
(example 2.2.6) and let x ∈ R be any point. A neighborhood basis Bx for x is given by

; if x = p
 {{p}}
Bx =
 {{x, p}}
; if x 6= p
Thus (R, Tp ) is first countable but it isn’t second countable according to Example 2.4.5.
Example 2.4.13. The real number line R with the finite complement topology
Tf c (Example 2.2.10) is not first countable. For suppose that Bp = {Ui | i ∈ N} were a
countable neighborhood basis at some point p ∈ R, with Ui = R−{ai1 , . . . , aini } for some
ai1 , . . . , aini ∈ R − {p} and some ni ∈ N. Additionally, let A = ∪i∈N {ai1 , . . . , aini } and
note that A is a countable set, and hence that R − A is infinite. For any neighborhood
V = R − {b1 , ..., bn } of p there would have to be some i ∈ N with Ui ⊂ V , that is
with {b1 , . . . , bn } ⊂ {ai1 , . . . , aini } ⊂ A. A contradiction is obtained by simply choosing
elements bj from R − (A ∪ {p}), showing that p has no countable neighborhood basis.
Compare to Exercise 2.5.22.
Example 2.4.14. A metric space (X, d) equipped with the metric topology Td
(Example 2.2.3) is always first countable. Namely, given a point x ∈ X we can define
Bx as
Bx = {Bx (r) | r ∈ Q+ }.
where, as before, Q+ is the set of positive rational number. Clearly Bx is a countable
set and if U is any neighborhood of x, then there must exist some real number r > 0
such that Bx (r) ⊂ U . Taking any r0 ∈ h0, ri ∩ Q gives an element Bx (r0 ) ∈ Bx with
x ∈ Bx (r0 ) ⊂ Bx (r).
We saw that second countability and separability are generally two distinct measures for the size of a topology (Lemma 2.4.9). However, for metric spaces, these two
notions agree. The reason for this lies in the first countability.
Proposition 2.4.15. A metric space (X, d) equipped with the metric topology Td
is separable precisely when it is second countable.
Proof. Let A = {ai | i ∈ N} be a countable dense subset of X and for each ai ∈ A
let Bi = {Bai (r) | r ∈ Q+ } be a countable neighborhood basis for ai . Let B = ∪∞
i=1 Bi .
Since B is a countable union of countable sets, it is itself a countable set. To see that B
is a basis for (X, T ), let U ⊂ X be an open set and let x ∈ U be any point. There has to
exist a rational number r > 0 such that Bx (r) ⊂ U (be definition of Td , Example 2.2.3).
Consider the open set Bx (r/2). By Corollary 2.3.7 the intersection A ∩ Bx (r/2) has to
be nonempty and so without loss of generality we can suppose that a1 ∈ A ∩ Bx (r/2).
But then x ∈ Ba1 (r/2) since a1 ∈ Bx (r/2), and additionally Ba1 (r/2) ⊂ Bx (r) for if
y ∈ Ba1 (r/2) then d(y, x) ≤ d(y, a1 ) + d(a1 , x) < r/2 + r/2 = r. Since Ba1 (r/2) ∈ B,
we have shown that for every point x ∈ U there is a set Ux ∈ B with x ∈ Ux ⊂ U .
Therefore U = ∪x∈U Ux showing that B is a basis.
2.5. EXERCISES
33
Remark 2.4.16. It is not true in general that a separable and first countable space
is second countable (Exercise 2.5.24).
Definition 2.4.17. We say that a topological space (X, T ) is metrizable if there is
a metric on X whose associated metric topology agrees with T .
Proposition 2.4.15 and the observation from Example 2.4.14 can be used to find
first examples of non-metrizable topologies (see Section 9.2 for more on metrizability).
For instance, the space (R, Tp ) is not metrizable as according to Example 2.4.10 it is
separable but not second countable (note however that it is a first countable space
according to Example 2.4.12). Similarly, (R, Tf c ) is not metrizable as it is not first
countable according to Example 2.4.13.
Theorem 2.4.18. Let (X, TX ) be a topological space and let Y ⊂ X be a subspace
of X. If X is either second countable or first countable, then so is Y . Separability of
X does not in general imply separability of Y .
Proof. Suppose that X is second countable and let B = {Ui | i ∈ N} be a basis
for the topology on X. Then BY = {Ui ∩ Y | i ∈ N} is a countable basis for the relative
topology on Y . To see this, let V ⊂ Y be an open set in Y and let U be an open set in
X such that V = U ∩ Y . Since B is a basis for the topology on X, then U = ∪i∈M Ui
for some subset M ⊂ N. Thus V = ∪i∈M (Ui ∩ Y ) proving the claim.
The case of first countability follows analogously. For a topological space showing
that separability is not necessarily inherited by a subspace, see Example 2.4.19.
Example 2.4.19. Consider the included point topology Tp on X = R. This is a
separable space with a dense subset given by {p}. Let Y = R − {p} be given the
subspace topology. Since the closed sets in X are those not containing p and X itself,
it follows that the closed subsets of Y are all subsets of Y . Accordingly, Ā = A for
any A ⊂ Y . Thus the only dense subset of Y is Y itself. Since Y is not countable, it
follows that it is not separable.
2.5. Exercises
Rb
2.5.1. Verify axiom 1 for the metric d(f, g) = a |f (t) − g(t)| dt from Example 2.2.3.
2.5.2. Verify that each of T1 , T2 , T3 , T4 from Example 2.2.5 is a topology on X =
{1, 2, 3, 4, 5}.
2.5.3. This exercise generalizes Examples 2.2.6 and 2.2.7. Let X be a topological
space and A ⊂ X a subset of X.
(a) Define TA as the collection of subset of X given by
TA = {U ⊂ X | U is the empty set, or A ⊂ U }.
Show that TA is a topology on X (called the Included subset topology).
34
2. TOPOLOGICAL SPACES
(b) Define T A as the collection of subset of X given by
T A = {U ⊂ X | U equals X, or A ⊂ (X − U )}.
Show that T A is a topology on X (called the Excluded subset topology).
2.5.4. Let X be a set and p1 , p2 ∈ X two distinct points. Define the collection Tpp12
of subsets of X as
Tpp12 = {U ⊂ X | p1 ∈ U, or p2 ∈ (X − U )}.
Show that Tpp12 defines a topology on X (called the Included-excluded points topology).
2.5.5. On R define the set T of subsets of R as
T = {U ⊂ X | U is the empty set, or R − U is a finite union of closed intervals}.
Show that T defines a topology on R (called the Compact complement topology).
2.5.6. Show that the functions dp from Definition 1.1.1 are each a metric for any
choice of p ∈ [1, ∞i ∪ {∞}.
2.5.7. Two metrics d and d0 on a set X are called equivalent if there exist positive
real numbers c1 , c2 such that
c1 · d(x, y) ≤ d0 (x, y) ≤ c2 · d(x, y),
for all x, y ∈ X.
(a) Show that the notion of equivalence between metrics on a set X is an equivalence
relation.
(b) Show that equivalent metrics induce the same metric topology, that is show
that if d and d0 are equivalent metrics, then Td = Td0 (Example 2.2.3).
(c) Show that for any pair p, p0 ∈ [1, ∞i ∪ {∞}, the two metrics dp and dp0 (Definition 1.1.1) are equivalent metrics on Rn . Conclude that all of the metrics dp ,
p ∈ [1, ∞i ∪ {∞} induce the Euclidean topology on Rn (Hint: Prove that for
any p ∈ [1, ∞i, the double inequality
1
√
· dp (x, y) ≤ d∞ (x, y) ≤ dp (x, y)
p
n
holds for all x, y ∈ Rn , and use parts (a) and (b) of the exercise.).
2.5.8. Show that the finite complement topology Tf c and the countable complement topology Tcc from Examples 2.2.10 and 2.2.11, satisfy the axioms of a topology
(Definition 2.1.1).
2.5.9. Verify axiom 3 of a topology (Definition 2.1.1) for the Fort topology TF ,p
from Example 2.2.12.
2.5.10. Show that X belongs to the order topology T≤ from Example 2.2.13, by
showing that X = Lx ∪ Ry for a pair of elements x, y ∈ X with y < x.
2.5.11. For the topological space (X, T ), determine the interior, closure and boundary of the given subset A ⊂ X.
2.5. EXERCISES
35
(a) X = R with the Euclidean topology TEu and A = Q.
(b) X = R with the Fort topology TF ,p (Example 2.2.12) with p = 0 and A = [0, 1].
(c) X = R2 with the lexicographic order topology T≤ (Example 2.2.13) and A being
the unit square A = [0, 1] × [0, 1].
(d) X = R with the lower limit topology Tll (Example 2.2.14) and with A = h0, 1i.
2.5.12. Are the irrational numbers dense in R equipped with the
(a) Euclidean topology TEu ?
(b) Countable complement topology Tcc (Example 2.2.11)?
(c) Fort topology TF ,p (Example 2.2.12)? Does the choice of p ∈ R matter?
2.5.13. Let X be a set and let T1 , T2 be two topologies on X and assume that
T1 ≤ T2 (Definition 2.1.3). For a subset A of X, write Inti (A), Cli (A) and ∂i (A) for
the interior, closure and boundary of A with respect to Ti . Show that
(a) Int1 (A) ⊂ Int2 (A).
(b) Cl1 (A) ⊂ Cl2 (A).
2.5.14. For a topological space X and subsets A, B ⊂ X, show the following relations: (a) • Cl(A ∪ B) = Cl(A) ∪ Cl(B).
(b) • Int(A) ∪ Int(B) ⊂ Int(A ∪ B).
• Cl(A ∩ B) ⊂ Cl(A) ∩ Cl(B).
• Int(A) ∩ Int(B) = Int(A ∩ B).
• Cl(Cl(A)) = Cl(A).
• Int(Int(A)) = Int(A).
Show by example that the inclusions from the second point of (a) and first point
of (b) may be proper inclusions.
2.5.15. Let X be a topological space and let Z ⊂ Y ⊂ X be subsets. Let TY ⊂X
and TZ⊂X be the relative topologies on Y and Z respectively, induced by the topology
on X. Furthermore, let TZ⊂Y be the relative topology on Z induced by the topology
TY ⊂X on Y . Show that TZ⊂Y = TZ⊂X .
2.5.16. Let X be a topological space and Y ⊂ X a subspace. Show that a subset
A ⊂ Y is closed in Y if and only if there exists a closed subset B ⊂ X of X such that
A=B∩Y.
2.5.17. Let X be a topological space and let A, B ⊂ X be two subspaces of X with
A ⊂ B.
(a) Show that if A is open in B and B is open in X then A is also open in X.
(b) Show that if A is closed in B and B is closed in X then A is also closed in X.
(c) Find an example of A, B and X with A open in B but not in X. Similarly,
find an example with A closed in B but not in X.
2.5.18. Let X be a topological space and Y ⊂ X a subspace. Write ClX (A) and
ClY (A) for the closure of A in X and Y respectively. Similarly, write IntX (A) and
IntY (A) for the interior of A in X and Y respectively. Show that for a subset A of Y ,
the following inclusions hold:
(a) ClY (A) ⊂ ClX (A).
36
2. TOPOLOGICAL SPACES
(b) IntX (A) ⊂ IntY (A).
Show by example that both inclusions may be proper.
2.5.19. Let X be a set and let T1 , T2 be two topologies on X with T1 ≤ T2 . Show
that if (X, T1 ) is separable, then so is (X, T2 ). Conclude that (R, Tll ) and (R, Tul )
(Example 2.2.14) are separable.
2.5.20. Show that Q with the discrete topology (Example 2.2.4) is second countable.
Show on the other hand that R with the discrete topology is not second countable.
2.5.21. Let X be a set and P = {Ui ⊂ X | i ∈ I} a partition of X. Show that X
equipped with the partition topology TP (Example 2.2.9) is second countable if and
only if the indexing set I is countable.
2.5.22. Show that (R, Tcc ) from Example 2.2.11 is neither first nor second countable.
2.5.23. Show that (R, Tf c ) from Example 2.2.10 is separable.
2.5.24. Show that (R, Tll ) from Example 2.2.14, is separable and first countable but
not second countable.
2.5.25. Show that any open subset U ⊂ R with respect to the lower limit topology
Tll (Example 2.2.14), is a countable union of intervals [a, bi with a, b ∈ R, a < b.
CHAPTER 3
C
Continuous functions and convergent sequences
ontinuous functions and convergent sequences are introduced, following the
intuition from Euclidean spaces introduced in Chapter 1 (see specifically Theorems 1.3.1 and 1.3.3). Many examples of continuous functions are presented,
and some of their basic properties are illuminated in Section 3.1. Section 3.2
focuses on convergent sequences in topological spaces, and heeds special attention to
the question of uniqueness of limit. The connection between continuous functions
and convergent sequences in general topological spaces holds in weakened form, and
Theorem 3.2.11 presents the generalization of Theorem 1.1.5 from the Euclidean case.
Section 3.3 presents only one result - the “Uniform convergence theorem”(Theorem
3.3.2) - which speaks to when a sequence of continues functions that converges point
wise, defines a continuous limit function. This theorem is used several times in the
remainder of the text, the first instance of which occurs in Section 3.4. The latter
discusses “space filling curves”, that is continuous surjections from the closed interval
[0, 1] to the closed cube [0, 1]n for any n ∈ N.
3.1. Continuous functions
Definition 3.1.1. Let (X, TX ) and (Y, TY ) be topological spaces and let f : X → Y
be a function.
(a) We say that f is continuous at x ∈ X if for every neighborhood V of f (x) there
exists a neighborhood U of x such that f (U ) ⊂ V .
(b) We say that f is continuous if f −1 (V ) ∈ TX for every V ∈ TY .
We shall use the term map as synonymous with “continuous function”.
This definition of both local continuity (part (a)) and global continuity (part (b))
are directly motivated by Theorem 1.3.1. In particular, we see that with the choices
of X = Rn and Y = Rm and with both TX and TY being the Euclidean topologies,
Definition 3.1.1 agrees with the usual definition of continuity on Euclidean spaces
familiar from analysis.
Global continuity of a function between Euclidean spaces (first encountered in Definition 1.1.3) is defined to simply mean local continuity at each point of the domain. In
contrast, Definition 3.1.1 gives separate meaning to both local and global continuity.
Nevertheless, the relation between the two remains the same.
Theorem 3.1.2. A function f : X → Y between two topological spaces is continuous if and only if it is continuous at every point x ∈ X.
37
38
3. CONTINUOUS FUNCTIONS AND CONVERGENT SEQUENCES
Proof. =⇒ Suppose that f : X → Y is continuous globally and let x be a
point in X. Pick an arbitrary neighborhood V of f (x) in Y , then U = f −1 (V ) is a
neighborhood of x in X with f (U ) ⊂ V .
⇐= Suppose that f : X → Y is continuous at every point x ∈ X, let V be an open
subset of Y and set U = f −1 (V ). We’d like to show that U is open. For that purpose,
let x ∈ U be an arbitrary point (if U is the empty set then it is automatically open)
and note that V is a neighborhood of f (x). By continuity of f at x, there must exist
a neighborhood Ux of x with f (Ux ) ⊂ V . This latter relation shows that Ux ⊂ U and
consequently we obtain U = ∪x∈U Ux . Being a union of open sets, U is itself open. Example 3.1.3. Let (X, TX ) = (R, Tf c ) and let f : X → X be the function
f (x) = x2 . Let U ∈ Tf c be any nonempty open set and assume that U = R −
{x1 , . . . , xk , y1 , . . . y` } with x1 , . . . , xk < 0 and y1 , . . . , y` ≥ 0. Then f −1 (U ) = R −
{−y1 , y1 , ..., −y` , y` } which has finite complement in R. Since in addition f −1 (∅) = ∅,
we see that f is continuous.
Example 3.1.4. Let (X, TX ) = (R, Tf c ), let (Y, TY ) = (R, Tp ) with p = 0 and let
f : X → Y be again the function f (x) = x2 . Then {0, 1} ∈ Tp but f −1 ({0, 1}) =
{−1, 0, 1} ∈
/ Tf c showing that f is not continuous.
Example 3.1.5. Let X be a non-empty set and let P1 and P2 be two partitions on
X and let T1 and T2 be the two associated partition topologies on X (Example 2.2.9).
Let f : X → X be the identity function f (x) = x whose domain is equipped with T1
and codomain with T2 . Then f is continuous if and only if every element in P2 is a
union of elements from P1 .
Example 3.1.6. The constant function f : X → Y , given by f (x) = p ∈ Y for all
x ∈ X, is always continuous since f −1 (U ) is either the empty set if p ∈
/ U or all of X
if p ∈ U .
Example 3.1.7. Let X be equipped with the discrete topology, then any function
f : X → Y is continuous. Conversely, if X is given the indiscrete topology, then a
function f : X → Y to a Hausdorff space Y is continuous if and only if it is constant.
For if f were not constant then we could find two points a, b ∈ X with f (a) 6= (b). The
Hausdorff property guarantees the existence of two open and disjoint sets U, V ⊂ Y
with f (a) ∈ U and f (b) ∈ V . But then f −1 (U ) 6= ∅ (since a ∈
/ f −1 (U )) and f −1 (U ) 6= X
−1
(since b ∈
/ f (U )), showing that f is not continuous.
Example 3.1.8. Let (X, dX ) and (Y, dY ) be two metric spaces and let TdX and TdY
be the associated metric topologies. Then a function f : X → Y is continuous at x ∈ X
if and only if for every ε > 0 there exists a δ > 0 such that x0 ∈ X and dX (x, x0 ) < δ
implies that dY (f (x0 ), f (x)) < ε. Thus continuous functions between metric spaces
satisfy the familiar (ε, δ)-rule for continuity from analysis. We leave the verification
of this claim as an exercise (Exercise 3.5.7), it follows along the lines of the proof of
Theorem 1.2.1.
3.1. CONTINUOUS FUNCTIONS
39
The next theorem provides alternative definitions of continuity. Part (a) is a generalization of Theorem 1.3.2 from the Euclidean case to general topological spaces.
Theorem 3.1.9. Let f : (X, TX ) → (Y, TY ) be a function between two topological
spaces. Then f is continuous if and only if any of the mutually equivalent conditions
below is met:
(a)
(b)
(c)
(d)
(e)
f −1 (B) is a closed subset of X for any closed subset B of Y .
For all subsets B ⊂ Y , the inclusion f −1 (B) ⊂ f −1 (B̄) holds.
For all subsets A ⊂ X the inclusion f (Ā) ⊂ f (A) holds.
For all subsets B ⊂ Y the inclusion f −1 (Int(B)) ⊂ Int(f −1 (B)) holds.
Given a basis B = {Vi ⊂ Y | i ∈ I} for TY , f −1 (Vi ) is open for every i ∈ I.
Proof. We will show that properties (a) and (e) are each equivalent to f being
continuous. We will then prove the implications (a)=⇒(b)=⇒(c)=⇒(a). Showing that
(d) is equivalent to the continuity of f is left as an exercise (Exercise 3.5.6).
(a) Suppose that f is continuous and let B ⊂ Y be a closed set. Then f −1 (B) =
X −f −1 (Y −B) is also closed since Y −B is open and continuity of f forces f −1 (Y −B)
to be open.
Conversely, suppose that f has property (a) and let V ⊂ Y be any open set. Then
−1
f (V ) = X − f −1 (Y − V ) is open since Y − V and f −1 (Y − V ) are both closed, the
latter by property (a).
(a)=⇒(b) Let B be any subset of Y . Since B ⊂ B̄, we obtain f −1 (B) ⊂ f −1 (B̄).
By property (a) of f , the set f −1 (B̄) is closed but since the set f −1 (B) is the smallest
closed set containing f −1 (B) (see Lemma 2.3.2), the inclusion f −1 (B) ⊂ f −1 (B̄) is
immediate.
(b)=⇒(c) Take A ⊂ X to be any subset of X and apply property (b) of f to
B = f (A) to obtain f −1 (f (A)) ⊂ f −1 (f (A)). Since A ⊂ f −1 (f (A)) we also get
Ā ⊂ f −1 (f (A)). Applying f to the inclusion Ā ⊂ f −1 (f (A)) yields the desired result.
(c)=⇒(a) Let B be a closed subset of Y and set A = f −1 (B). Pick a point x ∈ Ā.
Then, according to property (c) we must have f (x) ∈ f (A) = B̄ = B showing that
x ∈ A. This implies that Ā = A and thus that A is closed.
(e) The necessity of property (e) for a continuous function is obvious. Suppose then
that f possesses property (e) and let V ⊂ Y be an open set. Let J ⊂ I be such that
V = ∪j∈J Vj . Then f −1 (V ) = f −1 (∪j∈J Vj ) = ∪j∈J f −1 (Vj ) showing that f −1 (V ) is a
union of open sets and therefore open.
The reader may have noticed that parts (b), (c) and (d) of Theorem 3.1.9 express
continuity in terms of an image/preimage under f , combined with a choice of taking the
interior/closure of a set, see Table 3.1. Of the four possible combinations resulting in
this manner, one is noticeably absent, namely the one involving taking images under f
combined with taking interiors of set. The next example demonstrates that this fourth
possibility cannot be added to Theorem 3.1.9.
40
3. CONTINUOUS FUNCTIONS AND CONVERGENT SEQUENCES
Preimage Image
Closure
(b)
(c)
Interior
(d)
?
Table 1. Conditions (b), (c) and (d) from Theorem 3.1.9 each involve
a combination of taking an image/preimage under f and taking the closure/interior of a set.
Example 3.1.10. Consider the function f : R2 → R given by f (x, y) = x where
each of R2 and R are equipped with the Euclidean topology. Then f is clearly continuous, but, taking A = {(x, 0) ∈ R2 | x ∈ R}, we find that
Int(f (A) = Int(R) = R
while
f (Int(A)) = f (∅) = ∅.
Thus the inclusion Int(f (A) ⊂ f (Int(A)) fails in general for continuous functions.
On the other hand, consider the function g : R → R given by g(x) = 0 and assume
that both copies of R come with the Euclidean topology. Pick B = R, then
g(Int(B)) = g(R) = {0}
while
Int(g(B)) = Int({0}) = ∅.
We see that the inclusion g(Int(B)) ⊂ Int(g(B)) also fails in general for continuous
functions (see however part (a) of Proposition 3.1.22 below).
We next single out some simple functions that are always continuous. The reader
will no doubt recognize familiar properties of continuous functions from the Euclidean
case.
Proposition 3.1.11. Let X, Y , Z be topological spaces and let f : X → Y and
g : Y → Z be continuous functions. Additionally, let A ⊂ X be any subspace of X.
(a) The inclusion function ι : A → X is continuous. In particular, the identity
function id: X → X is always continuous.
(b) The composition function g ◦ f : X → Z is continuous.
(c) The restriction function f |A : A → Y is continuous.
(d) Let Ui ⊂ X, i ∈ I, be a collection of open subsets of X such that X = ∪i∈I Ui
and let h : X → Y be a function with h|Ui : Ui → Y continuous for each i ∈ I.
Then h is continuous.
Proof. (a) For an open subset U ⊂ X, the preimage ι−1 (U ) equals U ∩ A and is
therefore open in A with respect to its relative topology.
(b) Let W be an open subset of Z. Then V = g −1 (W ) is an open subset of Y and
thus U = f −1 (V ) must be open in X. Since U = f −1 (g −1 (W ) = (g ◦ f )−1 (W ), the
claim follows.
(c) This follows from parts (a) and (b) since f |A = f ◦ ι where ι : A → X is the
inclusion map.
(d) Set hi = h|Ui and let V ⊂ Y be an open set. Then h−1 (V ) = ∪i∈I h−1
i (V ) and,
−1
since each hi (V ) must be open in Ui (in its relative topology), there must be open
3.1. CONTINUOUS FUNCTIONS
41
subsets Wi ⊂ X with h−1
i (V ) = Ui ∩ Wi . As both Ui and Wi are open in X then so is
Ui ∩ Wi showing that h−1 (V ) is a union of open sets and therefore open.
The following simple lemma will prove a useful tool in subsequent sections.
Lemma 3.1.12. Let (X, TX ) be a topological space and A, B ⊂ X two subspaces with
X = A ∪ B. Assume that either both A and B are open or that both are closed. Let
(Y, TY ) be another topological space and let f : A → Y and g : B → Y be continuous
functions which agree on A ∩ B, that is assume that f (x) = g(x) for every x ∈ A ∩ B.
Then the function h : X → Y defined by
f (x)
; x ∈ A,
h(x) =
g(x)
; x ∈ B,
is continuous.
Proof. Let V ⊂ Y be any subset of Y . Then
h−1 (V ) = (h−1 (V ) ∩ A) ∪ (h−1 (V ) ∩ B) = f −1 (V ) ∪ g −1 (V ).
If A, B are both open, choose V to be open also. Continuity of f and g shows the sets
f −1 (V ) and g −1 (V ) to be open subsets of A and B respectively. Accordingly they are
also open in X since A and B are open subsets of X (part (a) of Exercise 2.5.17). Thus
h−1 (V ) is a union of two open sets and hence open, demonstrating the continuity of h.
If A, B are both closed, pick V ⊂ Y also closed. The same reasoning as in the
previous case (and by relying on part (b) of Exercise 2.5.17) shows that h−1 (V ) is a
closed set. Continuity of h now follows from part (a) of Theorem 3.1.9.
Definition 3.1.13. A function f : X → Y between topological spaces is called
a homeomorphism if f is a continuous bijection with a continuous inverse function
f −1 : Y → X. We say that two topological spaces X and Y are homeomorphic, and
write X ∼
= Y , if there exists at least one homeomorphism f : X → Y .
A function f : X → Y is called a local homeomorphism if it is surjective and every
point x ∈ X has a neighborhood U such that f |U : U → f (U ) is a homeomorphism,
where U and f (U ) come equipped with their relative topologies inherited from X and
Y respectively. Two spaces are called locally homeomorphic if there exists at least one
local homeomorphism between them.
Remark 3.1.14. The relation of “being homeomorphic to” between topological
spaces is an equivalence relation:
1. Reflexivity: X ∼
= X. A homeomorphism from X to X is given by the identity
map (see part (a) of Proposition 3.1.11).
2. Symmetry: X ∼
= Y implies Y ∼
= X. If f : X → Y is a homeomorphism, then
−1
f : Y → X is also a homeomorphism.
3. Transitivity: X ∼
= Y and Y ∼
= Z imply X ∼
= Z. If f : X → Y and g : Y → Z
are homeomorphisms, then g ◦ f : X → Z is also a homeomorphism (part (b)
of Proposition 3.1.11).
42
3. CONTINUOUS FUNCTIONS AND CONVERGENT SEQUENCES
In a similar vein one can also prove that the relation of “being locally homeomorphic”is
an equivalence relations among topological spaces (Exercise ??).
From an abstract point of view, that is, from a point of view where we disregard
the labels attached to the points of a set, two homeomorphic spaces (X, TX ) and
(Y, TY ) are indistinguishable. For all intents and purposes, two homeomorphic spaces
can be considered as being identical, a practice that we shall adopt from hereon out,
sometimes without specifically saying so. To imprint this point further, we turn to a
simple example.
Example 3.1.15. Let the sets X = {1, 2, 3} and Y = {a, b, c} be given the two
topologies
TX = {∅, X, {1, 2}}
and
TY = {∅, Y, {b, c}}.
The reader will have no trouble recognizing that the two spaces (X, TX ) and (Y, TY )
are homeomorphic with f : X → Y given by f (1) = c, f (2) = b and f (3) = a being
one possible homeomorphism. Thus, if we disregard that the elements in X are labeled
by 1, 2, 3 and those in Y by a, b, c, the spaces (X, TX ) and (Y, TY ) become identical.
We can think of the homeomorphism f : X → Y as a “relabeling” tool, one which
takes an element x from X and gives it a new label f (x). This relabeling has to be
done with care so as to respect the topologies on X and Y , the relabeling of an open
subset from X has to result in an open subset of Y .
Example 3.1.16. The function f : (R, T ) → (R, T ) given by f (x) = x3 is a
homeomorphism for every choice of T ∈ {TEu , Tp , T p , Tf c , Tcc , TF ,p }, with the choice of
p being either 0 or ±1. Proving this amounts to showing the f and its inverse are
continuous function with respect to the listed topologies. For instance, when T = Tp ,
and V ⊂ R is a nonempty open set, then p ∈ V and so p ∈ f −1 (V ) since f (p) = p for
p ∈ {−1, 0, 1}. Thus f −1 (V ) is open and hence f continuous. The verification of the
other claims is left as an exercise (Exercise 3.5.8).
Example 3.1.17. The function f : (R, TEu ) → (R, TEu ) given by f (t) = sinh t =
− e−t ) is a homeomorphism with inverse function f −1 (t) = sinh−1 t = ln(x +
+ 1).
1 t
(e
2
√
x2
Example 3.1.18. Let (X, dX ) and (Y, dY ) be two metric spaces. A function f :
X → Y is called an isometry if dY (f (a), f (b)) = dX (a, b, ) for any pair of points
a, b ∈ X. Considering X and Y equipped with their associated metric topologies TdX
and TdY respectively, any surjective isometry f : X → Y becomes an homeomorphism.
To see this we first verify that any isometry is continuous. Let x ∈ X be any
point and let V be any neighborhood of f (x). Then there exists an r > 0 such that
Bf (x) (r) ⊂ V . But then Bx (r) is a neighborhood of x with f (Bx (r)) ⊂ V for if
a ∈ Bx (r) then dX (x, a) = r = dY (f (x), f (a)) showing that f (x) ∈ Bf (x) (r).
It remains to see that f is a bijection and that its inverse function is also continuous.
These are deferred to Exercise 3.5.11.
3.1. CONTINUOUS FUNCTIONS
43
Example 3.1.19 (Stereographic projections). Consider Rn and Rn+1 (with n ∈ N)
both equipped with their Euclidean topologies and let S n ⊂ Rn+1 be the subspace
S n = {(x1 , ...xn+1 ) ∈ Rn+1 | x21 + x22 + ... + x2n+1 = 1}.
We shall refer to S n as the n-dimensional sphere or n-sphere for short. The goal of
this example is to show that S n − {p} is homeomorphic to Rn where p ∈ S n is an
arbitrarily chosen point. For convenience, we choose p = N = (0, 0, ..., 0, 1) to be the
“north pole”of S n .
Let ϕN : (S n − {N }) → Rn be the function, referred to as stereographic projection
from the north pole N , given by
(x1 , ..., xn )
ϕN (x1 , ..., xn+1 ) =
.
1 − xn+1
Note the ϕN is well defined since the only point (x1 , ..., xn+1 ) ∈ S n with xn+1 = 1
is the north pole N . It is easy to see that ϕN is a bijection with inverse function
n
n
ϕ−1
N : R → (S − {N }) given by
ϕ−1
N (y1 , ..., yn ) =
(2y1 , 2y2 , ..., 2yn , |y|2 − 1)
,
|y|2 + 1
where |y|2 stands for y12 + y22 + ... + yn2 .
Let a = (a1 , ..., an+1 ) ∈ S n − {N } be any point and let ε > 0 be chosen at will. Let
M = 1 − an+1 and define δ as any positive real number subject to the inequality
M
εM 2
√
δ < min
,
.
2 n(1 + M ) 2
Then ϕN (Ba (δ)) ⊂ BϕN (a) (ε) showing that ϕN is continuous at a and therefore continuous everywhere since a was arbitrary. To verify this inclusion, let a0 = (a01 , ..., a0n+1 ) ∈
Ba (δ) and note that this implies that |ai − a0i | < δ for each i = 1, ..., n + 1 and
1 − a0n+1 > M/2. Thus
2
n 0
X
a
a
i
2
i
(d2 (ϕN (a), ϕN (a0 ))) =
−
1 − an+1 1 − a0 n+1
i=1
=
n
X
|ai (an+1 − a0n+1 ) + (ai − a0i )(1 − an+1 )|2
i=1
< 4nδ 2
|1 − an+1 |2 · |1 − a0n+1 |2
(1 + M )2
M4
< ε2
as needed. Showing that ϕ−1
N is continuous follows a similar line of argument and is
left as an exercise.
The geometric meaning of the function ϕN , one from which its name derives, is
that ϕN (x) is the intersection point of the line through N and x ∈ S n − {N } with the
“equatorial plane ”Rn × {0} ⊂ Rn+1 as illustrated in Figure 1.
44
3. CONTINUOUS FUNCTIONS AND CONVERGENT SEQUENCES
N
x
ϕN (x)
Figure 1. The geometric interpretation of the stereographic projection
ϕN : (S n − {N }) → Rn . Given a point x ∈ S n − {N }, ϕN (x) is obtained
as the intersection point of the line through N and x, with the equatorial
plane Rn × {0} ⊂ Rn+1 .
Similar stereographic projection can be obtained from any point p ∈ S n to Rn
where Rn is characterized as those points in Rn+1 that have dot product zero with p,
i.e. Rn ∼
= {x ∈ Rn+1 | x · p = 0}. In Exercise 3.5.9 you will be asked to find formulas
describing the stereographic projection from the south pole of S n .
With these couple of examples in hand, we turn to what is one of the main questions
in topology:
Question 3.1.20. Given two topological spaces (X, TX ) and (Y, TY ), are they homeomorphic?
Given Example 3.1.15 and the discussion preceding it, this question can be interpreted as asking whether or not a pair of topological spaces (X, TX ) and (Y, TY ) are
the same topological space, from an abstract point of view. This is a very difficult
question and at best we can hope to answer it for specific categories of topological
spaces. The chief reason for the difficulty is that a given topological space (or better
said, a given homeomorphism class of a topological space) can have many different
descriptions, and it is often far from clear when two descriptions yield homeomorphic
spaces. Nevertheless, we shall have more to say about Question 3.1.20 as we progress
in our understudying of topology. The first partial answers we shall be able to provide,
will come after we develop one of the main tools of topology - “topological invariants”in Section 4.3.
Definition 3.1.21. Let f : X → Y be a map between topological spaces. We say
that f is open if f (U ) is an open subset of Y whenever U is an open subset of X.
Similarly, we say that f is closed if f (A) is closed in Y for every choice of a closed
subset A ⊂ X.
3.2. CONVERGENT SEQUENCES
45
Note that every homeomorphism is both an open and closed map. Indeed the
condition for a bijection f to be open is the same as the condition as for f −1 to be
continuous. Part (a) of Theorem 3.1.9 implies that a homeomorphism is also a closed
map. Perhaps more surprisingly, local homeomorphisms are also open maps.
Proposition 3.1.22. Let f : X → Y be a map between topological spaces.
(a) f is an open map if and only if f (Int(A)) ⊂ Int(f (A)).
b) f is a closed map if and only if f (A) ⊂ f (Ā) for every subset A ⊂ X.
(c) If f is a local homeomorphism then f is an open map.
Proof. (a) Suppose first that f is open and let A be any subset of X. Since
Int(A) ⊂ A we find that f (Int(A)) ⊂ f (A). Since f is open, the set f (Int(A)) is
an open subset of f (A). But Int(f (A)) is the largest open subset of f (A) forcing the
inclusion f (Int(A)) ⊂ Int(f (A)).
Conversely, suppose that f (Int(A)) ⊂ Int(f (A)) holds for all A ⊂ X. Pick an open
subset U ⊂ X, then f (Int(U )) = f (U ) ⊂ Int(f (U )). On the other hand, the inclusion
Int(f (U )) ⊂ f (U ) is trivially true showing that f (U ) = Int(f (U )), thus f (U ) is open.
(b) Assume that f is closed and that A ⊂ X is any set. Then the inclusion
A ⊂ Ā implies the inclusion f (A) ⊂ f (Ā). Since f (Ā) is closed and contains f (A),
the inclusion f (A) ⊂ f (Ā) follows since the closure of f (A) is the smallest closed set
containing f (A).
On the other hand, suppose that f (A) ⊂ f (Ā) for all subsets A of X, we’d like to
show that f is closed. Thus, let B ⊂ X be any closed set, then
f (B) ⊂ f (B̄) = f (B) ⊂ f (B),
showing that f (B) = f (B). Since f (B) is closed, then so is f (B).
(c) Assume that f is a local homeomorphism. To show that f is an open map, let
U ⊂ X and set V = f (U ). We’d like to show that V is an open subset of Y . Towards
this goal, pick a point x ∈ U . Then there exists a neighborhood Ux of x such that
f |Ux : Ux → Vx , with Vx = f (Ux ), is a homeomorphism. Without loss of generality
we can assume that Ux ⊂ U for if not, we simply replace Ux by Ux ∩ U . But then
Vx ⊂ V and therefore V = ∪x∈U Vx . Being a union of open sets, V is forced to be open
itself.
We will encounter open and closed maps again in subsequent chapters. Open maps
will play an important role in the study of quotient spaces (Chapter 8). Chapter 6 will
give a nice criterion for a map to be closed in terms of compactness of the domain of
the map (Corollary 6.1.10).
3.2. Convergent sequences
In this section we examine convergent sequences in topological spaces, first in the
their own right and then with regards to their relation to continuous functions. Familiar
properties of sequences from Euclidean spaces no longer hold in this more general
setting, perhaps most striking being the non-uniqueness phenomena for the limit of a
46
3. CONTINUOUS FUNCTIONS AND CONVERGENT SEQUENCES
convergent sequence. A characterization of continuity in terms of sequences is given in
Theorem 3.2.11.
Definition 3.2.1. Let (X, T ) be a topological space and {xk }k ⊂ X a sequence.
We say that the sequence {xk }k converges to x ∈ X, and write limk→∞ xk = x or just
lim xk = x, if for every neighborhood U of x there exists a k0 ∈ N such that for all
k ≥ k0 we obtain xk ∈ U .
Note that if one takes (X, TX ) = (Rn , TEu ) then this definition agrees with the
familiar definition of continuity in Euclidean spaces, compare to Theorem 1.3.3.
Example 3.2.2. Let X = R be equipped with the partition topology TP associated
to the partition
P = {[4a − 2, 4a + 2i | a ∈ Z}.
Then the sequence {xk }k with xk = (−1)k converges to any point x ∈ [−2, 2i since this
is the smallest non-empty open set containing both 1 and −1.
Example 3.2.3. If X is equipped with the indiscrete topology, then any sequence
in X is convergent and its limit is any point in X. Conversely, if X is given the discrete
topology, then a sequence {xk }k is convergent to x ∈ X if and only if xk = x for all
sufficiently large k.
Example 3.2.4. Let {xk }k ⊂ R be the sequence xk = 1/k. Determine whether or
not {xk }k converges in (R, T ) for the various choices of T below. If it converges, find
its limits.
(a) T = TEu . In the Euclidean topology, convergence in the sense of Definition 3.2.1
is the same as in the usual sense from analysis (Definition 1.1.4 and Theorem
1.3.3). Thus limk→∞ xk = 0.
(b) T = Tp . In the particular point topology Tp , the sequence {xk }k does not
converge at all. To see this, note that any sequence {yk }k converging to x 6= p
has to equal x or p for all sufficiently large indices k. This follows from the
fact that {x, p} is a neighborhood of x. On the other hand, any sequence {yk }k
converging to p has to be constant, and equal to p, for all sufficiently large k
since {p} is a neighborhood of p. Since xk = 1/k is not of these two types of
sequences, it cannot be convergent.
(c) T = T p . Here {xk }k is convergent and lim xk = p. Namely, given any sequence
{yk }k ⊂ R that converges to y 6= p, we must have yk = y for all large enough
k since {y} is a neighborhood of y. If {yk }k converges to p then, since the
only neighborhood of p is R, all that needs to be true is that yk ∈ R, a trivial
condition. Thus we see that in this example the limit of {xk }k exists and is
unique, though it may not be zero.
(d) T = T {p1 ,...,pn } . The excluded points topology T {p1 ,...,pn } is a variation of the
excluded point topology T p . A subset U ⊂ R belongs to T {p1 ,...,pn } if either
U = R or else neither of the points p1 , ..., pn lies in U (compare to part (b) of
Exercise 2.5.3). Arguing as in the previous case, we find that limk→∞ xk = pi
3.2. CONVERGENT SEQUENCES
47
for all i = 1, 2, ..., n. Thus, the sequence xk is convergent but has n different
limits.
(d) T = Tindis . As we already saw in Example 3.2.3, in the indiscrete topology we
find that limk→∞ xk = x for every x ∈ R showing that {xk }k is convergent and
has infinitely many limits.
Example 3.2.5. Let X = R be given the finite complement topology Tf c from
Example 2.2.10. Let {xk }k ⊂ R be a sequence with xk 6= x` whenever k 6= `. Then
{xk }k converges to any point x ∈ R since every neighborhood of x contains all but
finitely many elements of {x1 , x2 , x3 , ...}.
Quite to the contrary, if we equip R with the countable complement topology Tcc
instead (Example 2.2.11), then {xk }k is not a convergent sequence anymore. For if
we had lim xk = x we would need xk = x for all sufficiently large indices k (Exercise
3.5.15). As we assumed that xk 6= x` whenever k 6= `, the latter condition fails.
As the examples above show, in topological spaces limits of sequences may not be
unique and moreover, their number can vary wildly. Thus, in passing from our accustomed setting of Euclidean spaces to the much more general ambience of topological
spaces, we have paid the price of loosing a familiar and desirable property: uniqueness
of the limit of a convergent sequence {xk }k ⊂ Rn . However, with just a minor sacrifice
of generality, we can regain this attribute. To see how, here is first a definition.
Definition 3.2.6. A topological space X is called Hausdorff if every two points
have disjoint neighborhoods. Said differently, we require that for each pair of points
a, b ∈ X, a 6= b, there exist open sets Ua , Ub ⊆ X with a ∈ Ua , b ∈ Ub and Ua ∩ Ub = ∅.
Theorem 3.2.7. Let X be a topological space.
(a) If X Hausdorff space and xk ∈ X is a convergent sequence, then the limit lim xk
is unique.
(b) If X is fist countable and has the property that every convergent sequence xk ∈ X
has a unique limit, then X is Hausdorff.
Proof. (a) Suppose that there are two (or more) limits for {xk }k , say a and
b. Since X is Hausdorff, we can find disjoint neighborhoods Ua and Ub of a and b
respectively. Let ka ∈ N be such that xk ∈ Ua for all k ≥ ka and kb ∈ N have the
property that xk ∈ Ub for all k ≥ kb . Then for all k ≥ max{ka , kb } we have that
xk ∈ Ua ∩ Ub , a contradiction since Ua ∩ Ub = ∅.
(b) Given two arbitrary points a, b ∈ X with a 6= b, we need to find two disjoint
open sets of which one contains a and the other contains b. Let Ba = {Uia ⊂ X | i ∈ N}
and Bb = {Uib ⊂ X | i ∈ N} be countable neighborhood bases at a and b respectively.
We define new open sets Via and Vib as
Vka = U1a ∩ U2a ∩ ... ∩ Uka
and
Vkb = U1b ∩ U2b ∩ ... ∩ Ukb .
These sets are open since they are finite intersections of open sets. Furthermore, notice
that Vja ⊂ Uia and Vjb ⊂ Uib for every j ≥ i. Clearly, each Via is a neighborhood of a
and each Vib is a neighborhood of b.
48
3. CONTINUOUS FUNCTIONS AND CONVERGENT SEQUENCES
If Vka ∩ Vkb = ∅ for some k, we are done. So suppose instead that Vka ∩ Vkb 6= ∅ for all
k ∈ N. Let xk ∈ Vka ∩ Vkb be an arbitrary point. This yields a sequence in X which we
claim converges to a. To see this, let U be any neighborhood of a and find an Ukaa ∈ Ba ,
ka ∈ N, such that Ukaa ⊂ U . Then Vka ⊆ U for every k ≥ ka , in particular xk ∈ U for
every k ≥ ka , showing that lim xk = a. Repeating this same argument for b shows also
that lim xk = b. This is a contradiction since by assumption all convergent sequences
in X have a unique limit. We are thus forced to conclude that there is some k ∈ N for
which Vka ∩ Vkb = ∅, giving us disjoint neighborhoods of a and b, as needed.
The first countability condition from part (b) of Theorem 3.2.7 is necessary and
cannot be weakened as the next example shows.
Example 3.2.8. Consider X = R equipped with the countable complement topology. We claim that every convergent sequence {xk }k ⊂ X has a unique limit. Suppose
not, that is suppose that lim xk = a and lim xk = b with a 6= b. Let Ua be the open set
Ua = R − {xi | i ∈ N and xi 6= a}.
Clearly a ∈ Ua and so there must be some na ∈ N such that xn ∈ Ua for all n ≥ na .
But then xn = a for all n ≥ na since xn ∈ Ua ∩ {xi | i ∈ N} = {a}. A similar argument
shows that for some nb ∈ N all xn = b for n ≥ nb . But then xn = a and xn = b
for n ≥ max{na , nb } which is impossible since a 6= b. On the other hand, X is not
Hausdorff since every two non-empty open sets have nontrivial intersection. Compare
to Exercise 2.5.22.
Recall that a subset A ⊆ Rn (with the Euclidean topology) is closed if and only
if it contains the limits of all its convergent sequences (this was the original definition
of a closed subset of Rn , Definition 1.2.1). This characterization of closed sets is only
partly true in general topological spaces.
Theorem 3.2.9. Let X be a topological space and A a subset of X.
(a) If A is a closed then it contains the limits of all its convergent sequences.
(b) If X is first countable and A contains the limits of all of its convergent sequences, then A is closed.
Proof. (a) Let {xk }k ⊂ A be a sequence with limit x. If x ∈
/ A then x ∈ X − A
which is an open set. By convergence of {xk }k there must be some k0 ∈ N such that
xk ∈ X − A for all k ≥ k0 . This is impossible since xk ∈ A and A ∩ (X − A) = ∅.
(b) We will show that A is closed by exhibiting that A = Ā. Suppose this were not
true. Then Ā − A would be nonempty. Let x ∈ Ā − A and let Bx = {Ui ⊆ X | i ∈ N}
be a countable basis at x and define Vj as
Vj = U1 ∩ U2 ∩ ... ∩ Uj .
Note that the sets Vj are open, that the inclusion Vj ⊂ Ui holds for all j ≥ i and that
each Vj contains x. We claim next that each set Vi ∩ A must be non-empty. For if not
then Ā − Vj would be a closed set smaller than Ā and containing A, a contradiction.
Thus we can pick an element xk ∈ Vk ∩ A. But now {xk }k must converge to x since
3.2. CONVERGENT SEQUENCES
49
if V is any neighborhood of x then there must be an index k0 such that x ∈ Uk0 ⊂ V
and thus xk ∈ Vk ⊂ Uk0 ⊂ V for all k ≥ k0 . Since xk ∈ A and lim xk = x we conclude
that x ∈ A. Therefor A = Ā.
That the first countability condition from part (b) of the preceding theorem cannot
be dropped, is illustrated by the next example.
Example 3.2.10. Let X = R be equipped with the countable complement topology
and let A ⊆ X be the set A = X − {0}. Notice that A is not closed (since X − A = {0}
is not open). But A contains the limits of all of its convergent subsequences. To see
this we only need to show that no sequence {xk }k ⊂ A can converge to 0. This is
easy to see since the set U = X − {x1 , x2 , ...} is an open set which contains zero but
no element of the sequence xn . Thus A contains the limits of all of its convergent
sequences. Note that this, in conjunction with Theorem 3.2.9, shows that (R, Tcc )
cannot be first countable and hence neither second countable.
We conclude this section by examining the relation between continuous functions
and convergent sequences. As the reader may guess by now, with the assumption of
first countability, the relation between the two is just as in Euclidean space (Theorem
1.3.3).
Theorem 3.2.11. Let f : X → Y be a function between two topological spaces.
(a) If f is continuous at x ∈ X and if {xk }k ⊂ X is a convergent sequence with
lim xk = x, then {f (xk )}k ⊂ Y is also a convergent sequence with lim f (xk ) =
f (x).
(b) If X is first countable and f has the property that limf (xk ) = f (x) for all
sequences {xk }k ⊂ X converging to x, then f is continuous at x.
Proof. (a) Suppose that f is continuous at x and let {xk }k ⊂ X be a convergent
sequence with limit x. To show that {yk }k is a convergent sequence with limit y = f (x),
where yk = f (xk ), let V ⊆ Y be an arbitrary neighborhood of y. Since f is continuous,
the set U = f −1 (V ) ⊆ X is a neighborhood of x and by convergence of {xk }k to x,
there is some k0 ∈ N such that xk ∈ U for all k ≥ k0 . But then yk ∈ V for all k ≥ k0 ,
showing that lim f (xk ) = f (x).
(b) Let V be a neighborhood of f (x) and set U = f −1 (V ). We seek to show that
U contains a neighborhood of x. Suppose that this were not so. In that case, let
Bx = {Ui | i ∈ N} be a neighborhood basis around x and define, as in the proofs of
Theorem 3.2.7 and 3.2.9, the open sets Vj , j ∈ N as Vj = U1 ∩ · · · ∩ Uj . By assumption,
the sets (X − U ) ∩ Vk are nonempty and so we can pick points xk ∈ (X − U ) ∩ Vk .
The sequence {xk }k is easily seen to converge to x and thus by assumption {f (xk )}k
must converge to f (x). This however is impossible since V is open and f (x) ∈ V while
f (xk ) ∈
/ V . Consequently, U must contain a neighborhood Ux of x and hence f is
continuous at x.
50
3. CONTINUOUS FUNCTIONS AND CONVERGENT SEQUENCES
3.3. Uniform convergence of functions
This section is largely of a utilitarian nature in that its main result, Theorem 3.3.2,
shall be used as the basis for a number of arguments later in the book. The first
instance of this is in the next section where Theorem 3.3.2 is used in the construction
of space filling curves. For another instance, see Section 9.1 on the Tietze extension
theorem. The reader may skip this section on a first pass, and return to it as needed.
The main question we want to address here is the following: Suppose that (X, TX )
is a topological space and gm : X → R (where R is given the Euclidean topology) is a
sequence of functions indexed by m ∈ N such that for every fixed x ∈ X, the sequence
{gm (x)}m converges. In such a situation we can define a new function g : X → R by
setting g(x) = limm→∞ gm (x). We this in mind we ask: Under what conditions on the
functions gm is the function g continuous?
It is easy to see that even if all gm are continuous, the limiting function g need not
be. For example, taking X = [0, 1] and letting gm (x) = xm , we obtain
g(x) =
0
1
;
;
x ∈ [0, 1i,
x = 1,
which is not continuous. The extra condition needed to ensure continuity of g is one
that compares how gm (x) converges to g(x) for various x ∈ X.
Definition 3.3.1. Let (X, TX ) be a topological space and let gm : X → R, m ∈ N
be a sequence of continuous functions. We say that the sequence gm converges uniformly
to the function g : X → R if for every ε > 0 there exists an index m0 ∈ N so that all
m ≥ m0 imply the inequality |g(x) − gm (x)| < ε for all choices of x ∈ X.
Saying that gm (x) converges to g(x) for every x ∈ X means that for every fixed
x ∈ X and for every ε > 0, there is some integer m0 (x) ∈ N with |gm (x) − g(x)| < ε
for all m ≥ m0 (x). The extra condition incorporated in the definition of uniform
convergence of gm to g is that m0 (x) can be chosen to be the same for all x ∈ X (i.e.
we can take m0 (x) to be a constant function). With this understood, here is the main
result of this section.
Theorem 3.3.2. If gm : X → R, m ∈ N is a sequence of continuous functions
converging uniformly to the function g : X → R, then g is also continuous.
Proof. Let x0 ∈ X be any point and let V ⊂ R be any neighborhood of g(x0 ).
Find an ε > 0 so that hg(x0 ) − ε, g(x0 ) + εi ⊂ V . By the uniform convergence
property, we can find an index m0 so that for all m ≥ m0 and all x ∈ X we obtain
|g(x) − gm (x)| < 3ε . Pick any index m ≥ m0 , then by continuity of gm , there must exist
a neighborhood U of x0 such that gm (U ) ⊂ hgm (x0 ) − 3ε , gm (x0 ) + 3ε i. Given any x ∈ U ,
3.4. SPACE FILLING CURVES
51
we find
|g(x) − g(x0 )| = |g(x) − gm (x) + gm (x) − gm (x0 ) + gm (x0 ) − g(x0 )|
≤ |g(x) − gm (x)| + |gm (x) − gm (x0 )| + |gm (x0 ) − g(x0 )|
ε ε ε
< + + = ε,
3 3 3
showing that g(U ) ⊂ hg(x0 ) − ε, g(x0 ) + εi ⊂ V . From this, continuity of g at x0 follows
and since x0 ∈ X was arbitrary, g is continuous.
3.4. Space filling curves
In Section 3.1 we asked one of the main questions in topology, the questions of
finding effective ways to distinguish topological spaces up to homeomorphism (Question
3.1.20). While this question is much too difficult to address in full generality, one can
hope for a partial answer by restricting the choice of topological spaces considered. In
this section we ask such a more specific question, one which we shall return to on a
number of occasions later in the text and one which, unlike Question 3.1.20, we shall
be able to answer (the curious reader may skip ahead to Corollaries 7.2.8 and Theorem
??).
Question 3.4.1. For which values of m, n ∈ N can the n-dimensional Euclidean
space (Rn , TEu ) be homeomorphic to the m-dimensional Euclidean space (Rm , TEu )?
The goal of this section is to convince the reader that this question may have an
answer different from the expected one of “Only if n = m.” We shall attempt to do so
by demonstrating that there exist continuous surjective functions from the unit segment
[0, 1] ⊂ R to the unit n-dimensional cube [0, 1]n ⊂ Rn , each equipped with the relative
Euclidean topology. Such functions are referred to as space filling curves. Hence, if
a curve can fill out n-dimensional space, albeit only an n-dimensional cube, perhaps
there also exist homeomorphisms from R to Rn opening the possibility for unexpected
answers to Question 3.4.1. We’d like to emphasize that we do not claim that answers
besides m = n exist, we merely point out that one needs to exercise caution before
jumping to conclusions with regards to homeomorphisms between spaces of different
dimensions.
While there is a multitude of space filling curves, we shall utilize a particular construction and define only one such curve in detail. However, many others can be
obtained by varying the pattern of the curve below, some additional patterns are given
in Figure 5.
We will start by constructing a continuous surjection f : [0, 1] → [0, 1]2 . This is
done by defining piecewise linear and continuous functions fm : [0, 1] → [0, 1]2 for each
m ∈ N and then letting f (x) = limm→∞ fm (x). Of course, we will have to argue that
this latter limit exists and that the thus obtained function f is indeed continuous and
surjective. The case of surjective maps f : [0, 1] → [0, 1]n for n ≥ 3 is then easily
obtained from this example.
52
3. CONTINUOUS FUNCTIONS AND CONVERGENT SEQUENCES
To define the functions fm : [0, 1] → [0, 1]2 , m ∈ N, we proceed as follows: Start
with m = 1 and divide [0, 1] into four segements Ii1 , i = 1, ..., 4 of equal length. Thus
I11 = [0, 41 ],
I21 = [ 41 , 24 ],
I11 = [ 42 , 34 ],
I11 = [ 34 , 44 ].
Similarly divide the unit square [0, 1]2 into four congruent squares Ji1 , i = 1, ..., 4 by
defining
J11 = [0, 21 ] × [0, 12 ],
J21 = [0, 12 ] × [ 12 , 1],
J31 = [ 12 , 1] × [ 12 , 1],
J41 = [ 21 , 1] × [0, 21 ].
We then let f1 : [0, 1] → [0, 1]2 be any continuous function that maps Ii1 into Ji1 ,
I11 I21 I31 I41
I12
f1
......
2
I16
f2
J21
J31
J11
J41
J62
J72
2
J10
2
J11
J52
J82
J92
2
J12
J42
J32
2
J14
2
J13
J12
J22
2
J15
2
J16
Figure 2. The subdivision of [0, 1] and [0, 1]2 for the cases of m =
1, 2. The arrows inside of the two copies of [0, 1]2 indicate our choice
of ordering of the subdivision squares Jim (m = 1, 2) with respect to i.
Similar orderings are to be applied to Jim for m ≥ 3.
i = 1, . . . , 4 and whose image in [0, 1]2 looks as in Figure 4. We refer to the image of
f1 as the pattern of the curve to be constructed. Note that Im(f1 ) ∩ Ji1 6= ∅ for every
i = 1, . . . , 4.
To obtain f2 : [0, 1] → [0, 1]2 we proceed in much the same manner. This time we
divide [0, 1] into the 16 congruent segments Ii2 , i = 1, ..., 16 with Ii2 = [ i−1
, i ] and
16 16
likewise, we divide [0, 1]2 into 16 congruent squares Ji2 , i = 1, ..., 16. These subdivision
squares of [0, 1]2 are chosen as
J12 = [0, 14 ] × [0, 14 ],
J22 = [ 41 , 24 ] × [0, 14 ],
J32 = [ 41 , 24 ] × [ 41 , 24 ],
J42 = [0, 14 ] × [ 41 , 24 ],
J52 = [0, 14 ] × [ 24 , 43 ],
J62 = [0, 14 ] × [ 34 , 44 ],
J72 = [ 41 , 24 ] × [ 43 , 44 ],
J82 = [ 41 , 24 ] × [ 42 , 34 ],
2
2
2
J92 = [ 24 , 43 ] × [ 24 , 43 ], J10
= [ 42 , 34 ] × [ 34 , 44 ], J11
= [ 34 , 44 ] × [ 43 , 44 ], J12
= [ 34 , 44 ] × [ 42 , 34 ],
2
2
2
2
J13
= [ 34 , 44 ] × [ 14 , 42 ], J14
= [ 42 , 34 ] × [ 14 , 24 ], J15
= [ 24 , 43 ] × [0, 14 ], J16
= [ 34 , 44 ] × [0, 14 ].
3.4. SPACE FILLING CURVES
53
Figure 2 illustrates our choices of Ii1 , Ii2 , Ji1 and Ji2 for the various indices i. The
function f2 : [0, 1] → [0, 1]2 is chosen so that f2 (Ii2 ) ⊂ Ji2 for each i = 1, ..., 16 and so
that its image looks as in Figure 4. Notice again that our definition of f2 implies that
Im(f2 ) ∩ Ji2 6= ∅ for all i = 1, ..., 16.
From here on we proceed inductively on m to define fm : [0, 1] → [0, 1]2 , m ≥ 3.
We do so by subdividing [0, 1] into 4m congruent intervals Iim , i = 1, ..., 4m and by
subdividing [0, 1]2 into 4m congruent squares Jim , i = 1, ..., 4m . The function fm is then
chosen subject to the condition fm (Iim ) ⊂ Jim for each i = 1, ..., 4m and is otherwise
defined as in Figure 3. The images of the thus constructed functions fm are shown in
Figure 4 for m = 1, 2, 3, 4.
The two key features of the functions fm : [0, 1] → [0, 1]2 obvious from their construction, are:
(a) If fm (x) ∈ Jim for some i = 1, ..., 4m , then fm+k (x) ∈ Jim for all k ∈ N.
(3.1) (b) For every m ∈ N and for every i = 1, ..., 4m we obtain Im(fm ) ∩ Jim 6= ∅.
For the remainder of this section, we shall refer to these simply as “Properties (a) and
(b)”. The attentive reader will notice that the proofs of the Lemmas 3.4.2, 3.4.3 and
3.4.4 below, will be carried out only by relying on these two properties rather than on
any other specifics about the construction of fm .
Lemma 3.4.2. For every x ∈ [0, 1] and for every k ∈ N we obtain the inequality
√
2
d2 (fm (x), fm+k (x)) ≤ m ,
2
where d2 is the Euclidean metric from (1.1). Consequently, the sequence {fm (x)}m is
convergent for every x ∈ [0, 1].
Proof. The stated inequality for d√2 (fm (x), fm+k (x)) follows from Property (a) of
(3.1) and from the observation that 2m2 is the length of the diagonal of Jim and is
therefore the largest distance between any two points in Jim . This inequality shows
that the sequence {fm (x)}m is a Cauchy sequence in [0, 1]2 for any choice of x ∈ [0, 1]
and is thus convergent by Theorem 1.1.6.
We are now in the position to define our space filling function f : [0, 1] → [0, 1]2 as
(3.2)
f (x) = lim fm (x).
m→∞
√
Applying the limit as k goes to infinity to the inequality d2 (fm (x), fm+k (x)) ≤ 2m2 from
Lemma 3.4.2, we obtain the useful inequality
√
2
(3.3)
d2 (fm (x), f (x)) ≤ m
∀x ∈ [0, 1].
2
This step is justified by continuity of the function x 7→ d2 (y, x) for any choice of y ∈ R2
(Exercise 3.5.10) and by Theorem 1.1.5. Our first order of business is to show that f
is continuous.
3. CONTINUOUS FUNCTIONS AND CONVERGENT SEQUENCES
Im(fm )
Im(fm )
Im(fm )
Im(fm )
54
Im(fm )
Im(fm+1 )
Figure 3. The function fm+1 : [0, 1] → [0, 1]2 is constructed from the
function fm : [0, 1] → [0, 1]2 as follows: Take 4 copies of the unit square
[0, 1]2 and shrink each by a factor of 2. Arrange these 4 “shrunken
squares”into a new square [0, 1]2 as indicated on the right figure above.
Thus, 2 of the shrunken squares are placed side by side to fill the upper
half of [0, 1]2 while the remaining 2 shrunken squares are first rotated
by ±90◦ and the placed into the bottom half of [0, 1]2 . The image of
fm+1 in [0, 1]2 is then the union of the 4 shrunken copies of the image of
fm , connected suitably by small straight line segments (indicated in red)
so as to make fm+1 continuous. This figure also explains the indexing
of the subdivision squares Jim+1 , i = 1, ..., 4m+1 . Namely, the first 4m
of these squares are the 4m squares Jim from the previous stage of the
construction, shrunk by a factor of 2 and placed in the lower left corner
of [0, 1]2 . One repeats this process three more time placing three more
copies of the 4m squares Jim into the upper left, upper right and finally
lower right corner of [0, 1]2 . The indices i in Jim are increased by 4m ,
2 · 4m and 3 · 4m respectively, in this process.
Lemma 3.4.3. The function f : [0, 1] → [0, 1]2 constructed in (3.2), is continuous.
Proof. Inequality (3.3) shows the sequence {fm }m to converge uniformly to f .
The claim of the lemma is now immediate from Theorem 3.3.2.
Next we check that f is surjective.
Lemma 3.4.4. The function f : [0, 1] → [0, 1]2 defined in (3.2), is surjective.
3.4. SPACE FILLING CURVES
Im(f1 )
Im(f2 )
Im(f3 )
55
Im(f4 )
Figure 4. The images of the functions fm : [0, 1] → [0, 1]2 for m =
1, 2, 3, 4. These are constructed following the recipe described in Figure
3.
Proof. Let y0 ∈ [0, 1]2 be any point, we need to show that there exists a point
m
x0 ∈ [0, 1] with f (x0 ) = y0 . Since ∪4i=1 Jim = [0, 1]2 we can find for each m ∈ N, an index
im ∈ {1, ..., 4m } such that y0 ∈ Jimm . Let xm ∈ [0, 1] be any point with the property
that such a point xm has to exist by Property (b) from (3.1).
that fm (xm ) ∈ Jimm . Note
√
2
Then d(fm (xm ), y0 ) ≤ 2m .
Since the sequence {xm }m ⊂ [0, 1] is bounded, it must have a convergent subsequence {xk(m) }m with m 7→ k(m) an increasing function. Let x0 be the limit of
{xk(m) }m . By continuity of f , the sequence {f (xk(m) }m has to converge to f (x0 ) (Theorem 1.1.5). Thus, for any ε > 0 there exists an m0 ∈ N such that m ≥ m0 implies
that
√
2
ε
d(f (xk(m) ), f (x0 )) < 3 . For this same ε, let us chose a value m1 ∈ N such that
< 3ε .
2m1
√
2
Since k(m) ≥ m, choosing any m ≥ m1 will imply that the inequality 2k(m)
< 3ε still
holds. Putting these observations together, we find that for any m ≥ max{m0 , m1 },
the next inequality holds:
d(f (x0 ), y0 ) ≤ d(f (x0 ), f (xk(m) )) + d(f (xk(m) ), fk(m) (xk(m) )) + d(fk(m) (xk(m) ), y0 ),
ε ε ε
< + + ,
3 3 3
= ε.
The bound on the second term on the right hand side of the first line, comes from
Equation (3.3). Since d(f (x0 ), y0 ) < ε for every ε > 0, it must be that d(f (x0 ), y0 ) = 0
and therefore that f (x0 ) = y0 as desired.
Having constructed a continuous surjection f from [0, 1] onto [0, 1]2 , we can easily
obtain surjections from [0, 1] to [0, 1]n . To do this, let us write f (t) = (φ1 (t), φ2 (t))
with φ1 , φ2 : [0, 1] → [0, 1] being the coordinate functions of f . Let Fn : [0, 1] → [0, 1]n
be defined inductively as F2 = f and
(3.4)
Fn+1 (t) = (φ1 (t), Fn (φ2 (t)))
Since φ1 and φ2 are continuous, Fn+1 is also continuous given that Fn is continuous. An
induction on n then shows that all Fn are continuous since F2 is. Also by induction on
56
3. CONTINUOUS FUNCTIONS AND CONVERGENT SEQUENCES
n one can show that Fn surjects onto [0, 1]n . For example, when n = 3, let (t1 , t2 , t3 ) ∈
[0, 1]3 be any point. By surjectivity of F2 , we can find an s ∈ [0, 1] such that F2 (s) =
(t2 , t3 ). By surjectivity of (φ1 , φ2 ), we can then find a t ∈ [0, 1] such that (φ1 (t), φ2 (t)) =
(t1 , s). But then
F3 (t) = (φ1 (t), F2 (φ2 (t))) = (t1 , F2 (s)) = (t1 , t2 , t3 ),
showing that F3 is surjective. The case of n ≥ 3 is handled similarly and is left as an
exercise.
Our construction of the functions fm : [0, 1] → [0, 1]2 , m ∈ N started by choosing
the image of f1 in [0, 1]2 more or less arbitrary, and then constructing the functions
fm , m ≥ 2 according to the recipe in Figure 3. In this sense the functions fm , m ≥ 2
are pinned down by our choice of f1 , that is they are determined by the choice of
the pattern of the space filling curve. As long as the construction of the functions fm
satisfies Properties (a) and (b) from (3.1), we have complete freedom for choosing our
pattern for f1 . By varying the pattern, we may construct a multitude of space filling
curves. Figure 5 shows several alternative patterns.
3.5. Exercises
3.5.1. For a function f : X → Y between sets X and Y , and for any family of
subsets Vi ⊂ Y , i ∈ I, show that
(a) f −1 (∪i∈I Vi ) = ∪i∈I f −1 (Vi ).
(b) f −1 (∩i∈I Vi ) = ∩i∈I f −1 (Vi ).
3.5.2. Verify whether or not the given function f : R → R is continuous. Here the
domain of f is equipped with the topology T1 and the codomain with the topology T2 .
(a) f (x) = sin x, T1 = Tcc , T2 = TEu .
(b) f (x) = cos x, T1 = Tp with p = 0, T2 = Tp with p = 1.
(c) f (x) = x3 , T1 = T p with p = 0, T2 = TF ,p with p = 0.
3.5.3. For λ ∈ R let fλ : R → R be the function fλ (x) = x + λ. If both the domain
and codomain of fλ are equipped with the finite complement topology, for which λ ∈ R
is fλ continuous?
3.5.4. Consider the open ball Bx (r) ⊂ Rn equipped with the relative Euclidean
topology inherited from Rn . Show that for every choice of x ∈ Rn and radius r > 0,
Bx (r) is homeomorphic to (Rn , TEu ). (Hint: Show firstly that the tangent function
tan : h− π2 , π2 i → R is a homeomorphism and use this to solve the exercise for n = 1.
Apply this in conjunction with polar coordinates to the case of n ≥ 2.)
3.5.5. Construct an explicit homeomorphism f : Dn → [−1, 1]n . Here
Dn = {x ∈ Rn+1 | x21 + · · · + x2n ≤ 1},
[−1, 1]n = {x ∈ Rn | − 1 ≤ xi ≤ 1, i = 1, . . . , n},
and each of the spaces comes with the relative Euclidean topology.
3.5. EXERCISES
57
Figure 5. Each of the three rows of figures depicts the images of f1 , f2
and f3 for the choice of pattern given by the first picture in the row.
3.5.6. Show that a function f : X → Y is continuous if and only if f −1 (Int(B)) ⊂
Int(f −1 (B)) for every subset B ⊂ Y .
3.5.7. Verify the claim from Example 3.1.8. Specifically, let (X, dX ) and (Y, dY ) be
two metric spaces and let X and Y be equipped with the associated metric topologies.
Show that a function f : X → Y is continuous at x ∈ X if and only if for every ε > 0
there exists a δ > 0 such that dx (x0 , x) < δ implies dY (f (x0 ), f (x)) < ε.
58
3. CONTINUOUS FUNCTIONS AND CONVERGENT SEQUENCES
3.5.8. Verify the claims from Example 3.1.16: Show that f : (R, T ) → (R, T ) given
by f (x) = x3 is a homeomorphism for any choice of T ∈ {TEu , Tp , T p , Tf c , Tcc , TF ,p }
with p being either 0 or ±1.
3.5.9. Find a formula describing the stereographic projection ϕS : S n − {S} → Rn
from the “south pole S = (0, 0, . . . , 0, −1)”of the unit n-sphere in Rn+1 (see Example
3.1.19). Verify that it defines a homeomorphism between S n − {S} and Rn .
3.5.10. Let (X, d) be a metric space and Td the associated metric topology on X.
Show that for any fixed choice of y ∈ X, the function fy : X → [0, ∞i given by
fy (x) = d(x, y) is a continuous function. Here [0, ∞i carries the relative Euclidean
topology.
3.5.11. Let (X, dX ) and (Y, dY ) be two metric spaces. Recall from Example 3.1.18
that a function f : X → Y is called an isometry if dY (f (a), f (b)) = dX (a, b) for any
pair of points a, b ∈ X. To complete the claims from Example 3.1.18, prove that
(a) Every isometry is an injective function.
(b) Show that the inverse of a bijective isometry is again an isometry.
3.5.12. Show that the notions of “open map”and “closed map”(Definition 3.1.21)
are independent. Specifically, consider the functions f : R → R and g : R2 → R given
by
f (x) = x2 , and g(x, y) = x.
Show that f is a closed map that is not open, and show that g is an open map that is
not closed.
3.5.13. Show by example that a local homeomorphism is not necessarily a closed
map (compare to part (c) of Proposition 3.1.22).
3.5.14. Determine whether or not the sequence {xk }k ⊂ R converges when R is
given the topology T . If the sequence converges, find all of its limits.
(a) xk = (−1)k and T = TP with the partition P given by P = {[n, n + 1i | n ∈ Z}.
(b) xk = sin k and T = Tll .’
(c) xk = ln k and T is the topology generated by the subbases
S = {h−n, ni | n ∈ N}.
3.5.15. Consider R with the countable complement topology Tcc from Example
2.2.11. Show that a sequence {xk }k ⊂ R converges to x ∈ R if and only if there is an
integer k0 such that xk = x for all k ≥ k0 .
3.5.16. Show by example that the condition of first countability cannot be dropped
from part (b) of Theorem 3.2.11. Specifically, find an example of a function f : X → Y
that is not continuous at a point x ∈ X, but has the property that if {xk }k ⊂ X is a
sequence converging to x, then the sequence {f (xk )}k converges to f (x).
3.5. EXERCISES
59
3.5.17. Show that the maps Fn : [0, 1] → [0, 1]n defined in (3.4), are surjective for
each n ≥ 4. The case of n = 3 was addressed in the paragraph following equation
(3.4).
CHAPTER 4
S
Separation axioms
eparation axioms, introduced in this section, serve as a measure of how fine
or how coarse a topology T is in relation to the set X on which it is defined.
Section 4.1 formally introduces the 7 types of separation axioms and studies
some of the their most basic properties. Section 4.2 is devoted to examples,
and examines the topological spaces from Section 2.2 in light of the Separation Axioms.
Section 4.3 introduces the all important notion of a topological invariant, and give
several examples. Section 4.4 provides an illustration of how topological invariants are
typically applied to tell apart topological spaces up to homemorphism. The chapter
concludes with a proof of the Urysohn Lemma in Section 4.5.
4.1. Degrees of separation
In Chapter 2 we encountered two notions of “size” of a topology T on a set X, those
of separability (Definition 2.3.5) and second countability (Definition 2.4.8). These two
yardsticks measure the “size”of the topology T , without any regard to X itself. The
notions of “size”of a topological space that we shall encounter in this section, employs
an interplay between X and T . For instance, we shall ask whether T is “large enough”
so that any pair of distinct points in X has a pair of disjoint neighborhoods. If this is
not the case, we can think of T as being “small” in relation to X.
Definition 4.1.1. We say that two subsets A and B of the topological space (X, T )
are separated if Ā ∩ B = ∅ = A ∩ B̄.
For example, closed and disjoint sets A, B ⊂ X are separated.
Definition 4.1.2. Let (X, T ) be a topological space. Then
(a) X is called T0 or Kolmogorov if for any two points x, y ∈ X there exists an open
set U ⊂ X such that either x ∈ U and y ∈
/ U , or y ∈ U and x ∈
/ U.
(b) X is called T1 or Fréchet if for any two points x, y ∈ X there exist open sets
Ux , Uy ⊂ X such that x ∈ Ux − Uy and y ∈ Uy − Ux .
(c) X is called T2 or Hausdorff if for any two points x, y ∈ X there exist disjoint
open sets Ux , Uy ⊂ X such that x ∈ Ux and y ∈ Uy (Definition 3.2.6).
(d) X is called T3 if for any closed set A ⊂ X and any point x ∈ X − A there are
disjoint open sets UA , Ux ⊂ X such that A ⊂ UA and x ∈ Ux .
(e) X is called T3 1 if for any closed set A ⊂ X and any point x ∈ X − A there
2
exists a continuous function f : X → [0, 1] with f |A ≡ 0 and f (x) = 1.
61
62
4. SEPARATION AXIOMS
(f) X is called T4 if for any two disjoint closed sets A, B ⊂ X there are disjoint
open sets UA , UB ⊂ X with A ⊂ UA and B ⊂ UB .
(g) X is called T5 if for any two separated sets A, B ⊂ X (see definition 4.1.1 above)
there are disjoint open sets UA , UB ⊂ X with A ⊂ UA and B ⊂ UB .
From the properties T0 − T5 , we derive three additional definitions. Namely, we shall
that that
(h)
(i)
(j)
(k)
X
X
X
X
is
is
is
is
Regular if it is both T0 and T3 .
Completely Regular if it is both T0 and T3 1 .
2
Normal if it is both T1 and T4 .
Completely Normal if it is both T1 and T5 .
The various properties T0 −T5 are not entirely unrelated. Their various dependencies
are easily unraveled with the help of the next lemma.
Lemma 4.1.3. Let X be a T1 space, then for any point x ∈ X, the set {x} is closed.
Proof. For any point y ∈ X − {x} let Uy be a neighborhood of y not containing
x. Then
[
X − {x} =
Uy ,
y∈X−{x}
showing that X − {x} is a union of open sets and hence open.
This lemma and Definition 4.1.2 yield all but one of the following implications:
(4.1)
T1
T2
T1 and T3
T3 1
2
T1 and T4
T1 and T4
T5
=⇒
=⇒
=⇒
=⇒
=⇒
=⇒
=⇒
T0 ,
T1 ,
T2 ,
T3 ,
T3 1 (Corollary 4.5.4),
2
T3 ,
T4 .
For example, to see that T3 1 implies T3 , let X be a T3 1 space and let A ⊂ X be a
2
2
closed subset and x ∈ X − A any point. The T3 1 property guarantees the existence of
2
a map f : X → [0, 1] with f |A ≡ 0 and f (x) = 1. Then the sets UA = f −1 ([0, 1/2i) and
Ux = f −1 (h1/2, 1]) are open and disjoint sets containing A and x respectively, showing
that X is a T3 space.
The next theorem shows that the separation properties Ti are inherited by a subspace, with some subtleties in the case of T4 and T5 .
Theorem 4.1.4. Let X be a topological space and let Y ⊂ X be a subspace of X.
(a) If X is Ti for some i ∈ {0, 1, 2, 3, 3 12 } then so is Y .
(b) If Y is closed and X is Ti for some i ∈ {4, 5}, then Y is also Ti .
4.2. EXAMPLES
63
Proof. (a) For the cases of i = 0, 1, 2 let x, y ∈ Y be two distinct points. If Ux
and Uy are the neighborhoods of x and y in X showing up in the definition of Ti for
X, then Ux ∩ Y and Uy ∩ Y are neighborhoods of x and y in Y that can be taken to
verify property Ti for Y itself.
For property T3 , let A ⊂ Y be a closed subset and let x ∈ Y − A be any point.
Then there is a closed subset A0 ⊂ X with A = A0 ∩ Y and clearly x ∈
/ A0 . Let UA0 and
0
Ux0 be disjoint neighborhoods of A and x in X, then UA = Y ∩ UA0 and Ux = Ux0 ∩ Y
are disjoint neighborhoods of A and x in Y .
For property T3 1 , let again A ⊂ Y be a closed subset of Y and x ∈ Y −A any point.
2
Let A0 ⊂ X be a closed subset of X with A = A0 ∩ Y and note that x ∈
/ A0 . By the
T3 1 property of X, there exists a function f 0 : X → [0, 1] with f 0 |A0 ≡ 0 and f 0 (x) = 1.
2
But then f : Y → [0, 1] given by f = f 0 |Y is continuous (part (c) of Theorem 3.1.11)
and f |A ≡ 0 and f (x) = 1. This shows that Y also has the T3 1 property.
2
(b) For property T4 , let A, B ⊂ Y be two disjoint closed subsets of Y . Since Y is
assumed to be closed in X, we see that A, B are also closed subsets of X (Exercise
2.5.17). But then they can be separated by neighborhoods U, V in X and are therefore
separated by U ∩ Y and V ∩ Y in Y . A similar argument works for T5 , the details are
left to the reader (Exercise 4.6.1).
The hypothesis of Y being closed in part (b) of Theorem 4.1.4 cannot be removed,
as the next example shows.
Example 4.1.5. Consider X = R and for a choice of two distinct points p1 , p2 ∈ X,
let T be the topology
T = {U ⊂ X | U = ∅, or U = X, or p1 ∈ U and p2 ∈ (X − U )}.
Then X does not contain disjoint closed non-empty subsets rendering (X, T ) trivially
a T4 space. Let Y be the subspace of X given by Y = X − {p2 }, note that Y is not
closed in X. In Y there are many disjoint non-empty closed subsets, but there are no
disjoint non-empty open subsets, showing that Y cannot be T4 .
The following statement is a direct consequence of Theorem 4.1.4.
Corollary 4.1.6. A subspace of a (completely) regular space is (completely) regular. A closed subspace of a normal space is a normal space.
4.2. Examples
Example 4.2.1. A space that is T3 , T4 , T5 but not T0 , T1 , T2 . Let X = R be given
the partition topology TP associated to the partition P = {[2a, 2a + 2i | a ∈ Z}. This
space is not a Ti space for i = 0, 1, 2.
To see this, let x = 0 and y = 1. The smallest open set containing x is [0, 2i which
incidentally also contains y. Thus (R, TP ) is not T0 and hence neither T1 or T2 .
Observe that in this example the closed subsets and R agree with the open subsets.
Thus, given any closed subset A ⊂ R and a point x ∈ R − A, the open sets UA and Ux
64
4. SEPARATION AXIOMS
from separation axiom T3 can be taken to be UA = A and Ux = X − A. Accordingly,
(R, TP ) is T3 .
Let A, B ⊂ R be two separated sets. Then Ā ∩ B = ∅ implies that B ⊂ R − Ā. But
X − Ā is open and thus also closed in the partition topology and thus B̄ =⊂ R − Ā.
Given this, the two open sets UA and UB from the separation axiom for T5 can be taken
to be UA = Ā and UB = R − Ā. Recall that the T5 axiom implies the T4 axiom, cf.
(4.1).
Example 4.2.2. A space that is T0 , T1 but not T2 , T3 , T4 , T5 . Let X = R have the
finite complement topology and let x, y ∈ R be any two distinct points. Then the two
sets Ux = X − {y} and Uy = X − {x} are neighborhoods of x and y respectively with
y∈
/ Ux and x ∈
/ Uy . Thus (R, Tf c ) is T1 and hence T0 . It is not however T2 − T5 since
no disjoint non-empty open subsets of R exist in this topology, while there is a plenty
of non-empty closed and disjoint subsets of R.
Example 4.2.3. A space that is T0 but not T1 − T5 . Consider the included point
topology Tp on R with p = 0. There are no disjoint non-empty open sets in this
topology and so (R, Tp ) cannot be T2 − T5 . It is also not T1 since every non-empty open
set has to contain p. It is however T0 since given any two points x, y, with x 6= p, the
set {y, p} (or simply {p} if y = p) is an open set containing y but not x.
Example 4.2.4. A space that is T0 , T4 , T5 but not T1 , T2 , T3 . Consider X = R with
the excluded point topology T p with p = 0. Given any two distinct points x, y ∈ R,
one of them, say x, is not equal to 0. But then the set Ux = {x} is a neighborhood of
x not containing y. Thus (R, T p ) is T0 . However, it is not T1 (and thus also not T2 )
since taking y = 0, the only open set containing y is R which of course also contains x.
The closed subsets of R in this topology are those containing p = 0. Thus, given
any closed set A ⊂ R, the only neighborhood of A is R itself. This immediately shows
that (R, T p ) cannot be T3 . It also shows that it is T4 since no pair of non-empty disjoint
closed subsets of R exist.
Finally, let A, B ⊂ R be two separated sets. Note that Ā = A∪{0} and B̄ = B∪{0}.
Then Ā ∩ B = ∅ implies that 0 ∈
/ B while A ∩ B̄ = ∅ implies that 0 ∈
/ A But then A
and B are both already open sets and so we can set UA = A and UB = B showing that
(R, T p ) is T5 .
Example 4.2.5. A space that is T0 , . . . , T5 . Let X = R be given the lower limit
topology Tll . Then X is Ti for every i = 0, 1, ..., 5. To see this is suffices to show that it
is T2 and T5 since according to (4.1), all the other separation axioms follow from these.
To see that X is Hausdorff, pick two points x, y ∈ X with x < y. Define
Ux = [x, yi
and
Uy = [y, y + 1i
These are disjoint neighborhoods of x and y respectively showing that (R, Tll ) is T2 .
To verify separation axiom T5 , let A, B ⊆ X be two separated sets. Then X − B̄
is an open set and so we can, for each a ∈ A ⊂ X − B̄, find an xa ∈ X such that
[a, xa i ⊂ X − B̄ (since the half-open intervals are a basis for the topology). Define
4.3. TOPOLOGICAL INVARIANTS
65
the open set UA as UA = ∪a∈A [a, xa i and in a similar vein define also UB . These
are open sets (being a union of open sets) and contain A and B respectively. If we
had UA ∩ UB 6= ∅, then there would have to be some a ∈ A and some b ∈ B so that
[a, xa i ∩ [b, xb i 6= ∅. Suppose that a < b (the case b < a is treated analogously), then
b ∈ [a, xa i (since b is the smallest element of [b, xb i) but this would yield a contradiction
since b ∈ B and [a, xa i ⊆ X − B̄. Therefore UA and UB must be disjoint, showing that
(R, Tll ) is T5 .
Example 4.2.6. Metric spaces are always T0 , . . . , T5 . Let (X, d) be a metric space
and consider X equipped with the associated metric topology Td .
(X, Td ) is always T2 (and hence also T0 and T1 ) since, given any two distinct points
x, y ∈ X, the sets Bx (r/2) and By (r/2) are disjoint neighborhoods of x and y, where
r = d(x, y).
To see that it is also T3 − T5 , it suffices to show that it is T5 , see (4.1). Thus let
A, B ⊂ X be two separated set. Then for every point a ∈ A there exists some ra > 0
such that Ba (ra ) ∩ B = ∅. For if not, part (c) of Lemma 2.3.2 would imply that a ∈ B̄
contradicting the assumption A ∩ B̄ = ∅. Let Ua = Ba (ra /2) and define UA = ∪a∈A Ua .
In a similar vein define also rb , Ub and UB . It is clear that UA and UB are open sets
containing A and B respectively. It remains to show that they are disjoint. If not, we
could find a point p ∈ UA ∩ UB . But then there would be points a ∈ A and b ∈ B such
that p ∈ Ua ∩ Ub showing further that
d(a, b) ≤ d(a, p) + d(p, b) < ra /2 + rb /2.
For concreteness, suppose that ra ≤ rb . Then the above equality implies that d(a, b) <
rb showing that a ∈ Ub , a contradiction. We arrive at UA ∩ UB = ∅ verifying property
T5 for (X, Td ).
The results of this example can be used to show that certain topologies are not
metrizable (Definition 2.4.17). It was already pointed out that the included point
topology Tp and the finite complements topology Tf c on R are not metrizable (see
paragraph following Definition 2.4.17). Example 4.2.4 shows that the excluded point
topology T p is not metrizable either.
Example 4.2.7. (Rn , TEu ) is T0 , . . . , T5 . Recall that the Euclidean topology TEu
on Rn is a metric topology associated to the Euclidean metric d2 . Thus Example 4.2.6
shows that (Rn , TEu ) is T0 , . . . , T5 .
4.3. Topological invariants
We have thus far expanded most of our efforts towards the study of properties
of topological spaces and continuous functions between them. These are necessary
prerequisites for the more involved concepts studied in later chapters and will be used
throughout the remainder of the book. Before delving into these, we pause here to
reflect on a question that lies at the heart of most efforts in topology.
Question 4.3.1. Given a pair of topological spaces X and Y , are they homeomorphic?
66
4. SEPARATION AXIOMS
This is a difficult question in general, indeed its full resolution shall forever stay out
of grasp. Indeed, even restricting the choices of X and Y to particularly nice classes
of topological spaces, for example manifolds, this question is undecidable.
Nevertheless, we can try to obtain partial answers to Question 4.3.1. The main tool
in doing so is the use of topological invariants which we now introduce and then apply
to Question 4.3.1.
Definition 4.3.2. A topological invariant is a function
O : { Set of topological spaces } −→ { Set of objects },
such that if X and Y are homeomorphic than O(X) = O(Y ). The “Set of objects”
can be any set, examples include the set of integers, the set (ring) of polynomials over
a commutative ring, the set of (isomorphism classes of ) groups, etc.
Remark 4.3.3. It is important to note that the implication
X∼
=⇒
O(X) = O(Y ),
=Y
is one directional. It is generally not possible to conclude that X and Y are homeomorphic given O(X) = O(Y ). With regards to Question 4.3.1, topological invariants
are most frequently used to tell that two spaces X and Y are not homeomorphic, as
would follow from O(X) 6= O(Y ), rather than to conclude that they are.
Many of the properties of topological spaces encountered in the present chapter and
in Chapter 2, are topological invariants. We present one example with full details and
state others with less elaboration.
Example 4.3.4. Consider the set
V={
Is separable. ,
Is not separable. }
Thus V is a set of two elements, each being a phrase. We use V to define the function
O : { Set of topological spaces } → V,
by setting
O(X) =

 Is separable.

Is not separable.
;
If X is a separable topological space.
;
If X is not a separable topological space.
We claim that O is a topological invariant. To see this, suppose that X is a separable
space with A ⊂ X being a countable dense subset of X. Let f : X → Y be a
homeomorphism and let B = f (A) ⊂ Y . Since f is a bijection, it preserves cardinality
of sets and so B is also countable. To see that B is also dense, let V ⊂ Y be any open
set. Then U = f −1 (V ) is an open subset of X and so A ∩ U 6= ∅ according to Corollary
2.3.7. But then B ∩ V is also non-empty and so, again by Corollary 2.3.7, B must be
dense.
Similarly, if Y is separable then the same argument shows that X is also separable.
In conclusion, given two homeomorphic spaces X and Y , either both or neither is
separable and therefore O is a topological invariant.
4.4. A FIRST APPLICATION OF TOPOLOGICAL INVARIANTS
67
The preceding example of a topological invariant incorporated a greater than needed
number of details, in an effort to strictly adhere to Definition 4.3.2. Going forward,
we shall take a more relaxed approach. For instance, we may say that the “property
of being separable”is a topological invariant or, even shorter, that “separability”is a
topological invariant.
Example 4.3.5. Consider the two topological spaces: the Euclidean topology
(R, TEu ) and the excluded point topology (R, T p ) (Example 2.2.7). Then
O(R, TEu ) = Is separable.
O(R, T p ) = Is not separable.
The first claim is courtesy of Example 2.3.8 and the second follows from Example
2.4.10. We conclude that (R, TEu ) is not homeomorphic to (R, T p ).
Theorem 4.3.6. The properties of a topological space of being
· First Countable,
· Second Countable,
· Separable,
· Being Ti for i ∈ {0, 1, 2, 3, 3 12 , 4, 5},
are all topological invariants.
Proof. The ten proofs of the topological invariance of the ten properties listed are
very similar (the proof for separability was already given in Example 4.3.4). We shall
thus only give here the proof of the topological invariance of the Hausdorff property
T2 and defer there rest of the to the exercises (Exercise 4.6.5).
Let X be a Hausdorff space and f : X → Y be a homeomorphism. To show that
Y is also Hausdorff, pick a pair of distinct points y1 , y2 ∈ Y and let xi = f −1 (yi ) for
i = 1, 2. Then x1 and x2 are distinct points (since f is a bijection) and by the Hausdorff
property of X, there exist disjoint neighborhoods U1 and U2 of x1 and x2 respectively.
But then Vi = f (Ui ) for i = 1, 2 are disjoint neighborhoods of y1 and y2 , proving the
claim.
Example 4.3.7. Let TP be the partition topology on R associated to the partition
P = {[2a, 2a+2i | a ∈ Z}. Then the spaces (R, TP ) and (R, Tf c ) are not homeomorphic,
for the former is not T1 (Example 4.2.1) while the latter is (Example 4.2.2).
Example 4.3.8. The spaces (R, TEu ) and (R, T p ) are not homeomorphic as the
Euclidean line is a T3 space (Example 4.2.7) while (R, T p ) is not (Example 4.2.4).
4.4. A first application of topological invariants
Before proceeding with our exposition, we pause and give an application of the
topological invariants introduced in Section 4.3 to address Question 4.3.1 for the various
example of topological spaces presented in Section 2.2. Thus, we shall consider the
topological spaces obtained by endowing R with one of the nine topologies below, and
ask which, if any, resulting spaces are homeomorphic.
1. TEu - The Euclidean topology from Example 2.2.1.
68
4. SEPARATION AXIOMS
2.
3.
4.
5.
6.
7.
8.
9.
Tdis - The discrete topology from Example 2.2.4.
Tp - The included point topology from Example 2.2.6.
T p - The excluded point topology from Example 2.2.7.
Tf c - The finite complement topology from Example 2.2.10.
Tcc - The countable complement topology from Example 2.2.11.
TF,p - The Fort topology from Example 2.2.12.
Tll - The lower limit topology from Example 2.2.14.
Tul - The upper limit topology from Example 2.2.14.
We organize our results in the table below. Given the row containing the topology
T1 and the column containing T2 , the corresponding entry in the table is either a
checkmark X or a crossmark ×, indicating whether (R, T1 ) and (R, T2 ) are or are not
homeomorphic. The table entries are clearly symmetric about the subdiagonal and the
subdiagonal itself consists of X’s as each space is homeomorphic to itself.
TEu Tdis
TEu X
×
Tdis ×
X
Tp
×
×
p
T
×
×
Tf c
×
×
Tcc
×
×
TF,p ×
×
Tll
×
×
Tul
×
×
Table 1. Homeomorphism
troduced in Section 2.2.
T p T p Tf c
× × ×
× × ×
X × ×
× X ×
× × X
× × ×
× × ×
× × ×
× × ×
relationships
Tcc TF,p
× ×
× ×
× ×
× ×
× ×
X ×
× X
× ×
× ×
among the
Tll Tul
× ×
× ×
× ×
× ×
× ×
× ×
× ×
X X
X X
topologies on R in-
We explain some of the table entries and leave some for the reader to work out.
This is an instructive exercise that we highly recommend (Exercise 4.6.8). The red
checkmarks linking the finite complement with the countable complement topology is
there to emphasize that all of the topological invariants introduced in Theorem 4.3.6
agree for (R, Tf c ) and (R, Tcc ). To tell that these two spaces are not homeomorphic,
another topological invariant is used, see Exercise 4.6.6. We start by considering the
second countability invariant.
Lemma 4.4.1. Of the nine topologies TEu , Tdis , Tp , T p , Tf c , Tcc , TF,p , Tll and Tul
on R, only the Euclidean topology TEu is second countable.
Proof. The Euclidean topology TEu is second countable according to Example
2.4.4, while the included point topology Tp is not second countable according to Example 2.4.5. Example 2.3.9 shows that (R, Tdis ) is not separable and so by Lemma 2.4.9
it is not second countable either. Here are the remaining six cases:
4.4. A FIRST APPLICATION OF TOPOLOGICAL INVARIANTS
69
T p Every set {x} ⊂ R with x 6= p is open and cannot be written as the union of
any other non-empty open sets. Thus every basis B for (R, T p ) must at least
contain {x}, x ∈ R − {p} which is an uncountable set. Consequently, every
basis is uncountable.
Tf c Suppose a countable basis B = {Ui | i ∈ N} existed. Then we could pick an
arbitrary point x ∈ R and consider the countable non-empty set Bx = {Ui ∈
B | x ∈ Ui }. Since for every y 6= x, the set R − {y} is open, it must be a union of
elements from B showing that for every y 6= x there exists an element Ui ∈ Bx
with y ∈
/ Ui . But then ∩Ui ∈Bx Ui = {x} so that
R − {x} = R − ∩Ui ∈Bx Ui = ∪Ui ∈Bx (R − Ui ).
Tcc
TF,p
Tll
Tul
We have thus managed to write the uncountable set R − {x} as a countable
union of the finite sets R − Ui , a contradiction. Accordingly, (R, Tf c ) cannot be
second countable.
Since Tf c ⊂ Tcc and Tf c is not second countable then neither is Tcc .
The Fort topology is finer than the excluded point topology which we already
saw was not second countable, showing that neither is TF ,p .
Suppose there were a countable basis B = {Ui | i ∈ N} for Tll . Since each
open set in the lower limit topology is a countable union of intervals [a, bi
(Exercise 2.5.25), we can assume without loss of generality that Ui = [ai , bi i
for some ai , bi ∈ R, ai < bi . Since R is uncountable we can find a point
a ∈ R − {a1 , a2 , a3 , . . . }. Then the set [a, a + 1i cannot be gotten as a union of
the basis elements. Thus (R, Tll ) is not second countable.
Same argument as for the lower limit topology Tll .
The results of the preceding lemma suffice to fill out the first row and column of
Table 1. We next turn to the Hausdorff property.
Lemma 4.4.2. Of the nine topologies TEu , Tdis , Tp , T p , Tf c , Tcc , TF,p , Tll , Tul on R,
the Hausdorff ones are precisely TEu , Tdis , TF,p , Tll and Tul .
Proof. Examples 4.2.2, 4.2.3 and 4.2.4 show that (R, Tf c ), (R, Tp ) and (R, T p )
are not Hausdorff, while Examples 4.2.5 and 4.2.7 show that (R, Tll ) and (R, TEu ) are
Hausdorff. A straightforward adaption of Example 4.2.5 shows that (R, Tul ) is also
Hausdorff. Here are the remaining three cases.
Tdis This topology is Hausdorff since {x} and {y} are disjoint neighborhood for a
pair of distinct points x, y ∈ R.
Tcc As in the finite complement topology, no disjoint non-empty open sets exist.
TF,p Let x, y ∈ R be two distinct points with x 6= p. Then {x} and R − {x} are
two disjoint neighborhoods of x and y respectively, showing (R, TF,p ) to be
Hausdorff.
Lemma 4.4.2 contributes an additional 32 crossmarks × to Table 1.
70
4. SEPARATION AXIOMS
Lemma 4.4.3. The function f : R → R given by f (t) = −t is a homeomorphism
from (R, Tll ) to (R, Tul ).
Proof. This follows immediately from part (e) of Theorem 3.1.9 and the observation that f −1 (ha, b]) = [−b, −ai and f ([c, di) = h−d, −c].
The preceding lemma accounts for the two off-diagonal checkmarks X in Table 1.
The remaining 22 entries are deferred to Exercise 4.6.8.
4.5. The Urysohn Lemma
The following lemma gives an alternative characterization of T4 spaces. A similar
characterization can also be given for T3 spaces, see Lemma 6.3.5.
Lemma 4.5.1. A topological space (X, TX ) is T4 if and only if for every closed subset
A ⊂ X and any open set U ⊂ X containing A, there exists an open set V ⊂ X such
that
A ⊂ V ⊂ V̄ ⊂ U.
Proof. =⇒ Suppose firstly that X is T4 and let A ⊂ U ⊂ X be sets of which A
is closed and U open. Consider the two closed and disjoint sets A, X − U ⊂ X. By the
T4 property of X, there exist disjoint open subsets WA and WX−U containing A and
X − U respectively. But then
WA ∩ WX−U = ∅
=⇒
WA ⊂ X − WX−U ,
and since X − WX−U is a closed set, we see that W A ⊂ X − WX−U . On the other
hand, since X − U ⊂ WX−U then X − WX−U ⊂ U . Thus V = WA has the property
that A ⊂ V ⊂ V̄ ⊂ U , as desired.
⇐= Suppose that X has the property that for every pair A ⊂ U ⊂ X with A
closed and U open there is an open set V with A ⊂ V ⊂ V̄ ⊂ U and let A, B ⊂ X
be two disjoint and closed subsets of X. Then U = X − B contains A and so by
assumption there is an open set V with A ⊂ V ⊂ V̄ ⊂ X − B. Then V and X − V̄ are
disjoint neighborhoods of A and B respectively, showing X to be T4 .
Theorem 4.5.2 (The Urysohn Lemma). Let (X, TX ) be a T4 space and let A, B ⊂
X be two closed and disjoint subsets of X. Then there exists a continuous function
f : X → R such that f |A ≡ 0 and f |B ≡ 1. The codomain of f carries the Euclidean
topology.
Proof. Let Q[0,1] = Q ∩ [0, 1] be the set of rational numbers between 0 and 1. This
is a countable set and so we can list its elements as
Q[0,1] = {q0 , q1 , q2 , ...}
Any such enumeration of
choose q0 = 0 and q1 = 1.
0
1
1
Q[0,1] =
,
,
,
1
1
2
the elements of Q[0,1] will do but for convenience we shall
An example is
1 2
1 3
1 2 3 4
1 5
1 2 3 4 5 6
, ,
, ,
, , , ,
, ,
, , , , , ,... ,
3 3
4 4
5 5 5 5
6 6
7 7 7 7 7 7
4.5. THE URYSOHN LEMMA
71
in which, for instance, q5 = 1/4 and q9 = 3/5. We note that there is no relation
between the size of the indices i, j ∈ N ∪ {0} and the size of qi and qj .
In the first step of the proof, our strategy shall be to construct an open set Uq for
every point q ∈ Q[0,1] , subject to the conditions
(a) A ⊂ Uq0 and B ⊂ X − Ūq1 .
(b) If x, y ∈ Q[0,1] with x < y then Ūx ⊂ Uy .
In the second step of the proof, these sets Uq are then used to construct the function
f : X → R whose existence is claimed by the theorem.
Step 1 We shall define the sets Uqi , qi ∈ Q[0,1] by explicitly defining Uq1 and Uq0
and then proceed by induction for i ≥ 2. Thus, define Uq1 to be X − B and note
that A ⊂ Uq1 . According to Lemma 4.5.1, there is an open set Uq0 ⊂ X such that
A ⊂ Uq0 ⊂ Ūq0 ⊂ Uq1 . With these definitions of Uq0 and Uq1 , properties (a) and (b)
above are met.
Proceeding by induction, suppose that the sets Uq0 , , ..., Uqn , subject to conditions
(a) and (b) above, have been defined. Consider the finite set {q0 , q1 , ..., qn , qn+1 } and
suppose that, with respect to the ordering of rational numbers, qn+1 fits in between qi
and qj with i, j ∈ {0, 1, ..., n}. Thus, the indices i, j are chosen so that qi < qn+1 < qj
and so that for any other index k ∈ {0, ..., n} − {i, j} we either get qk < qi or qj < qk .
Such indices i, j must exist since q0 = 0 and q1 = 1 and qn+1 ∈ [0, 1]. By our inductive
assumption, the inclusion Ūqi ⊂ Uqj must hold and by Lemma 4.5.1, an open set
Uqn+1 ⊂ X with Ūqi ⊂ Uqn+1 ⊂ Ūqn+1 ⊂ Uqj must exist. This last double inclusion shows
that properties (a) and (b) continue to hold for the sets Uq0 , ..., Uqn , Uqn+1 , completing
the induction.
Step 2 Using the sets Uqi , qi ∈ Q[0,1] defined in Step 1, we are now in the position
to define a function f : X → R as
1
; x ∈ B,
f (x) =
inf{q ∈ Q[0,1] | x ∈ Uq }
; x∈
/ B.
Note that f is well defined since if x is not in B, then the set {q ∈ Q[0,1] | x ∈ Uq }
contains 0 and is therefore non-empty. We will prove that f satisfies the conditions
asserted by the theorem.
If x ∈ A then x ∈ U0 (and x ∈
/ B) so that inf{q ∈ Q[0,1] | x ∈ Uq } = 0 and hence
f |A ≡ 0. By definition of f , the relation f |B ≡ 1 is evident. To see that f is continuous
we will show that for every x ∈ X and every open set V ⊂ R containing f (x) ∈ V , there
exists a neighborhood U of x in X with f (U ) ⊂ V . This will show that f is continuous
at x and since x is arbitrary, we will have shown f to be continuous. Observe that
• If x ∈ Uqi then f (x) ≤ qi .
• If x ∈
/ Ūqj then qj ≤ f (x).
The first point above is obvious while the second point follows from the observation
that if f (x) = y < qj then x ∈ Uq for some rational number q in hy, qj i, an impossibility
since if x ∈
/ Uqj while Uq ⊂ Uqj .
72
4. SEPARATION AXIOMS
B
A
| {z }
U0
|
{z
U2
}
3
|
{z
U3
}
4
|
{z
U1
}
Figure 1. A depiction of some of the sets Uq , q ∈ Q ∩ [0, 1] used in the
proof of the Urysohn lemma. The values of q = 23 , 43 have been picked
randomly, however, there is less freedom in choosing the sets U0 and U1
and they have been portrayed as chosen in the proof of the Urysohn
lemma.
Pick any x ∈ X and let V ⊂ R be an open set with f (x) ∈ V . Find an ε > 0
so that hf (x) − ε, f (x) + εi ⊂ V . If f (x) = 1, find a rational number qj ∈ h1 − ε, 1],
then U = X − Ūqj has the property f (U ) ⊂ V . If f (x) = 0, find a rational number
qi ∈ [0, εi and set U = Uqi , then f (U ) ⊂ V . Finally, if 0 < f (x) < 1, find rational
numbers qi , qj with f (x) − ε < qj < x < qi < f (x) + ε and set U = Uqi − Ūqj . Then
again f (U ) ⊂ V .
Before considering some of the consequences of Urysohn’s Lemma, we’ll work through
an example.
Example 4.5.3. Consider the Euclidean line (R, TEu ) and let A, B ⊂ R be the
closed subsets A = [−2, −1] and B = [1, 2]. We shall follow, step by step, the proof of
Urysohn’s Lemma (Theorem 4.5.2) to construct a continuous function f : R → R with
4.5. THE URYSOHN LEMMA
73
f |A ≡ 0 and f |B ≡ 1. The sets Uq with q ∈ [0, 1] ∩ Q can be chosen as
U1 = h−∞, 1i ∪ h2, ∞i
Uq = h−∞, qi,
and
∀q ∈ [0, 1i ∩ Q.
Recall that the set U1 was defined as R − B in the proof of Theorem 4.5.2, while the
sets Uq for q 6= 1 had to be chosen so that A ⊂ U0 and whenever q1 < q2 then Ūq1 ⊂ Uq2 .
With our choices as above, Ūq1 = h−∞, q1 ] which is clearly contained in h−∞, q2 i if
q1 < q2 . The function f : R → [0, 1] is then defined as
1
; x∈B
f (x) =
inf{q ∈ Q[0,1] | x ∈ Uq }
; x∈
/B
showing that f takes on the form

 0
x
f (x) =
 1
x≤0
0≤x≤1
; 1≤x
;
;
This is clearly a continuous and A ⊂ f −1 (0) and B ⊂ f −1 (1), see Figure 2 for a graph
of f .
1
−2
|
{z
A
f
−1
1
}
|
2
{z
B
}
Figure 2. The graph of the function f (in green) constructed using the
proof of the Urysohn Lemma for the subsets A = [−2, −1] (in blue) and
B = [1, 2] (in red) of the Euclidean line.
Here are some immediate consequences of Urysohn’s Lemma.
Corollary 4.5.4. Every normal space is completely regular.
Proof. Suppose that X is normal, i.e. that it possesses the separation properties
T1 and T4 . To show that X is completely regular, we only need to demonstrate that X
is T3 1 . Let A ⊂ X be any closed set and let x ∈ X − A be any point. By Lemma 4.1.3,
2
74
4. SEPARATION AXIOMS
the set B = {x} is a closed set and so by the Urysohn Lemma (theorem 4.5.2), there
exists a map f : X → R with f |A ≡ 0 and f (x) = 1. Thus X is T3 1 , as needed.
2
The next statement follows from the Urysohn Lemma and from Example 4.2.6 by
which every metric space (X, d) is T0 ,. . . ,T5 .
Corollary 4.5.5. Let (X, d) be a metric space and equip X with the associated
metric topology Td . Then (X, Td ) is T3 1 and completely regular. Thus, given any two
2
closed and disjoint subsets A, B ⊂ X, there exists a continuous function f : X → [0, 1]
with f |A ≡ 0 and f |B ≡ 1.
In the case of metric spaces, the function f from the preceding corollary can be
made completely explicit. To see this, we first digress to make a definition and an
observation (Lemma 4.5.7).
Definition 4.5.6. Let (X, d) be a metric space and let A ⊂ X be a subset of X.
We define the function dA : X → [0, ∞i as
dA (x) = inf{d(x, a) | a ∈ A},
and call it the distance function from A (or simply the distance from A).
Lemma 4.5.7. Let (X, d) be a metric space.
(a) If A ⊂ X is closed subset, then d(x, A) = 0 if and only if x ∈ A.
(b) If A, B ⊂ X are two disjoint and closed subsets of X, then dA (x) + dB (x) > 0
for every x ∈ X.
(c) For every subset A ⊂ X, the function dA : X → [0, ∞i is continuous.
Proof. (a) If x ∈ A then clearly dA (x) = 0. Suppose conversely that x ∈ X is
some element with dA (x) = 0. Thus, for every n ∈ N there must exist an element
an ∈ A with d(x, an ) < 1/n (by the definition of the infimum). But then lim an = x
and since A is closed, x must lie in A (part (a) of Theorem 3.2.9).
(b) This is immediate from part (a) for if dA (x)+dB (x) = 0 then dA (x) = 0 = dB (x).
By part (a) these two equalities would imply x ∈ A and x ∈ B and hence x ∈ A ∩ B,
a contradiction since A and B are assumed to be disjoint.
(c) Let x ∈ X be any point and let V ⊂ [0, ∞i be any neighborhood of dA (x). If
dA (x) > 0, find and ε > 0 such that hdA (x) − 2ε, dA (x) + 2εi ⊂ V and if dA (x) = 0
then find an ε > 0 such that [0, 2εi ⊂ V . Let U = Bx (ε) ⊂ X, we claim that then
dA (U ) ⊂ V showing continuity of dA at x. Since x ∈ X was arbitrary, this proves
continuity of dA .
To see that dA (U ) ⊂ V , pick a point y ∈ U . For every n ∈ N there must exist a
point an ∈ A such that d(x, an ) < dA (x) + 1/n. By the triangle inequality, this implies
that d(y, an ) ≤ d(y, x) + d(x, an ) < ε + dA (x) + 1/n. Therefore, dA (y) − dA (x) ≤ ε.
Arguing the same way by swapping the roles of x and y one obtains the inequality
dA (x) − dA (y) ≤ ε showing that |dA (x) − dA (y)| ≤ ε and so dA (y) ∈ V . Since y ∈ U
was arbitrary, this shows that dA (U ) ⊂ V .
4.5. THE URYSOHN LEMMA
75
If the set A ⊂ X is not closed, part (a) of the lemma may fail. For instance, if
(X, d) = (R, d2 ) (the Euclidean metric on R) and A = h0, 1i then dA (0) = 0 = dA (1)
but neither of 0 or 1 belong to A.
Armed with Lemma 4.5.7, we return to the setting of Corollary 4.5.5. Let (X, d)
be a metric space and let A, B ⊂ X be a pair of disjoint, closed subsets of X. Define
f : X → [0, 1] as
dA (x)
(4.2)
f (x) =
.
dA (x) + dB (x)
By part (c) of Lemma 4.5.7, the denominator of the right hand side above, is never
zero showing that f is well defined. If x ∈ A then dA (x) = 0 so that f (x) = 0 while if
x ∈ B then dB (x) = 0 giving f (x) = 1 (by part (a) of Lemma 4.5.7). Thus A ⊂ f −1 (0)
and B ⊂ f −1 (1) showing that this function f has the properties of the function f from
Corollary 4.5.5. In fact, equality holds in both of these inclusions (Exercise 4.6.9) thus
yielding a slight strengthening of Corollary 4.5.5. With these findings, we return one
more time to the setting of Example 4.5.3.
Example 4.5.8. Consider the Euclidean line (R, TEu ) with the Euclidean metric
d2 (x, y) = |x − y| and let A, B ⊂ R be the two closed and disjoint subsets A = [−2, −1]
and B = [1, 2]. The two distance functions dA and dB take on the forms


; x ≤ 1,
; x ≤ −2,
 1−x
 −2 − x
0
; x ∈ [1, 2],
0
; x ∈ [−2, −1],
, dB (x) =
dA (x) =


x−2
; x ≥ 2.
x+1
; x ≥ −1,
Combining these, the function f : R → R from equation (4.2) becomes

2+x
; x ≤ −2


1+2x






0
; x ∈ [−2, −1]





x+1
; x ∈ [−1, 1]
f (x) =
2






1
; x ∈ [1, 2]






 x+1
; x≥2
2x−1
Figure 3 shows the graph of this function, the reader may find it instructive to compare
it to the graph of the function f constructed in Example 4.5.3, see Figure 2.
76
4. SEPARATION AXIOMS
1
−2
|
{z
A
f
−1
1
}
|
2
{z
B
}
Figure 3. The graph of the function f (in green) from equation (4.2) for
the subsets A = [−2, −1] (in blue) and B = [1, 2] (in red) of the Euclidean
line equipped with the standard metric d(x, y) = |x − y|. Compare this
to the graph of the function f from figure 2 constructed directly from
the proof of Urysohn’s lemma.
4.6. Exercises
4.6.1. Finish the proof of of part (b) of Theorem 4.1.4: Show that a closed subspace
Y of a T5 -space X is itself a T5 -space.
4.6.2. Consider the real line R equipped with the Fort topology TF ,p . Show that
(R, TF ,p ) is a Ti -space for all i ∈ {0, 1, 2, 3, 3 21 , 4, 5}.
4.6.3. Consider R2 equipped with the topology T generated by the basis
B = {{x} × U | x ∈ R, U ∈ TEu }.
Check that B is a basis of a topology, and then verify whether or not (R2 , T ) is
(a) Hausdorff.
(b) Normal.
4.6.4. Consider the rationals Q ⊂ R equipped with the relative Euclidean topology.
Show that Q is a T4 -space. Is it a T5 -space?
4.6.5. Prove the topological invariance of the properties of First and Second Countability, and of being a Ti -space for i ∈ {0, 1, 3, 3 1, 4, 5}. Recall that the topological
invariance of Separability and of being a T2 -space were addressed in Example 4.3.4 and
Theorem 4.3.6.
4.6. EXERCISES
77
4.6.6. Consider the functions Omax , Omin : { Set of topological spaces } → N∪{∞},
defined as

(
)
There
exists
a
conver


max n ∈ N gent sequence in X
; There exists M ∈ N so




with n limits.
that no convergent seOmax (X, T ) =
quence in X has more


than M limits.





∞
; Otherwise.
Omin (X, T ) =
)
(

There exists a conver



;
min n ∈ N gent sequence in X




with n limits.







∞
;
There is at least one
convergent sequence
in X with finitely
many limits.
Otherwise.
(a) Show that both Omax and Omin are topological invariants.
(b) Show that
Omax (R, Tf c ) = ∞ = Omin (R, Tf c )
and
Omax (R, Tcc ) = 1 = Omin (R, Tcc ).
Note that if (X, T ) is a Hausdorff space, then Omax (X, T ) = Omin (X, T ) = 1. The
converse is not true in general.
4.6.7. Show that the function O : { Set of topological spaces } → {0, 1}, defined
by
f (X, T ) =
0
1
;
;
There is no surjective map onto ({a, b}, Tdis ),
There is a surjective map onto ({a, b}, Tdis ),
is a topological invariant. Using this invariant, show that (R, TEu ) and (R − {0}, TEu )
are not homeomorphic spaces.
4.6.8. Fill in the eleven pairs of missing entries in Table 1 that are not accounted
for by Lemmas 4.4.1, 4.4.2 and 4.4.3. Specifically, show that of the nine topologies TEu ,
Tdis , Tp , T p , Tf c , Tcc , TF,p , Tll , Tul on R, the ones that are
(a) T1 are precisely TEu , Tdis , Tf c , Tcc , TF,p , Tll , Tul . Use this to add four pairs of
crossmarks × to Table 1.
(b) T4 are precisely TEu , Tdis , T p , TF,p , Tll , Tul . Use this to conclude that (R, Tp )
and (R, T p ) are not homeomorphic, and add an new pair of crossmarks × to
Table 1.
(c) separable are precisely TEu , Tp , Tll , Tul . Use this to add four pairs of crossmarks
× to Table 1.
78
4. SEPARATION AXIOMS
(d) first countable are precisely TEu , Tdis , Tp , T p , Tll , Tul . Conclude that (R, Tdis )
and (R, TF ,p ) are not homeomorphic, add a pair of crossmarks × to Table 1.
(e) Use the results of Exercise 4.6.6 to show that (R, Tf c ) and (R, Tcc ) are not
homeomorphic, completing Table 1.
4.6.9. Let (X, d) be a metric space and consider X equipped with the associated
metric topology Td . Let A, B ⊂ X be two closed and disjoint subsets so that the
function f : X → [0, ∞i in equation (4.2) is well defined. Show that f −1 (0) = A and
f −1 (1) = B.
4.6.10. Consider R2 with the Euclidean metric and let A, B ⊂ R2 be the sets
A = Cl(B(0,0) (1)) and B = Cl(B(4,0) (1)). Following the proof of the Urysohn Lemma
4.5.2, and the notation therein, show that possible choices of the sets Uq , q ∈ Q ∩ [0, 1],
are given by

; q 6= 1
 B(0,0) (2 + q)
Uq =
 R2 − B
; q = 1.
2
Determine the associated function f : R → [0, 1] and check that is possesses the
properties claimed by the Urysohn Lemma.
CHAPTER 5
P
Product spaces
roduct spaces are to topological spaces what Cartesian products are to sets.
Indeed, given a family of topological spaces, we discuss how to naturally
introduce a product topology on the Cartisan product of the sets underlying
the given topological spaces.
Section 5.1 addresses the case of finite products, and studies basic concepts such
as, for example, continuity of functions into and from a product. We also take up
the question of how properties of the factors, such a separability or the Ti -properties,
transfer to the product space, and vice versa (Theorem 5.1.10).
Section 5.2 turns to infinite products, where besides the already mentioned product
topology, there is a second natural choice of topology, referred to as the box topology.
The main goal of this section is to contrast the product and box topologies through
examples of sequences of functions.
5.1. Finite products
Definition 5.1.1. Let (X1 , T1 ),. . . ,(Xn , Tn ) be a collection of n ≥ 2 topological
spaces. Then their Cartesian product X1 ×...×Xn = {(x1 , . . . , xn ) | xi ∈ Xi , i = 1, ..., n}
becomes a topological space with the product topology TX1 ×...×Xn defined as the topology
generated by the basis
(5.1)
BX1 ×···×Xn = {U1 × · · · × Un ⊂ X1 × · · · × Xn | Ui ∈ Ti , i = 1, . . . , n}.
The space (X1 × ... × Xn , TX1 ×...×Xn ) is called the product of (X1 , T1 ), . . . , (Xn , Tn ) or
simply a product space. We shall refer to the spaces (Xi , Ti ), i = 1, . . . , n as the factors
of (X1 × · · · × Xn , TX1 ×···×Xn ). The functions
πi : X 1 × · · · × X n → X i ,
πi (x1 , . . . , xn ) = xi ,
shall be referred to as the projection maps or simply as the projections. Specifically πi
shall be called the i-th projection.
To see that the set BX1 ×···×Xn from Definition 5.1.1 forms the basis of a topology, we
invoke Lemma 2.4.2. Since Xi ∈ Ti for each i = 1, . . . , n, we see that X1 × · · · × Xn ∈
BX1 ×···×Xn so that the first condition of Lemma 2.4.2 is met. As for the second, pick
arbitrary elements Ui , Vi ∈ Ti for each i = 1, . . . , n and let (x1 , . . . , xn ) ∈ (U1 × · · · ×
Un ) ∩ (V1 × · · · × Vn ) be any point. Since
(U1 × · · · × Un ) ∩ (V1 × · · · × Vn ) = (U1 ∩ V1 ) × · · · × (Un ∩ Vn ),
79
80
5. PRODUCT SPACES
and because Ui ∩ Vi ∈ Ti , we see that (U1 × · · · × Un ) ∩ (V1 × · · · × Vn ) ∈ BX1 ×···×Xn
showing BX1 ×···×Xn to be the basis of a topology. Indeed, we have shown a bit more,
namely that the intersection of every two elements from BX1 ×···×Xn again belongs to
BX1 ×···×Xn .
For the most part, we shall be interested in Definition 5.1.1 with n = 2. This
causes no loss of generality since the n-fold product space (X1 × ... × Xn , TX1 ×...×Xn )
is homeomorphic to the space gotten by taking n − 1 successive twofold products (in
arbitrary order) of the spaces (X1 , T1 ), . . . , (Xn , Tn ) (Exercise 5.3.1).
Remark 5.1.2. It may be tempting, but incorrect, to assume that every open set in
(X × Y, TX×Y ) is a product of open sets from X and Y . Rather, according to Definition
5.1.1, every element W ∈ TX×Y is a union of such products:
[
W =
Ui × Vi with Ui ∈ TX , Vi ∈ TY , i ∈ I,
i∈I
where I is some indexing set.
Example 5.1.3. Consider the topological spaces (X, TX ) = (R, Tp ) and (Y, TY ) =
(R, Tq ) where Tp and Tq are the included point topologies associated to the points
p, q ∈ R. An open set W ⊂ X × Y is a union of open set U × V with U ∈ Tp and
V ∈ Tq . In particular, if U and V are non-empty, then p ∈ U and q ∈ V so that
(p, q) ∈ W .
The converse of this is false, i.e. a subset W of X × Y that contains (p, q) may not
be open. An example is any set W = {(p, q), (x, y)} with x 6= p and y 6= q. Thus, the
product topology of Tp and Tq on R2 is not equal to the included point topology T(p,q)
on R2 .
The result of the following lemma is certainly expected though not completely
immediate from Definition 5.1.1.
Lemma 5.1.4. Let X and Y be two topological spaces and let A ⊂ X and B ⊂ Y
be two non-empty subsets.
(a) A × B is an open subset of X × Y with respect to the product topology TX×Y if
and only if A and B are both open.
(b) A × B is a closed subset of X × Y with respect to the product topology TX×Y if
and only if A and B are both closed.
Proof. (a) =⇒ If A and B are both open then A×B is open simply by definition
of TX×Y .
⇐= Suppose conversely that A×B is open. Then we can write A×B = ∪i∈I Ui ×Vi
with Ui ⊂ X and Vi ⊂ Y each open. But then A = ∪i∈I Ui and B = ∪i∈I Vi showing
that they are both unions of open sets and thus open themselves.
(b) =⇒ Suppose that A ⊂ X and B ⊂ Y are closed subsets and note that
(X × Y ) − (A × B) = X × (Y − B) ∪ (X − A) × Y.
5.1. FINITE PRODUCTS
81
Both of the sets on the right hand side of the above are open subsets of X × Y showing
that (X × Y ) − (A × B) is open and thus that A × B is closed.
⇐= Suppose that A × B ⊂ X × Y is closed and let x ∈ X − A be any point. Pick
y ∈ B arbitrarily and observe that then (x, y) lies in (X × Y ) − (A × B) which is an
open set. Thus there must be neighborhoods Ux ⊂ X of x and Vy ⊂ Y of y such that
Ux × Vy ⊂ (X × Y ) − (A × B). We claim that Ux is disjoint from A. If it were not,
we could pick a point z ∈ Ux ∩ A giving us the relation (z, y) ∈ (Ux × Vy ) ∩ (A × B),
a contradiction since (Ux × Vy ) ∩ (A × B) = ∅. Therefore Ux ∩ A = ∅ showing that
X − A = ∪x∈X−A Ux making X − A an open and hence A a closed set. The proof that
B is closed is analogous.
Definition 5.1.5. Let fi : Xi → Yi , i = 1, . . . , n be continuous functions and
define the function
f1 × · · · × fn : X1 × · · · × Xn → Y1 × · · · × Yn
by f1 × ... × fn (x1 , . . . , xn ) = (f1 (x1 ), . . . , fn (xn )). We shall refer to f1 × · · · × fn as
the product of the functions f1 , . . . , fn and shall call each fi a factor function.
Theorem 5.1.6. Let (X, TX ) and (Y, TY ) be two topological spaces.
(a) The projection maps πX : X × Y → X and πY : X × Y → Y are continuous,
open and surjective functions.
(b) For any two points x0 ∈ X and y0 ∈ Y , the two inclusion maps ιy0 : X → X ×Y
and ιx0 : Y → X × Y given by ιy0 (x) = (x, y0 ) and ιx0 (y) = (x0 , y), are
continuous functions.
(c) Let (Z, TZ ) be a topological space and let f : Z → X × Y be a function with
coordinate functions f = (fX , fY ). Then f is continuous if and only if each of
fX : Z → X and fY : Z → Y is continuous.
(d) A product of finitely many functions is continuous if and only if each of the
factors is continuous.
Proof. (a) Surjectivity of πX and πY are obvious. To prove their continuity, let
U ⊂ X and V ⊂ Y be open sets. Then
−1
πX
(U ) = U × Y
and
πY−1 (V ) = X × V,
which are both open sets showing that πX and πY are continuous.
Let W ⊂ X × Y be an open set and write W = ∪i∈I Ui × Vi with Ui ∈ TX and
Vi ∈ TY . Then
πX (W ) = πX (∪i∈I Ui × Vi ) = ∪i∈I Ui ∈ TX ,
showing that πX is an open map. By the same token, so is πY .
(b) To show that ιy0 : X → X ×Y is continuous we shall rely on part (e) of Theorem
3.1.9, by which it suffices to show that ι−1
y0 (W ) is open for W ranging through a basis
B of TX×Y . Choosing B = BX×Y from Equation (5.1), and thus picking W = U × V
82
5. PRODUCT SPACES
with U ∈ TX and V ∈ TY , we find that
ι−1
y0 (U
×V)=
U
∅
;
;
y0 ∈ V,
y0 ∈
/ V.
In either case ι−1
y0 (U × V ) ∈ TX , as needed. The case of ιx0 follows analogously.
(c) =⇒ Suppose first that f = (fX , fY ) : Z → X × Y is continuous. Then clearly
so are fX and fY since fX = πX ◦f and fY = πY ◦f (by part (a) of the present theorem,
and part (b) of Proposition 3.1.11).
⇐= Assume that fX : Z → X and fX : Z → Y are continuous and let U ⊂ X and
V ⊂ Y be arbitrary open sets. Relying again on part (e) of Theorem 3.1.9, continuity
of f follows by showing that f −1 (U × V ) = fX−1 (U ) × fY−1 (V ) is open. This in turns
follows plainly from the continuity of fX and fY and the definition of TX×Y .
(d) The coordinate functions of the product function f = f1 ×f2 : X1 ×X2 → Y1 ×Y2 ,
are the two functions f1 ◦ πX1 and f2 ◦ πX2 . If f1 and f2 are continuous, then parts (a)
and (c) of the theorem imply that f is also continuous. Conversely, if f is continuous,
then so are f1 ◦πX1 and f2 ◦πX2 (by part (c) of the present theorem, with Z = X1 ×X2 ).
But then, for any open set U ⊂ Y1 , the preimage f1−1 (U ) equals πX1 ((f1 ◦ πX1 )−1 (U ))
which is open since f1 ◦ πX1 is continuous and πX1 is open (by part (a)). This implies
continuity of f1 , a similar argument gives continuity of f2 .
The next example demonstrates that projection maps πX , πY don’t need to be
closed maps.
Example 5.1.7. Consider the topological spaces (X, TX ) = (Y, TY ) = (R, TEu )
and let Γf ⊂ R × R be the graph of the function f : R → R defined as f (x) =
2
sin(x)(1 − e−x ). We leave it as an easy exercise to show that Γf is a closed subset of
R × R. On the other hand, πY (Γf ) = h−1, 1i which is not a closed subset of (R, TEu ).
Thus, πY is not a closed map.
Inclusion maps on product spaces can fail to be open or closed maps.
Example 5.1.8. Let (X, TX ) and (Y, TY ) be two topological spaces and consider
the inclusion map ιy0 : X → X × Y (remember that ιy0 (x) = (x, y0 )). For any subset
A ⊂ X we obtain ιy0 (A) = A × {y0 }. If A is an open set then, according to Lemma
5.1.4, A × {y0 } is an open set if and only if {y0 } is open. On the other hand, if A is
closed then, again using Lemma 5.1.4, the set A × {y0 } is closed if and only if {y0 } is
closed.
Examples of spaces (Y, TY ) in which {y0 } fails to be both open and closed, abound.
An easy example is to take (Y, TY ) = (R, Tindis ), the indiscrete topology on R. For this
example, the inclusion map ιy0 is neither open nor closed for any choice of y0 ∈ R.
Corollary 5.1.9. Let X and Y be two topological spaces. Then, for any choice
of y0 ∈ Y , the inclusion map ιy0 : X → X × Y is a homeomorphism onto its image.
This allows us to view X as a subspace of X × Y . If additionally Y is T1 , then X is a
closed subspace of X × Y (for any choice of y0 ∈ Y ). Identical conclusions hold for Y .
5.1. FINITE PRODUCTS
83
Proof. Clearly ιy0 is injective and it is continuous by Theorem 5.1.6, part (b). Its
inverse function ι−1
y0 = πX |X×{y0 } is also continuous since the restriction of a continuous
function is continuous (Theorem 3.1.11). If Y is T1 then {y0 } is a closed subspace of
Y (Lemma 4.1.3) rendering X × {y0 } a closed subspace of X × Y by Lemma 5.1.4. We next explore how properties of topological spaces, such as second countability
and the Hausdorff property, transfer from factors to the product space and vice versa.
Theorem 5.1.10. Let X, Y be two topological spaces and let X ×Y be equipped with
the product topology. Let A and B be chosen from the sets of properties of topological
spaces


Separabiltiy,






Normality ( T1 and T4 ),
First countability,
.
, B∈
A∈
Complete Normality ( T1 and T5 ).
Second countability, 





T ,T ,T ,T .
0
1
2
3
(a) X × Y possesses property A if and only if both X and Y possess property A.
(b) If X × Y possesses property B, then so do X and Y (the converse is generally
not true, see Example 5.1.12).
Proof. With the exception of separability, each A-property is inherited by a subspace (Theorems 2.4.18 and 4.1.4). As Corollary 5.1.9 allows us to view X and Y as
subspaces of X × Y , it follows that if X × Y has any of the A-properties (other than
separability), then so do X and Y .
Similarly, the B-properties are inherited by a closed subspace (Theorem 4.1.4).
Lemma 4.1.3 shows that a singleton in a T1 space is a closed set, from which it follows
(with the help of Corollary 5.1.9) that X and Y are closed subspaces of X × Y . We
conclude that if X × Y possesses any of the B-properties, then so do X and Y .
The remaining claims of the theorem are each addressed separately.
• A is “Separability”
=⇒ Assume that X and Y are separable and let A ⊂ X and B ⊂ Y be countable
dense subsets. Then A × B is a countable subset of X × Y and A × B = Ā × B̄
(Exercise 5.3.2), showing that X × Y is separable.
⇐= Assume X × Y is separable with C ⊂ X × Y a countable dense subset. Let
A = πX (C) ⊂ X and B = πY (C) ⊂ Y . For any open set U ⊂ X we must have
A ∩ U 6= ∅ for otherwise we would obtain C ∩ (U × Y ) = ∅. Since U × Y is open in
X ×Y , this latter equality of sets produces a contradiction to the denseness property of
C according to Corollary 2.3.7. Thus A ∩ U 6= ∅ and so this same corollary guarantees
the denseness of A, and thus the second countability of X. The case of showing that
B̄ = Y follows similarly.
• A is “Second Countability”
Assume that X and Y are second countable and let BX = {Ui | i ∈ N} and BY =
{Vj | j ∈ N} be two countable bases for X and Y respectively. Define B as B =
{Ui × Vj | i, j ∈ N}. This is a countable set of open subsets of X × Y . To see that it is
84
5. PRODUCT SPACES
a basis, let W ⊂ X × Y be any open set. By Definition 5.1.1 of the product topology,
we can find open sets Ai ⊂ X and Bi ⊂ Y , i ∈ I, such that W = ∪i∈I Ai × Bi . For
each i ∈ I, let Ji and Ki be indexing sets such that Ai = ∪j∈Ji Uj and Bi = ∪k∈Ki Vk ,
then
W = ∪i∈I Ai × Bi = ∪i∈I [(∪j∈Ji Aj ) × (∪k∈Ki Bk )] = ∪i∈I,j∈Ji ,k∈Ki Aj × Bk .
This shows B to be a basis and thus X × Y to be second countable.
• A is “First Countability”
This proof mirrors the one just given for second countability and the details are left to
the reader (Exercise 5.3.3).
• A is the “ T0 property”, the “ T1 property”or the “ T2 property”.
We address the T0 − T2 properties together since their proofs are very similar. Let
(x1 , y1 ) and (x2 , y2 ) be two distinct points in X × Y .
T0 : If x1 6= x2 , find either a neighborhood U1 ⊂ X of x1 that doesn’t contain x2 or
else a neighborhood U2 ⊂ X of x2 that doesn’t contain x1 . Then either U1 × Y
is a neighborhood of (x1 , y1 ) not containing (x2 , y2 ) or U2 × Y is a neighborhood
of (x2 , y2 ) not containing (x1 , y1 ). If x1 = x2 then y1 6= y2 and one proceeds in
a similar fashion.
T1 : If x1 6= x2 , let U1 , U2 ⊂ X be neighborhoods of x1 and x2 respectively with
x2 ∈
/ U1 and x1 ∈
/ U2 . Then U1 × Y and U2 × Y are neighborhoods of (x1 , y1 )
and (x2 , y2 ) respectively with (x2 , y2 ) ∈
/ U1 × Y and (x1 , y1 ) ∈
/ U2 × Y . The case
of y1 6= y2 goes similarly.
T2 : If x1 6= x2 , let U1 , U2 ⊂ X be disjoint neighborhoods of x1 and x2 . Then U1 × Y
and U2 × Y are disjoint neighborhoods of (x1 , y1 ) and (x2 , y2 ) respectively. The
case of y1 6= y2 goes accordingly.
• A is the “T3 property”
Assume that X and Y are T3 . Let C ⊂ X × Y be a closed subset and pick a point
(x, y) ∈ X × Y − C. Consider X × Y − C as a neighborhood of (x, y) and find
neighborhoods U of x and V of y such that U × V ⊂ X × Y − C. By regularity of X
we can separate x and the closed set X − U by neighborhoods Wx and WX−U . Note
that
Wx ⊂ X − WX−U =⇒ W x ⊂ X − WX−U ⊂ U,
since X − WX−U is a closed set. In the same vein we find neighborhoods Wy and
WY −V of y and Y − V with W y ⊂ V . Then Wx × Wy is a neighborhood of (x, y) with
W x × W y ⊂ U × V ⊂ X × Y − C so that X × Y − W x × W y becomes a neighborhood
of C. Clearly Wx × Wy and X × Y − W x × W y are disjoint.
Corollary 5.1.11. The product X × Y is a regular space ( T0 and T3 ) if and only
if each of the factors X, Y is regular.
The following example shows even when X and Y are normal spaces, X × Y need
not be, demonstrating that part (b) of Theorem 5.1.10 cannot strengthened.
5.2. INFINITE PRODUCTS
85
Example 5.1.12. Consider the topological space (X, TX ) = (R, Tll ) where Tll is the
lower limit topology from Example 2.2.14 generated by the subbasis Sll = {[a, bi | a, b ∈
R, a < b}. It was already shown in Example 4.2.5 that X is T0 − T5 and thus both
(completely) regular and (completely) normal. To see that X × X is not T4 with
the product topology consider the subdiagonal ∆0 = {(x, −x) | x ∈ X}. Since ha, bi =
1
0
∪∞
n=1 [a+ n , bi is an open subset of X, it is easy to see that ∆ is a closed subset of X ×X.
0
0
The relative topology on ∆ is the discrete one because ∆ ∩ ([a, a+1i× [−a, −a +1i) =
{(a, −a)}. Therefore, the two disjoint sets
AQ = {(a, −a) ∈ ∆0 | a ∈ Q}
and
AI = {(a, −a) ∈ ∆0 | a ∈ I = R − Q},
are both closed in ∆0 and thus also in X × X.
If AQ and AI were separated by disjoint open sets UQ , UI ⊂ X × X, then for
every point (a, −a) ∈ AQ there would have to a basis element [α, βi × [γ, δi with
(a, −a) ∈ [α, βi × [γ, δi ⊂ UQ , and similarly for points (b, −b) ∈ AI . But the only basis
elements [α, βi × [γ, δi that contain (a, −a) ∈ AQ and do not intersect AI , are those
of the form [a, βi × [−a, δi (with a < β and −a < τ ). Similarly, basis elements that
contain (b, −b) ∈ AI and do not intersect AQ are of the form [b, σi × [−b, τ i (with b < σ
and −b < δ). From this it follows that for every (a, −a) ∈ AQ there exist βa , δa and for
every (b, −b) ∈ AI there exist σb , τb , such that
[
[
UQ =
[a, βa i × [−a, δa i and UI = [b, σb i × [−b, τb i.
a∈Q
b∈I
This relation however forces UQ ∩ UI 6= ∅, a contradiction, showing that X × X is not
a normal space.
Since X is regular, Corollary 5.1.11 implies that X × X is also regular. We have
thus found an example of a regular space that isn’t normal.
5.2. Infinite products
In this section we briefly turn to the infinite product spaces. While in the case
of finite products, the product topology is a natural choice, in the case of infinite
products there are two such natural choices. Depending on context, one may prove
more advantageous than the other (see for example Tychonoff’s compactness theorem
in Section 9.4).
Let (X, TX ) and (Y, TY ) be two topological space. The product topology TX×Y on
X × Y , generated by the basis {U × V | U ∈ TX , V ∈ TY }, can be characterized in two
ways:
1. TX×Y is the smallest topology which contains all of the products of open sets
from X and Y .
2. TX×Y is the smallest topology on X × Y for which the two projection maps
πX : X × Y → X
and
πY : X × Y → Y,
are continuous. Indeed, if we assume their continuity, then for any U ∈ TX and
any V ∈ TY , the sets π −1 (U ) = U × Y and πY−1 (V ) = X × V would have to be
86
5. PRODUCT SPACES
open in X × Y . But then their intersection (U × Y ) ∩ (X × V ) = U × V is also
forced to be open.
A similar characterization holds for the product topology of finitely many factors. It
is of course easy to see that these two descriptions lead to the same topology. This,
however, is no longer the case when infinitely many factors are involved.
Let I be an indexing set, possibly infinite, and for each i ∈ I let (Xi , Ti ) be a
topological space. Let X = ×i∈I Xi denote their Cartesian product, that is let X be
X = {x : I → ∪i∈I Xi | x(i) ∈ Xi }.
As customary, we shall denote x(i) simply by xi . For example, if Xi = Y for all i ∈ I,
then X is simply the set of functions from I to Y .
Definition 5.2.1. Let (Xi , Ti ), i ∈ I be a collection of topological space and let
X = ×i∈I Xi .
(a) The box topology TBox on X is the topology generated by the basis
BBox = {×i∈I Ui | Ui ∈ Ti }.
(b) The product topology TP rod on X is the topology generated by the basis
BP rod = {×i∈I Ui | Ui ∈ Ti and Ui = Xi for all but finitely many i ∈ I}.
The verification that the sets BBox and BP rod are indeed bases of a topology, is easy
and is left as an exercise (Exercise 5.3.4). It should be noted that TBox is strictly finer
than TP rod whenever I is infinite. When I is finite, the two topologies agree.
Example 5.2.2. Choose I = N and let (Xi , Ti ) = (R, TEu ). To tell copies of R
apart we shall label them as Ri , i ∈ I. Denote the product ×i∈N R by RN and note
that RN is simply the set of sequences in R. We shall consider the subset Y ⊂ X of
bounded sequences in R. On Y , consider the metric d : Y × Y → [0, ∞i (Exercise
5.3.5) given by
∞
X
|xi − yi |
d(x, y) =
.
2i
i=1
The associated metric topology is the relative topology on Y induced by Tbox . On the
other hand, the relative topology on Y induced by Tprod is not a metric topology since
it is not Hausdorff. For example, there are no disjoint neighborhoods of xi = 1/i and
yi = −1/i in Tprod .
Example 5.2.3. Pick I = R and let (Xλ , Tλ ) equal the Euclidean line for each
λ ∈ I. To distinguish the copy of R corresponding to λ ∈ I, we shall denote it by Rλ .
We’ll denote the Cartesian product ×λ∈R Rλ by RR . As in the previous example, we
can interpret RR as a set of functions, namely functions from R to R. We will examine
when a sequence of functions fi ∈ RR converges to a function f ∈ RR with respect to
the two topologies Tprod and Tbox .
5.2. INFINITE PRODUCTS
87
• Consider first RR with the product topology, let {fk }k ⊂ RR be a sequence with
limit f . Let λ ∈ R be arbitrary and let U ⊂ R be a neighborhood of f (λ). Consider
the open set in the product topology
U = (×µ<λ Rµ ) × U × (×µ>λ Rµ ).
Then U is a neighborhood of f and so there must exist an integer kλ such that fk ∈ U
for all k ≥ kλ . This last statement is equivalent to saying that fk (λ) ∈ U for all k ≥ kλ
which is simply saying that the sequence fk (λ) converges to f (λ) in the Euclidean
topology on R.
Conversely, suppose that for each λ ∈ R, the equation lim fk (λ) = f (λ) holds with
respect to the Euclidean topology on R. Pick any neighborhood of V of f and suppose
that
V = ×µ∈R−{λ1 ,...,λn } Rµ × (Vλ1 × ... × Vλn ) ,
where λ1 , ..., λn ∈ R are chosen arbitrarily and where Vλ1 , ..., Vλn are arbitrary neighborhoods of f (λ1 ), ..., f (λn ) respectively. Then there exists an integer k0 such that
k ≥ k0 implies that fk (λj ) ∈ Vλj for all k ≥ k0 and for any choice j ∈ {1, ..., n}. This
of course is equivalent to saying that fk ∈ V showing that lim fk = f with respect to
the product topology. In conclusion, we have shown that convergence of a sequence
of functions in RR with respect to the product topology, is equivalent to pointwise
convergence with respect to the Euclidean topology
lim fk = f in (RR , Tprod )
⇔
lim fk (λ) = f (λ) in (R, TEu ) for every λ ∈ R.
• Next consider RR equipped with the box topology and suppose again that {fk }k ⊂
RR is a sequence with limit f . For each λ ∈ R, let Uλ be a neighborhood of f (λ) with
respect to the Euclidean topology on Rλ and let U be the neighborhood of f (with
respect to the box topology) defined as
U = ×λ∈R Uλ .
By assumption, there exists an integer k0 such that k ≥ k0 implies that fk ∈ U.
The latter inclusion is equivalent to fk (λ) ∈ Uλ for every λ ∈ R. If we choose Uλ =
hf (λ) − ε, f (λ) + εi for each λ ∈ R and for some fixed ε > 0, then we find that
|fk (λ) − f (λ)| < ε for all k ≥ k0 , and for all values of λ ∈ R. The latter says that the
sequence of functions fk : R → R converges uniformly to the function f : R → R (with
respect to the Euclidean topology on R). Thus
(5.2)
lim fk = f in (RR , Tbox )
⇓
lim fk = f uniformly as functions (R, TEu ) → (R, TEu ).
For instance, define fk : R → R as fk (t) = et−k . Then, for every fixed but arbitrary
t ∈ R, the relation lim fk (t) = 0 holds in (R, TEu ) so that lim fk = 0 in (RR , Tprod ).
However, the convergence lim fk (t) = 0 in (R, TEu ) is not uniform (since for every fixed
88
5. PRODUCT SPACES
k, the quantity |fk (t)| can be arbitrarily large as we vary t) so that lim fk 6= 0 in
(RR , Tbox ).
The implication in (5.2) is in one direction only, as evidenced by the next example,
which exhibits a sequence of functions fk : R → R that converges uniformly to a
function f : R → R, but such that the sequence {fk }k does not converge to f in
(RR , TBox ).
2
Example 5.2.4. Consider the sequence of functions fk : R → R, fk (λ) = e−(λ +k) ,
k ∈ N. Then fk uniformly converges to f ≡ 0 because for every ε > 0 and any k0 with
e−k0 < ε, we obtain for each k ≥ k0 and for every λ ∈ R, the inequality
|fk (λ) − f (λ)| = e−(λ
2 +k)
≤ e−k ≤ e−k0 < ε.
On the other hand, {fk }k does not converge to f in (RR , TBox ) for if it did, then for
4
the neighborhood U = ×λ∈R B0 (e−λ ) of f , we should be able to find an integer k0
with k ≥ k0 implying fk ∈ U. The latter inclusion is tantamount to the inclusions
4
fk (λ) ∈ B0 (e−λ ) for every k ≥ k0 and every λ ∈ R. This in turn amounts to having
2
4
the inequality e−(λ +k) < e−λ hold for all k ≥ k0 and all λ ∈ R, clearly an impossibility.
5.3. Exercises
5.3.1. Prove the following two claims about finite product spaces.
(a) Given topological spaces (X1 , T1 ), . . . , (Xn , Tn ) show that
(X1 × . . . Xn , TX ×···×X ) ∼
= (X1 × . . . Xn−1 , TX ×···×X ) × (Xn , Tn ).
1
n
1
n−1
Said differently, an n-fold product space can be constructed by taking the product of two factors at a time.
(b) For topological spaces (X, TX ) and (Y, TY ), show that
(X × Y, TX×Y ) ∼
= (Y × X, TY ×X ).
Thus, up to homeomorphism, the order of the factors in a product spaced does
not matter.
5.3.2. Let (X, TX ) and (Y, TY ) be two topological spaces, and let A ⊂ X and B ⊂ Y
be a pair of given subsets. Viewing A × B as a subset of (X × Y, TX×Y ), show that
(a) A × B = A × B.
(b) (A × B)◦ = Å × B̊.
(c) ∂(A × B) = (∂A × B̄) ∪ (Ā × ∂B).
5.3.3. Show that if X and Y are first countable spaces, then X × Y , equipped with
the product topology, is also a first countable space.
5.3.4. Prove that BBox and BP rod from Definition 5.2.1, are each a basis of some
topopology.
5.3.5. Fill in the missing details from Example 5.2.2. Namely, let Y ⊂ RN be the
subset of bounded sequences. Show that
5.3. EXERCISES
89
P
i|
(a) The function d(x, y) = i∈N |xi2−y
defines a metric on Y .
i
(b) The associated metric topology Td on Y agrees with the relative box topology
TBox inherited from RN .
(c) Show that the relative product topology on Y is not Hausdorff, and thus not
metrizable.
5.3.6. Let (X, TX ) be a topological space and consider X × X equipped with the
product topology TX×X .
(a) Show that the function ∆ : X → X × X given by ∆(x) = (x, x) is continuous.
(b) Show that X is Hausdorff if and only if the image of ∆ is a closed subset of
X × X.
5.3.7. Consider [0, 1i and [0, 1] each equipped wit the relative Euclidean topology.
Show that with respect to the product topologies, the space [0, 1i × [0, 1] is homeomorphic to [0, 1i × [0, 1i. Does the same claim hold for the spaces [0, 1] × [0, 1] and
[0, 1i × [0, 1i?
5.3.8. For any m, n ∈ N, show that (Rn , TEu ) × (Rm , TEu ) ∼
= (Rn+m , TEu ).
CHAPTER 6
C
Compactness
compactness of a space is one of the most fundamental and important properties associated to a topological space, and has far reaching consequences
both for subspaces and maps on the compact space. As the name suggests,
a compact space can be thought of as being “small”. In that sense, compactness is not unlike the notions of separability and second countability introduced in
Chapter 2, but its impact on properties of the underlying space far surpasses those of
the latter two.
Section 6.1 introduces compactness, provides examples and establishes first properties of compact spaces. Section 6.2 heeds special attention to compactness on metric
spaces, culminating with the explicit characterization of compactness of Theorem 6.2.10
(see Corollary 6.2.11 for compactness in Euclidean spaces). Section 6.3 studies more
in depth properties of compact spaces, some of which rely on results from Section 6.2.
Section 6.4 studies the “One Point Compactification”, a procedure for exhibiting any
non-compact space X as a subspace of a compact space obtained by adding just a
single point to X. Finally, Section 6.5 touches on other flavors of compactness, namely
“Local Compactness”, “σ-compactness” and “Paracompactness”.
6.1. Definition and first examples
Definition 6.1.1. Let (X, TX ) be a topological space.
(a) An open cover F of X is a collection of open subsets of X whose union is all of
X.
(b) A subcover F 0 of F is a subset of F which is still an open cover of X. A proper
subcover of F is any subcover of F not equal to F.
(c) A refinement F 00 of F is an open cover of X with the property that for every
U ∈ F 00 there exists a V ∈ F such that U ⊂ V .
Definition 6.1.2. A topological space (X, TX ) of compact if every open cover F
of X has a finite subcover.
Example 6.1.3. Any topological space (X, TX ) with TX a finite set, is compact. In
particular, if X is a finite set then (X, TX ) is compact for any choice of a topology TX .
Example 6.1.4. The Euclidean line (R, TEu ) is not compact. This can be seen by
considering the open cover
F = {ha − 1, a + 1i | a ∈ Z}
91
92
6. COMPACTNESS
There are no proper subcovers of F at all, let alone finite subcovers, since each integer
a ∈ Z lies in precisely one element of F, namely ha − 1, a + 1i. A similar argument can
be used to show that the n-dimensional Euclidean space (Rn , TEu ) is also not compact
(Exercise 6.6.1).
Example 6.1.5. Consider the finite complement topology Tf c on R and let F be
an open cover. Pick any U ∈ F with U 6= ∅. Then U = R − {x1 , ..., xn } for some points
xi ∈ R, i = 1, ..., n. For each xi let Vi ∈ F be such that xi ∈ Vi (such a Vi has to exist
since F is an open cover). But then {U, V1 , ..., Vn } is a finite subcover of F showing
that (R, Tf c ) is a compact space.
Example 6.1.6. Consider the included point topology Tp on R. The infinite open
cover
F = {{p, x} | x ∈ R − {p} }
has no proper subcovers. Consequently (R, Tp ) is not compact.
Example 6.1.7. Consider the excluded point topology T p on R and let F be any
of its open covers. Since R is the only non-empty open set that contains p, R must be a
member of every open cover. Given this observation, the cover {R} is a finite subcover
of any open cover. Therefore, (R, T p ) is compact.
We turn to some simple properties of compact spaces.
Theorem 6.1.8. Let f : X → Y be a map. If X is compact then so is f (X). In
particular, compactness is a topological invariant.
Proof. Let Ff (X) = {Vi | i ∈ I} be an open cover of f (X) and consider the open
cover FX = {f −1 (Vi ) | i ∈ I} of X. By compactness of X, this latter cover has a
finite subcover FX0 = {f −1 (Vi1 ), ..., f −1 (Vin )} for some indices i1 , ..., in ∈ I. But then
Ff0 (X) = {Vi1 , ..., Vin } is a finite subcover of Ff (X) .
If f : X → Y is a homeomorphism and X is compact, then so if Y , being equal
to f (X). If X is not compact then neither is Y since compactness of Y would imply
that of X, being equal to f −1 (Y ). Thus, two homeomorphic spaces X and Y are either
both compact or both non-compact.
Theorem 6.1.9. Let X and Y be two topological spaces.
(a) If X is compact and A is a closed subspace of X then A is also compact.
(b) If X is Hausdorff and A ⊂ X is a compact subspace then A is closed in X.
Proof. (a) Let FA = {Vi | i ∈ I} be an open cover of A and let Ui be open subsets
of X such that Vi = A ∩ Ui . Then FX = {X − A, Ui | i ∈ I} is an open cover of X
and thus has a finite subcover FX0 that looks like FX0 = {Ui1 , ..., Uin , X − A} or FX0 =
{Ui1 , ..., Uin } for some choice of indices i1 , ..., in ∈ I. In either case, FA0 = {Vi1 , ..., Vin }
is a finite subcover of FA showing that A is compact.
(b) Fix a point a ∈ A. The Hausdorff condition guarantees that for every point
x ∈ X − A there exist disjoint neighborhoods Ua,x and Vx,a of a and x respectively. The
6.1. DEFINITION AND FIRST EXAMPLES
93
collection {Ua,x ∩ A | a ∈ A} is an open cover of A and so by compactness, it has a finite
subcover {Ua1 ,x , ..., Uan ,x } for some points a1 , ..., an ∈ A. Let Vx = ∩ni=1 Vx,ai . Clearly
Vx is a neighborhood of x but moreover, Vx and A are disjoint. For if not, we could
find a point y ∈ Vx ∩ A. Then we would have y ∈ Uaj ,x for some j ∈ {1, ..., n}. But
on the other hand y ∈ Vx ⊂ Vx,aj which is impossible since Uaj ,x ∩ Vx,aj = ∅. Finally,
we note that X − A = ∪x∈X−A Vx showing that X − A is open and therefore that A is
closed.
Corollary 6.1.10. Let X be a compact space and Y a Hausdorff space and let
f : X → Y be a continuous function. Then f is a closed map. In particular, if f is a
bijection then it is a homeomorphism.
Proof. Let A ⊂ X be any closed subset of X. According to part (a) of Theorem
6.1.9, A is compact. According to part (b) of Theorem 6.1.9, f (A) is compact, which
renders it a closed subset of Y by part (c) of Theorem 6.1.9. Thus f is a closed map.
If f is a bijection, it suffices to show that f −1 : Y → X is continuous in order to
demonstrate that f is a homeomorphism. According to part (a) of Theorem 3.1.9, the
latter follows from the closedness property of f which has already been established. Theorem 6.1.11. Let (X, TX ) and (Y, TY ) be two topological spaces and let X × Y
be equipped with the product topology TX×Y . Then X × Y is compact if and only if each
of X and Y are compact.
Proof. =⇒ Suppose firstly that X and Y are both compact and let W = {Wi | i ∈
I} be an open cover of X × Y . For each Wi ∈ W, find open sets Ui,j ⊂ X and
Vi,j ⊂ Y , j ∈ Ji such that Wi = ∪j∈Ji Ui,j × Vi,j . If we can show that the open cover
f = {Ui,j × Vi,j | j ∈ Ji , i ∈ I} has a finite subcover, then clearly so does W. Thus,
W
without loss of generality, we shall assume that each Wi has the form Ui × Vi , i ∈ I to
begin with.
Pick a point y ∈ Y and consider the subspace X × {y} ⊂ X × Y . As we saw in
Corollary 5.1.9, X × {y} is homeomorphic to X and so X × {y} must also be compact
(Proposition 6.1.8). Therefore, the induced open cover
Wy = {(Ui × Vi ) ∩ (X × {y}) | i ∈ I},
of X × {y}, must have a finite subcover Wy0 = {Uj × {y} | j ∈ Jy } for some finite subset
of indices Jy ⊂ I. Let Vy = ∩j∈Jy Vj and note that Vy is a neighborhood of y ∈ Y with
the property that Uj × Vy ⊂ Uj × Vj for all j ∈ Jy .
Perform the above procedure for all y ∈ Y to arrive at the open cover {Vy | y ∈ Y }
of Y . By compactness of Y , this has a finite subcover {Vy1 , ..., Vyn } for some points
y1 , ..., yn ∈ Y . We claim that then the set
W 0 = {Uj × Vj | j ∈ Jyi , i = 1, ..., n}
is a finite subcover of W . Given any point (x, y) ∈ X × Y , there exists an index
i ∈ {1, ..., n} such that y ∈ Vyi . Since the sets Uj , j ∈ Jyi cover X, we find that x ∈ Uj
for some j ∈ Jyi . But then (x, y) ∈ Uj × Vyi ⊂ Uj × Vj ∈ W 0 . So W 0 is an open cover.
94
6. COMPACTNESS
⇐= Since the projections maps πX : X × Y → X and πY : X × Y → Y are both
continuous and surjective (part (a) of Theorem 5.1.6), and since the continuous image
of a compact space is compact (part (b) of Theorem 6.1.9), the compactness of X × Y
immediately implies that of both X and Y .
By induction, the result of the preceding theorem extends without difficulty to
arbitrary finite products to show that X = X1 × ... × Xn is compact (with respect
to the product topology) if and only if each of the factors Xi is compact. The case
of infinite products is more subtle. On an infinite product Y = ×i∈I Yi there are
two natural choices of topologies available: the product topology TP rod and the box
topology TBox (Definition 5.2.1). Regardless of the choice, the compactness of Y still
implies the compactness of the factors Yi (by the same argument as in the second half
of the proof of Theorem 6.1.11, which only relied on the surjectivity of the projection
maps πi : Y → Yi ). Conversely however, the compactness of Y is only implied by the
compactness of the Yi when Y is given the product topology, but not with respect to
the box topology. See Section 9.4 for more details on this.
Before proceeding to study more in depth properties of compact spaces in Section
6.3, we digress first to take a closer look at compactness of metric spaces.
6.2. Compactness for metric spaces
Compactness for metric spaces is much easier to characterize than in the case of
general topological spaces (Theorems 6.2.6 and 6.2.10). An important special case of
these findings is a completely explicit descriptions of compact subspaces of Euclidean
space (Rn , TEu ) (Corollary 6.2.11).
Definition 6.2.1. Let (X, d) be a metric space.
(a) (X, d) is called totally bounded if for S
every ε > 0 there exist finitely many points
xi ∈ X, i = 1, . . . , n such that X = ni=1 Bxi (ε).
(b) (X, d) is said to be bounded if there exists some r > 0 such that d(x, y) < r for
all x, y ∈ X, otherwise we call (X, d) unbounded.
(c) If (X, d) is bounded, we define its diameter diam(X) as
diam(X) = sup{d(x, y) | x, y ∈ X}.
If (X, d) is unbounded we define diam(X) to be infinite.
Using the triangle inequality (relation (3) from the axioms of a metric, Example
2.2.3), it is easy to show that a a totally bounded metric space is always bounded. The
converse is not generally true (Exercise 6.6.2).
For the remainder of this chapter we shall always assume, without further mention,
that a metric space (X, d) has been made into a topological space with the metric
topology Td (Example 2.2.3).
Lemma 6.2.2. A totally bounded metric space (X, d) is separable.
6.2. COMPACTNESS FOR METRIC SPACES
95
Proof. For every n ∈ N and for the choice of εn = n1 , there is a finite set of points
An = {xn1 , . . . , xn`n } ⊂ X such that X = ∪x∈An Bx n1 . Note that A = ∪n∈N An is a
countable subset of X. We claim that A is also dense. To see this, let U ⊂ X be any
non-empty open set and let x ∈ U be any element. Then there exists an ε > 0 such
that Bx (ε) ⊂ U . Find an integer n large enough so that n1 < ε and let x ∈ Bxni ( n1 ) for
some i ∈ {1, . . . , `n }. Then d(x, xni ) < n1 < ε showing that xni ∈ Bx (ε) ⊂ U and thus
A ∩ U 6= ∅. Since U was arbitrary it follows that Ā = X (Corollary 2.3.7).
Remark 6.2.3. Recall from Proposition 2.4.15 that a metric space is second countable if and only if it is separable. In particular, a totally bounded metric space must
be second countable.
Lemma 6.2.4. Let (X, d) be a metric space with the property that every sequence
{xk }k ⊂ X has a convergent subsequence. Then (X, d) is totally bounded.
Proof. Suppose X were not totally bounded. Then there would be some ε > 0
so that X cannot be covered by a finite number of balls of radius ε. Using this,
we can construct a sequence {xk }k ⊂ X as follows: let x1 be arbitrary and choose
consecutive elements of the sequence so that xk+1 ∈
/ ∪kj=1 Bxj (ε). Notice that it follows
that xm ∈
/ ∪`j=1 Bxj (ε) whenever ` < m.
Let yn = xkn be some convergent subsequence of xk with lim yn = y ∈ X. Then
there must exist an n0 ∈ N such that n ≥ n0 implies yn ∈ By (ε/2). Let `, m be integers
such that n0 ≤ ` < m, then
d(xk` , xkm ) ≤ d(xk` , y) + d(y, xkm ) = d(y` , y) + d(y, ym ) < ε/2 + ε/2 = ε.
This implies that xkm ∈ Bxk` (ε) which is a contradiction since k` < km and by definition,
/ ∪k≤k` Bxk (ε). Thus X must be totally bounded.
xkm ∈
Lemma 6.2.5 (Lindelöf’s Theorem). If X is a second countable topological space
then every open cover F of X has a countable subcover.
Proof. Let B = {Un | n ∈ N} be countable basis for X and let F = {Vi | i ∈ I}
be an open cover of X. Since each Vi is an open set, there must exist a subset Ji ⊂ N
such that Vi = ∪j∈Ji Uj . Define F 00 = {Uj ∈ B | j ∈ Ji for some i ∈ I}. Clearly F 00 is
countable and it is also easy to see that F 00 is an open cover of X. For if x ∈ X is an
arbitrary point then there is some i ∈ I with x ∈ Vi and thus x ∈ Uj for some j ∈ Ji .
On the other hand, F 00 is a refinement of F by definition. For each Uj ∈ F 00 , let j 0 ∈ I
be such that Uj ⊂ Vj 0 . Then F 0 = {Vj 0 | Uj ∈ F 00 } is a countable subcover of F.
Theorem 6.2.6. A metric space (X, d) is compact if and only if every sequence
{xk }k ⊂ X has a convergent subsequence.
Proof. =⇒ Let X be a compact space and let {xk }k ⊂ X be an arbitrary
sequence. Suppose that there exists a point y ∈ X such that every neighborhood of
y contains infinitely many distinct points of the sequence {xk }k . For such a point y
we define a subsequence yn = xkn of {xk }k as follows: Pick y1 to be any point in
96
6. COMPACTNESS
By (1) ∩ {xk | k ∈ N}. With y1 , . . . , yn−1 chosen, pick yn to be an arbitrary point in the
set By ( n1 ) ∩ {xk | k > kn−1 }. It is easy to see that {yn }n is a convergent subsequence of
{xk }k with limit y, proving the theorem.
If this doesn’t occur, then every y ∈ Y possesses a neighborhood Uy that contains
only finitely many distinct points of the sequence {xk }k . Then F = {Uy | y ∈ X}
is an open cover of X and so by compactness of X, it has a finite subcover F 0 =
{Uy1 , . . . , Uym }| for some points y1 , . . . , ym ∈ X. But then the infinite sequence {xk }k
is contained in the finite union Uy1 ∪ ... ∪ Uy1 showing (by the “Pigeon Hole Principle”)
that some point y must equal xk for infinitely many distinct values k1 , k2 , k3 , . . . of
k. Then the sequence {xk1 , xk2 , xk3 , . . . }, being a constant sequence, is a convergent
subsequence of {xk }k .
⇐= Suppose that every sequence {xk }k ⊂ X has a convergent subsequence and
let F be an arbitrary open cover of X. Lemma 6.2.4 shows that X is totally bounded,
and hence by Lemma 6.2.2 it is also separable. By virtue of Proposition 2.4.15 (see
Remark 6.2.3) and using Lemma 6.2.5, we see that there must be a countable subcover
G = {U1 , U2 , U3 , . . . } of F.
We will prove the theorem by showing that G has a finite subcover. Assume to
the contrary that G has no finite subcovers. Form a sequence {xk}k ⊂ X by choosing
k−1
x1 ∈ X arbitrarily and by picking xk from the set X − ∪i=1
Ui for k ≥ 2. By the
assumption, there must be a subsequence yn = xkn with converges to some point y ∈ X.
Since G is an open cover of X, there must be some index m ∈ N with y ∈ Um . But
then, by construction, yn ∈
/ Um for all n > m, a contradiction given that lim yn = y.
Therefore G, and hence also F, must have a finite subcover. Since the cover F was
arbitrary, it follows that X is compact.
Corollary 6.2.7. Every compact metric space X is totally bounded, separable and
second countable.
Proof. This follows from Theorem 6.2.6 and Lemmas 6.2.2 and 6.2.4 and Remark
6.2.3.
While Theorem 6.2.6 provides a necessary and sufficient condition for a metric
space (X, d) to be compact, it is often tedious in practice to check that X has the
property that every of its sequences has a convergent subsequence. We provide a
second characterization of compactness on metric spaces, one which involved easier to
verify conditions. To state that theorem, we first introduce two additional concepts.
Recall that a sequence {xk }k ⊂ Rn is said to be a Cauchy sequence (with respect
to the Euclidean topology) if for every ε > 0 there exists a natural number n0 such
that for all m, n ≥ n0 the inequality d2 (xn , xm ) < ε holds. This definition involves only
the Euclidean metric d2 on Rn and none of the other more advances structures on Rn .
Hence, it easily transfers to any metric space.
Definition 6.2.8. Let (X, d) be a metric space. A sequence {xk }k ⊂ X is said to
be a Cauchy sequence if for every ε > 0 there exists a natural number n0 such that for
6.2. COMPACTNESS FOR METRIC SPACES
97
all m, n ≥ n0 we obtain d(xn , xm ) < ε. The metric space (X, d) is said to be a complete
metric space (or simply complete) if every Cauchy sequence is convergent.
Example 6.2.9. Consider Rn equipped with the Euclidean topology associated to
the Euclidean metric d2 . For a subset A ⊂ Rn , the metric space (A, d2 |A×A ) is complete
if and only if A is closed. In particular, (Rn , d2 ) itself is complete.
To see this, suppose that A is closed in Rn and let {xk }k ⊂ A be a Cauchy sequence.
Every Cauchy sequence is bounded and every bounded sequence in Rn has a convergent
subsequence yn = xkn (these are standard results in analysis whose proofs we omit)
with limit lim yn = y. By Exercise 6.6.3 the sequence {xk }k is then also convergent
with lim xk = y. Since A is closed it must contain the limits of all its convergent
sequences, and thus y ∈ A and A is complete.
Conversely, suppose that A is complete. If A weren’t closed then Rn − A wouldn’t
be open and we could find a point y ∈ Rn − A all of whose neighborhoods intersect A.
We could then define a sequence {xk }k ⊂ A by choosing xk arbitrarily from By ( k1 ) ∩ A.
As d2 (xk , x` ) < k1 + 1` , the sequence {xk }k is Cauchy, creating a contradiction to the
assumption that A is complete and the fact that lim xk = y ∈
/ A. Thus A must be
closed.
The following theorem is the main result of this section.
Theorem 6.2.10. A metric space (X, d) is compact if and only if is complete and
totally bounded.
Proof. =⇒ Suppose that X is compact. To see that X is complete, let {xk }k ⊂
X be any Cauchy sequence. By Theorem 6.2.6 there is a convergent subsequence
yn = xkn of {xk }k with limit y. But then xk itself has to converge to y (Exercise 6.6.3)
showing that X is complete.
To see that X is totally bounded, pick an arbitrary ε > 0 and consider the open
cover Fε = {Bx (ε) | x ∈ X} of X. By compactness of X, Fε has a finite subcover,
showing that X is totally bounded.
⇐= Assume that X is complete and totally bounded. We will show that X is
compact by relying on Theorem 6.2.6. Specifically, we will show that every sequence
{xk }k ⊂ X has a convergent subsequence.
Since X is totally bounded, there is a finite cover of X with balls of radius 1, say
F1 = {Bx1,1 (1), ..., Bxn1 ,1 (1)}. Since the sequence {xk }k has infinitely many terms,
one of the balls from F1 must contain infinitely many elements of the said sequence.
Without loss of generality, assume that Bx1,1 (1) is such a ball. Using total boundedness
of X once more, we find that there is a finite cover F2 = {Bx1,2 ( 12 ), ..., Bxn2 ,2 ( 21 )} of X
by balls of radius 12 . Since Bx1,1 (1) contains infinitely many elements of the sequence
{xk }k , there must be a ball in F2 whose intersection with Bx1,1 (1) contains infinitely
many elements of the sequence {xk }k . Again, and without loss of generality, we assume
that Bx1,2 ( 21 ) is that ball so that Bx1,1 (1) ∩ Bx1,2 ( 12 ) contains infinitely many elements
of the sequence xk . Proceeding inductively, we obtain a family of balls Bx1,k ( k1 ) of
98
6. COMPACTNESS
shrinking radii such that for any n ∈ N the set ∩nj=1 Bx1,j ( 1j ) contains infinitely many
elements of the sequence {xk }k .
To find a convergent subsequence of {xk }k , pick indices k1 < k2 < k3 < . . . such that
xkn ∈ ∩nj=1 Bx1,j ( 1j ) and set yn = xkn . Then yn is a Cauchy sequence since d(yn , ym ) <
max{ n1 , m1 }. Since X is assumed to be complete, yn must be a convergent sequence and
so we have generated a convergent subsequence of {xk }k , as needed. This completes
the proof.
Corollary 6.2.11. Let A be a subspace of n-dimensional Euclidean space Rn .
Then A is compact if and only if it is closed and bounded.
Proof. If A is bounded then it is also totally bounded (Exercise 6.6.2) and if in
addition it is also closed, then it is complete by Example 6.2.9. Theorem 6.2.10 then
shows that A is compact.
Conversely, if A ⊂ Rn is compact, then by Theorem 6.2.10 is must be complete and
hence closed by Example 6.2.9. Theorem 6.2.10 also shows that A has to be totally
bounded and therefore bounded (Exercise 6.6.2).
Example 6.2.12. The n-dimensional sphere S n is the subspace of Rn+1 consisting
of vectors of norm 1:
S n = {(x1 , ..., xn+1 ) ∈ Rn+1 | x21 + ... + x2n+1 = 1}.
It is quite obviously bounded and it is also closed. An easy way to verify closedness
of S n is to consider the continuous function f : Rn+1 → R given by f (x1 , ..., xn+1 ) =
x21 + ... + x2n+1 . Then S n = f −1 (1) and {1} ⊂ R is a closed set. According to Corollary
6.2.11, S n is compact.
6.3. Properties of compact spaces
In this section we return to the investigation of properties of compact spaces already
initiated in Section 6.1. Some of the properties established in this section rely on the
results about metric spaces obtained in the preceding section.
Theorem 6.3.1. Let X be a compact topological space and let f : X → R be
a continuous function to the Euclidean line. Then f attains both its maximum and
minimum value.
Proof. By part (b) of Theorem 6.1.9, the image f (X) is a compact subspace of
R. By Corollary 6.2.11, f (X) must be closed and bounded and thus f (X) = [a1 , b1 ] ∪
· · · ∪ [an , bn ] for some ai , bi ∈ R with ai ≤ bi . This shows that both the supremum
sup f (X) = max{b1 , . . . , bn } and the infimum inf f (X) = min{a1 , . . . , an } are contained
in f (X) and are therefore attained at some points xmax , xmin ∈ X.
Theorem 6.3.2. Let (X, d) be a metric space and let F be an open cover of X.
If X is compact then there exists a real number L > 0, called the Lebesgue number of
F, such that for every open ball Bx (L) of radius L (and with x ∈ X arbitrary), there
exists an element U ∈ F with Bx (L) ⊂ U .
6.3. PROPERTIES OF COMPACT SPACES
99
Proof. Let y ∈ X be an arbitrary element. Since F is an open cover of X, there
must be some element Uy ∈ F such that y ∈ Uy . Let ry > 0 be chosen so that
By (ry ) ⊂ Uy . Consider then the open cover G of X given by G = {By ( r2y ) | y ∈ X}.
r
Since X is compact, G has a finite subcover G 0 = {By1 ( y21 ), . . . , Byn ( ry2n )} for some
points y1 , . . . , yn ∈ X. Let L > 0 be chosen that
nr
rx o
x1
,..., n .
L < min
2
2
We claim that this L has the property specified by the theorem. To see this, let
x ∈ X be any point. Since G 0 is a cover, there is some index i ∈ {1, . . . , n} such that
r
x ∈ Byi ( 2yi ). But then, if x0 ∈ Bx (L), we obtain
ry
ry
ry
d(x0 , yi ) ≤ d(x0 , x) + d(x, yi ) < L + i < i + i = ryi ,
2
2
2
showing that Bx (L) ⊂ Byi (ryi ) ⊂ Uyi ∈ F. Since x was arbitrary, the theorem is
proved.
The following has already been established for a compact metric space (Theorem
6.2.6). We show here that it remains valid in general compact spaces provided they
are first countable (which metric spaces always are by Example 2.4.14).
Proposition 6.3.3. Every sequence {xk }k in a first countable, compact space
(X, TX ) has a convergent subsequence.
Proof. Case 1. Suppose that we can find a point x ∈ X with the property that
every neighborhood U of x contains infinitely many elements of the sequence {xk }k .
Let Bx = {Un | n ∈ N} be a countable neighborhood basis of x. Define the subsequence
yn = xkn of {xk }k by picking y1 from U1 ∩ {xk | k ∈ N} at random, and by picking yn
from the non-empty set U1 ∩ ... ∩ Un ∩ {xk | k > kn−1 } in an arbitrary fashion. Then,
given any neighborhood V of x, we can find a natural number n0 such that Un0 ⊂ V
and consequently, yn ∈ V for all n ≥ n0 . In particular, lim yn = x.
Case 2. If the hypothesis from Case 1 fails, then for every point x ∈ X there has
to exist a neighborhood Ux containing only finitely many points of the sequence {xk }k .
By compactness of X, we can reduce the open cover {Ux | x ∈ X} of X to a finite
subcover {Ua1 , . . . , Uan } with a1 , . . . , an ∈ X. But then, by the pigeonhole principle,
one of the sets Uai must contain infinitely many elements of the sequence xk , producing
a contradiction. Therefore Case 2 cannot occur while in Case 1 we were able to find a
convergent subsequence of {xk }k .
The next set of theorems shows that the compactness of X “enhances”the separation axioms T0 − T5 that X possesses. By this we mean, for example, that if a compact
space X is T2 then it is automatically T3 , something that if of course false in general
topological spaces. The chief reason for this is that closed subsets of X behave in
many ways like points, given that every open cover of a closed subset of X has a finite
subcover. How exactly this plays out can be seen in the proof of the next theorem.
Theorem 6.3.4. Let X be a compact topological space.
100
6. COMPACTNESS
(a) If X is Hausdorff then X is also T3 and T4 .
(b) If X is T3 then it is also T4 .
The next lemma is used in the proof of Theorem 6.3.4, it is the analogue for T3
spaces of what Lemma 4.5.1 is for T4 spaces.
Lemma 6.3.5. A topological space (X, TX ) has the T3 property if and only if for
every closed set A ⊂ X and every point x ∈ X − A there exists an open set V ⊂ X
such that
x ∈ V ⊂ V̄ ⊂ X − A.
Proof. =⇒ Let X be a T3 space and let A ⊂ X be a closed set and x ∈ X − A
any point. By the T3 property there are disjoint open sets Ux , UA ⊂ X with x ∈ Ux
and A ⊂ UA . From Ux ∩ UA = ∅ it follows that Ux ⊂ X − UA . Since X − UA is a closed
set, we find that Ūx ⊂ X − UA and since X − UA ⊂ X − A we get the desired relation
x ∈ Ux ⊂ Ūx ⊂ X − A. Choosing V = Ux establishes one direction of the theorem.
⇐= Suppose that for every closed subset A ⊂ X and for every point x ∈ X − A
there exists an open set V with x ∈ V ⊂ V̄ ⊂ X − A. We define the open sets Ux and
UA as V and X − V̄ respectively. Then A ⊂ UA and Ux ∩ UA = ∅ and so X is a T3
space.
Proof of Theorem 6.3.4. (a) The proof of this part of the theorem follows
closely the proof of part (c) of Theorem 6.1.9. Let A ⊂ X be a closed set and x ∈ X −A
any point. The Hausdorff condition guarantees for every a ∈ A, the existence of disjoint
neighborhoods Ua,x and Vx,a of a and x respectively. The collection {Ua,x ∩ A | a ∈ A}
is an open cover of A and, since A is compact (part (a) of Theorem 6.1.9), it has a
finite subcover {Ua1 ,x , . . . , Uan ,x } for some points a1 , . . . , an ∈ A. Let Vx = ∩ni=1 Vx,ai .
Clearly Vx is a neighborhood of x but moreover, Vx and A are disjoint. For if not, we
could find a point y ∈ Vx ∩ A. Then we would have y ∈ Uaj ,x for some j ∈ {1, . . . , n}
and on the other hand y ∈ Vx ⊂ Vx,aj , a contradiction since Uaj ,x ∩ Vx,aj = ∅. Let
UA = Ua1 ,x ∪ · · · ∪ Uan ,x . Then A ⊂ UA and clearly UA ∩ Vx = ∅, showing that X is T3 .
To verify the T4 property of X, let A, B ⊂ X be two disjoint closed subsets of
X. For a fixed point b ∈ B, let Vb,A and UA,b be disjoint open sets with b ∈ Vb,A and
A ⊂ UA,b . The set FB = {Vb,A | b ∈ B} is an open cover of B which, by compactness of
B, (again, part (a) of Theorem 6.1.9), has a finite subcover FB0 = {Vb1 ,A , . . . , Vbn ,A } for
some points b1 , . . . , bn ∈ B. Set VB = ∪ni=1 Vbi ,A and UA = ∩ni=1 UA,bi . These are open
sets containing B and A respectively. Moreover, VB ∩ UA = ∅ for if we had a point
y ∈ VB ∩ UA then y would lie in some Vbj ,A and at the same time UA,bj , a contradiction
since these two sets are disjoint by construction. Therefore X is T4 .
(b) Let A, B ⊂ X be two disjoint and closed subsets of X. According to Lemma
6.3.5, for every point a ∈ A there exists a neighborhood Ua of a such that
a ∈ Ua ⊂ Ūa ⊂ X − B.
6.4. THE ONE POINT COMPACTIFICATION
101
Then {X − A, Ua | a ∈ A} is an open cover of X and has a finite subcover {X −
A, Ua1 , Ua2 , . . . , Uan }. In a similar vein, every point b ∈ B has a neighborhood Vb with
b ∈ Vb ⊂ V̄b ⊂ X − A,
and the open cover {X − B, Vb | b ∈ B} has a finite subcover {X − B, Vb1 , Vb2 , . . . , Vbn }
(for convenience and without loss of generality, we assume that this cover has as many
elements as the cover {X − A, Ua1 , Ua2 , ..., Uan }). We now define the open sets
U10 = Ua1 − V̄b1 ,
U20 = Ua2 − (V̄b1 ∪ V̄b2 ),
..
.
V10 = Vb1 − Ūa1 ,
V20 = Vb2 − (Ūa1 ∪ Ūa2 ),
..
.
Un0 = Uan − (V̄b1 ∪ · · · ∪ V̄bn ),
Vn0 = Vbn − (Ūa1 ∪ · · · ∪ Ūan ).
Observe that Ui0 ∩ Vj0 = ∅ for all indices i, j ∈ {1, . . . , n}. Since V̄b1 ∪ · · · ∪ V̄bn ⊂ X − A
we see that UA = ∪ni=1 Ui0 is an open set containing A and likewise, VB = ∪nj=1 Vj0 is an
open set containing B. Moreover, the set UA and VB are disjoint for if they were not
then we would obtain Ui0 ∩ Vj0 6= ∅ for some indices i, j, an impossibility. Thus X is
T4 .
6.4. The one point compactification
Definition 6.4.1. A compactification of a topological space (X, TX ) is a compact
topological space (Y, TY ) that contains X as a dense subspace.
The goal of this section is to show that compactifications always exist for every
space X. We will describe one method for obtaining a compactification called the One
Point Compactification.
If (X, TX ) is already compact then X is its own compactification. If X is not
compact, let p be an abstract point not contained in X and define Y (as a set) by
Y = X ∪ {p}. Define TY to be the following collection of subsets of Y :


Either
p
∈
/
U
and
U
∈
T
,


X
U ⊂ Y or
(6.1) TY =


p ∈ U and X − U is a compact closed subset of X.
Remark 6.4.2. Note that if X −U is a closed subset of X for any choice of U ∈ TY .
Lemma 6.4.3. TY is a topology on Y .
Proof. This is a direct verification of the three axioms of a topology.
1. ∅ ∈ TY since p ∈
/ ∅ and ∅ ∈ TX .
Y ∈ TY since p ∈ Y and X − Y = ∅, the latter being closed and compact in X.
2. Let Ui ∈ TY , i ∈ I and let U = ∪i∈I Ui . If for all i ∈ I we hve p ∈
/ Ui and
Ui ∈ TX then p ∈
/ U and U ∈ TX showing that U ∈ TY . On the other hand, if
p ∈ Uj for some j ∈ I, then X − Uj is a closed and compact subset of X. But
then p ∈ U and
X − U = X − ∪i∈I Ui = ∩i∈I (X − Ui ) ⊂ X − Uj .
102
6. COMPACTNESS
Since each X − Ui is closed in X (Remark 6.4.2) we see that X − U is a closed
subset of the compact space X − Uj and hence itself compact (part (a) of
Theorem 6.1.9). It follows that U ∈ TY .
3. Let V1 , ..., Vn ∈ TY and set V = V1 ∩ ... ∩ Vn . If there is at least one index
j ∈ {1, ..., n} with p ∈
/ Vj and Vj ∈ TX then p ∈
/ V and V ∈ TX . If p ∈ Vi for
all i = 1, ..., n then X − Vi is closed and compact for all i. But then
X − V = X − ∩ni=1 Vi = ∪ni=1 (X − Vi ),
showing X − V to be a finite union of closed and compact subspaces of X.
Accordingly, X − V itself is a closed and compact subspace of X showing that
V ∈ TY .
We remark that the definition of TY makes it immediate that the subspace topology
on X induced by TY is simply TX . Therefore we can view (X, TX ) as a subspace of
(Y, TY ).
We next turn to the verification of the density of X in Y . Since Y = X ∪ {p},
it is clear that X̄ must either be X or Y (since these are the only two subsets of Y
containing X). But X̄ = X could happen only if X were closed in Y . By definition of
TY , this would mean that {p} would have to be open in Y which in turn would mean
that Y − {p} = X would need to be a closed and compact subspace of X. But our
assumption was that X is not compact, and so {p} is not open in Y and consequently
X is not closed in Y . This excludes the possibility X̄ = X leaving only X̄ = Y .
It remains to be seen that (Y, TY ) is a compact space. Thus, let F be an open cover
of Y . There must be some element U0 ∈ F which contains p and so by the definition
of TY , X − U0 is a compact subspace of X. Thus, X − U0 is covered by finitely many
U1 , ..., Un ∈ F showing that Y is covered by the finite subcover {U0 , U1 , ..., Un } ⊂ F of
F. Ergo, Y is compact. We summarize our findings in the next theorem.
Theorem 6.4.4. Let (X, TX ) be a non-compact space and let Y = X ∪ {p} for
some abstract point p ∈
/ X. Then (Y, TY ), with TY defined as in (6.1), is a compact
space that contains X as a dense subspace. The space Y is referred to as the one point
compactification of X.
Example 6.4.5. Let (X, TX ) = (Rn , TEu ) be n-dimensional Euclidean space and let
p = ∞ be the symbol for an abstract point not in Rn . Consider also the n-dimensional
sphere S n defined as
S n = {(x1 , . . . , xn+1 ) ∈ Rn+1 | x21 + · · · + x2n+1 = 1},
equipped with the relative Euclidean topology on Rn+1 . We will show that the one point
compactification of Rn is homeomorphic to S n . Define the function f : Rn ∪ {∞} → S n
6.4. THE ONE POINT COMPACTIFICATION
103
as
 2x1
2xn
|x|2 − 1


,..., 2
,
; x 6= ∞,

|x|2 + 1
|x| + 1 |x|2 + 1
f (x1 , ..., xn ) =



(0, 0, ..., 0, 1)
; x = ∞,
p
where |x| = x21 + ... + x2n . We note that f is a bijection with inverse function
 y1
yn


,...,
; y 6= (0, . . . , 0, 1),

1 − yn+1
1 − yn+1
f −1 (y1 , ..., yn+1 ) =



∞
; y = (0, . . . , 0, 1).
To see that f and f −1 are continuous we will show that they are open maps. If
U ∈ TRn ∪{∞} is an open set with ∞ ∈
/ U or if V ⊂ S n is an open set with (0, ..., 0, 1) ∈
/ V,
−1
then f (U ) and f (V ) are open since f |Rn : Rn → S n − {(0, ..., 0, 1)} is quite obviously
a homeomorphism (since the coordinate functions of f −1 |S n −{(0,...,0,1)} and of f |Rn are
continuous functions).
Next, pick a set U ∈ TRn ∪{∞} with ∞ ∈ U . According to (6.1), Rn − U is a compact
subspace of Rn and so, according to corollary 6.2.11, it is also bounded. This allows
us to pick an R > 0 so that Rn − B0 (R) ⊂ U (where 0 ∈ Rn is the origin). But then
B(0,...,0,1) (ρ) ∩ S n ⊂ f (U )
with
ρ= √
2
R2
+1
,
showing that (0, ..., 0, 1) ∈ f (U ) has a neighborhood contained in f (U ). By continuity
of f |Rn , every point y ∈ f (U ) − {(0, ..., 0, 1)} also has a neighborhood contained in
f (U ) and so f (U ) is open and therefore f −1 is continuous.
A neighborhood basis for S n around (0, ..., 0, 1) is given by {B(0,...,0,1) (ρ)∩S n | ρ > 0}.
To show that f −1 is open, it suffices to show that f −1 (B(0,...,0,1) (ρ) ∩ S n ) is an open
subset of Rn . But this is a completely explicit computation and one finds that
ρ
p
f −1 (B(0,...,0,1) (ρ) ∩ S n ) = Rn − B0 (R) with R =
.
1 − 1 − ρ2
Note that Rn − B0 (R) ∈ TRn ∪{∞} . For any open subset V ⊂ S n − {(0, ..., 0, 1)}, f −1 (V )
is obviously open and so f −1 is an open map and therefore f is continuous and thus a
homeomorphism.
Example 6.4.6. Consider the included point topology Tp on R. The space (R, Tp )
is not compact (Example 6.1.6) and so we can define Y = R ∪ {∞} to be the one
point compactification of (R, Tp ). To describe the elements of TY , note that a subset
A ⊂ R is compact with respect to Tp if and only if it is finite. For if A = {a1 , a2 , . . . }
is infinite and p ∈
/ A, then the set {{a1 }, {a2 }, . . . } is an open cover of A with no
proper subcovers at all. On the other hand, if A is infinite and p ∈ A, then the set
{{a1 , p}, {a2 , p}, . . . } is again an open cover of A with no proper subcovers. A subset
104
6. COMPACTNESS
A ⊂ R is closed if it doesn’t contain p. Consequently, closed and compact subsets of
(R, Tp ) are precisely finite subsets of R that don’t contain p. From this we obtain
TY = {U ⊂ Y | (p ∈ U and ∞ ∈
/ U ) or (p, ∞ ∈ U and R − U is finite)}.
We see that the one point compactification of (R, Tp ) contains features of the finite
complement topology. Observe that every sequence in Y converges to ∞.
6.5. Flavors of compactness
Definition 6.5.1. Let (X, TX ) be a topological space and F an open cover of X.
We shall say that F is locally finite if every point x ∈ X has a neighborhood Ux that
intersects only finitely many elements from F.
For later intentions, it is not sufficient to define local finiteness to be the property
that every point x ∈ X intersects only finitely many sets in F. A priori, it is not
clear that this is actually a weaker than Definition 6.5.1, though Example 6.5.3 below
confirms that it is. Note that every finite cover F is locally finite.
Example 6.5.2. Consider the cover F = {ha + 1, a − 1i | a ∈ Z} of (R, TEu ). This
is locally finite since for every point x ∈ R, the interval hx − 1, x + 1i intersects at most
2 elements from F.
On the other hand, G = {ha, bi | a, b ∈ Q, a < b} is not locally finite since, given
any point x ∈ R and any neighborhood Ux of x, there are infinitely many rational
numbers contained in Ux . Let q1 , q2 , q3 , · · · ∈ Ux ∩ Q be an infinite sequence chosen so
that qi < qi+1 , then hq1 , qi i intersects Ux for every i = 2, 3, 4, . . . .
Example 6.5.3. Consider the excluded point topology T p on R and let
F = {U ⊂ R | U = R or U = {x} for some x ∈ R − {p} }.
Then F is an open cover of R with the property that every point x ∈ R is contained
in either exactly two elements of F (if x 6= p) or in only one such element (if x = p).
But p ∈ R has no neighborhood that intersects only finitely many elements from F.
We conclude that F is not a locally finite cover even though every point in R only
intersects finitely many sets in F.
Definition 6.5.4. Let (X, TX ) be a topological space.
(a) We shall say that X is locally compact if for every point x ∈ X and every
neighborhood U of x, there exists a neighborhood V of x such that x ∈ V ⊂
V̄ ⊂ U and with V̄ compact.
(b) We call X σ-compact if there exists a sequence A1 , A2 , A3 , . . . of compact subsets
of X such that X = ∪∞
i=1 Ai .
(c) The space X is said to be paracompact if every open cover F of X has a locally
finite refinement F 0 .
Remark 6.5.5. We comment briefly on each point from the previous definition
before moving on. Regarding part (a) from Definition 6.5.4, some authors define local
6.6. EXERCISES
105
compactness of X as meaning that every point x ∈ X has a neighborhood with compact
closure. While this is implied by our definition, it is not equivalent to it (Example ??).
If X is a compact space then we can take each Ai from part (b) of Definition 6.5.4
to be X. Thus a compact space is σ-compact.
Similarly, if X is compact, then every of its open covers F has a finite subcover F 0 .
Since every finite cover is necessarily locally finite, we see that compactness implies
paracompactness.
Example 6.5.6. Euclidean space (Rn , TEu ) is σ-compact since each of Ai = B0 (i),
n
i ∈ N is compact and clearly Rn = ∪∞
i=1 Ai . As we already saw, R is not compact.
Example 6.5.7. The exluded point topology T p on R is not locally compact.
Namely, the point p itself has only one neighborhood, R itself. But we saw in example 6.1.7 that (R, T p ) is not compact
Theorem 6.5.8. Let (X, TX ) be a locally compact, non-compact, Hausdorff space
and let (Y, TY ) be its one point compactification (with Y = X ∪ {∞}, see (6.1)). Then
Y is a compact Hausdorff space.
Proof. Let x, y ∈ Y be two distinct points. If x, y ∈ X then, since X is Hausdorff
and since TX ⊂ TY , x and y possess disjoint neighborhoods Ux and Uy .
On the other hand, if y = ∞ and thus x 6= ∞, then x lies in X and local compactness
of X guarantees the existence of a neighborhood Ux of x in X with Ūx compact. But
then Uy = Y − Ūx belongs to TY since Ūx is compact and closed (the latter because
it is a compact subspace of a Hausdorff space, see part (c) of Theorem 6.1.9). Clearly
Ux ∩ Uy = ∅ demonstrating that Y is Hausdorff.
Corollary 6.5.9. Let X be a locally compact, Hausdorff space. Then X is completely regular (part (b) of Definition 4.1.2).
Proof. If X is compact, let (Y, TY ) = (X, TX ) and if X is not compact, let (Y, TY )
be the one point compactification of (X, TX ). Then Y is a compact, Hausdorff space
(either by assumption on X or by Theorem 6.5.8) and so Y is a normal space according
to part (a) of Theorem 6.3.4. The Hausdorff condition of Y implies that Y has the
T0 property while Corollary 4.5.4 shows that Y is a T3 1 space. The conclusion of
2
the corollary now follows from part (a) of Theorem 4.1.4 by which the subspace of a
completely regular space is again completely regular.
Example 6.5.10. Let X be a locally compact, Hausdorff space. Then for any two
distinct points x, y ∈ X there is a continuous function f : X → R such that f (x) = 0
and f (y) = 1.
6.6. Exercises
6.6.1. Show that (Rn , TEu ) is not compact by finding an explicit open cover of Rn
that has no finite subcovers.
6.6.2. Referring to Definition 6.2.1, prove that
106
6. COMPACTNESS
(a) A totally bounded metric space is bounded.
(b) Show by example that there exist bounded metric spaces that are not totally
bounded.
(c) Consider Rn with the Euclidean metric d2 . Show that a subset A ⊂ Rn is
bounded if and only if it is totally bounded.
6.6.3. Let (X, d) be a metric space and {xk }k ⊂ X a Cauchy sequence. Show that
if {xk }k possesses a convergent subsequence yn = xkn with lim yn = y, then {xk }k is
also convergent and lim xk = y.
6.6.4. Show that the conclusion of part (b) of Theorem 6.3.4 remains true under
a weaker hypothesis. Namely, show that every second-countable space X that is T3 is
also T4 . Hint: Emulate the proof of part (b) of Theorem 6.3.4 and use Lemma 6.2.5
instead of compactness of X, to reduced a cover to a countable subcover.
6.6.5. Prove the following converse of Theorem 6.5.8: If the one point compactification (Y, TY ) of (X, TX ) (with Y = X ∪ {p} and X non-compact) is a Hausdorff space,
then X is locally compact.
6.6.6. Consider the “extended real line ” R̄ = R ∪ {±∞} with ±∞ being two
abstract points, not belonging to R. Consider the collection TR̄ of subsets of R̄ defined
as
±∞ ∈
/ U and U ∈ TEu , or
TR̄ = U ⊂ R̄
.
{±∞} ∩ U 6= ∅ and R − U is closed and compact in R.
Show that TR̄ is a topology on R̄.
Show that (R̄, TR̄ ) is a compact space.
Show that (R, TEu ) is a dense subspace of (R̄, TR̄ ).
Show that (R̄, TR̄ ) is homeomorphic to [0, 1] (with the relative Euclidean topology on [0, 1]).
Conclude that the two-point compactification of the real line is a segment (of positive
length).
(a)
(b)
(c)
(d)
6.6.7. Let (X, d) be a metric space and F1 ⊃ F2 ⊃ F3 ⊃ . . . a sequence of nested,
non-empty compact subspaces of X. Show that if lim diam(Fi ) = 0, then ∩i∈N Fi
i→∞
consists of a single point.
6.6.8. Which of the following subspaces of (Rn , TEu ) is compact?
(a) The sphere S n−1 ⊂ Rn with finitely many points removed.
(b) The graph Γf of a map f : [0, 1]n−1 → R.
(c) The Cantor set C ⊂ R (part (d) of Example 1.5.1).
6.6.9. Find an example of an open cover of (R, TEu ) that has no (positive) Lebesgue
number (compare to Theorem 6.3.2).
6.6.10. Let (X, d) be a compact metric space.
(a) Show that X has finite diameter.
6.6. EXERCISES
(b) Show that there exist points a, b ∈ X with d(a, b) = diam(X).
107
CHAPTER 7
I
Connectedness and Path-Connectedness
tuitively speaking, a disjoint union of two abstract topological spaces consists
of (at least) two “components”- the constituent topological spaces. A prime
examples of this phenomenon is the two point space obtained by taking the
disjoint union of two singletons, forcing the two point space to carry the
discrete topology. Indeed, this archetypical example of a “disconnected”space can be
used to characterized “disconnected”spaces as being precisely those that continuously
surject on it.
The notions of connectedness and disconnectedness of a topological space are explored in Section 7.1. Section 7.2 explores the notion of path-connectedness of a space,
and compares and contrasts it to that of connectedness. While connectedness is implied
by path-connectedness, the converse if false in general. The final Section 7.3 introduces
local version of connectivity and path-connectivity. The main goal of introducing these
concepts is to understand when connected and path-connected components of a space
are open and closed (Theorem 7.3.6 and Corollary 7.3.7).
7.1. Connected topological spaces
Definition 7.1.1. A topological space (X, TX ) is said to be connected if there is
no continuous surjection f : X → {0, 1} where the two point set {0, 1} is given the
discrete topology. Otherwise X is said to be disconnected.
Theorem 7.1.2. Let (X, TX ) be a topological space. Then the following are equivalent:
(a) X is connected.
(b) The only open and closed subsets of X are the empty set and X itself.
(c) If X = A ∪ B with A, B disjoint and open sets, then A = ∅ or B = ∅.
Proof. (a)=⇒(b) If A ⊂ X is both open and closed, then so is X − A. If both
of these were nonempty, as would be the case if A 6= ∅, X, then we could define the
surjective function f : X → {0, 1} as
0
; if x ∈ A,
f (x) =
1
; if x ∈ X − A.
An easy check reveals that f is continuous, contradicting the assumption from part (a)
that X is connected.
109
110
7. CONNECTEDNESS AND PATH-CONNECTEDNESS
(b)=⇒(c) If X = A ∪ B with A and B open, then A and B are also closed (each
being the complement of the other). According to part (b), one of A or B is then the
empty set.
(c)=⇒(a) If f : X → {0, 1} is a surjective map, then f −1 ({0}) and f −1 ({1}) are
disjoint subsets of X that are open and closed (since f is continuous and since {0}
and {1} are open and closed subsets of {0, 1}). By surjectivity of f , they are also both
non-empty contradicting part (c). Therefore, no continuous surjection f : X → {0, 1}
can exist.
Theorem 7.1.3. Let f : X → Y be a continuous function between topological
spaces. If X is connected then so is f (X). In particular, connectedness is a topological
invariant.
Proof. Suppose that f (X) = A ∪ B with A, B ⊂ f (X) disjoint and open sets.
Then f −1 (A) and f −1 (B) are open subsets of X with X = f −1 (A) ∪ f −1 (B). By
connectedness of X, one of f −1 (A) or f −1 (B) has to be the empty set, say f −1 (B).
But then B = ∅ showing that f (X) is connected.
If X and Y are homeomorphic, each of them is the image of the other under a
continuous map and so they are both connected or both disconnected.
An important corollary of the previous result is the Intermediate Value Theorem.
Corollary 7.1.4 (Intermediate Value Theorem). Let X be a connected space and
f : X → R a continuous function and let a, c ∈ Im(f ) be two distinct points with a < c.
If b ∈ ha, ci, then there exists and x ∈ X with f (x) = b (that is b ∈ Im(f )).
Proof. If we could find an element b ∈ ha, ci − Im(f ), we could write
f (X) = (f (X) ∩ h−∞, bi) ∪ (f (X) ∩ hb, ∞i) .
This would imply disconnectedness of f (X), contradicting the previous theorem and
the assumption that X is connected.
Example 7.1.5. If X = {x} is a set with only a single point, then it is connected
for there is no surjection from a set with one element onto a set with two elements.
Example 7.1.6. The excluded point topology makes R into a connected space.
This follows from the observation that if R = A ∪ B with A, B open, then if p ∈ A we
get A = R and if p ∈ B then B = R. In either case, A and B can only be disjoint if
one of them is the empty set.
Example 7.1.7. The set of rational number Q ⊂ R with its √
relative Euclidean
topology,
is
disconnected.
This
can
be
seen
by
setting
A
=
h−∞,
2i ∩ Q and B =
√
h 2, ∞i ∩ Q. Then A, B ⊂ Q are two open and disjoint sets and Q = A ∪ B.
Example 7.1.8. If X is any set with at least two elements, then (X, Tdis ) is disconnected. On the other hand, for any set Y , the space (Y, Tindis ) is connected (here
Tdis is the discrete and Tindis is the indiscrete or trivial topology).
7.1. CONNECTED TOPOLOGICAL SPACES
111
Example 7.1.9. With the finite complement topology, R is a connected space
since there are no non-empty disjoint open subsets of R. The same holds true with the
countable complement topology.
In the statement of the next theorem we shall use the word interval to mean any
of ha, bi, ha, b], [a, bi or [a, b], including intervals of infinite length. Said differently,
intervals are precisely the convex subsets of R. The next theorem characterizes the
connected subsets of the Euclidean line.
Theorem 7.1.10. Consider the Euclidean line (R, TEu ). The nonempty connected
subspaces of R are precisely the intervals in R.
Proof. =⇒ Let X be a nonempty connected subspace of R. If X contains only
one point then it is an interval and the claim of the theorem is proved. Suppose that
X contains at least two points, say x and y, and suppose that x < y. If z ∈ R is any
point with x < z < y then z has to belong to X. For if not, then X could be written
as the union
X = (h−∞, zi ∩ X) ∪ (hz, ∞i ∩ X),
with both sets on the right hand side being open and nonempty (and clearly disjoint)
contradicting the connectedness assumption of X. Therefore z ∈ X. Let ` = inf X and
L = sup X, then X equals an interval from ` to L, the boundaries may be included or
not and may be infinite or not.
⇐= Let X = ha, bi be an open interval for some a < b and assume that X can be
written as X = A ∪ B with A, B disjoint, open subsets of X and with A non-empty.
We shall show that then B is forced to equal the empty set.
Let x ∈ A be any element. Since X is open in R and A is open in X, then A is also
open in R. Thus there exists an open interval around x that is contained in A, making
the following definitions meaningful:
` = inf{r ∈ h−∞, xi | hr, x] ⊂ A}
and
L = sup{s ∈ hx, ∞i | [x, si ⊂ A}.
For each n ∈ N, there exist rn , sn ∈ R such that
1
1
0 < rn − ` <
and 0 < L − sn < ,
n
n
and hrn , sn i ⊂ A. Thus h`, Li = ∪n∈N hrn , sn i is also contained in A. However, since A
is an open interval, neither ` or L can be continued in A.
If we had a < `, it would follow that ` ∈ B and since B is open in R, an interval
around ` would be contained in B. This in turn would force B to intersect A nontrivially, a contradiction, showing that ` = a. In a similar vein it follows that L = b and
thus that A = X, implying B = ∅.
The cases of X being one of the other types of intervals follow similarly, and we
leave them as an exercise (Exercise 7.4.1). We add that these additional cases are also
implied by the next theorem using the current theorem with X = ha, bi.
Theorem 7.1.11. Let (X, TX ) be a topological space and let Y be a subspace of X.
If Y is connected and Z ⊂ X is any set with Y ⊂ Z ⊂ Ȳ , then Z is also connected.
112
7. CONNECTEDNESS AND PATH-CONNECTEDNESS
Proof. Let Z ⊂ Ȳ be any set with Y ⊂ Z and write Z = A ∪ B with A, B open
and disjoint subsets of Z. Then AY = A ∩ Y and BY = B ∩ Y are open and disjoint
subsets of Y with Y = AY ∪ BY . By connectedness of Y , one of AY or BY is the empty
set, say BY = ∅. Let B 0 ⊂ Ȳ be an open set with B = Z ∩ B 0 , then Y ⊂ Ȳ − B 0 .
Since B 0 is open, Ȳ − B 0 is closed and contains Y and so Ȳ ⊂ Ȳ − B 0 . This shows that
B 0 = ∅ and therefore also B = Z ∩ B 0 = ∅. It follows that Z is connected.
Corollary 7.1.12. The closure of a connected subspace is connected. If a space
X possesses a connected dense subset, then X itself is connected.
Example 7.1.13. Consider the included point topology Tp on R. The closure of
{p} is all of R (since R is the only closed set containing p) and since a set with only
one element is always connected (Example 7.1.5), it follows from Theorem 7.1.11 that
(R, Tp ) is connected.
Lemma 7.1.14. Let (X, TX ) be a topological space and Yi ⊂ X, i ∈ I be a family of
connected subspaces. If ∩i∈I Yi 6= ∅ then ∪i∈I Yi is connected.
Proof. Let p ∈ ∩i∈I Yi be any point and let A, B ⊂ ∪i∈I Yi be two open and
disjoint subsets of ∪i∈I Yi with ∪i∈I Yi = A ∪ B. Then Ai = A ∩ Yi and Bi = B ∩ Yi are
open and disjoint subsets of Yi with Yi = Ai ∪ Bi . By the connectedness assumption
of Yi , one of Ai or Bi needs to be the empty set. Suppose that for a given index i0 ∈ I
we had Bi0 = ∅ and notice that then p ∈ Ai0 ⊂ A. For any other index i ∈ I, we
cannot have Ai = ∅ since that would imply that p ∈ Bi ⊂ B, a contradiction since we
already found that p ∈ A. Therefore we obtain that Bi = ∅ for every i ∈ I. But then
B = ∪i∈I Bi = ∅ showing that ∪i∈I Yi is connected.
Definition 7.1.15. Let (X, TX ) be a topological space.
(a) A connected component U of X is any maximal connected subset of X. Said
differently, U is a connected component of X if, whenever V is a connected
subspace of X and U ⊂ V , then U = V .
(b) We say that X is totally disconnected if every connected component of X consists of just a single point.
Theorem 7.1.16. Let (X, TX ) be a topological space.
(a) X is the disjoint union of its connected components.
(b) Every connected component of X is a closed subset of X. In particular, if X
has finitely many components, then each component is both open and closed.
(c) X is connected if an only if it has precisely one connected component.
Proof. (a) Let x ∈ X be any point and let Ux be the subset of X obtained by
Ux = ∪U ∈Ux U
with
Ux = {W ⊂ X | x ∈ W and W is connected}.
By Lemma 7.1.14 Ux is connected and it is clearly a maximal connected subset of X.
Thus every point x ∈ X belongs to a connected component of X showing that X is
the union of its connected components.
7.1. CONNECTED TOPOLOGICAL SPACES
113
Let U, V ⊂ X be two different connected components of X. If we had U ∩ V 6= ∅
then U ∪ V would be a connected space (by Lemma 7.1.14), a contradiction to the
maximal property of U and V . Thus U ∩ V = ∅.
(b) If U is a component of X then Ū is also connected (Theorem 7.1.11) and so by
maximality, U = Ū . Accordingly, U is closed.
(c) This follows directly from the definition. If the only component of X is U then,
since every component is connected, so is X. If X is connected, let Ui , i ∈ I be
its components. Then X is a connected set containing Ui and so by the maximallity
property of Ui , we must have X = Ui for all i ∈ I.
We note that part (c) of the above theorem cannot in general be improved by
claiming that connected components of X are always open subsets (however, compare
to Corollary 7.3.7). Here is an example demonstrating this.
Example 7.1.17. Consider the set of rational numbers Q ⊂ R equipped with the
relative Euclidean topology. If Y ⊂ Q is any subset with at least two elements, say
y1 , y2 ∈ Y with y1 < y2 , then
A = h−∞, λi ∩ Y
and
B = hλ, ∞i ∩ Y,
with λ ∈ hy1 , y2 i ∩ (R − Q) arbitrary, are two disjoint, open and non-empty subsets of
Y with Y = A ∪ B. Thus, any such Y is disconnected showing that the only connected
subsets of Q are sets with only one element. Accordingly, Q is totally disconnected
with closed (but not open!) components {q}, q ∈ Q.
Example 7.1.18. Consider the lower limit topology Tll on R. The connected components (R, Tll ) are the single point sets {x}, x ∈ R since whenever A ⊂ R has at least
two elements a, b ∈ A with a < b, then A can be written as the union of two disjoint
and non-empty open subsets of A:
A = (h−∞, bi ∩ A) ∪ ([b, ∞i ∩ A)
This shows that any such A is disconnected. Consequently, (R, Tll ) is completely disconnected.
Example 7.1.19. Consider the Euclidean plane (R2 , TEu ) and let Y be its subspace
given by
Y = {(x, y) ∈ R2 | xy = 0 and (x, y) 6= (0, 0)}
Geometrically, Y is the union of the x-axis with the y-axis and with the origin removed.
Then Y has 4 connected components and they are:
U+ = {(x, 0) ∈ R2 | x > 0} = Positive half of the x-axis.
U− = {(x, 0) ∈ R2 | x < 0} = Negative half of the x-axis.
V+ = {(0, y) ∈ R2 | y > 0} = Positive half of the y-axis.
V− = {(0, y) ∈ R2 | y < 0} = Negative half of the y-axis.
114
7. CONNECTEDNESS AND PATH-CONNECTEDNESS
To see that these are indeed the components of Y , note firstly that each of U± and V±
is homeomorphic to R and therefore connected (Theorem 7.1.10). To see that U+ is
a maximal connected subspace of Y , let A, B ⊂ R2 be the open and disjoint subsets
given by
A = {(x, y) ∈ R2 | x < |y|}
and
B = {(x, y) ∈ R2 | x > |y|}
Then U+ ⊂ A and U− ∪ V+ ∪ V− ⊂ B. If U+ were not a maximal connected subspace
of Y and were contained in a larger connected subset Ũ ⊂ Y , then A ∩ Ũ and B ∩ Ũ
would be disjoint, open and non-empty subsets of Ũ with Ũ = (A ∩ Ũ ) ∪ (B ∩ Ũ ), a
contradiction to the connectedness of Ũ . Consequently, U+ is a component of Y , the
cases of U− , V+ , V− are handled analogously.
7.2. Path-connectedness
Definition 7.2.1. Let (X, TX ) be a topological space.
(a) A path in X is a continuous function α : [0, 1] → X where [0, 1] is equipped with
the relative Euclidean topology. The points α(0), α(1) are called the initial (or
starting) and terminal (or ending) points of α. We will say that α joins x to y
or that x is joined to y by α.
(b) If α : [0, 1] → X is a path in X then we shall call the path ᾱ : [0, 1] → X given
by ᾱ(t) = α(1 − t), the inverse path of α.
(c) Given two paths α, β : [0, 1] → X with α(1) = β(0), we define the product path
α · β : [0, 1] → X of α and β as

; t ∈ [0, 21 ],
 α(2t)
α · β(t) =

β(2t − 1)
; t ∈ [ 21 , 1].
A couple of remarks are in order. If x is joined to y by means of a path α, then
y is joined to x by the path ᾱ. The function α · β from part (c) of Definition 7.2.1 is
continuous by Lemma 3.1.12. If α joins x to y and β joins y to z then α · β joins x to
z. With this in mind, we are ready to introduce the main concept of this section.
Definition 7.2.2. A topological space X is called path-connected if for every two
points x, y ∈ X there exists a path α in X with α(0) = x and α(1) = y.
Theorem 7.2.3. Let (X, TX ) and (Y, TY ) be two topological space and let f : X → Y
be a continuous function. If X is path-connected then so is f (X). In particular, pathconnectedness is a topological invariant.
Proof. Given two points x, y ∈ f (X), let a, b ∈ X be points with f (a) = x and
f (b) = y. Let α : [0, 1] → X be a path connecting a to b, then f ◦ α is a path in f (X)
connecting x to y.
If X and Y are homeomorphic spaces then they are each the image of the other.
Consequently they are either are both path-connected or neither of them is.
7.2. PATH-CONNECTEDNESS
115
Theorem 7.2.4. If X is path connected then it is connected. The converse is
generally not true (Example 7.2.7).
Proof. Suppose X were not connected. Then we could write X = A ∪ B with
A, B two open, disjoint and non-empty sets. Let a ∈ A and b ∈ B be any two points
and let α : [0, 1] → X be a path joining a to b. Then the sets
A0 = α−1 (A)
and
B 0 = α−1 (B),
are both open (since α is continuous), non-empty (since 0 ∈ A0 and 1 ∈ B 0 ) and disjoint
(since A and B are disjoint) subsets of [0, 1]. This implies that [0, 1] is disconnected, a
contradiction to Theorem 7.1.10. Therefore X must be connected.
Example 7.2.5. Euclidean n-dimensional space (Rn , TEu ) is path-connected and
hence also connected. Given any two points x, y ∈ Rn , the path α : [0, 1] → Rn given
by α(t) = x + t(y − x) starts at x and ends at y.
Example 7.2.6. For any point x ∈ Rn and for any r > 0, the ball Bx (r) is
path-connected (and thus also connected). This is seen by observing that every point
y ∈ Bx (r) can be connected to x via the path αy : [0, 1] → Bx (r) defined by αy (t) =
y + t(x − y). Given any pair of points y1 , y2 ∈ Bx (r), the path αy1 · ᾱy2 connects y1 to
y2 .
Example 7.2.7. (Of a connected but not path connected space) Let X be the topologists sine curve from Example 2.2.15, that is let X be the subspace of the Euclidean
plane (R2 , TEu ) given by
X = {(x, sin x1 ) | x ∈ h0, 1]} ∪ ({0} × [0, 1]) ⊂ R2 .
Note that the set Y1 = {(x, sin( x1 )) | x ∈ h0, 1]} ⊂ X is homeomorphic to h0, 1] and is
thus connected by Theorem 7.1.10. According to Theorem 7.1.11, the closure Ȳ1 of Y1
is then also closed as is any set Z ⊂ R2 with Y1 ⊂ Z ⊂ Ȳ1 . We claim that Z = X
satisfies this double inclusion showing that it is connected. To show that X ⊂ Ȳ1 , it
suffices to show that every element in Y2 = {0} × [0, 1] is the limit of a convergent
sequence {xk }k ⊂ Y1 . This is easily seen for if (0, y) ∈ Y2 , let xk = (tk , y) ∈ Y1 be any
sequence with tk a convergent sequence with limit 0. For example, choosing tk as
1
tk =
with k ∈ N and t1 = arcsin |[ π , 3π ] (y),
2 2
t1 + 2πk
will do.
We will now show that X is not path-connected. If we suppose to the contrary
that X is path-connected, then we can find a path α : [0, 1] → X connecting (0, 0) to
(1, sin 1). Since the y-axis is a closed subset of R2 , the set
A = α−1 (y-axis ∩ X),
is a closed subset of [0, 1] and it is nonempty since it contains 0. Let t0 ∈ A be any
point and suppose that α(t0 ) = (0, b0 ) for some b0 ∈ [0, 1]. Consider the neighborhood
116
7. CONNECTEDNESS AND PATH-CONNECTEDNESS
V of (0, b0 ) in X given by
V = h− 12 , 12 i × hb0 − 21 , b0 + 12 i ∩ X.
By continuity of α, there exists a δ > 0 so that α(ht0 − δ, t0 + δi ∩ [0, 1]) ⊂ V . Since
ht0 − δ, t0 + δi ∩ [0, 1] is path-connected, so is its image (Theorem 7.2.3), and it must
therefore lie in the path-connected component of V that contains α(t0 ). This however
is simply V ∩ y-axis (see Figure 1) and so ht0 − δ, t0 + δi ∩ [0, 1] is a subset of A. This
shows that A is both open and closed and so is therefore B = [0, 1] − A. Likewise, both
A and B are nonempty seeing as they contain 0 and 1 respectively. This contradicts
the connectedness of [0, 1] and so our supposed path α cannot exist. Thus X is not
path-connected.
0.5
.....
−0.5
0.5
.....
−0.5
Figure 1. The path-connected components of V for the case of α(t0 ) =
(0, 0). The picture looks similar for lather values of α(t0 ).
With our investigations of connectedness and path-connectedness thus far, we can
now partially answer Question 3.4.1 about when Euclidean n-space (Rn , TEu ) and Euclidean m-space (Rm , TEu ) can be homeomorphic.
Corollary 7.2.8. If the Euclidean line (R, TEu ) is homeomorphic to Euclidean
n-dimensional space (Rn , TEu ), then n = 1.
Proof. Suppose that f : R → Rn is a homeomorphism. Then f |R−{0} : R − {0} →
R − {f (0)} is also a homeomorphism. However, R − {0} is not an interval and
is therefore not connected by Theorem 7.1.10. On the other hand, we claim that
Rn − {f (0)} is even path-connected, and therefore connected, when n ≥ 2. To see this,
let x, y ∈ Rn be any two points and consider the straight line path α(t) = x + t(y − x).
n
7.2. PATH-CONNECTEDNESS
117
If f (0) does not lie on α, then α is a path in Rn − {f (0)} from x to y. If f (0) does
lie on α, let z ∈ Rn be any point not collinear with x and y and consider the straight
line paths β(t) = x + t(z − x) and γ(t) = z + t(y − z) connecting x to z and z to y
respectively. Then β ·γ is a path from x to y in Rn −{f (0)} showing that Rn −{f (0)} is
path-connected. Since connectedness is a topological invariant, it follows that n must
be 1.
Theorem 7.2.9. An open subset U of Euclidean space Rn is connected if and only
if it is path connected.
Proof. In view of Theorem 7.2.4, we only need to show that if U is connected
then it is also path-connected. Let x ∈ U be any point and define
A = {y ∈ U | x and y can be joined by a path in U },
B = {y ∈ U | x and y can not be joined by a path in U }.
Clearly X = A ∪ B and A 6= ∅ since x ∈ A. We will show that both A and B are open
subsets of U . Since U is connected, this will imply that B = ∅ and A = U , as desired.
To see that A is open, let y ∈ A be any point and let α : [0, 1] → U be a path joining
x to y. Let ε > 0 be such that By (ε) ⊂ U and for any z ∈ By (ε) let βz : [0, 1] → U
be the radial path from y to z, that is βz (t) = (1 − t) · y + t · z. Then α · βz is a path
from x to z showing that By (ε) ⊂ A. Since y ∈ A was arbitrary, we conclude that A
is open.
To see that B is open, we proceed similarly. Let y ∈ B be any point and let ε > 0
be such that By (ε) ⊂ U . If there were a point z ∈ By (ε) ∩ A, there would have to be
a path α : [0, 1] → U from z to x. Letting βz be as in the previous paragraph, the
product path βz · α would be a path from y to x contradicting our choice of y ∈ B.
Therefore By (ε) is contained in B and hence B is open.
Theorem 7.2.10. Let (X, TX ) and (Y, TY ) be topological spaces and let X × Y be
given the product topology. Then X × Y is (path-)connected if and only if each of X
and Y are (path-)connected.
Proof. Since X and Y get surjected on by the projection maps πX : X × Y → X
and πY : X ×Y → Y , (path-)connectedness of X ×Y implies that of X and Y (Theorem
7.1.3 and Theorem 7.2.3).
In the other direction, assume that X and Y are connected. Note that each of
X × {y0 },
{x0 } × Y
and Z = X × {y0 } ∪ {x0 } × Y
is a connected subspace of X × Y . The connectedness of the first two these follows
from Corollaries 5.1.9 and 7.1.3, for the third it follows from Lemma 7.1.14 (since
X × {y0 } ∩ {x0 } × Y = {(x0 , y0 )} =
6 ∅). To show that X × Y is connected, write X × Y
as the union A ∪ B with A, B open and disjoint subsets of X × Y . By connectedness,
each (X × {y0 }) ∪ ({x0 } × Y ) with x0 ∈ X and y0 ∈ Y arbitrary, has to be contained
entirely in either A or B. The same is true of (X × {y1 }) ∪ ({x0 } × Y ) for every other
118
7. CONNECTEDNESS AND PATH-CONNECTEDNESS
y1 . But
[(X × {y0 }) ∪ ({x0 } × Y )] ∩ [(X × {y1 }) ∪ ({x0 } × Y )] = {x0 } × Y,
so that both of (X × {y0 }) ∪ ({x0 } × Y ) and (X × {y1 }) ∪ ({x0 } × Y ) lie in the same
set, either A or B. Since y0 , y1 ∈ Y were completely arbitrary, we see that all of X × Y
lies in either A or B. Thus, one of A or B has to be the empty set, as claimed.
Finally, if X and Y are path-connected then so is X × Y . Namely, given two points
(x1 , y1 ), (x2 , y2 ) ∈ (X × Y ), let α : [0, 1] → X and β : [0, 1] → Y be paths joining x1 to
x2 and y1 to y2 respectively. Then α × β : [0, 1] → X × Y is a path joining (x1 , y1 ) to
(x2 , y2 ).
The following lemma is the “path-connected”version of Lemma 7.1.14.
Lemma 7.2.11. Let (X, TX ) be a topological space and Yi ⊂ X, i ∈ I be a family
of path-connected subspaces. If ∩i∈I Yi 6= ∅ then ∪i∈I Yi is a path-connected subspace of
X.
Proof. Let p ∈ ∩i∈I Yi be any point and for x ∈ Yi , let αx be a path in Yi
connecting x to p. Given any x, y ∈ ∪i∈I Yi , the path αx · ᾱy connects x to y.
Definition 7.2.12. Let (X, TX ) be a topological space. A path-connected component of X is any maximal path-connected subspace U of X. Maximallity here refers
to the property that if V is a path-connected subspace of X with U ⊂ V then U = V .
Example 7.2.13. The topologists sine curve defined in Example 2.2.15 and considered again in Example 7.2.7, has two path connected components, namely Y1 =
{(x, sin x1 ) ∈ R2 | x ∈ h0, 1]} and Y2 = {0} × [0, 1], but only has one connected component.
The following is the analogue of Theorem 7.1.16 for the path-connected case.
Theorem 7.2.14. Let (X, TX ) be a topological space.
(a) X is the disjoint union of its path-connected components.
(b) X is path-connected if and only if it has a single path-connected component.
Proof. (a) For a given x ∈ X, the path-connected component Vx containing it
can be obtained as
Vx = ∪V ∈V V
with
V = {W ⊂ X | x ∈ W and W is path connected}.
The set Vx is path-connected by Lemma 7.2.11, and it is clearly maximal with respect
to this property. Two path-connected components U, V ⊂ X are disjoint for if they
were not, Lemma 7.2.11 would imply that U ∪ V is also path-connected, contradicting
the maximallity of both U and V .
(b) This proof is the complete analogue of the proof of part (c) of Theorem 7.1.16
and is left as an exercise.
Unlike in the case of connectedness (part (b) of Theorem 7.1.16), a path-connected
component of X need not be closed nor open.
7.3. LOCAL CONNECTIVITY
119
Example 7.2.15. Consider the topologists sine curve X from Examples 2.2.15 and
7.2.7). The two subsets Y1 , Y2 ⊂ X defined as
Y1 = {(x, sin 1/x) | x ∈ h0, 1]},
Y2 = {0} × [0, 1],
are path-connected seeing as they are homeomorphic to h0, 1] and [0, 1] respectively.
Since X = Y1 ∪ Y2 and X is not path connected (Example 7.2.7), Y1 and Y2 are the
only two path-connected components of X. It is easy to see that Y2 is closed in X for
Y2 = X ∩ ({0} × R). However, Y2 is not open in X for if it were we could find an open
set U ⊂ R2 with Y2 = U ∩ X. But then, given any point (0, y) ∈ Y2 , we would need to
be able to find an ε > 0 with B(0,x) (ε) ⊂ U , an impossibility since each such B(0,y) (ε)
will intersect Y1 . We conclude that
The path connected component Y1 is open but not closed in X.
The path connected component Y2 is closed but not open in X.
The root for this phenomenon lies in the lack of a path-connected analogue of part (b)
of Theorem 7.1.11.
7.3. Local connectivity
Definition 7.3.1. Let (X, TX ) be a topological space.
(a) We say that X is locally connected if every point x ∈ X has a neighborhood
basis Bx consisting of connected open sets. Said differently, we require that for
every point x ∈ X and every neighborhood U of x there exists a connected
open set V ⊂ X with x ∈ V ⊂ U .
(b) We say that X is locally path connected if every point x ∈ X has a neighborhood
basis Bx consisting of path connected sets. Said differently, we require that
for every point x ∈ X and every neighborhood U of x there exists an pathconnected open set V ⊂ X with x ∈ V ⊂ U .
The next lemma is the local analogue of Theorem 7.2.4. The proof is omitted as it
is largely identical to that of Theorem 7.2.4.
Lemma 7.3.2. A locally-path connected space is locally connected.
Example 7.3.3. Every open subset U of Euclidean space (Rn , TEu ) is locally-path
connected and hence locally connected. This is obvious since, given any point x ∈ Rn
and given any neighborhood of U of x, there exists an ε > 0 with Bx (ε) ⊂ U . Any
such Bx (ε) is path-connected by Example 7.2.6.
The notions of path-connectedness and of local path-connectedness are independent
in that neither implies the other. Here are examples to substantiate this claim.
Example 7.3.4 (A path-connected but not locally path-connected space). Let X ⊂
R be the infinite broom from Example 2.2.16, that is let X be the subspace of the
2
120
7. CONNECTEDNESS AND PATH-CONNECTEDNESS
Euclidean plane given by
X=
∪∞
n=0 In
with
In =

 {(x, n1 · x) ∈ R2 | x ∈ [0, 1]} ;

{(x, 0) ∈ R2 | x ∈ [0, 1]}
;
n ∈ N,
n = 0.
It is easy to see that X is path-connected (and hence also connected), for is x ∈ X is
any point then αx : [0, 1] → X, given by αx (t) = (1−t)·x, is a path connecting x to the
origin. Given any two points x1 , x2 ∈ X, the path αx1 · ᾱx2 connects the former to the
latter. However, X is not locally connected and hence also not locally path connected.
To see this, consider the point x0 = (1, 0) ∈ X and its neighborhood U = Bx0 ( 21 ) ∩ X.
Then U contains no connected neighborhood of x0 . For if V ⊂ U were a connected
neighborhood of x0 , we could find some ε > 0 so that Bx0 (ε) ∩ X ⊂ V . But every
such set has infinitely many path connected (and hence connected) components as is
evident from Figure 2.
Bx0 (ε)
X
..
..
..
..
.
U
..
.
| {z }
Bx0 (ε)
Figure 2. The two neighborhoods U and Bx0 (ε) from Example 7.3.4.
The intersection of the infinite broom X with Bx0 (ε) is an infinite disjoint
union of intervals and is therefore not connected.
Example 7.3.5 (A locally path-connected but not path-connected space). The
subspace X = [0, 1] ∪ [2, 3] ⊂ R of the Euclidean line is both locally connected and
locally path-connected by not connected or path-connected.
Every discrete space (X, Tdis ) with at least two points is locally connected and
locally path-connected by not connected or path-connected.
Theorem 7.3.6. Let (X, TX ) be a topological space.
(a) X is locally connected if and only if every connected component of every open
set U ⊂ X is an open subset of X.
(b) X is locally path-connected if and only if every path-connected component of X
is an open subset of X.
7.4. EXERCISES
121
Proof. (a) =⇒ Suppose that X is locally connected, let U ⊂ X be a an open
subset and let U0 ⊂ U be a connected component of U . Then every point x ∈ U0
has a connected neighborhood Ux,0 . It is necessary that Ux,0 be contained in U0 for
if not, then U0 ∪ Ux,0 would be a connected subspace of U (Lemma 7.1.14), violating
the maximallity condition from the definition of a connected component (Definition
7.1.15). Given this, we can write
U0 = ∪x∈U Ux,0
showing that U0 is an open set.
⇐= Suppose the connected components of open subsets of X are open sets. Let
x ∈ X be any point and U any neighborhood of x. Then Ux , the connnected component
of U that contains x, is a connected, open set with x ∈ Ux ⊂ U , showing that X is
locally connected.
(b) This part of the proof is a verbatim repeat of the proof of part (a), but substituting “connected” with “path-connected”.
The next corollary follows from Theorems 7.1.16 and 7.3.6.
Corollary 7.3.7. In a locally connected space, the connected components are both
open and closed. Similarly, in a locally path-connected space, the path connected components are both open and closed.
Proof. The connected case is immediate from Theorems 7.1.16 and 7.3.6.
For the path-connected case, let X be a path-connected space and let Ui , i ∈ I be
its path-connected components. If x ∈ Ui and Ux is a path-connected neighborhood of
x, then Ux ⊂ Ui by Lemma 7.2.11. Thus Ui = ∪x∈Ui Ux showing that Ui is open. But
then ∪j∈I−{i} Uj is also open and equal to X − Ui showing that Ui is closed.
7.4. Exercises
7.4.1. Using an approach as in the second half of the proof of Theorem 7.1.10, show
that intervals of the type [a, bi, ha, b] and [a, b] are connected (without appealing to
Theorem 7.1.11).
7.4.2. For n ≥ 2, let A ⊂ Rn be the subspace consisting of all points with at least
one rational coordinate. Is A (path-)connected?
7.4.3. Show that each of the included and excluded point topologies Tp and T p ,
render R a path-connected space.
7.4.4. Examine whether the irrational numbers in R are (path-)connected with
respect to the topology T ∈ {Tp , T p , Tcc , TF ,p }. Determine its (path-)connected components in each case.
7.4.5. Let X be a topological space and An ⊂ X, n ∈ N, a family of (path)connected subspaces. If An ∩ An+1 6= ∅, show that A = ∪∞
n=1 An is a (path-)connected
subspace of X.
122
7. CONNECTEDNESS AND PATH-CONNECTEDNESS
7.4.6. Find an example of a topological space X with a disconnected subspace Y
such that the closure Ȳ is connected.
7.4.7. Show that no two among the spaces h0, 1i, [0, 1i and [0, 1] are homeomorphic.
7.4.8. Let A ⊂ R2 be a countable subset, and consider R2 equipped with the
Euclidean topology. Show that R2 − A is a path-connected space.
7.4.9. Find an example of a topological space X and a connected subspace Y such
that
(a) Y̊ is disconnected.
(b) ∂Y is disconnected.
7.4.10. Show that a connected and locally path-connected space is path-connected.
CHAPTER 8
T
Quotient spaces
his chapter introduces a powerful method for constructing new topological
spaces from given ones by means of taking “quotients”. This construction is
commonplace in set theory and we extend it here to the topological setting
by endowing the quotient set with a natural topology - the quotient topology.
Section 8.1 introduces quotient spaces, presents first examples and explores some of
the properties of quotient spaces, and maps from and into quotient spaces. Of the
remaining sections, each focuses on a particular set of examples.
8.1. Quotient spaces: Definitions, properties and first examples
We start this section by reviewing quotient sets first, postponing the relevant topology for a couple of paragraphs. We introduce three mutually equivalent but slightly
different viewpoints of quotient sets. We shall use each of these approaches below for
dealing with quotient spaces.
Definition 8.1.1. Let X be a non-empty set.
(a) The quotient set X/π associated to a surjective function π : X → Y onto a
non-empty set Y is defined to be X/π = Y .
(b) The quotient set X/ ∼ associated to an equivalence relation ∼ on X is the set
of equivalence classes: (X/ ∼) = {[x] | x ∈ X} with [x] = {x0 ∈ X | x0 ∼ x}.
(c) The quotient set X/P associated to a partition P = {Pi | i ∈ I} of X is defined
as X/P = I.
The three notions of quotient sets from Definition 8.1.1 coincide with one another,
provided the surjection π, the equivalence relation ∼ and the partition P are properly
related. Here are the details:
π induces ∼ Let π : X → Y be a surjective function and define the relation ∼ on
X as
x 1 ∼ x2
if and only if
π(x1 ) = π(x2 ).
Then the function (X/π) → (X/ ∼) given by x 7→ [x] is a bijection of quotient sets.
∼ induces P Given an equivalence relation ∼ on X, we define the partition P on
X by
P = {[x] | x ∈ X},
123
124
8. QUOTIENT SPACES
where [x] is as in part (b) of Definition 8.1.1. The function (X/ ∼) → (X/P) given
by [x] 7→ [x] (where [x] is thought of an equivalence class and as an element of the
partition P respectively) is again a bijection between quotient sets.
P induces π Let P = {Pi | i ∈ I} be a partition of X with Pi 6= ∅ for all i ∈ I.
We define π : X → Y by setting Y = I and
π(x) = i
if
x ∈ Pi .
The quotient sets (X/P) and (X/π) are equal by Definition 8.1.1.
Notice that our considerations establish the commutativity of the diagram below
in which each map is a bijection.
/ {Equivalence relations ∼ on X.}
Surjections π : X → Y .
O
g
gggg
gggg
g
g
g
gggg
gs ggg
Partitions P = {Pi | i ∈ I}
of X with Pi 6= ∅, ∀i ∈ I.
Example 8.1.2. Let X = {1, 2, 3, 4, 5, 6}, let Y = {e, o} and define π : X → Y by
π(1) = π(3) = π(5) = o
and
π(2) = π(4) = π(6) = e.
Define the relation ∼ on X by
a∼b
if and only if
b − a is even.
Finally, let P be the partition of X given by P = {{1, 3, 5}, {2, 4, 6}}. Then π, ∼ and
P are related as above and each of X/π, X/ ∼ and X/P, is a set of two elements.
With these preliminaries out of the way, we now turn to the topology of quotient
sets. We state our definition in terms of a surjective function π : X → Y though we
could equally well use equivalence relations or partitions.
Definition 8.1.3. Let (X, TX ) be a topological space, let Y be a any non-empty
set and let π : X → Y be a surjective function.
(a) The quotient topology TX/π on Y , induced by X and π (or simply the quotient
topology), is defined by
TX/π = {U ⊂ Y | π −1 (U ) ∈ TX }.
(b) For topologies TX and TY on X and Y respectively, the surjective function
π : (X, TX ) → (Y, TY ) is called a quotient map if it has the property that
U ∈ TY
if and only if
π −1 (U ) ∈ TX .
It is quite straightforward to verify that TX/π is a topology on Y . Clearly the empty
set and Y belong to TX/π since π −1 (∅) = ∅ and π −1 (Y ) = X. The second and third
axiom of a topology (Definition 2.1.1) follow from the equalities
π −1 (∪i∈I Ui ) = ∪i∈I π −1 (Ui )
and
π −1 (∩i∈I Ui ) = ∩i∈I π −1 (Ui ),
where Ui ∈ Tx/π , i ∈ I (these two equalities were the subject of Exercise 3.5.1).
8.1. QUOTIENT SPACES: DEFINITIONS, PROPERTIES AND FIRST EXAMPLES
125
Remark 8.1.4. While Definition 8.1.3 seemingly defines two new terms, those of
a quotient space and of a quotient map, each of these determines the other. It is easy
to see (Exercise 8.7.1) that π : (X, TX ) → (Y, TY ) is a quotient map if and only if
TY = TX/π .
We postpone the exploration of concrete examples of quotient spaces for the moment
and turn to some of their general properties instead.
Proposition 8.1.5. Let (X, TX ) be a topological space, let π : X → Y be a surjective function onto the set Y and let TX/π be the associated quotient topology.
(a) The quotient map π : (X, TX ) → (Y, TX,π ) is continuous.
(b) The quotient topology TX/π on Y is the finest topology (Definition 2.1.3) for
which π is continuous. Said differently, if TY is any topology on Y for which
π : (X, TX ) → (Y, TY ) continuous, then TY ⊂ TX,π .
(c) If f : (X, TX ) → (Y, TY ) is a continuous surjection that is additionally also
either open or closed, then f is a quotient map.
Proof. (a) Immediate from Definition 8.1.3.
(b) If TY is a topology on Y for which the map π : X → Y is continuous, then for
every U ∈ TY the set π −1 (U ) belongs to TX . Thus any such U automatically lies in
TX/π verifying the claim.
(c) Let f : (X, TX ) → (Y, TY ) be a continuous surjection and let U ∈ TY be any
set. By continuity of f , f −1 (U ) lies in TX . Suppose that f is also an open map and
that U ∈ TY is a set such that f −1 (U ) ∈ TX . The openness property of f implies then
that the set f (f −1 (U )) = U is open in Y . Thus TY = {U ⊂ Y | f −1 (U ) ∈ TX } is the
quotient topology associated to X and f .
If f is closed rather than open and U ⊂ Y is again a set with f −1 (U ) ∈ TX , then
X − f −1 (U ) is a closed set and hence so is f (X − f −1 (U )) = Y − U . Thus U is again
an open set and we arrive at the same conclusion.
We next investigate when functions on a quotient space are continuous. For this
purpose, let (X, TX ) be topological space, let π : X → Y be a surjection and assume
that Y is given the quotient topology TX/π . Let (Z, TZ ) be another topological space
and let f : Y → Z be a function. Associated to f is the function fˆ : X → Z defined
by fˆ = f ◦ π, see commutative diagram below.
π /
Y
X
fˆ
~
~~
~~
~
~~~ f
Z
ˆ
Notice that if x1 , x2 ∈ π (y) then f (x1 ) = fˆ(x2 ), a fact that we will express by saying
that “fˆ is constant on π −1 (y) for each y ∈ Y ”. Conversely, given a function fˆ : X → Z
that is constant on π −1 (y) for each y ∈ Y , we can define f : Y → Z as
−1
f (y) = fˆ(xy )
for any choice
xy ∈ π −1 (y).
126
8. QUOTIENT SPACES
The constancy of fˆ on π −1 (y) guarantees that f is well defined. Thus, there is a
bijective correspondence between functions f : Y → Z and functions fˆ : X → Z, the
latter of which need to be constant on π −1 (y) for all y ∈ Y . The next theorem shows
that this correspondence remains bijective after restricting to the subset of continuous
functions in each of these sets.
Theorem 8.1.6. Let π : (X, TX ) → (Y, TX/π ) be a quotient map and let (Z, TZ ) be
a topological space. A function f : Y → Z is continuous if and only if the function
fˆ : X → Z given by fˆ = f ◦ π, is continuous.
Proof. If f : Y → Z is continuous then clearly so is fˆ : X → Z, being the
composition of two continuous functions. If on the other hand fˆ is continuous, then,
for any open set V ⊂ Z, consider f −1 (V ). This is open in Y if and only if π −1 (f −1 (V ))
is open in X. But
π −1 (f −1 (V )) = (f ◦ π)−1 (V ) = fˆ−1 (V ),
and fˆ−1 (V ) is open in X by continuity of fˆ. Thus f too is continuous.
The situation of functions to a quotient space is somewhat more precarious. Certainly given a continuous function ĝ : Z → X, the induced function g : Z → Y given
by g = π ◦ ĝ is again continuous, see commutative diagram below.
π /
XO
>Y
ĝ
~~
~~
~
~ g
~~
Z
But given a continuous function g : Z → Y , there need not be a continuous function
ĝ : Z → X with g = π ◦ ĝ. Instances of this type abound, for illustration see Examples
?? and ??.
In the concrete examples of quotient spaces considered in Section 8.2, it will sometimes prove advantageous to construct a quotient space in several stages. What is
meant by this is that we first form an “intermediate”quotient space from a given space
X, say by moding out by a partition P1 of X, and then further moding out by a partition P2 of X/P1 to arrive at (X/P1 )/P2 . The next theorem show that the latter can
be arrived at by just a single quotient space construction.
Theorem 8.1.7. Let (X, TX ) be a topological space, let P1 = {Ui ⊂ X | i ∈ I1 } be
a partition of X and let P2 = {Vi ⊂ X/P1 | i ∈ I2 } be a partition of X/P1 . Then
(X/P1 )/P2 ∼
= X/P3 ,
where P3 is the partition of X given by P3 = {Wi ⊂ X | Wi = π1−1 (Vi ), i ∈ I2 }. Here
π1 : X → X/P1 is the quotient map.
Proof. Let π2 : (X/P1 ) → (X/P1 )/P2 and π3 : X → X/P3 be the additional two
quotients maps. Note that as sets, each of (X/P1 )/P2 and X/P3 are defined to equal
the indexing set I2 . Thus we define f : (X/P1 )/P2 → X/P3 to simply be the identity.
The four maps π1 , π2 , π3 and f fit into the commutative diagram:
8.1. QUOTIENT SPACES: DEFINITIONS, PROPERTIES AND FIRST EXAMPLES
π1
X
π3
f =id
X/P3
/
/
127
X/P1
π2
(X/P1 )/P2
Showing that f is a homeomorphism is equivalent to showing that a subset W ⊂ X/P3
is open if and only if that same set W is also open when regarded as a subset of
(X/P1 )/P2 . But, W ⊂ (X/P1 )/P2 is open if and only if π2−1 (W ) is an open subset of
X/P1 which in turn is true if and only if π1−1 (π2−1 (W )) = (π2 ◦ π1 )−1 (W ) is an open
subset of X. Since f ◦ π3 = π2 ◦ π1 , the conclusion follows.
Theorem 8.1.8. Let fˆ : X1 → X2 be a homeomorphism and let πi : Xi → Yi ,
i = 1, 2 be two quotient maps such that there exists a injective function f : Y1 → Y2 for
which the diagram
X1
π1
Y1
fˆ
f
/
/
X2
π2
Y2
commutes. Then f is a homeomorphism between Y1 and Y2 .
Proof. The function f : Y1 → Y2 is surjective since both π2 and fˆ are surjective,
showing that f is a bijection. According to Theorem 8.1.6, f : Y1 → Y2 is continuous if
and only if π2 ◦ fˆ : X1 → Y2 is continuous. The latter is a composition of two continuous
functions and therefore continuous. A symmetric argument shows that f −1 : Y2 → Y1
is likewise continuous.
The next theorem points to which topological properties are inherited by quotient
spaces.
Theorem 8.1.9. Let π : X → Y be a quotient map. If X is either compact,
connected or path-connected, then so is Y .
Proof. This follows from the fact that Y = f (X) and that each of the properties of compactness, connectedness and path-connectedness is preserved under taking
continuous images (Theorem 6.1.8, Theorem 7.1.3 and Theorem 7.2.3 respectively). With some of the general properties of quotient spaces out of the way, we turn
to examples. Specific groups of examples are considered in sections ?? – ??. In the
present section we only look at some very basic examples to illustrate Definition 8.1.3.
We do so after first singling out two special scenarios of surjective functions π : X → Y
that play a prominent role in the remainder of this section.
Construction 8.1.10 (Quotienting out by a subset). Let (X, TX ) be a topological
space and let A ⊂ X be a subset of X. Let Y be the set Y = (X − A) ∪ {a} where a
128
8. QUOTIENT SPACES
is some abstract element not in X. Define the function π : X → Y by
x
; x ∈ X − A,
π(x) =
a
; x ∈ A,
and note that π is surjective. The space (Y, TX/π ) is typically denoted by (X/A, TX/A )
and referred to as the quotient of X by A. Note that it is the quotient space X/PA
associated to the partition PA = {A, {x} | x ∈ X − A} of X.
Construction 8.1.11 (Gluing of topological spaces). Let (X1 , T1 ) and (X2 , T2 ) be
two disjoint topological spaces, let A ⊂ X1 be a given subspace and let f : A → X2 be
a continuous function. Let X = X1 t X2 equipped with the topology TX = TX1 t TX2
(that is U ∈ TX if and only if U ∩ Xi ∈ TXi , i = 1, 2) and consider the partition P of
X defined by
P = {{x}, {y} ∪ f −1 (y) | x ∈ X1 − A, y ∈ X2 }.
The space Y = X/P is denoted by Y = X1 ∪f X2 and is said to have been obtained by
“gluing X1 to X2 along A by using the function f : A → X2 . ” (which we will typically
abbreviate by simply saying “X1 ∪f X2 is obtained by gluing X1 to X2 ”). Note that Y
is constructed by identifying points x ∈ A to their image f (x) ∈ Y . Points x ∈ X1 − A
and points y ∈ X2 − f (A) do not get identified with other points.
Example 8.1.12. Let Dn = {x ∈ Rn | ||x|| ≤ 1} and S n−1 = {x ∈ Rn | ||x|| = 1}
be the n-dimensional closed ball and the (n − 1)-dimensional sphere, each equipped
with the relative Euclidean topology. Since S n−1 ⊂ Dn , it makes sense to form the
quotient space Dn /S n−1 (Construction 8.1.10).
The main goal of this example is to establish the existence of a homeomorphism
f : (Dn /S n−1 , TDn /S n−1 ) → (S n , TEu ).
To define the function f , we first introduce a couple of auxiliary functions. Let g1 :
Int(Dn ) → Rn be the homeomorphism given by (Exercise 8.7.2)
 1+||x||
x

· ||x||
; x 6= 0,
 ln 1−||x||
g1 (x) =


0
; x = 0,
and let g2 : Rn → S n − {(1, 0, ..., 0} be the homeomorphism
2y1
2y2
2yn
||y||2 − 1
g2 (y1 , ..., yn ) =
,
,...,
,
.
||y||2 + 1 ||y||2 + 1
||y||2 + 1 ||y||2 + 1
Write Dn /S n−1 = (Dn −S n−1 )∪{s} where s ∈ Dn /S n−1 represents the image of S n−1 in
Dn /S n−1 under the quotient map π : Dn → Dn /S n−1 . Then we define f : Dn /S n−1 →
S n as

; x ∈ Dn − S n−1 ,
 g2 (g1 (x))
f (x) =

(0, ..., 0, 1)
; x = s.
8.1. QUOTIENT SPACES: DEFINITIONS, PROPERTIES AND FIRST EXAMPLES
129
Clearly f is a bijection so we only need to verify that it is continuous and open. To
show that f is continuous, it suffices to show that fˆ = f ◦ π : Dn → Rn is continuous
(Theorem 8.1.6). From its definition, it is evident that fˆ is continuous at all points
x ∈ Int(Dn ) so we are left with verifying that this remains true for x0 ∈ S n−1 . Noting
that fˆ(x0 ) = (0, 0, ..., 0, 1) for each x0 ∈ S n−1 , let V ⊂ S n be any neighborhood of
(0, .., 0, 1). Continuity of fˆ at x0 will be demonstrated if we can find a neighborhood U
of x0 in Dn such that fˆ(U ) ⊂ V . Firstly, find an r > 0 such that B(0,...,0,1) (r) ∩ S n ⊂ V .
Without loss of generality we can assume that r ≤ 2 for otherwise V = S n and then
any neighborhood U of x0 maps into V . But if r ≤ 2, then U = {x ∈ Dn | ||x|| > R}
with
q
2 2
exp
−1 −1
r
q
R=
where
exp(t) = et ,
2
2
exp
−1 +1
r
is a neighborhood of x0 with fˆ(U ) = B(0,...,0,1) (r) showing that fˆ is continuous at x0
and consequently that f is continuous. The verification of the continuity of f −1 is left
as an exercise.
Example 8.1.13. Let X1 and X2 be two disjoint copies of the closed unit ball
D = {(x1 , ..., xn ) ∈ Rn | x21 + ... + x2n ≤ 1} equipped with the relative Euclidean
topology and let A ⊂ X1 be the subset A = S n−1 = {(x1 , ..., xn )Rn | x21 + .... + x2n = 1}.
Define f : A → X2 be f (x) = x. We will show that X1 ∪f X2 (Construction 8.1.11) is
homeomorphic to S n .
As usual, let π : X1 t X2 → X1 ∪f X2 denote the quotient map and let the function
ĝ : X1 t X2 → S n be defined as
p

; (x1 , ..., xn ) ∈ X1 ,
 (x1 , ..., xn , 1 − x21 − ... − x2n )
ĝ(x1 , ..., xn ) =
p

(x1 , ..., xn , − 1 − x21 − ... − x2n )
; (x1 , ..., xn ) ∈ X2 .
n
Note that ĝ induces a function g : X1 ∪f X2 → S 1 since ĝ(x) = ĝ(f (x)) for all x ∈ ∂X1 .
According to Theorem 8.1.6, to show that g is continuous it suffices to show that ĝ
is continuous. The continuity of ĝ, however, is immediate and follows from Lemma
3.1.12 after observing that the subsets X1 and X2 are closed in X1 t X2 . Finally, g is
a bijection since it has in inverse function, namely

; yn+1 ≥ 0,
 (y1 , ..., yn ) ∈ X1
−1
g (y1 , ..., yn+1 ) =
 (y , ..., y ) ∈ X
; yn+1 ≤ 0.
1
n
2
Since S n is compact and since X1 ∪f ∪X2 is Hausdorff (which we leave as an easy exercise), it follows from Corollary 6.1.10 that g is a homeomorphism. Figure 1 illustrates
the formation of X1 ∪f X2 from X1 and X2 .
130
X1
X2
8. QUOTIENT SPACES
X1
X2
X1 ∪f X2
Figure 1. An illustration of the creation of the identification space
X1 ∪f X2 from Example 8.1.13. Each of Xi is a copy of the closed unit
ball Dn (n being 2 in this illustration) and the gluing map f : ∂X1 → ∂X2
is the identity.
Example 8.1.14. Consider the Euclidean line (R, TEu ) and let A ⊂ R be the
subset of all integers A = Z. Using the notation from Construction 8.1.10, we form
the quotient space R/Z. Visually, we think of R/Z as having been gotten by collapsing
Z ⊂ R to a single point, as in Figure 2. This representation of R/Z is not unlike that
of the Hawaiian earrings H from Example 2.2.17 (also shown in Figure 2). Recall that
H is the union ∪∞
n=1 Cn of the circles
Cn = {(x, y) ∈ R2 | (x −
1 2
1
) + y 2 = 2 },
n
n
and as a subset of R2 , H is equipped with the relative Euclidean topology.
The goal of this example is to show that in fact R/Z and H are not homeomorphic,
despite the similarities in their appearances. We first observe that R/Z is not compact.
Namely, let
1
1
1
3 FR =
a − ,a +
a∈Z ,
, a + ,a +
2
2
4
4
be an open cover of R and let FR/Z = {π(U ) | U ∈ FR } be the associated open cover
of R/Z where π : R → R/Z is the quotient map. This open cover of R/Z has no finite
subcovers at all since every point π(a + 21 ), a ∈ Z lives in a unique open set π(ha +
1
, a + 34 i) while every point π(a + 15 ), a ∈ Z lives in a unique open set π(ha − 12 , a + 12 i).
4
On the other hand, H is a compact space. Being a subspace of R2 , it suffices to
show that H is bounded and closed. The boundedness of H is obvious and to verify
the closedness of H, let p ∈ R2 − H be any point. Then there exists an integer n0 ∈ N
such that p lies inside of Cn for all n < n0 and p lies outside of Cn for all n ≥ n0
(if p lies outside of C1 we take n0 = 1). If n0 = 1 let r = dC (p) and if n0 > 1 let
r = min{dCn0 (p), dCn0 +1 (p)}. In either case, the ball Bp (r) lies entirely in R2 − H
showing that R2 − H is open and thus that H is closed.
8.1. QUOTIENT SPACES: DEFINITIONS, PROPERTIES AND FIRST EXAMPLES
h−2, −1i
131
(x − 1)2 + y 2 = 1
h−1, 0i
(x − 21 )2 + y 2 =
1
4
h0, 1i
z
o
h4, 5i
h1, 2i
h3, 4i
h2, 3i
(a)
(b)
Figure 2. (a) A visual representation of R/Z. Each of the infinitely
many loops represents an open interval ha, a + 1i for some integer a ∈ Z.
Several such intervals are explicitly labeled. The distinguished point z is
marked by a black dot in the middle. (b) The Hawaiian earrings. Each
of the circles corresponds to {(x, y) ∈ R2 | (x − n1 )2 + y 2 = n12 } for some
integer n ∈ N. The distinguished point o = (0, 0) is labeled by a black
dot.
Finally, since compactness is a topological invariant (Theorem 6.1.8), R/Z and H
cannot be homeomorphic.
Example 8.1.15. Consider again the Euclidean line (R, TEu ) and let A ⊂ R be the
set of rational number Q. By considering the quotient space R/Q, we will demonstrate
that the Hausdorff property of a topological space may disappear when passing to a
quotient space.
To see that R/Q is not Hausdorff, let x, y ∈ R be any two distinct irrational elements
in R and let π(x), π(y) be their images in R/Q under the quotient map π : R → R/Q.
Note that π(x) 6= π(y). Suppose that Uπ(x) and Uπ(y) are neighborhoods of π(x) and
π(y) in R/Q. Then Vx = π −1 (Uπ(x) ) and Vy = π −1 (Uπ(y) ) are neighborhoods of x and
y in R and as such, both contain a rational number, say qx ∈ Vx ∩ Q and qy ∈ Vy ∩ Q.
But then we can conclude that π(qx ) ∈ Uπ(x) and π(qy ) ∈ Uπ(y) and since clearly
π(qx ) = π(qy ), we find that Uπ(x) and Uπ(y) cannot be disjoint. Accordingly, R/Q is
not Hausdorff even though R clearly is.