Download Spring 2009 Topology Notes

Contents
1 Introduction and Set Theory
1
1.1
Cardinality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.2
Restrictions and Compositions of Functions . . . . . . . . . . . .
5
1.3
Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
1.4
Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
2 Topological Spaces
11
2.1
Bases of Topologies . . . . . . . . . . . . . . . . . . . . . . . . . .
14
2.2
Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
2.3
Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . .
20
3 New Spaces from Old
25
3.1
Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
3.2
Quotient Spaces
. . . . . . . . . . . . . . . . . . . . . . . . . . .
27
3.3
Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29
3.4
Weak Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
4 Separation Axioms
37
4.1
T0 , T1 , and T2 Axioms . . . . . . . . . . . . . . . . . . . . . . . .
37
4.2
Regularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
4.3
Complete Regularity . . . . . . . . . . . . . . . . . . . . . . . . .
42
4.4
Normality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43
4.5
Lindelöf Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
i
ii
CONTENTS
5 Convergence and Metric Spaces
51
5.1
First Countable Spaces . . . . . . . . . . . . . . . . . . . . . . . .
51
5.2
Convergence of Nets . . . . . . . . . . . . . . . . . . . . . . . . .
53
5.3
Complete Metric Spaces . . . . . . . . . . . . . . . . . . . . . . .
57
6 Compactness
63
6.1
Local Compactness . . . . . . . . . . . . . . . . . . . . . . . . . .
68
6.2
Compactifications . . . . . . . . . . . . . . . . . . . . . . . . . . .
70
6.3
Paracompactness . . . . . . . . . . . . . . . . . . . . . . . . . . .
72
7 Connectedness
75
7.1
Basic Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
75
7.2
Components and Local Connectedness . . . . . . . . . . . . . . .
78
7.3
Path Connectedness . . . . . . . . . . . . . . . . . . . . . . . . .
81
8 Homotopy And Fundamental Groups
85
8.1
Fundamental Group . . . . . . . . . . . . . . . . . . . . . . . . .
85
8.2
Homotopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
86
0
CONTENTS
Chapter 1
Introduction and Set
Theory
This chapter should be mostly review, although the particular presentation may
be new in many cases. While we do not do a thorough axiomatic treatment of
sets, we will try to point out where naive set theory has difficulties and hint at
how to solve them.
A set is a collection of mathematical objects. Sets, in turn, are mathematical
objects. Hence sets can have other sets as members. When X is a set and x
is anything, we write x ∈ X to denote the fact that x is a member (also called
an element) of X. The negation of this is written x 6∈ X. Sets are typically
defined by giving some property that every element of the set satisfies and that
nothing outside of the set satisfies. Hence, if P (x) is any property of and object
x (in other words, P (x) can be either true or false for each x), then the set,
A, of those things which satisfy property P (x) is written A = {x : P (x)}. In
this case, x ∈ A if and only if P (x) is true. Another way of defining a set is
simply by listing the elements in the set. Hence, with A = {a, b}, x ∈ A if and
only if x = a or x = b. Notice that a = b is allowed in this. Similarly, with
A = {a1 , a2 , · · · , an }, x ∈ A if and only if x = a1 or x = a2 or · · · or x = an .
If X and Y are sets, we say that X is a subset of Y and write X ⊆ Y if
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CHAPTER 1. INTRODUCTION AND SET THEORY
x ∈ X implies x ∈ Y for all objects x. If X ⊆ Y and X 6= Y , we write X ( Y
and say that X is a proper subset of Y . A particularly important set is the
empty set ∅ = {x : x 6= x}. It is always the case that ∅ ⊆ X for any set X.
We also have the intersection X ∩ Y = {x : x ∈ X and x ∈ Y } and the union
X ∪ Y = {x : x ∈ X or x ∈ Y }. Also, we define X \ Y = {x ∈ X : x 6∈ Y }, the
set difference.
S
More generally, if X is a set, we can define the union X = {x : x ∈
T
A for some A ∈ X} and the intersection X = {x : x ∈ A for all A ∈ X}.
This is typically done when X is a collection of sets, but the definitions work
even in the general case.
It turns out that inconsistencies arise unless some restriction is placed on
those properties P (x) which can be used to define sets. In particular, forming
the Russell ’set’ R = {x : x 6∈ x} leads quickly to a contradiction: is R ∈ R or
is R 6∈ R? This is avoided by not allowing the construction of very ’large’. In
spite of this, there are some important sets that are allowed. For example, if
S
X is a set, the power set P(X) = {A : A ⊆ X} is always allowed. Also, X
T
is allowed no matter what set X is, but X is only allowed if X is non-empty
(why?).
Whenever x and y are objects, we can define the ordered pair (x, y) =
{{x}, {x, y}}. This has the property that (x, y) = (z, w) if and only if x = z
and y = w. In fact, this is the only important property of ordered pairs from
our point of view. This can be generalized by setting, say (x, y, z) = (x, (y, z)),
so (x1 , y1 , z1 ) = (x2 , y2 , z2 ) if and only if x1 = x2 , y1 = y2 , and z1 = z2 . Clearly,
this can be generalized to any n−tuple.
If X and Y are sets, we define the cross product set X × Y = {(x, y) : x ∈
X, y ∈ Y }. In a similar way, we can define X × Y × Z by using ordered triples.
Some important sets:
1. The set
N of natural numbers, N = {1, 2, 3, · · · }.
2. The set
R of real numbers.
3. The set
Rn of n−tuples of real numbers.
3
4. The set
Q of rational numbers.
n+1
5. The set S n = {x ∈ R
: kxk = 1}. the n−sphere.
A function from X to Y is now a subset f ⊆ X × Y such that a) for each
x ∈ X, there is a y ∈ Y with (x, y) ∈ f and b) if (x, y1 ), (x, y2 ) ∈ f , then
y1 = y2 . When we think of functions actvely, we say that x ∈ X ’goes to’ y ∈ Y
if (x, y) ∈ f . To be a function, then, each x ∈ X has to ’go to’ something in
Y and only one such thing in Y . If (x, y) ∈ f , we write y = f (x). We write
f : X → Y when f is a function from X to Y . The set {y ∈ Y : there is an x ∈
X with y = f (x) is called the image of f and denoted by im f . Similarly, we
say that X is the domain of f . The collection of all functions from X to Y is a
set denoted by Y X = {f |f : X → Y }.
Now let I be a set and suppose there is a function whose domain is I that
assigns each α ∈ I a non-empty set Aα . We call {Aα }α∈I an indexed family of
sets. In this case, we define the union and intersections
[
Aα = {x : x ∈ Aα for some α ∈ I},
α∈I
\
Aα = {x : x ∈ Aα for all α ∈ I}.
α∈I
These are special cases of the union and intersection defined above.
Theorem 1.1 Let {Aα }α∈I be an indexed family of subsets of X and suppose
that A ⊆ X. Then
a) A \
S
α∈I
Aα =
T
b) A \
T
α∈I
Aα =
S
c) A ∪
S
α∈I
Aα =
S
d) A ∩
T
α∈I
Aα =
T
α∈I
A \ Aα ,
α∈I
A \ Aα ,
α∈I
A ∪ Aα ,
α∈I
A ∩ Aα ,
Definition 1.2 Let f : X → Y and A ⊆ X and B ⊆ Y . Define
a) f [A] = {y ∈ Y : there exists x ∈ X such that y = f (x)}
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CHAPTER 1. INTRODUCTION AND SET THEORY
b) f −1 [B] = {x ∈ X : f (x) ∈ B}
Definition 1.3 Let f : X → Y . We say that f is
a) one-to-one (or injective) if x1 , x2 ∈ X with f (x1 ) = f (x2 ) implies that
x1 = x2 .
b) onto (or surjective) if for each y ∈ Y , there is an x ∈ X with f (x) = y.
c) a one-to-one correspondence (or bijective ) if f is both one-to-one and
onto.
Notice that f is onto exactly when im f = Y .
If f is bijective, then there is an inverse function f −1 : Y → X defined by
setting f −1 (y) = x exactly when f (x) = y. For any set, there is an identity
function 1X : X → X defined by 1X (x) = x. This is clearly a one-to-one
correspondence from X to itself.
1.1
Cardinality
If X and Y are sets, we say that X and Y are equivalent (in cardinality) if there
is a one-to-one correspondence from X to Y . We think of X and Y as being the
’same size’ when this happens. In particular, a set X is said to be countably
infinite if X is equivalent to
N, the set of natural numbers. In particular, Q is
countably infinite. We say that X is countable if it is either finite or countably
infinite. Any subset of a countable set is countable. Also, a countable union of
countable sets is again countable. Finally, X is said to be uncountable if X is
not countable. In particular,
R is an uncountable set.
Theorem 1.4 (Schröder-Bernstein) Suppose that f : X → Y and g : Y →
X are one-to-one functions. Then, there exists a one-to-one correspondence
between X and Y .
Proof: Let C0 = X \g[Y ] and Cn+1 = g ◦f [Cn ]. Then let C =
S∞
n=0
Cn . Notice
that if x 6∈ C, then x ∈ g[Y ]. Since g is one-to-one, there is then a unique y ∈ Y
such that g(y) = x.
1.2. RESTRICTIONS AND COMPOSITIONS OF FUNCTIONS
5
Define h : X → Y by h(x) = f (x) if x ∈ C and h(x) = g −1 (x) if x 6∈ C. By
the last paragraph, h IS well-defined. We claim that h is one-to-one and onto.
Suppose that x1 , x2 ∈ with h(x1 ) = h(x2 ). There are three cases: if x1 , x2 ∈
C, then f (x1 ) = f (x2 ), so x1 = x2 since f is one-to-one. If x1 , x2 6∈ C, then
g −1 (x1 ) = g −1 (x2 ), so applying g gives x1 = x2 . Finally, if x1 ∈ C and x2 6∈ C
(the other case is the same), f (x1 ) = g −1 (x2 ), so g ◦ f (x1 ) = x2 . But, if x1 ∈ C,
then x1 ∈ Cn for some n and so g ◦ f (x1 ) ∈ Cn+1 , so x2 ∈ Cn+1 ⊆ C, a
contradiction. Thus h is one-to-one.
Next, if y ∈ Y , there are two cases. If y ∈ f [C], then there is an x ∈ C with
h(x) = f (x) = y. Otherwise, let x = g(y). Since x ∈ g[Y ], x 6∈ C0 . But, now,
if x ∈ Cn+1 for some n ≥ 0, there is a z ∈ Cn with g ◦ f (z) = x = g(y). Since
g is one-to-one, y = f (z) ∈ f [C], a contradiction. Hence x 6∈ ∪∞
n=0 Cn = C, so
h(x) = g −1 (x) = y. Hence h is onto.
1.2
Restrictions and Compositions of Functions
Theorem 1.5 If f : X → Y , the following hold:
S
S
a) If {Aα }α∈I is any indexed family of subsets of X, then f [ Aα ] = f [Aα ].
T
T
b) If {Aα }α∈I is any indexed family of subsets of X, then f [ Aα ] ⊆ f [Aα ].
In this case, equality may not hold.
S
c) If {Bα }α∈I is any indexed family of subsets of Y , then f −1 [ Bα ] =
S −1
f [Bα ].
T
d) If {Bα }α∈I is any indexed family of subsets of Y , then f −1 [ Bα ] =
T −1
f [Bα ].
Now, suppose that f : X → Y and g : Y → Z are functions where the
codomain of f is the same as the domain of g. We can then define the composition g ◦ f : X → Z by setting g ◦ f (x) = g(f (x)). It is easy to see that
(g ◦ f )−1 [C] = f −1 [g −1 [C]].
Next, if f : X → Y and A ⊆ X, then we can define the restriction of f
to A, f |A : A → Y by letting f |A (x) = f (x) for x ∈ A. The domain of this
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CHAPTER 1. INTRODUCTION AND SET THEORY
restriction is then A and the image is now f [A]. This is actually a special case
of a composition, as we can see as follows: Define the inclusion map iA : A → X
by iA (x) = x. Then f |A = f ◦ iA .
In the case when f : X → Y is a one-to-one correspondence and f −1 is the
inverse function, we have f ◦ f −1 = 1Y and f −1 ◦ f = 1X .
1.3
Products
Given two sets X and Y , there are natural projections πX : X × Y → X and
πY : X × Y → Y defined by πX (x, y) = x and πY (x, y) = y. We will generalize
the idea of products so that infinitely many factors will be allowed.
If {Aα } is an indexed family of sets, we can define the product set via
Y
α∈I
Aα = {x : A →
[
Aα |x(α) ∈ Aα for all α ∈ I}.
α∈I
An element of the product is called a choice function. It is quite common to
write xα instead of x(α) and call xα the α coordinate of x.
Y
For each α0 ∈ I, there is a natural projection πα0 :
Aα → Aα0 defined
by πα0 (x) = x(α0 ). This generalizes the projections defined above for finite
products.
The Axiom of Choice states that if {Aα }α∈I is an indexed family of nonY
empty sets, then the product
Aα is non-empty. We will assume this axiom
and use it without comment when needed.
The most important property of products is the following:
Theorem 1.6 Suppose that {Aα }α∈I is an indexed family of sets and that X is
any set. Suppose also that there is an indexed family of functions fα : X → Aα .
Y
Then there is a unique function f : X →
Aα such that fα = πα ◦ f for all
α ∈ I.
Proof: For x ∈ X and α ∈ I, define f (x)(α) = fα (x). It is easy to verify both
that this works and that it is unique.
1.4. RELATIONS
1.4
7
Relations
Let X be a set. A relation on X is a subset, R, of X × X. We generally write
xRy in place of (x, y) ∈ R when dealing with relations.
An equivalence relation on a set X is a relation R satisfying the following
properties:
a) For all x ∈ X, xRx (Reflexivity)
b) Whenever xRy, we have yRx (Symmetry)
c) If xRy and yRz, then xRz (Transitivity)
If R is an equivalence relation on X and x ∈ X, we define the equivalence
class of x by [x] = {y ∈ X : xRy}. It is then easy to show that [x] = [y] if
and only if xRy and that [x] ∩ [y] = ∅ when xRy fails. The collection of all
equivalence classes is denoted by X/R = {[x] : x ∈ X}. There is a natural
quotient map qR : X → X/R defined by qR (x) = [x]. This map is always onto.
We have the following property of such quotients:
Theorem 1.7 Suppose that R is an equivalence relation on X and f : X → Y
is such that x1 Rx2 implies that f (x1 ) = f (x2 ). Then, there is a unique function
fR : X/R → Y such that f = fR ◦ qR .
Proof:
For x ∈ X, define fR to be the collection of ordered pairs fR =
{([x], f (x)) : x ∈ X}. Then, if [x1 ] = [x2 ], we have that x1 Rx2 , so f (x1 ) =
f (x2 ), so fR is a function. It is clear that dom f = X/R and that f = fR ◦ qR .
Definition 1.8 A partition of a set X is a family P ⊆ P(X) such that
a)
S
P =X
b) If A, B ∈ P and A 6= B, then A ∩ B = ∅.
From above, if R is an equivalence relation on X, the collection of equivalence
classes X/R is a partition of X. Conversely, if P is a partition of X, we define
xRy if there is an A ∈ P with x, y ∈ A. This gives an equivalence relation
whose equivalence classes are exactly the elements of P .
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CHAPTER 1. INTRODUCTION AND SET THEORY
Another important type of relation is that of a partial order. For this, we
usually use the symbol ≤ to denote the relation and we assume the following
properties:
a) For each x ∈ X, x ≤ x.
b) If x ≤ y and y ≤ x, then x = y.
c) If x ≤ y and y ≤ z, then x ≤ z.
So a partial order is symmetric, transitive, but anti-symmetric. There are many
types of partial orders, most of which we will not have occasion to investigate.
We say that x0 ∈ X is a maximal element if y ∈ X with x0 ≤ y implies that
y = x0 . If A ⊆ X, we say that x0 ∈ X is an upper bound for A if y ≤ x0
for all y ∈ A. Notice that it can happen that x0 is a maximal element of X
while not being an upper bound for X. An upper bound for X is also called a
largest element for X. Of course, similar definitions are made for the concepts
of minimal elements, lower bounds, and smallest elements.
But we will be interested in two types: linear orders and well-orderings. We
say that a partial order ≤ is a linear order if, for each x, y ∈ X, we have x ≤ y
or y ≤ x. In other words, every two elements of X are comparable. The usual
order on the real line is a linear order.
If X is a partially ordered set, we say that C ⊆ X is a chain if C is linearly
ordered via ≤. In particular, every subset of a linearly ordered set is a chain.
A well-ordering on a set X is a partial order ≤ with the property that every
non-empty subset A ⊆ X has a smallest element in A. In particular, using
A = {x, y}, we see that well ordered sets are linearly ordered. The usual order
on the natural numbers,
N, is a well-ordering, but that on the real line is NOT
a well-ordering: the set (0, 1) does not have a smallest element.
The following result gives two important equivalences of the Axiom of Choice.
Both are heavily used in this course.
Theorem 1.9 The following are equivalenct to the Axiom of Choice:
a) Every set can be given a well-ordering. (The Well-Ordering Principle).
1.4. RELATIONS
9
b) Suppose that X is a partially ordered set. Suppose that every chain C in
X has an upper bound in X. Then X has a maximal element. (Zorn’s
Lemma).
Theorem 1.10 There exists an uncountable well-ordered set (Ω0 , ≤) with the
property that {x : x ≤ x0 } is countable for each x0 ∈ Ω0 .
Proof:
Let ≤ be a well-ordering of any uncountable set X. There are two
possibilities: either the set {x : x ≤ y} is countable for each y ∈ X or there is
some y where this set is uncountable. In the first case, let Ω0 = X.
Otherwise, since ≤ is a well-ordering, in the second case, there is a smallest
y with {x : x ≤ y} uncountable. Let Ω0 = {x : x < y}. Then Ω0 is a wellordered, uncountable set with the property that {x : x ≤ x0 } is countable for
every x0 ∈ Ω0 .
We call Ω0 the first uncountable ordinal. Notice that Ω0 does not have a
largest element. We can extend it, however, by picking any ω1 6∈ Ω0 and setting
Ω = Ω0 ∪ {ω1 } with the order extended by defining y ≤ ω1 for all y ∈ Ω0 . Then,
Ω is a well-ordered, uncountable set with a largest element.
The crucial propert of Ω0 is the following:
Theorem 1.11 Suppose that A ⊆ Ω0 is a countable subset. Then A has an
upper bound in Ω0 .
Proof: For each x ∈ A, the set Ix = {y ∈ Ω0 : y ≤ x} is countable. Since A is
S
countable, so is x∈A Ix . Since Ω0 is uncountable, there is some y that is not
in this union. Let y0 be the least such y. Then, for each x ∈ A, we must have
x < y0 , so y0 is an upper bound for A.
Exercises:
1. Let A = {a, b, c}. List all elements of P(A).
2. What happens with Y X when X = ∅? When Y = ∅?
3. Let X = {a, b, c} and Y = {x, y}. How many elements does Y X have?
How about X Y ?
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CHAPTER 1. INTRODUCTION AND SET THEORY
4. Let X be any set. Show that X {0,1} and X × X are equivalent via the
function f (x) = (x(0), x(1)).
5. Let X be a set. Show that {0, 1}X and P(X)are equivalent via the function
Φ(f ) = {x ∈ X : f (x) = 1}.
6. Let X be a set. Show that P(X) and X are not equivalent. Hint: Suppose
that f : X → P(X) and consider A = {x ∈ X : x 6∈ f (x)} to show that f
cannot be onto.
7. Prove Theorem 1.5.
8. Suppose that f : X → Y is one-to-one. Show that equality obtains in part
b) of Theorem 1.5
9. Suppose that A = {1, 2} and let A1 and A2 be non-empty sets. Show that
Y
Aα is equivalent to A1 × A2 via the function f (x) = (x(1), x(2)).
α∈A
10. Suppose that Aα = X for all α ∈ I. Show that
Y
Aα = X I .
α∈I
11. By assuming the Axiom of Choice, show that the projections πα are all
onto if every set Aα is non-empty.
12. Suppose that R is an equivalence relation on X and S is an equivalence
relation on Y . Define a relation R × S on X × Y by setting (x1 , y1 )(R ×
S)(x2 , y2 ) if and only if x1 Rx2 and y1 Sy2 . Show that R × S is an equivalence relation on X × Y and that (X × Y )/(R × S) is equivalent to
(X/R) × (Y /S) via the function f ([(x, y)]) = ([x], [y]).
13. Suppose that R, S are equivalence relations on X with R ⊆ S. Show that
there is an equivalence relation, E on X/R defined by [x]R E[y]R if and
only if xSy.
Chapter 2
Topological Spaces
Let X be a set. A collection τ ⊆ P(X) of subsets of X is called a topology if
the following hold:
a) ∅ ∈ τ and X ∈ τ .
b) If U, V ∈ τ , then U ∩ V ∈ τ .
c) If A ⊆ τ , then
S
A ∈ τ.
A topological space is a set X with a topology τ on X. We call elements of τ
open sets in X.
Example: Let X be any set. The discrete topology on X is defined by setting
τ = P(X). Hence, every subset of X is open.
Example:
Let X be a set. The indiscrete topology on X is defined by τ =
{∅, X}.
Example:
Let X = {a, b}. Define τ = {∅, {a}, X}. Then τ is a topology on
X.
Example:
Let X be a set. Define τ = {U ⊆ X : U = ∅ or X \ U is finite }.
By DeMorgan’s laws and the fact that subsets of finite sets are finite, property
c) is satisfied. Since unions of finite sets are finite, property b) is also satisfied.
11
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CHAPTER 2. TOPOLOGICAL SPACES
We call τ the co-finite topology on X. Notice that this topology is the same as
the discrete topology if X is finite, but differs otherwise.
Example:
We can define the co-countable topology on a set X by replacing
the word finite by the word countable in the previous definition. In this case,
the topology is discrete exactly when X is countable.
We will find many more examples of topologies, but we first have to find
better ways of defining them. Before we do that, we define some important
concepts.
Definition 2.1 If τ1 and τ2 are both topologies on the same set, we say that
τ1 is coarser (alternatively, weaker) than τ2 if τ1 ⊆ τ2 . In this case, we also
say that τ2 is finer (alternatively, stronger) than τ1 . In particular, the discrete
topology on X is the strongest topology on X and the indiscrete topology is the
weakest.
Suppose that X is a topological space. We say that A ⊆ X is closed in X if
X \ A is open. Let F denote the collection of closed sets of X. Then, we have
the following from an easy application of DeMorgan’s laws:
a) ∅, X ∈ F..
b) If C, K ∈ F, then C ∪ K ∈ F.
c) If A ⊆ F, then
T
A ∈ F.
It is clear that we can define a topology by declaring which sets are closed
instead of which sets are open as long as these three properties are satisfied.
In fact, if F is a collection of subsets of X which satisfy these properties, then
{U : X \ U ∈ F} is a topology on X.
Now, let A ⊆ X be any subset. We define the interior of A by letting
S
A = {U : U ⊆ A, U ∈ τ }. In other words, A◦ is the union of all open subsets
◦
of A. Since unions of open sets are open, A◦ is an open set of A. In fact, because
of our definition, it is the em largest open set contained in A. Another common
notation for the interior of A is int(A).
13
In a similar way, we can define the closure of A, A =
T
{F : A ⊆ F, F ∈ F}.
Then the closure is the intersection of all closed sets containing A as a subset.
Since intersections of closed sets are closed, A is a closed set containing A. In
fact, it is the smallest closed set containing A. Another notation for the closure
of A is cl(A).
The following holds:
Theorem 2.2 For subsets A, B of the topological space X, we have
a) A◦ ⊆ A.
b) A ⊆ A.
c) (A◦ )◦ = A◦ .
d) (A ∩ B)◦ = A◦ ∩ B ◦
e) X \ A◦ = X \ A.
f) X \ A = (X \ A)◦ .
g) cl(cl(A)) = cl(A).
h) A ∪ B = A ∪ B.
i) A◦ = A if and only if A is open.
j) A = A if and only if A is closed.
Definition 2.3 We say a subset A ⊆ X is dense if A = X
Proposition 2.4 Let A ⊆ X. Then the following are equivalent:
a) A is dense in X
b) Whenever U 6= ∅ is open, we have A ∩ U 6= ∅
This will be left as an exercise.
Definition 2.5 If A ⊆ X and x ∈ A, we say that A is a neighborhood of x if
x ∈ A◦ .
14
CHAPTER 2. TOPOLOGICAL SPACES
It is clear that the intersection of two neighborhoods of x is again a neigh-
borhood of x. Notice that neighborhoods do not have to be open sets, but we
do have the following.
Proposition 2.6 A set is open if and only if it is a neighborhood of each of its
points.
2.1
Bases of Topologies
The first difficulty in defining a topology is that, in general, topologies are too
large to define easily. Our first goal is find an easier way to obtain topologies
that makes them easier to work with also. This is done by considering collections
of open sets that do not satisfy property c) in the definition of topology, but
allow the full topology to be obtained by use of this property.
Definition 2.7 Let X be a topological space with topology τ . A collection B ⊆ τ
is called a base for the topology if whenever ∅ 6= U ∈ τ , there is a collection
S
{Bα } ⊆ B such that U = α Bα . In other words, B is a collection of open sets
and all non-empty open sets can be obtained by taking unions from B.
Theorem 2.8 Let X be a topological space. In order for B ⊆ P(X) to be a
base for the topology τ , the following must hold.
(i) Each B ∈ B must be open. In other words, B ⊆ τ .
(ii) If U is a non-empty open set and x ∈ U , there exists a B ∈ B such that
x ∈ B ⊆ U.
Conversely, if B satisfies these properties, then B is a base for the topology of
X.
Proof: Since sets in a base have to be open, i) is clear. So suppose that U is a
S
non-empty open set and x ∈ U . If B is a base, then we can write U = α Bα for
some indexed family {Bα }. But then x ∈ Bα for some α. Hence, x ∈ Bα ⊆ U .
For the converse, we need to show that every non-empty open set is a union
of elements from B. So let U be a non-empty open set. For each x ∈ U , there
2.1. BASES OF TOPOLOGIES
is a Bx ∈ B such that x ∈ Bx ⊆ U . But then
15
S
x
Bx = U since containment is
true both ways.
In this result, we already have a topology on the set X and wish to determine
if a collection of open sets is a base for that topology. In the next result, we
want to define the topology by declaring a certain collection of sets to be a base.
Of course, there are conditions on the collection for this to work.
Theorem 2.9 Suppose that B ⊆ P(X) satisfies
(i)
S
B = X,
(ii) If B1 , B2 ∈ B and x ∈ B1 ∩ B2 , then there is a B3 ∈ B with x ∈ B3 ⊆
B1 ∩ B2 .
Then there is a unique topology τ such that B is a base for τ .
Notice that if (i) fails, we may simply replace B by B ∪ {X}
Proof: Define τ to contain the empty set and all sets of the form
S
Bα where
{Bα } is an indexed family of elements of B. Clearly, X ∈ τ and ∅ ∈ τ . This
is also the only topology for which B could possibly be a base. Since unions of
elements of τ are again simply unions of elements of B, τ is closed under unions.
We need only show that τ is closed under intersections.
First, suppose that B1 , B2 ∈ B. For each x ∈ B1 ∩ B2 , by assumption, there
S
is a Bx ∈ B such that x ∈ Bx ⊆ B1 ∩ B2 . Then, x Bx = B1 ∩ B2 . Hence,
B1 ∩ B2 ∈ τ .
S
S
Now suppose that U, V ∈ τ and write U = α Bα and V = β Bβ0 where
S S
Bα , Bβ0 ∈ B. Then U ∩ V = α β (Bα ∩ Bβ0 ). By the previous paragraph, each
Bα ∩ Bβ0 ∈ τ . Since we already know τ is closed under unions, U ∩ V ∈ τ ,
showing that τ is, in fact, a topology on X.
Example:
Let X =
R and set B = {(a, b) : a < b}. Since intersections of
open intervals are open intervals, there is a unique topology making B a base.
This is called the usual topology on
R. Open sets in R are then unions of open
intervals. Unless we state otherwise, we will always assume that R has this
topology.
16
CHAPTER 2. TOPOLOGICAL SPACES
Example:
Again, let X =
R, but this time let B = {[a, b) : a < b}. Again,
intersections of these half-open intervals are again half-open intervals, so B
defines a base for a topology. We call this the half-open topology and call
R with this topology the Sorgenfry line.We will denote it by L.
It can be important to know when two different bases manage to define the
same topology. The next result gives an easy criterion for this determination.
It actually is stronger than this, giving a criterion for determining when one
topology is finer than another.
Theorem 2.10 Suppose that B1 and B2 are bases for the topologies τ1 and τ2
on the set X, respectively. In order that τ1 ⊆ τ2 it is necessary and sufficient
that whenever x ∈ B1 ∈ B1 there exists B2 ∈ B2 with x ∈ B2 ⊆ B1 .
Definition 2.11 Let X be a topological space with topology τ . We say that S ⊆
τ is a sub-base for the topology of X if the collection of all finite intersections
{S1 ∩ · · · Sn : S1 , · · · Sn ∈ S} ∪ {X} is a base for the topology of X.
Proposition 2.12 Any collection of subsets of X defines a sub-base for some
topology on X.
Proof:
Just show that the collection of finite intersections satisfies the condi-
tions of Theorem 2.9.
Example: Let X be any partially ordered set and define (←, x) = {y : y < x}
and (x, →) = {y : x < y}. Let S = {(←, x) : x ∈ X} ∪ {(x, →) : x ∈ X}. Then
S is a subbase for a topology, called the order topology on X. In particular,
notice that the order topology of
R agrees with the usual topology. Unless
otherwise stated, Ω and Ω0 will be given their order topologies.
Definition 2.13 Let X be a topological space and x ∈ X. A base of neighborhoods at x is a collection N of neighborhoods at x such that whenevewr A is any
neighborhood at x, there is an N ∈ N such that x ∈ N ⊆ A.
In particular, if B is a base for the topology, N = {B : x ∈ B ∈ B} is a base
of neighborhoods at x.
2.2. METRIC SPACES
17
Definition 2.14 Let X be a topological space. We say that X is
a) second countable if there is a countable base for the topology of X.
b) first countable if every point in X has a countable neighborhood base.
c) X is separable if there is a countable set D ⊆ X that is dense in X.
Proposition 2.15 If X is second countable, then X is first countable and separable.
We will learn later that the converse of this is not true.
Proof: Let B be a countable base for the toplogy of X. Then for each x ∈ X,
the collection {B : x ∈ B ∈ B} is a countable neighborhood base at x, so X is
first countable. Now, write B = {Bn : n ∈
N}. For each n ∈ N, pick xn ∈ Bn
and let D = {xn : n ∈ N}. We use the result from Exercise 23 to show that D
is dense. So, let U be a non-empty open set in X. Since B is a base, there is a
Bn ∈ B such that Bn ⊆ U . But then xn ∈ U ∩ D, showing U ∩ D is non-empty.
Since D is clearly countable, X is separable.
Proposition 2.16 Suppose that A, B ⊆ Ω0 are closed uncountable sets. Then
A ∩ B 6= ∅.
Proof: Pick any x1 ∈ A. Since B is uncountable and {y : y ≤ x1 } is countable,
there is a y1 ∈ B with x1 < y1 . Similarly, there is an x2 ∈ A with x1 < y1 < x2 .
Continuing in this way, we find sequences x1 < y1 < x2 < y2 < · · · where
xn ∈ A and yn ∈ B for all n. Now let a = sup{xn : n ∈
N} = sup{yn ; n ∈ N}.
By exercise 9, a ∈ A ∩ B.
2.2
Metric Spaces
A very important class of topological spaces derive from spaces having a concept
of distance. These are called metric spaces.
Definition 2.17 Let X be a set. A metric on X is a function d : X × X → R
such that
18
CHAPTER 2. TOPOLOGICAL SPACES
a) For every x, y ∈ X, d(x, y) ≥ 0.
b) We have d(x, y) = 0 if and only if x = y.
c) For every x, y ∈ X, d(x, y) = d(y, x).
d) For every x, y, z ∈ X, we have d(x, z) ≤ d(x, y) + d(y, z).
A metric space is a set X together with a metric d on X.
Example: Let X = R and set d(x, y) = |x − y|.
n
Example: Let X = R and set
d(x, y) =
n
X
! 21
|xk − yk |2
k=1
This is called the usual metric on
n
R .
n
Example: Let X = R and set
d1 (x, y) =
n
X
|xk − yk |
k=1
This is called the taxi-cab metric on
Rn .
n
Example: Let X = R and set
d∞ (x, y) = max{|xk − yk | : 1 ≤ k ≤ n}
This is called the max metric or box metric on
Rn .
Definition 2.18 Let X be a metric space with metric d and let x ∈ X and
r > 0. We define the open ball around x or radius r by
Bd (x, r) = {y ∈ X : d(x, y) < r}.
When the metric is understood, we shall often omit the subscript.
Theorem 2.19 Let X b e a metric space with metric d. Define B = {B(x, r) :
x ∈ X, r > 0}. Then B defines a base for a topology on X. We call this topology
the metric topology. Moreover, for each x ∈ X, the collection {B(x, r) : r > 0}
is a neighborhood base at x.
2.2. METRIC SPACES
Proof: Clearly, for each x ∈ X, x ∈ B(x, 1), so X =
19
S
x
B(x, 1). This gives the
first condition in Theorem 2.9. For the second, suppose that z ∈ B(x1 , r1 ) ∩
B(x2 , r2 ). Let r = min{r1 −d(z, x1 ), r2 −d(z, x2 ). If y ∈ B(x, r), then d(x1 , y) ≤
d(x1 , x)+d(z, y) < d(x1 , x)+r ≤ r1 . So, y ∈ B(x1 , r1 ). Similarly, y ∈ B(x2 , r2 ).
Thus x ∈ B(x, r) ⊆ B(x1 , r1 ) ∩ B(x2 , r2 ) as required to invoke Theorem 2.9.
Now assume that x ∈ X and x ∈ U with U open. By the definition of a
base, there is a y ∈ X and an r1 > 0 such that x ∈ B(y, r1 ) ⊆ U . The subtlety
here is that we want an open ball with center x instead of one with center y.
But, if we let r = r1 − d(x, y), we have that B(x, r) ⊆ B(y, r1 ) ⊆ U , as required
for a neighborhood base.
Unless otherwise stated, we will always give a metric space the corresponding
metric topology.
Proposition 2.20 Every metric space is first countable.
Proof:
If X is a metric space and x ∈ X, then {B(x, n1 ) : n ∈
N} is a
neighborhood base at x.
Proposition 2.21 A separable metric space is second countable.
Proof: Let D be a countable set with D = X. Write D = {xn : n ∈
B = {B(xn , k1 : k, n ∈
N}. Let
N}. Then B is a countable collection of open sets in X.
We show that B is a base for the topology of X. So, let x ∈ U where U is open
in X. There is a k ∈
N such that B(x, k1 ) ⊆ U . Since D is dense, there is an
1
1
1
xn ∈ B(x, 2k
). But then x ∈ B(xn , 2k
) ⊆ B(x, k1 ) ⊆ U . Since B(xn , 2k
) ∈ B,
this shows that B is a base for X. Hence, X is second countable.
It is instructive to see where this argument breaks down for general topological spaces. The reader is encouraged to work through this.
Definition 2.22 We say a topology on X is metrizable if it is induced by some
metric on X. We say that two metrics are topologically equivalent if they induce
the same topology on X.
20
CHAPTER 2. TOPOLOGICAL SPACES
Example: The metrics d∞ , d2 , andd1 are all topologically equivalent on
Rn
Example: Suppose that d is a metric on X and define

 d(x, y) d(x, y) ≤ 1
d1 (x, y) =
 1
d(x, y) ≥ 1
Then d1 is a metric on X that is topologically equivalent to d. The important
thing about d1 is that it is bounded, yet defines the same topology as d.
2.3
Continuous Functions
Definition 2.23 Let X and Y be two topological spaces with topologies τX and
τY respecitvely. We say that a function f : X → Y is continuous at the point
x ∈ X if, whenever V is open in Y with f (x) ∈ V , there is an open set U in X
with x ∈ U and f [U ] ⊆ V . A function that is continous at every point of X is
said to be continuous on X.
The following shows that continuity can be tested using only elements from
bases defining the respective topologies.
Proposition 2.24 Suppose that BX and BY are bases for the topologies of X
and Y respectively. Then f : X → Y is continous at x ∈ X if and only if
whenever f (x) ∈ B ∈ BY , there is a B 0 ∈ BX with x ∈ B 0 and f [B 0 ] ⊆ B.
Example:
Let (X, d) and (Y ρ) be metric spaces. A function f : X → Y is
continous at x ∈ X in the metric topologies if and only if for each ε > 0 there
is a δ > 0 such that f [B(x, δ)] ⊆ B(f (x), ε). In more detail, whenever x0 ∈ X
with d(x, x0 ) < δ, we can conclude that ρ(f (x), f (x0 )) < ε. This should be the
familiar definition for the case where X and Y are Euclidean spaces.
Theorem 2.25 Let X and Y be topological spaces and f : X → Y . The following are equivalent:
a) f is continuous from X to Y .
2.3. CONTINUOUS FUNCTIONS
21
b) Whenever V is open in Y , f −1 [V ] is open in X.
c) Whenever C is closed in Y , f −1 [C] is closed in X.
d) For every set A ⊆ X, f [A] ⊆ f [A].
Proof:
a) implies b): Suppose that f is continous and that V ⊆ Y is open.
For each x ∈ f −1 [V ], f (x) ∈ V , so by continuity, there is an open set Ux such
that x ∈ Ux and f [Ux ] ⊆ V . But this implies that x ∈ Ux ⊆ f −1 [V ]. Hence,
S
f −1 [V ] = x Ux is a union of open sets, so f −1 [V ] is open in X.
b) implies c): Suppose that b) holds and let C ⊆ Y be closed. Then Y \ C
is open in Y , so by assumption f −1 [Y \ C] is open in X. But f −1 [Y \ C] =
X \ f −1 [C], so f −1 [C] is closed in X.
c) implies d): Suppose that c) holds and that A ⊆ X. Then f [A] is closed
in Y , so f −1 [f [A]] is closed in X. Notice that A ⊆ f −1 [f [A]]. Since A is the
smallest closed set containing A, we have that A ⊆ f −1 [f [A]. This, in turn,
implies f [A] ⊆ f [A].
d) implies a): Suppose that d) holds, x ∈ X, V ⊆ Y is open, and f (x) ∈ V .
Let A = f −1 [Y \ V ] = X \ f −1 [V ].Then, f [A] ⊆ f [A] ⊆ Y \ V = Y \ V since
Y \ V is closed. Since f (x) 6∈ Y \ V , this shows that x 6∈ A. Let U = X \ A.
Then x ∈ U . Furthermore, U ⊆ X \ A = f −1 [V ]. Hence f [U ] ⊆ V .
An easy, but important, consequence is the following:
Theorem 2.26 Suppose that f : X → Y and g : Y → Z are continuous
functions between topological spaces. Then g ◦ f : X → Z is continuous.
Proof: Let V ⊆ Z be open. Then (g ◦ f )−1 [V ] = f −1 [g −1 [V ]] is open in X.
Example: If X has the discrete topology, then any function f : X → Y into a
topological space is continuous. Similarly, if Y has the indiscrete topology, then
any function f : X → Y into Y is continuous.
Definition 2.27 Let X and Y be topological spaces. A homeomorphism from
X to Y is a continous, one-to-one correspondence f : X → Y such that the
function f −1 : Y → X is also continuous. We say that two topological spaces
are homeomorphic if there is a homeomorphism from one to the other.
22
CHAPTER 2. TOPOLOGICAL SPACES
Since compositions and inverses of homeomorphisms are again homeomor-
phisms, the following hold:
(i) Every topological space X is homeomorphic to itself.
(ii) If X is homeomorphic to Y , then Y is homeomorphic to X.
(iii) If X is homeomorphic to Y and Y is homeomorphic to Z, then X is
homeomorphic to Z.
Hence, the property of being a homeomorphism is almost an equivalence
relation. The only problem is that the collection of all topological spaces is too
large to be a set, so we do not technically have a relation at all. However, for
every set of topological spaces, we do get an equivalence relation. We generally
regard homeomorphic spaces as being the same topologically.
Definition 2.28 A function f : X → Y between topological spaces is said to be
an open map if f [U ] is open in Y whenever U is open in X. Similarly, we say
that f is a closed map if f [C] is closed for every closed set C ⊆.
In particular, a continuous, one-to-one onto function is a homeomorphism if
and only if it is also either an open map or a closed map. Note that if B is a
base for the topology of X, it is enough to check that f [B] is open in Y for each
B ∈ B to show that f is an open map by Theorem 1.5.
Exercises:
1. Find all topologies on the three element set X = {a, b, c}.
2. What are the closed sets in the co-finite topology? The co-countable
topology?
3. Suppose that τ1 and τ2 are both topologies on the set X. Show that τ1 ∩τ2
is a topology on X. What about τ1 ∪ τ2 ?
4. Suppose that {τα }α∈I is a collection of topologies on X.
T
α∈I τα is a topology on X.
Show that
2.3. CONTINUOUS FUNCTIONS
23
5. Suppose that S ⊆ P(X) is any collection of subsets of X. Show that there
is a weakest topology on X containing S.
6. Show that [a, b) is both open and closed in the half-open topology on
R.
7. Prove Theorem 2.10.
8. Show that the half-open topology on
R is finer than the usual topology.
9. Suppose that A ⊆ Ω0 is closed and {xn }n∈N is a sequence in A. Let
x = sup{xn : n ∈ N}. Show that x ∈ A.
10. Suppose that A, B ⊆ Ω0 are closed, uncountable sets. Show that A ∩ B is
uncountable.
11. Suppose that {An }∞
n=1 is a countable family of closed uncountable subsets
T∞
of Ω0 . Show that n=1 An is uncountable.
12. Let B be a bse for the topology of X and suppose that A ⊆ X. Show that
S
A◦ = {B : B ∈ B, B ⊆ A}. In other words, we may obtain the interior
of a set using only basic open sets in our definition.
13. What do the open balls look like in
R2 with the taxi-cab metric? R3 ?
14. What do the open balls look like in
R2 in the max metric? R3 ?
15. Show that the max metric and the taxi-cab metric define the same topology on
R2 .
16. Let X be a metric space with metric d. The closed ball of radius r centered
at x ∈ X is defined by B(x, r) = {y : B(x, y) ≤ r}. Show that B(x, r) is
a closed set in the metric topology.
17. Let X = [0, 1]∪[2, 3] with distance d(x, y) = |x−y|. Show that the closure
of B(2, 1) is NOT B(2, 1).
18. Let (X, d1 ) and (Y, d2 ) be metric spaces. Define d : (X ×Y )×(X ×Y ) → R
by d((x1 , y1 ), (x2 , y2 )) = d1 (x1 , x2 ) + d2 (y1 , y2 ). Show that d is a metric
on X × Y .
24
CHAPTER 2. TOPOLOGICAL SPACES
19. Let (X, d) be a metric space and give X × X the metric from the previous
exercise (use d for both coordinates). Show that the function d : X × X →
R is continuous on X × X.
20. Let τ1 and τ2 be topologies on the set X. When is the identity function
1X : (X, τ1 ) → (X, τ2 ) continuous?
21. Let (X, d) be a metric space and A ⊆ X. Define d(x, A) = inf{d(x, a) :
a ∈ A}. Show that f : X → R defined by f (x) = d(x, A) is continuous.
22. Let (X, d) be a metric space and A ⊆ X. Show that d(x, A) = 0 if and
only if x ∈ A.
23. Show that D is dense if and only if whenever U is a non-empty open set,
we have D ∩U 6= ∅. If B is a base, this is equivalent to D ∩B 6= ∅ whenever
B ∈ B is non-empty.
Chapter 3
New Spaces from Old
Our next concern is to find ways of creating new topological spaces from ones
we have already found.
3.1
Subspaces
So, suppose that A ⊆ X and that X is a topological space with topology τ .
Define the subspace or relative topology on A by setting τA = {A ∩ U : U ∈ τ }.
It is easily verified that τA is, in fact, a topology on the set A. We say that
elements of τA are open in A. Unless otherwise stated, we will always give
subsets of a topological space the subspace topology.
Proposition 3.1 Let X be a topological space and A ⊆ X. The following hold:
a) A subset V ⊆ A is open in A if and only if there is an open set U in X
with V = A ∩ U .
b) A subset C ⊆ A is closed in A if and only if there is a closed set K in X
with C = A ∩ K.
c) If B is a base for the topology of X, then {A ∩ B : B ∈ B} is a base for
the topology of A.
d) The inclusion map iA : A → X is continuous.
25
26
CHAPTER 3. NEW SPACES FROM OLD
e) If f : X → Y is a continuous map, then the restriction f |A : A → Y is
continuous.
It should be pointed out that τA is the weakest topology on A that makes
the inclusion map iA continuous. This follows since i−1
A [U ] = A ∩ U for U ⊆ X.
This is one motivation for defining the topology on A as we have done.
Notice that if A is an open subset of X and U is open in the subspace
topology of A, then U is open in X. Similarly, if A is closed in X and C ⊆ A is
closed in the subspace topology of A, we can conclude that C is cloased in X.
It is often the case that we want to define a continuous function on a large
set X by saying what it does on certain subsets. The following allows us to do
this:
Theorem 3.2 Suppose that X = A1 ∪ A2 is a topological space with A1 , A2
closed subsets. Suppose that f : X → Y is a function such that f1 = f |A1 and
f2 = f |A2 are continuous on A1 and A2 , resp. Then, f is continuous on X.
Proof: Let K ⊆ Y be closed in Y . Then f −1 [K] = f1−1 [K] ∪ f2−1 [K]. Since
f1 and f2 are assumed to be continuous, f1−1 [K] is closed in A1 and f2−1 [K] is
closed in A2 . Since A1 and A2 are closed in X, these are both closed sets in X,
so f −1 [K] is closed in X. Hence f is continuous.
Clearly, this can be extended to the case where there are finitely many closed
subsets {Ai } of X. Also, if we use open sets instead of closed sets, we may use
arbitrarily many sets and obtain the same result.
Definition 3.3 Let X and Y be topological spaces. We define the em dijoint
union of X and Y by X ⊕ Y = (X × {0}) ∪ (Y × {1}). A subset U is declared
to be open in X ⊕ Y if {x ∈ X : (x, 0) ∈ U } and {y ∈ Y : (y, 1) ∈ U } are both
open.
Notice that the map iX : X → X ⊕ Y defined by iX (x) = (x, 0) is a
homeomorphism of X onto the subspace X × {0}. Similarly, the map iY :
Y → X ⊕ Y defined by iY (y) = (y, 0) is a homeomorphism of Y onto the
subspace Y × {0}. Because of this, we will usually identify X and Y with their
corresponding subspaces of X ⊕ Y .
3.2. QUOTIENT SPACES
27
Proposition 3.4 Suppose that X, Y and Z are topological spaces and that fX :
X → Z and fY : Y → Z are continuous maps. Then the map f : X ⊕ Y :→ Z
defined by setting f (x, 0) = fX (x) for x ∈ X and f (y, 0) = fY (y) for y ∈ Y is
a continous function.
3.2
Quotient Spaces
Suppose that X is a topological space and that R is an equivalence relation
on X. There is a quotient map qR : X → X/R defined by qR (x) = [x], the
equivalence class of x. Our goal is to put a topology on X/R that will make
this map continuous. Of course, the indiscrete topology will work for this, but
it doesn’t have anything to do with the topology of X. The problem is that it
is too small. We want to find the finest topology on X/R which will make the
quotient map continuous.
Definition 3.5 Let X be a toplogical space and R an equivalence relation on
X. Define the quotient topology on X/R by declaring U ⊆ X/R to be open in
−1
X/R exactly when qR
[U ] is open in X.
Proposition 3.6 The quotient topology is, in fact, a topology on X/R. Furthermore, it is the finest topology on X/R making qR continuous.
Proposition 3.7 Let X be a topological space and R and equivalence relation
on X. Then, a function f : X/R → Y is continuous if and only if f ◦ qR is
continuous.
−1 −1
Proof: Just notice that for an open set U ⊆ Y , (f ◦ qR )−1 [U ] = qR
[f [U ]] is
open in X if and only if f −1 [U ] is open in X/R by the definition of the quotient
topology.
More generally, we have the following definition:
Definition 3.8 A continuous map between topological spaces, f : X → Y is
said to be a quotient map if f is surjective and U ⊆ Y is open in Y if and only
if f −1 [U ] is open in X.
28
CHAPTER 3. NEW SPACES FROM OLD
Example: Suppose that X and Y are topological spaces, R is an equivalence
relation on X and f : X → Y is a continous function such that x1 Rx2 implies
f (x1 ) = f (x2 ). Then the function fˆ : X/R → Y defined by fˆ([x]) = f (x) is
well-defined and continuous.
A very common situation is that of ’collapsing a set to a point.’ For this, let
X be a topological space and A ⊆ X. Define a partition of X by P = {{x} :
x 6∈ A} ∪ {A}. We let R be the corresponding equivalence relation on X. Then
we will write X/A instead of X/R and say that X/A is the space obtained from
X by collapsing A to a point.
Example: Let X = [0, 1] and A = {0, 1}. Let S 1 = {(x, y) ∈
R2 : x2 + y2 =
1}. Define f : X → S 1 by f (t) = (cos t, sin t). Then f is continous and
f (0) = (1, 0) = f (1), so there is an induced map on the quotient fˆ : X/A → S 1 .
Notice that f |[0,1/2] and f |[1/2,1] are both homeomorphisms, so are closed. This
shows that f is closed, so is a quotient map, so fˆ is a homeomorphism.
Now let X and Y be topological spaces and f : A → Y be a continuous
function where A ⊆ X. We define the attachment space of X onto Y via f to
be (X⊕Y )/R where the equivalence classes for R are sets of the form {y}∪f −1 [y]
for y ∈ Y . We denote this attachment space as X ⊕f Y .
Definition 3.9 We say the equivalence relation R is upper semi-continuous if
whenever y ∈ X/R with q −1 [y] ⊆ U open, there is a V ⊆ X/R open with y ∈ V
and q −1 [V ] ⊆ U .
Theorem 3.10 The quotient map q : X → X/R is a closed map if and only if
R is upper semi-continuous.
Proof: Let Y = X/R for convenience. Suppose q is closed. Suppose y ∈ Y and
U is open in X with q −1 [y] ⊆ U . Then X \ U is closed in X, so V = Y \ q[X \ U ]
is open in Y and y ∈ V . Notice that q −1 [V ] ⊆ U . Hence R is upper semicontinuous.
Conversely, suppose that R is upper semi-continuous and suppose that C ⊆
X is closed. Let y 6∈ q[C]. Then q −1 [y] ⊆ U = X \ C, so there is an open set
3.3. PRODUCT SPACES
29
V ⊆ Y with y ∈ V and q −1 [V ] ⊆ X \ C. But then, V ∩ q[C] = ∅, showing that
q[C] is closed in Y .
3.3
Product Spaces
Definition 3.11 Let X and Y be topological spaces. We define the product
topology on X × Y by declaring B = {U × V : U open in X, V open in Y } to be
a base for the topology.
Proposition 3.12 The above is a base for a topology on X × Y . Furthermore,
it is the weakest topology such that both πX and πY are continuous.
Proof:
Since (U1 × V1 ) ∩ (U2 × V2 ) = (U1 ∩ U2 ) × (V1 ∩ V2 ), B satisfies the
conditions in Theorem 2.9 to be a base for a topology on X × Y . Next, if U
−1
is open in X, then πX
[U ] = U × Y is open in X × Y , so πX is continuous.
Similarly, πY is continuous. Conversely, suppose that τ is any topology making
both πX and πY continuous. Then, if U is open in X and V is open in Y ,
−1
U × V = πX
[U ] ∩ πY−1 [V ] will be in τ . Hence, τ contains B and so is stronger
than the product topology.
Proposition 3.13 Suppose that X, Y , and Z are topological spaces with f :
Z → X × Y . Then f is continuous if and only if both πX ◦ f and πY ◦ f are
continuous.
Proof: If f is continuous, then πX ◦f and πY ◦f are compositions of continuous
functions so are also continuous. For the converse, assume that the compositions
are continuous. To show that f is continuous, it is enough to show that f −1 [U ×
V ] is open in Z for every basic open set U ×V . But f −1 [U ×V ] = (πx ◦f )−1 [U ]∩
(πY ◦ f )−1 [V ], so this follows.
Example:
We show that L is not second countable. If it were, then L × L
would be second countable by exercise 14. In that case, every subspace of
L × L would be second countable by exercise 4. But consider the subspace
A = {(x, −x) : x ∈ L}. Every singleton is open in the subspace since A∩([x, x+
30
CHAPTER 3. NEW SPACES FROM OLD
1) × [−x, −x + 1)) = {(x, −x)}. But an uncountable discrete space is definitely
NOT second countable. Hence L cannot be second countable. This gives an
example of a first countable, separable space which is not second countable. It
also shows that L is not metrizable.
We now want to extend the idea of product topologies to the case of infinite products. So, suppose that {Xα }α∈I is a collection of topological spaces.
Q
We want to put a topology on
Xα . The natural goal would be to require
each projection πα to be continuous. The product topology will be the weakest
topology such that this is the case.
Definition 3.14 The product topology on
Q
Xα is defined by declaring a sub-
basic open set to be of the form πα−1 [Uα ] where Uα is open in Xα , α ∈ I.
From this, a basic open set in the product topology is a set of the form
Tn
i=1
[Uαi ] where each Uαi is open in Xαi . Clearly, we may assume that
πα−1
i
the αi are all distinct. But now, let Uα = Xα unless α = α1 , · · · , αn , then
Q
Tn
Uα = i=1 πα−1
[Uαi ]. In other words, products of open sets where all but
i
finitely many factors are trivial are open in the product topology and such sets
form a base for the product topology.
Q
Note: if we allowed every product of the form
Uα where Uα is open in
Q
Xα , we get a different topology on
Xα known as the box topology. This
topology is, in general, a finer topology than the product topology. Of the two,
the product is the nicer, as we shall see.
Proposition 3.15 The product topology on
Q
Xα is the weakest topology mak-
ing all projections πα continuous. Furthermore, each πα is an open map.
Proof: Since πα−1 [Uα ] is open in the product topology whenever Uα is open in
Xα , each πα is continuous. Also, πα [B] is open for each basic open set, so πα
Q
is an open map. On the other hand, if Xα has a topology τ making each πα
T
continuous, then all sets of the form πα−1 [Uα ] with finitely many terms will be
open. Hence every basic open set in the product topology would be open in τ .
Thus τ is stronger than the product topology.
3.3. PRODUCT SPACES
31
Proposition 3.16 Suppose that Z is a topological space. Then, f : Z →
Q
Xα
is continuous if and only if each πα ◦ f is continuous.
Proof: Since each πα is continuous, the continuity of f would imply that of all
πα ◦ f . For the converse, suppose that B = ∩ni=1 πα−1
[Uαi ] is a basic open set.
i
Then f −1 [B] = ∩ni=1 (παi ◦ f )−1 [Uαi ], showing f to be continuous.
Lemma 3.17 If U is a non-empty open set in the product
Q
Xα , then {α :
πα (U ) = Xα } is finite.
Proof:
Let x ∈ U and pick a basic open set
Q
Uα such that x ∈
Q
Uα ⊆ U .
Then {α : Uα 6= Xα } is finite and Uα ⊆ πα [U ].
Theorem 3.18 Suppose that {Xα } is a collection of topological space. Then
Q
Xα is second countable if and only if each Xα is second countable and {α :
Xα is not indiscrete} is countable. This also holds with first countable in place
of second countable everywhere.
Proof: First suppose that
Q
Xα is second countable. By an exercise, each Xα
is second countable. Let B = {Bn : n ∈ N} be a countable base for the product.
Let An = {α : πα [Bn ] 6= Xα }. Then each An is finite, so A = ∪∞
n=1 An is
countable. Now suppose that α 6∈ A and that Xα is not indiscrete. Then there
is a non-empty open set Uα 6= Xα . Then U = πα−1 [Uα ] is open in the product
and πα [U ] 6= Xα . Since α 6∈ A, this shows that no Bn is a subset of U . THis
contradicts that B is a base for the product.
Now, suppose that Xn is second countable for each n ∈
N. Next, let
Bn = {Bnm : m ∈ N} be a countable base for Xn for each n ∈ N. Let
Tk
B = { i=1 πi−1 [Bimi ] : k ∈ N, Bimi ∈ Bi }. Then B is a countable base for
Q
Xn .
Similar proofs hold for first countability.
Theorem 3.19 Suppose that (Xn , dn ) for n ∈
N are metric spaces such that
Q
dn (xn , yn ) ≤ 1 for all xn , yn ∈ Xn . For x, y ∈ X =
Xn , define d(x, y) =
P
n
n dn (xn , yn )/2 . Then d is a metric on X inducing the product topology.
Hence, countable products of metrizable spaces are metrizable.
32
CHAPTER 3. NEW SPACES FROM OLD
Proof: Clearly the sum defining d(x, y) converges and gives a metric on X. We
need to show that the metric topology induced by d is the same as the product
topology on X.
We first show that every open set in the product topology is open in the
metric topology. It is sufficient to show that every sub-basic open set in the
product is open in the metric topology. So, suppose that Un is open in Xn
and that x ∈ πn−1 [Un ]. Then xn ∈ Un . Since Xn has the topology induced
by the metric dn , there is an ε > 0 such that B(xn ; ε) ⊆ Un . Let δ = ε/2n .
Then, if d(x, y) < δ, dn (xn , yn ) < ε, so yn ∈ Un , so y ∈ πn−1 [Un ]. Hence
B(x, δ) ⊆ πn−1 [Un ], showing this subbasic open set is open in the metric topology.
Conversely, consider a basic open set B(x, ε) in the metric topology. Choose
P
an N ∈ N such that n≥N 21n < ε/2. Next, let Un = B(xn ; ε/2). Then Un
is open in Xn . Let U = U1 × · · · UN −1 × XN · · · . Then U is a basic open
set in the product topology and x ∈ U . Furthermore, suppose that y ∈ U .
Then, for n < N , dn (xn , yn ) < ε/2 and for n ≥ N , dn (xn , yn ) ≤ 1. Hence
P
P
P
d(x, y) = n dn (xn , yn )/2n < (ε/2) · n<N 21n + n≥N 21n < ε/2 + ε/2 = ε.
So, U ⊆ B(x; ε). Hence B(x; ε) is open in the product topology.
3.4
Weak Topologies
We can generalize the construction of the product topology as follows. Suppose
that {Xα } is a collection of topological spaces and X is a set. Suppose we are
given functions fα : X → Xα . We can give X a topology be defining a subbase
for X to be S = {fα−1 [Uα ] : Uα open in Xα }. We call this topology the weak
topology induced by the functions {fα }.
Proposition 3.20 The weak topology on X is the weakest topology making each
fα continuous.
Proposition 3.21 Let X have the weak topology induced by the functions {fα }.
If X is a topological space, then f : Z → X is continuous if and only if fα ◦ f
is continuous for each α.
3.4. WEAK TOPOLOGIES
33
Definition 3.22 We say a function f : X → Y is an embedding if it is a
homeomorphism of X to f [X] where f [X] is regarded as a subspace of Y .
Notice that every inclusion map is an embedding.
Definition 3.23 We say a collection of functions fα : X → Xα separates points
of X if, whenever x 6= y in X, there is an α such that fα (x) 6= fα (y).
Recall, from Theorem 1.6 that a collection of functions fα : X → Xα induces
Q
a function f : X → Xα such that fα = πα ◦ f . The following shows when
this map is an embedding into the product topological space. This is ultimately
the basis for many important results.
Theorem 3.24 Let {Xα } be a collection of topological spaces, X a topological
Q
space, and fα : X → Xα functions. In order for the induced map f : X → Xα
to be an embedding it is necessary and sufficient for X to have the weak topology
induced by {fα } and for this collection of functions to separate points of X.
Proof: First assume that f is an embedding. The collection of all f [X]∩πα−1 [Uα ]
is a subbase for the topology of f [X]. Hence the collection of f −1 [f [X]∩πα−1 [Uα ]]
is a subbase for the topology of X. But this set is exactly fα−1 [Uα ], so X has
the weak topology induced by {fα }. Also, if x 6= y in X, then f (x) 6= f (y), so
there is an α such that πα ◦ f (x) 6= πα ◦ f (y). Hence fα (x) 6= fα (y), so {fα }
separates points of X.
For the converse, suppose that X has the weak topology induced by {fα }
which separates points of X. Then, if x 6= y in X, there is an α so that
fα (x) 6= fα (y). Since fα = πα ◦ f , this implies that f (x) 6= f (y), so f is oneto-one. Since each fα is continuous, the function f is also continuous. We only
need to show that f : X → f [X] is an open map. But a subbasic open set in X
is one of the form fα−1 [Uα ] whose image in f [X] is f [X] ∩ πα−1 [Uα ], which is a
subbasic open set in f [X].
Because of this result, it is important to be able to identify when a space X
has the weak topology induced by a family of functions. The crucial concept is
that of separation of points and closed sets.
34
CHAPTER 3. NEW SPACES FROM OLD
Definition 3.25 We say that a family of functions {fα } separates points from
closed sets in X if, whenever C ⊆ X is closed and x ∈
/ C, there is a function
fα such that fα (x) 6∈ fα [C].
Theorem 3.26 Suppose that {fα : X → Xα } is a collection of continuous
functions which separates points from closed sets in X. Then the collection of
all sets of the form fα−1 [Uα ] where Uα is open in Xα is a base for the topology of
X. In particular, X then has the weak topology induced by the functions {fα }.
Proof:
Certainly all sets of this form are open in X. So, suppose that U is
open in X and x ∈ U . Let C = X \ U , so x 6∈ C. By assumption, there is an
α such that fα (x) 6∈ fα [C]. Let Uα = Xα \ fα [C], which is an open set in Xα .
Then x ∈ fα−1 [Uα ] ⊆ U , proving the result.
Exercises:
1. Let (X, d) be a metric space with the metric topology. Let A ⊆ X and let
dA : A × A → R be the restriction of d. Show that the subspace topology
on A induced from X is the same as the metric topology induced by dA .
2. Prove Proposition 3.1.
3. Suppose that X is first countable and A ⊆ X. Show that A is first
countable.
4. Suppose that X is second countable and A ⊆ X. Show that A is second
countable.
5. Suppose that X and Y are topological spaces, A ⊆ X and that f : X → Y
is continous with f |A a constant map. Show that there is a continuous
function fˆ : X/A → Y such that fˆ ◦ qA = f where qA : X → X/A is the
quotient map.
6. Suppose that X and Y are topological spaces and f : X → Y is an onto
continuous function. Define an equivalence relation on X by setting x1 Rx2
if and only if f (x1 ) = f (x2 ). Show that the induced map fˆ : X/R → Y is
a continuous, one-to-one, onto map. Show that fˆ is a homeomorphism if
and only if f is a quotient map.
3.4. WEAK TOPOLOGIES
35
7. Suppose that f : X → Y is a continous, closed, onto map. Show that f is
a quotient map.
8. Suppose that f : X → Y is a continuous, open, onto map. Show that f is
a quotient map.
9. Let R, S be equivalence relations on X such that R ⊆ S. Show that there
is a continous map qRS : X/R → X/S such that qRS ◦ qR = qS . Show
that qRS is a quotient map.
10. Suppose that X and Y are topological spaces, A ⊆ X and f : A → Y
is continuous. Suppose also that Z is a topological space and that the
functions g : X → Z and h : Y → Z are continous with g|A = h ◦ f .
Show that H : X ⊕A Y → Z defined by H([(x, 0)]) = g(x) for x ∈ X
and H([(y, 0)]) = h(y) for y ∈ Y is well-defined and gives a continuous
function H.
11. Suppose that (X, d) and (Y, ρ) are metric spaces. Define d∞ on X × Y by
d∞ ((x1 , y1 ), (x2 , y2 )) = max{d(x1 , x2 ), ρ(y1 , y2 )}. Show that d∞ induces
the product topology on X × Y .
12. As in the previous exercise, define d1 ((x1 , y1 ), (x2 , y2 )) = d(x1 , x2 ) +
ρ(y1 , y2 ). Show that d1 also induces the product topology on X × Y .
13. Suppose that BX is a base for the topology of X and BY is a base for the
topology of Y . Show that {B1 × B2 : B1 ∈ BX , B2 ∈ BY } is a base for the
topology of X × Y .
14. Suppose that X and Y are both second countable spaces. Show that X ×Y
is second countable.
15. Suppose that X and Y are first countable spaces. Show that X × Y is
first countable.
16. Show that A × B = A × B for A ⊆ X and B ⊆ Y .
17. Suppose that X and Y are separable spaces. Show that X ×Y is separable.
36
CHAPTER 3. NEW SPACES FROM OLD
18. Suppose that f : X → Y is a continuous, onto function which is also either
open or closed. Show that Y is homeomorphic to X/R where x1 Rx2 if
and only if f (x1 ) = f (x2 ).
19. Suppose that f : X → Y is an onto, continuous, open map. Suppose that
B is a base for the topology of X. Show that {f [B] : B ∈ B} is a base for
the topology of Y .
20. Suppose that f : X → Y is an onto continuous map and D ⊆ X is dense.
Show that f [D] is dense in Y .
21. Suppose that
Q
Xα is second countable. Show that each Xα is second
countable.
22. Suppose that
Q
Xα is separable. Show that each Xα is separable for each
α.
23. Let T = {0, 1, 2} with the topology τ = {∅, {0, 1}, T }. Let X be any
topological space and A the collection of continuous functions f : X → T .
Show that A separates points of X and separates points from closed sets.
Conclude that X can be embedded into T A .
Chapter 4
Separation Axioms
The concept of a topology is a very general one. One difficulty that may arise
is that a given topology may not have very many open sets. We will find it
convenient to look at additional axioms that will guarantee a good collection of
open sets. The most useful such axioms are the separation axioms.
4.1
T0 , T1 , and T2 Axioms
Definition 4.1 We say that a topological space satisfies the T0 separation axiom
(usually, we just say that X is T0 ) if, whenever x 6= y in X, there is an open
set U such that either x ∈ U and y 6∈ U or vice versa.
Definition 4.2 We say that a topological space X satisfies the T1 separation
axiom if whenever x 6= y in X, there is an open set U such that x ∈ U and
y 6∈ U . Note that there will also be an open set V such that y ∈ V and x 6∈ V .
Theorem 4.3 For a topological space X, the following are equivalent:
a) X is T1 ,
b) For each x ∈ X, the set {x} is closed in X,
c) For any A ⊆ X, A =
T
{U : A ⊆ U open}.
37
38
CHAPTER 4. SEPARATION AXIOMS
Proof: Suppose that X is T1 and let x ∈ X. Then, for each y 6= x, there is an
S
open set Uy such that y ∈ Uy and x 6∈ Uy . But then, X \ {x} = y Uy is open,
so {x} is closed.
Now suppose that each {x} is closed in X and let A ⊆ X. Then A =
T
x
X \ {x} where x runs over the x 6∈ A. Thus, A is the intersection of open
sets.
Finally, suppose that the last condition holds and suppose x 6= y in X. Let
A = {x}. By assumption, there is an open set U such that A ⊆ U and y 6∈ U .
Hence, X is T1 .
Theorem 4.4 Products and subspaces of T1 spaces are T1 .
Proof:
Suppose that X is T1 and A ⊆ X. Let x 6= y in A. Then, there is an
open set U in X such that x ∈ U and y 6∈ U . But then A ∩ U is open in A and
x ∈ A ∩ U while y 6∈ A ∩ U .
Now suppose that {Xα } is a collection of T1 spaces and x 6= y in
Q
Xα .
Then, there is a coordinate α such that xα 6= yα . Hence, there is an open set
Uα in Xα such that xα ∈ Uα but yα 6∈ Uα . Then, with U = πα−1 [Uα ], we have
Q
x ∈ U and y 6∈ U . Hence, Xα is T1 .
Perhaps the most important separation axiom is the next one.
Definition 4.5 We say that a topological space satisfies the T2 separation axiom
(or say that X is Hausdorff ) if whenever x 6= y in X, there are open sets U and
V such that x ∈ U , y ∈ V and U ∩ V = ∅.
Example:
Let (X, d) be a metric space and suppose that x 6= y in X. Let
ε = d(x, y)/2 > 0 and set U = B(x, ε) and V = B(y, ε). Then U and V are
open and U ∩ V = ∅. Hence, metric spaces are always T2 .
Theorem 4.6 Products and subspaces of T2 spaces are T2 .
Proof: Suppose that X is T2 and A ⊆ X. Let a1 , a2 ∈ A with a1 6= a2 . Then,
there are open sets U and V in X such that a1 ∈ U , a2 ∈ V and U ∩ V = ∅.
4.1. T0 , T1 , AND T2 AXIOMS
39
But then a1 ∈ A ∩ U , a2 ∈ A ∩ V where A ∩ U and A ∩ V are open sets in A
with (A ∩ U ) ∩ (A ∩ V ) = ∅.
Now suppose that {Xα } is a family of T2 spaces. Suppose that x = (xα ) 6=
Q
(yα ) = y in Xα . Then, there is an index α such that xα 6= yα . Since Xα is
T2 , there are disjoint open sets Uα and Vα such that xα ∈ Uα and yα ∈ Vα . But
Q
then U = πα−1 [Uα ] and V = πα−1 [Vα ] are disjoint open sets in Xα with x ∈ U
and y ∈ V .
Theorem 4.7 A space X is T2 if and only if the diagonal ∆ = {(x, x) : x ∈ X}
is closed in X × X.
Proof: Suppose that X is T2 and (x, y) 6∈ ∆. Then x 6= y, so there are disjoint
open sets U and V with x ∈ U and y ∈ V . But then (x, y) ∈ U × V and
(U × V ) ∩ ∆ = ∅. Hence the complement of ∆ is open in X × X.
Conversely, Suppose that ∆ is closed in X × X and suppose that x 6= y in
X. Then (x, y) 6∈ ∆, so there is a basic open set U × V with (x, y) ∈ U × V ⊆
(X × Y ) \ ∆. But then, x ∈ U , y ∈ V and U ∩ V = ∅.
Theorem 4.8 Suppose that f X → Y is continuous and Y is T2 . Then A =
{(x1 , x2 ) : f (x1 ) = f (x2 )} is closed in X × X.
Proof: Suppose that (x1 , x2 ) 6∈ A. Then f (x1 ) 6= f (x2 ). Then there are disjoint
open sets U and V with f (x1 ) ∈ U and f (x2 ) ∈ V . But then x1 ∈ f −1 [U ],
x2 ∈ f −1 [V ]. Furthermore, if (a, b) ∈ f −1 [U ] × f −1 [V ], then f (a) ∈ U and
f (b) ∈ V . Since U and V are disjoint, f (a) 6= f (b), so (a, b) 6∈ A. Hence
f −1 [U ] × f −1 [V ] ∩ A = ∅, so A is open in X × X.
Theorem 4.9 Suppose that f, g : X → Y are continuous and Y is T2 . Then
{x ∈ X : f (x) = g(x)} is closed in X.
Proof: Let A = {x ∈ X : f (x) = g(x)}. Suppose that x0 6∈ A, so f (x0 ) 6= g(x0 ).
Since Y is T2 , there are disjoint open sets U and V in Y such that f (x0 ) ∈ U
and g(x0 ) ∈ V . Then x0 ∈ f −1 [U ] ∩ g −1 [V ] is open and if x ∈ f −1 [U ] ∩ g −1 [V ],
then f (x) ∈ U and g(x) ∈ V , so f (x) 6= g(x), so x 6∈ A. Hence X \ A is open.
40
CHAPTER 4. SEPARATION AXIOMS
Corollary 4.10 Suppose that f, g : X → Y are continuous and D ⊆ X is dense
in the space X and that Y is T2 . Suppose that f (x) = g(x) for all x ∈ D. Then
f = g.
The following should be compared to the corresponding result for second
countablility where the number of factors must be countable.
Theorem 4.11 Let {Xα } be a collection of T2 spaces. Then
Q
Xα is separable
if and only if each Xα is separable and card{α : cardXα 6= 1} ≤ card[0, 1].
Proof: Let A = {α : cardXα 6= 1}. For each α ∈ A, let Uα and Vα be disjoint
open sets in Xα .
First suppose that
Q
Xα is separable and let D be a countable dense set.
Then πα [D] is dense in Xα for each α. In fact, if Uα is open in Xα , then
πα−1 [Uα ] ∩ D 6= ∅, so πα [D] ∩ Uα 6= ∅.
Next, for each α ∈ A, consider the subset πα−1 [Uα ] ∩ D = Dα of D. If α 6= β
in A, then πα−1 [Uα ] ∩ πβ−1 [Vβ ] 6= ∅ so the intersection of this set with D is nonempty. Hence Dα ∩ πβ−1 [Vβ ] 6= ∅ while Dβ ∩ πβ−1 [Vβ ] = ∅. Thus Dα 6= Dβ . In
other words, the map that takes α ∈ A to Dα ∈ P(D) is one-to-one. Hence,
cardA ≤ cardP(D) = card[0, 1].
Conversely, suppose that each Xα is separable and cardA ≤ card[0, 1]. We
may regard A as a subset of [0, 1]. Let Dα = {d1α , d2α , · · · } be a countable
dense subset of Xα . For each finite sequence of natural numbers {n1 , · · · , nk }
and each finite disjoint collection of closed intervals {F1 , · · · , Fk } in [0, 1] with
Q
rational endpoints, let d(n1 , · · · , nk , F1 , · · · , Fk ) ∈ Xα be defined by setting

 d
if α ∈ Fi
ni α
d(n1 , · · · , nk , F1 , · · · , Fk )(α) =
 d1α
if α 6∈ ∪Fi
There are only countably many such sequences of natural numbers and intervals
with rational endpoints, so the collection, D, of all d(n1 , · · · , nk , F1 , · · · , Fk ) is
a countable subset of the product.
Q
Q
Now, suppose that U =
Uα is a basic open set in
Xα with Uα =
Xα unless α = α1 , · · · , αk .
Since A ⊆ [0, 1], we may find disjoint closed
4.2. REGULARITY
41
intervals F1 , · · · , Fk with rational endpoints such that αi ∈ Fi if αi ∈ A.
Next, we may find dni αi ∈ Uαi ∩ Dαi since Dα is dense in Xα . But then
d(n1 , · · · , nk , F1 , · · · , Fk ) ∈ U ∩ D. Thus D is dense in the product.
4.2
Regularity
Definition 4.12 We say that a topological space X is regular if whenever C ⊆
X is closed and x 6∈ C, there are disjoint open sets U and V such that x ∈ U
and C ⊆ V . In other words, we can separate points from closed sets via open
sets. We say that X is T3 if it is both regular and T1 .
Since singletons are closed in T1 spaces, a T3 space is T2 . It is possible,
however, that a regular space is not T1 and hence will not be T2 .
Proposition 4.13 For a topological spaces X, the following are equivalent:
a) X is a regular space.
b) Whenever U ⊆ X is open and x ∈ U , there is an open set V such that
x ∈ V ⊆ V ⊆ U.
Theorem 4.14 Products and subspaces of regular spaces are regular.
Proof:
Suppose that X is regular and A ⊆ X. Suppose that x 6∈ C, which is
closed in A. Then C = A ∩ K for some closed subset K ⊆ X. Since x 6∈ K,
there are disjoint open sets U and V in X such that x ∈ U and K ⊆ V . But
then A∩U and A∩V are disjoint open sets in A and x ∈ A∩U while C ⊆ A∩V .
Theorem 4.15 Suppose that q : X → Y is an open, closed, continuous, onto
map and that X is T3 . Then Y is T2 .
Proof: Since q is closed, continuous and onto, q is a quotient map. Since X is
T1 , singletons are closed in X, so singletons are closed in Y , so Y is T1 .
Now suppose that y1 = q(x1 ) 6= q(x2 ) = y2 in Y . Then x1 ∈ X \ q −1 [y2 ].
Since X is T3 , there are disjoint open sets U and V so that x1 ∈ U and q −1 [y2 ] ⊆
42
CHAPTER 4. SEPARATION AXIOMS
V . By theorem 3.10, there is an open set W of Y such that y2 ∈ W and
q −1 [W ] ⊆ V . Now, q[U ] is open in Y and y1 ∈ q[U ]. Finally, q[U ] ∩ W = ∅,
showing that Y is T2 .
4.3
Complete Regularity
The next separation axiom seems rather different than the others at first glance.
It is, however, an important step.
Definition 4.16 We say that a topological space X is completely regular if
whenever C ⊆ X is closed and x 6∈ C, there is a continuous function f : X →
[0, 1] such that f (x) = 0 and f [C] = {1}. We say that X is T3 12 if it is both T1
and completely regular.
Theorem 4.17 Products and subspaces of completely regular spaces are completely regular.
Proof: Suppose that X is completely regular and A ⊆ X. Suppose that C ⊆ A
is closed in A and that x ∈ A with x 6∈ C. Then, we may write C = A ∩ K
where K is closed in X. Notice that x 6∈ K, so there is a continuous function
f : X → [0, 1] such that f (x) = 0 and f [K] = {1}. Then f |A : A → [0, 1] is
continuous and f |A [C] = {1}. Hence subspaces of completely regular spaces are
completely regular.
Now suppose that {Xα } is a collection of completely regular spaces, C ⊆
Q
X = Xα is closed and x ∈ X \ C. Then, there are finitely many α1 , · · · αn
Tn
and open sets Uαk ⊆ Xαk such that x ∈ U = k=1 πα−1
[Uαk ] ⊆ X \ C. Then,
k
xαk ∈ Uαk , so there is a continuous function fαk : Xαk → [0, 1] such that
f (xαk ) = 1 and fαk [Xαk \ Uαk ] = {0}. Let f : X → [0, 1] be defined to be the
product of fαk ◦ παk where 1 ≤ k ≤ n. Then f is continuous on X, f (x) = 1
and if y 6∈ U , some yαk 6∈ Uαk , so f (y) = 0. Hence f [X \ C] = {0}.
Theorem 4.18 If a space X is completely regular then it has the weak topology
induced by the collection of continuous functions into [0, 1].
4.4. NORMALITY
Proof:
43
If X is completely regular, the collection of all continuous functions
from X into [0, 1] separates points from closed sets. Hence, by theorem 3.26, X
has the weak topology induced by these functions.
Theorem 4.19 A space can be embedded into a product of [0, 1] if and only if
it is T3 21 .
Proof: This now follows easily from theorem 3.24 and the previous results for
subspaces and products of T3 21 spaces.
In essence, this shows that T3 12 spaces are those with enough continuous
real-valued functions to define the topology. These are then the natural spaces
on which to do analysis of continuous functions.
It is known that there are T3 spaces with more than two points where every
continuous function is constant. Such spaces are not completely regular.
4.4
Normality
Definition 4.20 We say that a topological space X is normal if whenever C
and D are disjoint closed sets, there are disjoint open sets U and V such that
C ⊆ U and D ⊆ V . We say that X is T4 if it is both T1 and normal.
Clearly T4 spaces are T3 . It is not as clear, althought it is true, that T4
spaces are T3 21 .
Theorem 4.21 Metric spaces are T4 .
Proof: Let C and D be disjoint closed sets in X. Define
f (x) =
d(x, C)
.
d(x, C) + d(x, D)
Then f : X → [0, 1] is defined for every x ∈ X since C and D are disjoint and
gives a continuous function. Furthermore, f (x) = 0 if and only if x ∈ C and
f (x) = 1 if and only if x ∈ D. Let U = f −1 [[0, 14 )] and V = f −1 [( 34 , 1]]. Then
C ⊆ U , D ⊆ V and U and V are disjoint open sets in X.
44
CHAPTER 4. SEPARATION AXIOMS
This proof shows, by the way, that metric spaces are T3 12 .
Example:
The Sorgenfry line L is normal. In fact, suppose that C and D
are disjoint closed sets in L. For each x ∈ C, there is a εx > 0 such that
[x, x + εx ) ∩ D = ∅. Similarly, for each y ∈ D there is a δy > 0 such that
S
S
[y, y + δy ) ∩ C = ∅. Now, let U = [x, x + εx ) and V = [y, y + δy ). Clearly
U and V are open with C ⊆ U and D ⊆ V . But, if U ∩ V 6= ∅, there would
be x ∈ C and y ∈ D such that [x, x + εx ) ∩ [y, y + δy ) 6= ∅. But then either
x ∈ [y, y + δy ) or y ∈ [x, x + εx ). Either one is a contradiction.
Unfortunately, it is not the case that subspaces and products of T4 spaces
are T4 . So, while this separation axiom is very nice in some ways, it is not so
nice in others.
Theorem 4.22 Suppose that X is a T4 space. Suppose that D ⊆ X is dense and
that S ⊆ X is closed and discrete (in the subspace topology). Then 2|S| ≤ 2|D| .
Proof: For each A ⊆ S, both A and S \ A are closed in X, so there are disjoint
open sets U (A) and V (A) such that A ⊆ U (A) and S \ A ⊆ V (A). Consider the
subset U (A) ∩ D of D. This gives us a function from P(S) to P(D). I claim
that this map is one-to-one, showing the result.
In fact, if A 6= B ⊆ S, we may assume that there is x ∈ A \ B. Then
x ∈ U (A) ∩ V (B) so this open set has non-empty intersection with D, that is
U (A) ∩ V (B) ∩ D 6= ∅. But, since U (B) ∩ V (B) = ∅, this shows that U (A) ∩ D 6=
U (B) ∩ D, giving the result.
Example:
The product L × L is not a T4 space. In fact, let D =
Q × Q,
a countable dense subset and A = {(x, −x) : x ∈ L}, a discrete, closed subset
of L × L. Then |A| = 2|D| , so 2|A| > 2|D| . Hence L × L cannot be T4 . Thus
product of T4 spaces need not be T4 .
On the other hand, we have the following crucial result about normal spaces.
Theorem 4.23 (Urysohn’s Lemma) A topological space is normal if and only
if whenever C and D are disjoint closed sets, there is a continuous function
f : X → [0, 1] such that f [C] = {0} and f [D] = {1}.
4.4. NORMALITY
45
Proof: If such a function exists, then U = f −1 [[0, 41 )] and V = f −1 [( 34 , 1]] are
disjoint open sets containing C and D. So this condition implies normality. The
converse is more difficult because we have to construct a continuous function on
X ’from scratch’.
First, let U1 = X \ D. Then C ⊆ U1 , so there is an open set U0 such that
C ⊆ U0 ⊆ U0 ⊆ U1 . Next, there is an open set U 12 such that U0 ⊆ U 12 ⊆ U 12 ⊆
U1 . Continuing in this fashion we can find open sets Uq for every dyadic rational
q = m/2n such that if q < r, we have Uq ⊆ Ur .
Now, define f : X → [0, 1] by f (x) = inf{q : x ∈ Uq } if x ∈ U1 while
f (x) = 1 otherwise. Clearly, if x ∈ C, f (x) = 0 and if x ∈ D, f (x) = 1. It
remains to prove that f is continuous. To see this, notice that
f −1 [[0, a)] =
[
Uq
q<a
and
f −1 [(b, 1]] =
[
X \ Uq
b<q
are both open sets in X. The second equality follows since if f (x) > b, there are
dyadic rationals with b < q < r < f (x), so x 6∈ Ur , which implies that x 6∈ Uq .
Conversely, if x 6∈ Uq , then f (x) ≥ q > b.
It follows that T4 spaces are T3 12 . This also shows that every T4 space can
be embedded in a product of [0, 1]. The next result is another crucial result,
allowing the extension of continuous functions from subspaces to the whole
space.
Theorem 4.24 (Tietze’s Extension Theorem). A topological space X is normal
if and only if whenever C ⊆ X is closed and f : C → [0, 1] is continuous, there
is a continuous F : X → [0, 1] with F |C = f .
Proof: It is easier to use [−1, 1] in place of [0, 1] for this proof. So suppose that
f : A → [−1, 1] is continuous. Let C = f −1 [−1, 1 31 ] and D = f −1 [ 13 , 1]. Then
C and D are disjoint closed sets in A and hence in X. But Urysohn’e Lemma,
there is a function g1 : X → [− 13 , 31 ] such that g1 [C] = − 13 and g1 [D] =
Then, for every a ∈ A, we have |f (a) − g1 (a)| ≤
2
3
and |g1 (x)| ≤
1
3
1
3.
for all x ∈ X.
46
CHAPTER 4. SEPARATION AXIOMS
Now consider the function ( 23 )(f − g1 ) : A → [−1, 1]. By the previous
paragraph, there is a function g2 : X → [− 13 , 13 ] such that |( 32 (f − g1 ) − g2 | ≤
on A and |g2 | ≤
1
3
2
3
on X. But then |f − (g1 − 23 g2 )| ≤ ( 23 )2 on A.
Continue in this way to find a sequence gn : X → [− 13 , 13 ] of continuous
functions such that
|f −
n
X
2
2
( )k−1 gk | ≤ ( )n
3
3
k=1
Notice that the sum in this is convergent on X and the limit will be a
continuous F : X → [−1, 1]. By construction, f = F on A.
The same result is true if we replace [0, 1] by (0, 1). In fact, suppose that
f : A → (−1, 1) is continuous. Use Tietze’s Extension theorem to extend to
F : X → [−1, 1]. Now, A and B = f −1 [{−1, 1}] are disjoint closed sets in
X, so there is a continuous function g : X → [0, 1] such that g[A] = {1} and
g[B] = {0}. Now let F 0 (x) = F (x)g(x). Then F 0 : X → (−1, 1) is continous
and extends f .
4.5
Lindelöf Spaces
Definition 4.25 An open cover of a topological space X is a collection of open
S
sets {Uα }α∈A in X such that X = α∈A Uα . A subcover of {Uα }α∈A is a cover
{Uβ }β∈B where B ⊆ A. We say the subcover is finite if the set B is finite.
Similarly, the subcover is said to be countable if B is countable.
Definition 4.26 A topological space X is said to be Lindelöf if every open cover
of X has a countable subcover.
Proposition 4.27 In the definition of Lindelöf, is is sufficient to consider only
basic open sets.
Proof: Suppose that B is a base for the topology of X and that U is an open
cover of X. For each x ∈ X, there is a Ux ∈ U and a Bx ∈ B such that
x ∈ Bx ⊆ Ux . But then {Bx }x∈X is a cover of X by basic open sets. If {Bxn }
is a countable subcover, then {Uxn } is a countable subcover of U.
In particular, second countable spaces are Lindelöf.
4.5. LINDELÖF SPACES
47
Theorem 4.28 Closed subspaces and continuous images of Lindelöf spaces are
Lindelöf.
Proof:
Suppose that X is Lindelöf and A ⊆ X is closed. Suppose also that
{Vα } is an open cover of A in the subspace topology. Then, there are open sets
{Uα } in X with Vα = A ∩ Uα . Then, the collection of open sets {X \ A} ∪ {Uα }
is an open cover of X. Since X is Lindelöf, there is a countable subcover, say
{X \ A} ∪ {Uαn }. Notice that it does no harm to assume that X \ A is in this
subcover. But then {Vαn } is a countable subcover of {Vα } for A.
Theorem 4.29 A regular Lindelöf space is normal.
Proof: Suppose that C and D are disjoint closed subsets of the Lindelöf space
X. For each x ∈ C, there is an open set Ux with x ∈ Ux and D ∩ Ux = ∅.
Similarly, for each y ∈ D, there is a Vy open with y ∈ Vy and C ∩ Vy = ∅.
Since closed subspaces of Lindelöf spaces are Lindelöf, there are countably many
S∞
S∞
{Un } and {Vn } with C ⊆ n=1 Un and D ⊆ n=1 Vn . Now, let W1 = U1 ,
Sn
Z1 = V1 \ W1 , W2 = U2 \ Z1 , Z2 = V2 \ (W1 ∪ W2 , Wn+1 = Un+1 \ k=1 Zk ,
Sn+1
Zn+1 = Vn+1 \ k=1 Wk .
S∞
S∞
Finally, let W = n=1 Wn and Z = n=1 Zn . Then W and Z are open,
C ⊆ W and D ⊆ Z. Finally, W ∩ Z = ∅.
Example: The Sorgenfry line L is Lindelöf.
Proof: Suppose that {[aα , bα )} is a cover of L by basic open sets. Let A =
S
α (aα , bα ). Since R is second countable, the subset A of R is second countable,
S∞
and so it is Lindelöf. Hence, we can write A = n=1 (an , bn ).
Now, suppose that x ∈ L \ A. Then x = aα for some α and we may pick
a rational qx ∈ (aα , bα ) ⊆ A. Suppose that y 6= x in L \ A. Write y = aβ .
Then, for qx = qy , we would need that (aα , bα ) ∩ (aβ , bβ ) 6= ∅. But then, either
x = aα ∈ (aβ , bβ ) ⊆ A or y = aβ ∈ (aα , bα ). Either case is a contradiction.
Since there are only countably many rational numbers, the set L \ A must be
countable. Thus L \ A can be covered by countably many of the original open
sets. Since A is already covered by countably many, we have that L is covered
by countably many of the basic open sets.
48
CHAPTER 4. SEPARATION AXIOMS
Since L × L is not normal, but is T3 , the product of Lindelöf spaces need
not be Lindelöf.
Finally, we give an incredible result which identifies purely topological properties which imply metrizability, at least in the case of separablility.
Theorem 4.30 (Urysohn Metrization Theorem) The following are equivalent
for a topological space X.
a) X is a T3 , second countable space.
b) X is separable and metrizable.
c) X can be embedded into a countable product of [0, 1].
Proof: First notice that a countable product of [0, 1] is metrizable and second
countable, so every subspace will be metrizable and second countable, hence separable. Hence c) implies b). Next, a separable metric space is second countable
and T4 , hence T3 , so b) implies a).
So, now assume that X is T3 and second countable. Then X is Lindelöf, so
we know that X is T4 . Let B be a countable base for the topology of X and
consider the collection {(B1 , B2 ) : B1 , B2 ∈ B, B1 ⊆ B2 }. This is a countable
0
collection of pairs, so we may re-write this set as {(Bn , Bn ) : n ∈ N}.
0
Now, for each pair, (Bn , Bn ), Urysohn’s lemma gives a continuous function
0
fn : X → [0, 1] such that fn [Bn ] = {0}. and fn [X \ Bn ] = {1}. Notice that the
collection {fn } separates points from closed sets. In fact, if C ⊆ X is closed
and x 6∈ C, then there is a B ∈ B such that x ∈ B and B ∩ C = ∅. But, since
X is T3 , there is a B 0 ∈ B such that x ∈ B 0 and B 0 ⊆ B. Then, the pair (B 0 , B)
corresponds to some n ∈ N and fn (x) = 0 while fn [C] = {1}.
Q
This now shows that the function f : X → n [0, 1] such that fn = πn ◦ f is
an embedding. Hence a) implies c).
Exercises:
1. Suppose that fn : X →
R are continuous functions which converge uni-
formly to f : X → R. Show that f is continuous on X.
4.5. LINDELÖF SPACES
49
2. Prove Proposition 4.13
3. Suppose that X is a T3 space and x 6= y in X. Show that there are open
sets U and V with x ∈ U , y ∈ V and U ∩ V = ∅.
4. Suppose that X is a T3 space that A ⊆ X is an infinite set. Show that
there are open sets Un such that Un ∩A 6= ∅ for all n ∈ N and Un ∩Um = ∅
for all n 6= m.
5. A subset A of a topological space is called a retract of X if there is a
continuous map r : X → A such that r(a) = a for all a ∈ A. Suppose that
X is T2 and that A is a retract of X. Show that A is closed in X.
6. Show that A ⊆ X is a retract if and only if whenever f : A → Y is a
continuous function, there is an F : X → Y which extends f .
7. We call a space Y an absolute retract if whenever A ⊆ X is a closed
subspace of a T4 space and f : A → Y is continuous, there is an extension
F : X → Y . Note the Tietze’s Extension theorem says exactly that [0, 1] is
an absolute retract. Show that if Y is an absolute retract and f : Y → X
is an embedding as a closed subspace of a T4 space, then f [Y ] is a retract
of X.
8. Show that a product of absolute retracts is again an absolute retract.
9. Suppose that A ⊆
R is uncountable. Show that there is an x ∈ R such
that for every ε > 0, the set A ∩ [x, x + ε) is uncountable.
10. Suppose that A ⊆ X is a closed subset. Show that the quotient map
q : X → X/A is a closed map.
11. Suppose that X is a T3 space and A ⊆ X is a closed subset. Show that
X/A is T2 .
12. Suppose that both X and Y are T1 (resp, T2 , T3 , T3 12 , T4 ). Show that
X ⊕ Y is T1 (resp, T2 , T3 , T3 12 , T4 ).
50
CHAPTER 4. SEPARATION AXIOMS
13. A space X is said to be completely normal if every subspace of X is normal.
Show that X is completely normal if and only if whenever A, B ⊆ X with
A ∩ B = ∅ = B ∩ A, there are disjoint open sets U and V with A ⊆ U and
B ⊆ V . Hint: for one direction consider the subspace X \ (A ∩ B).
Chapter 5
Convergence and Metric
Spaces
5.1
First Countable Spaces
Definition 5.1 A sequence {xn } in a topological space X is said to converge to
the point x ∈ X if, whenever U is an open set with x ∈ U , there is an N ∈
N
such that n ≥ N implies that xn ∈ U . We will write xn → x.
Notice that it is enough to check this condition for open sets U in a neighborhood base at x. In particular, if (X, d) is a metric space, then xn → x if
and only if for each ε > 0, there is an N ∈
N such that n ≥ N implies that
xn ∈ B(x; ε), in other words, n ≥ N implies that d(xn , x) < ε.
Theorem 5.2 Suppose that X is a first countable space. Then the following
hold.
a) Suppose that A ⊆ X. Then x ∈ A if and only if there is a sequence xn ∈ A
with xn → x.
b) A subset A ⊆ X is closed if and only if whenever xn is a sequence in A
and xn → x ∈ X, then x ∈ A.
51
52
CHAPTER 5. CONVERGENCE AND METRIC SPACES
c) A subset U ⊆ X is open if and only if whenever xn is a sequence with
xn → x ∈ U , then there is an N inN such that n ≥ N implies xn ∈ U .
d) Suppose that Y is a topological space and f : X → Y . Then f is continuous if and only if whenever xn → x ∈ X, we have that f (xn ) → f (x) ∈ Y .
Proof:
a) Suppose that x ∈ A and let N = {Un : n ∈
N} be a countable
neighborhood base at x. We may assume that Un+1 ⊆ Un for all n by replacing
Un by U1 ∩ · · · Un . For each n, A ∩ Un 6= ∅, so pick xn ∈ A ∩ Un . By construction
n ≥ N implies that xn ∈ UN , so xn → x.
Conversely, suppose that xn ∈ A and xn → x ∈ X but that x 6∈ A. Then
x ∈ U = X \ A, so by definition of convergence there is an N ∈
N such that
n ≥ N implies that xn ∈ U . But this contradicts that xn ∈ A.
b) This follows immediately from a).
c) This follows by letting A = X \ U and applying b).
d Suppose that f : X → Y is continuous, that xn → x ∈ X and that
f (x) ∈ V , open in Y . Then x inf −1 [V ] is open in X, so there is an N ∈ N such
that n ≥ N implies that xn ∈ f −1 [V ]. But then n ≥ N implies that f (xn ) ∈ V ,
so f (xn ) → f (x). Notice that this direction does not require that X be first
countable.
Conversely, suppose that whenever xn → x ∈ X, we can conclude that
f (xn ) → f (x) ∈ Y . We need to show that f is continuous. So, suppose that
V ⊆ Y is open and consider the set U = f −1 [V ]. We use the criterion in c
to show that U is open in X. But, if xn → x ∈ U , then f (xn ) → f (x) ∈ V .
Hence, there is an N ∈
xn ∈ f
−1
N such that n ≥ N implies that f (xn ) ∈ V . But then
[V ] = U for n ≥ N . Hence, U is open in X and so f is continuous.
One way to think of this theorem is that convergence of sequences is enough
to define the topology of any first countable space. Unfortunately, this is not
the case for general topological spaces and the following example shows.
Example: Let X = [0, 1][0,1] as a product space with uncountably many factors.
Let A be the collection of f : [0, 1] → [0, 1] such that {x : f (x) 6= 0} is a finite
set. Let g be the function which is identically 1. The claim is that g ∈ A but
5.2. CONVERGENCE OF NETS
53
that no sequence fn from A converges to g. For the first, suppose that U is
Q
a basic open set with g ∈ U . Write U = x∈[0,1] Ux where Ux = [0, 1] except
for finitely many x = x1 , · · · xn . Let f (x) = 1 if x = x1 , · · · xn and f (x) = 0
otherwise. Then, clearly f ∈ A ∩ U . Thus A ∩ U 6= ∅, showing that g ∈ A.
However, if fn is a sequence in A, we can let Fn = {x : fn (x) 6= 0}. Then
each Fn is a finite set, so F = ∪n Fn is a countable set. Pick any x 6∈ F . Then
πx : [0, 1][0,1] → [0, 1] is continuous, πx (fn ) = 0 for all n ∈
N, but πx (g) = 1.
Hence, πx (fn ) does not converge to πx (g). This fn does not converge to g.
Example:
Another illustrative example is found in Ω, the first uncountable
ordinal. Recall that subbasic open sets are those of the form (←, ω0 ) = {ω :
ω < ω0 } and (ω0 , →) = {ω : ω0 < ω}. Recall that ω1 is the largest element of
Ω and that Ω0 = {ω ∈ Ω : ω < ω1 }. Then ω1 ∈ Ω0 , but if {ωn } is a sequence in
Ω0 , there is an upper bound in Ω0 , so ωn does not converge to ω1 .
5.2
Convergence of Nets
Definition 5.3 We say that Λ is a directed set if it has a relation ≤ such that
a) λ ≤ λ for each λ ∈ Λ
b) If λ1 ≤ λ2 and λ2 ≤ λ3 , then λ1 ≤ λ3 for λi ∈ Λ
c) If λ1 , λ2 ∈ Λ, then there is a λ3 ∈ Λ such that λ1 , λ2 ≤ λ3 .
Notice that ≤ does not have to be a partial order on Λ since we do not
assume anti-symmetry. A very important directed set for our purposes is Nx =
{U ∈ τ : x ∈ U } where x ∈ X, a topological space and we define U ≤ V if and
only if V ⊆ U .
Definition 5.4 A net in a set X is a function x : Λ → X from a directed set
Λ to X. We often denote the image of λ ∈ Λ by xλ . If X is a topological space
and x0 ∈ X, we say that the net (xλ ) converges to x0 if whenever U is an open
set with x ∈ U , there is a λ0 ∈ Λ such that λ0 ≤ λ implies that xλ ∈ U . In this
case, we write xλ → x.
54
CHAPTER 5. CONVERGENCE AND METRIC SPACES
If (xλ ) is a net and λ0 ∈ Λ, we call a set of the form {xλ : λ0 ≤ λ} a tail of
the net (xλ ). Convergence of a net to x means that every neighborhood around
x contains a tail of the net.
Notice that sequences are particular types of nets and our notion of convergence for ners generalizes that for sequences.
The following should be compared to theorem 5.2) in both the statement
and the proof.
Theorem 5.5 Let X be a topological space. Then, the following hold.
a) Let A ⊆ X. Then x ∈ A if and only if there is a net (xλ ) in A such that
xλ → x.
b) A subset A ⊆ X is closed if and only if whenever (xλ ) is a net in A with
xλ → x ∈ X, we can conclude that x ∈ A.
c) A subset U ⊆ X is open if and only if whenever (xλ ) is a net with xλ →
x ∈ U , then some tail of (xλ ) is contained in U .
d) Let f : X → Y be a function between topological spaces. Then f is
continuous if and only if whenever (xλ ) is a net in X with xλ → x ∈ X,
then f (xλ ) → f (x) ∈ Y .
Proof: a) Suppose that (xλ ) is a net in A with xλ → x ∈ X. Let U be an open
set with x ∈ U . Then, there is a λ0 ∈ Λ such that λ ≥ λ0 implies that xλ ∈ U .
Hence A ∩ U 6= ∅. This shows that x ∈ A.
Conversely, suppose that x ∈ A. Let Λ = {U : x ∈ U open} with U1 ≤ U2 if
and only if U2 ⊆ U1 . Then, for each U ∈ Λ, A ∩ U 6= ∅, so there is a xU ∈ A ∩ U .
Now, (xU ) is a net in A which convwerges to x. In fact, if x ∈ U is open, just
take λ0 = U .
b) This follows immediately from a).
c) Again, this follows immediately from b) by letting A = X \ U .
d) Suppose that f is continuous and that xλ → x ∈ X. Let f (x) ∈ V , an
open set in Y .Then x ∈ f −1 [V ], an open set in X, so there is a λ0 ∈ Λ such
5.2. CONVERGENCE OF NETS
55
that λ ≥ λ0 implies that xλ ∈ f −1 [V ]. But then f (xλ ) ∈ V , so f (xλ ) → f (x)
in Y .
Converely, we use c) to show that f −1 [V ] is open in X when V is open in
Y . In particular, suppose that xλ → x ∈ f −1 [V ]. Then f (xλ ) → f (x) ∈ V , so
there is a λ0 ∈ Λ such that λ ≥ λ0 implies that f (xλ ) ∈ V . But then the tail
{xλ : λ ≥ λ0 } is contained in f −1 [V ]. Thus, f −1 [V ] is open in X and so f is
continuous.
Theorem 5.6 Suppose that X is a T2 space and (xλ ) is a net in X that converges to both x and y. Then x = y.
Proof:
If not, there are disjoint open sets U and V with x ∈ U and y ∈ V .
There are then λ0 , λ1 ∈ Λ such that λ ≥ λ0 implies that xλ ∈ U and λ ≥ λ1
implies that xλ ∈ V . Since there is a λ ≥ λ0 , λ1 , this is a contradiction.
Theorem 5.7 Let {Xα } be a collection of topological spaces and let X =
Q
Xα .
Then, a net (xλ ) in X converges if and only if πα (xλ ) → πα (x) for every index
α.
Proof: Since each πα is continuous, one implication is automatic. Now suppose
[Uαi ] be a basic open set
that πα (xλ ) → πα (x) for every α. Let U = ∩ni=1 πα−1
i
containing x. Then παi (x) ∈ Uαi for each 1 ≤ i ≤ n. Hence, there are λi ∈ Λ
such that λ > λi implies that παi (xλ ) ∈ Uαi . Since Λ is a directed set, there is
a λ0 ≥ λi for all 1 ≤ i ≤ n. But then, if λ > λ0 , we have that παi (xλ ) ∈ Uαi for
all i, which shows that xλ ∈ U . Thus, xλ → x.
Example:
Let’s reconsider example 5.1. In that example, A is the set of
functions which are 0 except on a finite subset of [0, 1] and g is the constant
function 1. It was shown there that g ∈ A but that no sequence from A converges
to g. Now, let Λ consist of all finite subseteq of [0, 1] and define F1 ≤ F2 if
F1 ⊆ F2 . For F ∈ Λ, let fF be the function which is 1 on F and 0 elsewhere.
Q
We claim that the net (fF ) converges to g. In fact, suppose that U = Uα is
a basic open set containing g. Then F0 = {α : Uα 6= [0, 1]} ∈ Λ and if F0 ≤ F ,
we clearly have that fF ∈ U .
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CHAPTER 5. CONVERGENCE AND METRIC SPACES
Definition 5.8 A subnet of a net (xλ ) is the composition x◦φ where φ : M → Λ
is a function from a directed set M into Λ such that m1 ≤ m2 in M implies
that φ(m1 ) ≤ φ(m2 ) in Λ (we say that φ is increasing) and such that whenever
λ0 ∈ Λ, there is an m ∈ M such that λ0 ≤ φ(m) (we say that φ is cofinal).
Definition 5.9 Let (xλ ) be a net in a topological space X and let x ∈ X. We
say that x is a cluster point of (xλ ) if whenever x ∈ U , an open set and λ0 ∈ Λ,
there is a λ ≥ λ0 such that xλ ∈ U . We say that xλ is in U frequently.
Theorem 5.10 Let (xλ ) be a net in a topological space X and let x ∈ X. Then
the following are equivalent:
a) x is a cluster point of (xλ )
b) There is a subnet of (xλ ) which converges to x
c) x ∈
Proof:
T
λ∈Λ
{xµ : µ ≥ λ}.
First assume that x is a cluster point of (xλ ). Define M = {(λ, U ) :
x ∈ U open , xλ ∈ U }. Define (λ1 , U1 ) ≤ (λ2 , U2 ) if and only if λ1 ≤ λ2 and
U2 ⊆ U1 . Then M is a directed set. Now define φ(λ, U ) = λ. Clearly φ is
increasing. Also, φ(λ, X) ≥ λ, so φ is also cofinal. We show that the subnet
x ◦ φ converges to x. For this, suppose that x ∈ U open. By assumption,
there is a λ ∈ Λ such that xλ ∈ U . Hence (λ, U ) ∈ M . Now suppose that
(µ, V ) ≥ (λ, U ). Then, we have that x ◦ φ(µ, V ) = xµ ∈ V ⊆ U , as required.
Next, suppose that the subnet x ◦ φ converges to x, λ ∈ Λ and let U be
open with x ∈ U . Then, there is a m0 ∈ M such that whenever m ≥ m0 , we
have x ◦ φ(m) ∈ U . Next, there is an m1 ∈ M such that φ(m1 ) ≥ λ. Finally,
since M is a directed set, there is an m ∈ M such that m ≥ m0 , m1 . But
then x ◦ φ(m) ∈ U and φ(m) ≥ φ(m1 ) ≥ λ, so U ∩ {xµ : µ ≥ λ} 6= ∅. Hence
x ∈ {xµ : µ ≥ λ}. This holds for all λ ∈ Λ.
T
Finally, suppose that x ∈ λ∈Λ {xµ : µ ≥ λ}. Then, for each open set U
such that x ∈ U and for each λ ∈ Λ there is a µ ≥ λ such that xµ ∈ U . Hence
x is a cluster point of (xλ ).
5.3. COMPLETE METRIC SPACES
5.3
57
Complete Metric Spaces
Definition 5.11 Let (X, d) be a metric space. A sequence {xn } in X is said to
be a Cauchy sequence if whenever ε > 0, there is a N ∈
N such that whenever
n, m ≥ N , we have d(xn , xm ) < ε. We say that (X, d) is a complete metric
space if every Cauchy sequence in X converges.
Proposition 5.12 In any metric space, any convergent sequence is Cauchy.
Example:
The metric space
R with the usual metric is complete. However,
the homeomorphic space (0, 1) is not complete in the usual metric. This shows
that completeness is a metric property and not a topological property.
Example: If (X, d) is complete if and only if (X, d0 ) is complete where d0 (x, y) =
max{d(x, y), 1}. Hence every complete metric space has a complete, bounded
metric.
Definition 5.13 A subset A of a metric space (X, d) is said to be bounded if
there is an M ∈
R such that d(x, y) ≤ M for all x, y ∈ A. in this case, we
define the diameter of A by diam(A) = sup{d(x, y) : x, y ∈ A}.
Proposition 5.14 Cauchy sequences are bounded.
Proof: There is an N ∈ R such that whenever n, m ≥ N , we have d(xn , xm ) ≤
1. Let M = 1 + max{d(xn , xm ) : n, m ≤ N }. Then, for every n, m ∈
N,
d(xn , xm ) ≤ M . In particular, if n < N and m ≥ N , we have d(xn , xm ) ≤
d(xn , xN ) + d(xN , xm ) ≤ 1 + d(xn , xN ) ≤ M .
Proposition 5.15 If a Cauchy sequence has a convergent subsequence, then
the Cauchy sequence converges.
Proof: Suppose that the subsequence xnk converges to x. Let ε > 0 and choose
K ∈ N so that k ≥ K implies that d(xnk , x) < ε/2. Now choose N ∈ N so that
n, m ≥ N implies that d(xn , xm ) < ε/2. Then, if n ≥ max{nK , N }, we have
d(xn , x) ≤ d(xn , xnK ) + d(xnK , x) < ε/2 + ε/2 = ε.
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CHAPTER 5. CONVERGENCE AND METRIC SPACES
Theorem 5.16 A subspace of a complete metric space is complete (in the restricted metric) if and only if it is closed.
Proof: Suppose that A ⊆ X is complete and suppose that xn → x0 ∈ X. Then
{xn } is Cauchy in X, so is Cauchy in A, so it converges in A. Since metric
spaces are T2 , this implies that x0 ∈ A, so A is closed.
Conversely, if A is closed and {xn } is a Cauchy sequence in A, then it is a
Cauchy sequence in X, so it converges to some point x0 ∈ X. But since A is
closed, x0 ∈ A, so {xn } converges in A, so A is complete.
Theorem 5.17 Suppose that X is a complete metric space and f : X → X
satisfies d(f (x), f (y) ≤ α · d(x, y) for all x, y ∈ X with α < 1. Then, there is a
unique x0 ∈ X such that f (x0 ) = x0 .
Proof:
Pick any x1 ∈ X and define, inductively, xn+1 = f (xn ). Let A =
d(x1 , x2 ). Then, d(xn+1 , xn+2 ) ≤ α · d(xn , xn+1 ) ≤ · · · ≤ αn d(x1 , x2 ) = αn A.
Thus, for m < n, we have d(xm , xn ) ≤ d(xm , xm+1 ) + d(xm+1 , xm+2 ) + · · · +
d(xn−1 , xn ) ≤ αm−1 A + αm A + · · · + αn−2 A ≤ Aαm−1 /(1 − α). Since αm−1 → 0
as m → ∞, this shows that {xn } is a Cauchy sequence. Suppose that xn → x0 .
Then xn+1 = f (xn ) → f (x0 ), but xn+1 → x0 also. Thus f (x0 ) = x0 .
To show uniqueness, suppose that f (y0 ) = y0 also.
Then d(x0 , y0 ) =
d(f (x0 ), f (y0 )) ≤ α · d(x0 , y0 ). Since α < 1, this implies that d(x0 , y0 ) = 0,
so x0 = y0 .
Theorem 5.18 Suppose that (Xn , dn ) is a sequence of complete metric spaces
Q∞
with dn (xn , yn ) ≤ 1 for all xn , yn ∈ Xn . Let X = n=1 Xn with the metric
d(x, y) =
∞
X
dn (xn , yn )
2n
n=1
Then (X, d) is a complete metric space.
Proof: To reduce confusion, we will let the index for sequences in X be a superscript rather than a subscript. Hence, we assume that x(n) is a Cauchy sequence
(n)
(m)
in X. Then, for each subscript, we have that dk (xk , xk ) ≤ 2k d(x(n) , x(m) ).
5.3. COMPLETE METRIC SPACES
59
Thus, the sequence of k th coordinates is a Cauchy sequence in Xk . Let this
sequence converge to x ∈ X . Let x ∈ X have the k th coordinate x . Then
k
k
k
πk (xn ) converges to πk (x) for all n, so xn converges to x.
Example:
Let (X, d) and (Y, ρ) be metric spaces and let Cb (X, Y ) denote
all continuous bounded functions from X to Y . For f, g ∈ Cb (X, Y ), define
d∞ (f, g) = sup{ρ(f (x), g(x)) : x ∈ X}. Then d∞ gives a metric for Cb (X, Y ).
We claim that d∞ is complete if and only if ρ is complete. So suppose that ρ is
complete and that {fn } is a Cauchy sequence in Cb (X, Y ).
For each x ∈ X, the sequence {fn (x)} is a Cauchy sequence in Y since
ρ(fn (x), fm (x)) ≤ d∞ (fn , fm ).
Hence, this sequence converges in Y .
Let
fn (x) → f (x) define f : X → Y . We need to show that f ∈ Cb (X, Y ) and
that fn → f using the d∞ metric.
Let ε > 0 and choose N ∈
N so that n, m ≥ N implies that d∞ (fn , fm ) <
ε/3. For every x ∈ X, there is an m ≥ N so that ρ(f (x), fm (x)) < ε/3. But then,
whenever n ≥ N , we have ρ(f (x), fn (x)) ≤ ρ(f (x), fm (x)) + ρ(fm (x), fn (x)) <
2ε/3. This holds for every x ∈ X and n ≥ N , so d∞ (f, fn ) ≤ 2ε/2 < ε for
n ≥ N . Hence, if f ∈ Cb (X, Y ), then fn → f .
But now, if ε > 0 and x0 ∈ X are given, choose N ∈ N so that n ≥ N implies
that d∞ (f, fn ) < ε/3. Next, choose δ > 0 so that d(x, x0 ) < δ implies that
ρ(fN (x), fN (x0 )) < ε/3. But then, for such x, ρ(f (x), f (x0 )) ≤ ρ(f (x), fN (x))+
ρ(fN (x), fN (x0 ) + ρ(fN (x0 ), f (x0 )) < ε. Hence f is continuous.
Finally, choose N so that n ≥ N implies that d∞ (f, fn ) < 1 and let
M ∈
R be such that x1 , x2 ∈ X implies that ρ(fN (x1 ), fN (x2 )) ≤ M . Then,
ρ(f (x1 ), f (x2 )) ≤ M + 2 for all x1 , x2 ∈ X, so f is bounded.
Definition 5.19 Let (X, d) and (Y, ρ) be metric spaces. A function f : X → Y
is said to be an isometry into Y if ρ(f (x), f (y)) = d(x, y) for all x, y ∈ X. If,
in addition, f can be found to be onto, we say that X and Y are isometric.
Theorem 5.20 Let (X, d) be a metric space. Then, there is a complete metric
space (Y, ρ) along with an isometry Φ : X → Y into Y such that Φ[X] is dense
in Y . Furthermore (Y, ρ) is unique in the sense that if (Z, η) is another complete
60
CHAPTER 5. CONVERGENCE AND METRIC SPACES
metric space and g : X → Z is an isometry with g[Z] dense in Z, then Y and
Z are isometric via an isometry h : Y → Z such that h ◦ Φ = g.
Proof:
First, fix x0 ∈ X. For each a ∈ X, define φa (x) = d(x, x0 ) − d(x, a).
Then φa is a continuous function from X into
R. Also, we have for each x ∈ X
that |φa (x)| ≤ d(x0 , a), so φa is a bounded function.
Now define Φ : X → Cb (X, R) by Φ(a) = φa . We show that Φ is an isometry.
In other words, d∞ (Φ(a), Φ(b)) = d∞ (φa , φb ) = d(a, b) for all a, b ∈ X.
Now let x ∈ X. We have that |φa (x) − φb (x)| = |(d(x, x0 ) − d(x, a)) −
(d(x, x0 ) − d(x, b))| = |d(x, a) − d(x, b)| ≤ d(a, b). Thus, d∞ (φa , φb ) ≤ d(a, b).
On the othe rhand, when x = a, |φa (x) − φb (x)| = d(a, b), so the inequality is
not strict.
Now, let Y = Φ[X]. Then Y is closed in the complete metric space Cb (X, R)
so is complete. Clearly Φ[X] is dense in Y . This shows the existence in the
theorem.
Now suppose that g : X → Z is an isometry, Z is complete and g[X] is
dense in Z. Let y ∈ Y . By density, there is a sequence {xn } in X such that
Φ(xn ) → y. Then {Φ(xn )} is Cauchy in Y and since Φ is an isometry, {xn } is
Cauchy in X. But this implies that {g(xn )} is Cauchy in Z. Hence, there is a
z ∈ Z such that g(xn ) → z. We want to define h(y) = z. To do this, we have
to show that h is well defined.
0
0
0
So, suppose that Φ(xn ) → y also. Then, d(xn , xn ) = d∞ (Φ(xn ), Φ(xn )) ≤
0
0
0
d∞ (φ(xn ), y) + d∞ (y, Φ(xn )) → 0. Thus η(g(xn ), g(xn )) → 0, so η(g(xn ), z) ≤
0
0
η(g(xn ), g(xn )) + η(g(xn ), z) → 0. Hence {g(xn )} also converges to z.
We leave it as an exercise to show that h is an isometry between Y and Z
and that h is onto.
Theorem 5.21 (Baire Category Theorem) Suppose that (X, d) is a complete
metric space and {Un } is a sequence of dense open sets in X. Then ∩∞
n=1 Un is
dense in X.
Proof: Let D = ∩∞
n=1 Un . Let U be any non-empty open set in X. We wish to
show that D ∩ U is non-empty.
5.3. COMPLETE METRIC SPACES
61
Since U1 is dense, U ∩ U1 is non-empty, so there is an x1 ∈ U ∩ U1 . Since
U ∩ U1 is open, there is a ε1 < 1 such that B(x1 ; ε1 ) ⊆ U ∩ U1 .
Now, B(x1 ; ε1 ) is open and U2 is dense, so B(x1 ; ε1 ) ∩ U2 is non-empty. As
before, there is a x2 and a ε2 <
1
2
such that B(x2 ε2 ) ⊆ B(x1 ; ε1 ) ∩ U2 .
Continue in this way to find a sequence {xn } and εn <
1
2n
such that
B(xn+1 ; εn+1 ) ⊆ B(xn , εn ) ∩ Un+1 ⊆ U ∩ U1 ∩ · · · Un+1 .
Now, if ε > 0, find N so that
xn , xm ∈ (xN , εN ), so d(xn , xm )
1
2N −1
< 21N
< ε. Then, if n, m ≥ N , we have that
< ε, so the sequence {xn } is Cauchy.
Since X is complete, xn → x for some x ∈ X.
But now, if m ≥ n, xm ∈ B(xn ; εn ) ⊆ Un . Since B(xn ; εn ) is closed, x ∈ Un .
This holds for every n ∈ N, so x ∈ U ∩ ∩∞
n=1 Un , as desired.
Example: In particular, we can see that
Q is not the intersection of a countable number of open sets in R. For, if it were, Q = ∩∞
n=1 Un , then each Un would
be dense and open. Also, for each x ∈ Q, R \ {x} is dense and open. So we
would then have ∩∞
n=1 Un ∩ ∩x∈Q R \ {x} = ∅ is an intersection of a countbable
family of dense open sets, but is no longer desne.
Example:
We can no show explicitly that L × L is not normal. Let A =
{(x, −x) : x ∈
Q} and B = {(x, −x) : x 6∈ Q}. Then A and B are disjoint
closed subsets of L × L. Suppose that U and V are open with A ⊆ U and
B ⊆V.
For each n ∈
Bn =
N define An = {x ∈ R : (x + 1/n, −x + 1/n) ∈ U } and
◦
∪∞
k=n An .
Q ⊆ Bn for every n.
In fact, if x ∈ Q, then there is an ε > 0 such that [x, x+ε)×[−x, −x+ε) ⊆ U .
Claim:
Choose N ≥ n such that 2/N < ε. Then, if y ∈ (x − 1/N, x + 1/N ), we have
(y + 1/N, −y + 1/N ) ∈ U , so y ∈ AN . Thus, (x − 1/N, x + 1/N ) ⊆ AN , so
x ∈ A◦N .
But this shows that Bn is a dense open set. By the previous example,
∩∞
n=1 Bn
6=
Q. Pick some x ∈ ∩∞
n=1 Bn \ Q. Then (x, −x) ∈ B, so there is an
ε > 0 such that [x, x + ε) × [−x, −x + ε) ⊆ V . Now, pick 1/N < ε and n ≥ N
such that x ∈ A◦n . Then (x + 1/n, −x + 1/n) ∈ U ∩ V , so there are no disjoint
62
CHAPTER 5. CONVERGENCE AND METRIC SPACES
open sets containing A and B.
Exercises:
1. Show that h as defined in the proof of theorem 5.20 is an isometry and is
onto.
2. We say that a metric space (X, d) is totally bounded if for each ε > 0,
there are finitely many ε-balls that cover X. Show that a totally bounded
metric space is bounded.
3. Show that a totally bounded metric space is separable.
4. Show that every sequence in a totally bounded space has a Cauchy subsequence.
5. Show that every subspace of a totally bounded space is totally bounded.
6. Suppose that A is a dense, totally bounded subspace of the metric space
X. Show that X is totally bounded.
7. Show that
Q is not a Gδ in R.
8. We say that a subset of a topological space X is nowhere dense if (A)◦ = ∅.
Show that an open set is dense if and only if X \ U is nowhere dense.
9. We say that a subset A of X is of first category if it is the union of
countably many nowhere dense subsets. Show that a complete metric
space is not of first category in itself.
Chapter 6
Compactness
Definition 6.1 A collection A of subsets of X is said to have the finite intersection property if whenever A1 · · · , An ∈ A, then ∩nk=1 Ak 6= ∅.
Definition 6.2 A topological space is said to be compact if every open cover of
X has a finite subcover.
Theorem 6.3 For a topological space X, the following are equivalent:
a) X is compact.
b) Whenever C is a collection of closed sets with the finite intersection propT
erty, then C 6= ∅.
c) Every net has a cluster point.
d) Every net has a convergent subnet.
Proof: Suppose that X is compact and that C is a collection of closed sets of X
with the finite intersection property. Then, {X \ C}C∈C is a collection of open
sets such that no finite sub-collection covers X by DeMorgan’s theorem. Hence,
This collection cannot cover X, so, again by DeMorgan, ∩C∈C C 6= ∅. Hence a)
implies b). The converse is evidence again by an argument using DeMorgan.
By our general results on convergence of nets, we know that c) and d) are
equivalent.
63
64
CHAPTER 6. COMPACTNESS
Next, suppose that (xλ ) is a net and let Fλ = {xα : α ≥ λ}. Then C = {Fλ }
is a collection of closed sets. Since Λ is a directed set, given any λ1 · · · , λn ,
there is a λ ≥ λ1 · · · λn , the collection C has the finite intersection property. If
we assume b), the intersection of all Fλ is non-empty. But this shows that the
net (xλ ) has a cluster point.
Finally, assume c) and let C be a collection of closed sets with the finite intersection propery. Let Λ consist of all finite sets F = {F1 · · · , Fn } ⊆ C ordered
by inclusion. For each such F, there is an xF ∈ F1 ∩ · · · Fn by assumption.
Now, if x is a cluster point of the net (xF ), we must have x ∈ F = F for each
F ∈ C. Thus, the intersection of all elements of C is non-empty, so b) follows.
Theorem 6.4 A closed subspace of a compact space is compact. A compact
subspace of a T2 space is closed.
Proof:
Suppose that A ⊆ X is a closed subspace of the compact space X.
Suppose that {Vα } is an open cover of A in the subspace topology. Then, we can
write Vα = A ∩ Uα where each Uα is open in X. Then {X \ A} ∪ {Uα } is an open
cover of X. Since X is compact, this has a finite subcover, {X\A, Uα1 , · · · , Uαn }.
Then Vα1 · · · , Vαn } is a cover of A.
Now suppose that X is T2 and A ⊆ X is compact. Let x 6∈ A. We find an
open set containing x that is disjoint from A. This will show that A is closed.
For each y ∈ A, the T2 property gives disjoint open sets Uy and Vy such that
x ∈ Uy and y ∈ Vy . Now, {A ∩ Vy }y∈A is an open cover of A so there are
y1 , · · · yn so that A ⊆ Vy1 ∪ · · · ∪ Vyn . Let U = Uy1 ∩ · · · Uyn . Then x ∈ U and
U ∩ A = ∅.
Theorem 6.5 A compact T2 space is T4 .
Proof: Let X be compact and T2 We first show that X is T3 . Let x 6∈ C where
C is closed in X. For each y ∈ C, there are disjoint open sets Uy and Vy such
that x ∈ Uy and y ∈ Vy . But then {Vy }y∈Y is an open cover of C, which is
compact since it is closed in a compact space. Hence, there are y1 , · · · , yn such
Sn
T
that C ⊆ k=1 Vyk = V . Let U = Uyk . Then x ∈ U , C ⊆ V and U ,V are
disjoint open sets in X.
65
Now we show that X is T4 . Suppose that C and D are disjoint closed sets
in X. For each x ∈ C, there are disjoint open sets Ux and Vx such that x ∈ Ux
and D ⊆ Vx . Again, the collection {Ux } covers the compact set C, so there are
Tn
Sn
x1 , · · · , xn such that C ⊆ k=1 Uxk = U . Let V = k=1 Yxk . Then C ⊆ U ,
D ⊆ V and U ,V are disjoint open sets in X.
Theorem 6.6 The continuous image of a compact space is compact.
Proof:
Suppose that X is compact and f : X → Y is continuous and onto.
if {Uα } is an open cover of Y , then {f −1 [Uα ]} is an open cover of X. By
S
compactness, there are α1 , · · · , αn such that X = f −1 [Uαk ]. But then Y =
S
Uαk , so we have a finite subcover of the original cover.
Theorem 6.7 (Alexander Sub-base Theorem) Suppose that S is a subbase for a
topological space X. In order for X to be compact, it is necessary and sufficient
that every cover of X by elements of S have a finite subcover.
Proof:
Clearly, if X is compact, every cover by elements of S has a finite
subcover.
Conversely, suppose that every cover by elements of S has a finite subcover,
but that X is not compact. In particular, there is an open cover with no finite
subcover. Let O denote the collection of all open covers without finite subcovers.
If U1 , U2 ∈ O are open covers, we say U1 ≤ U2 if U1 ⊆ U2 ; in other words, U1
has fewer open sets than U2 . In this way, O becomes a partially ordered set.
S
Now, suppose that C ⊆ O is a chain. Let U = C. Then U is an open cover
of X since every element of C is an open cover. Also, U has no finite subcover.
In fact, if U1 , U2 , · · · Un are in U, then there are Ui ∈ C such that Ui ∈ Ui for
each 1 ≤ i ≤ n. But, since C is a chain, there is a U0 ∈ C such that Ui ⊆ U0
for each 1 ≤ i ≤ n. But then U1 , U2 , · · · , Un cannot be a cover since U0 has no
finite subcover. Thus, U ∈ O is an upper bound for C.
By Zorn’s lemma, there is a maximal element V ∈ O. Then V is an open
cover of X with no finite subcover, but if U is an open set, U 6∈ V, then V ∪ {U }
has a finite subcover. Now consider S ∩ V. There is no finite subcover from this
set, so by our assumption, S ∩ V is not a cover of X.
66
CHAPTER 6. COMPACTNESS
Suppose that x 6∈
S
(S ∩ V). Since V is a cover, there is a V ∈ V such
that x ∈ V . Also, since S is a subbase for X, there are S1 , · · · , Sn ∈ S such
Tn
that x ∈ k=1 Sk ⊆ V . But no Sk is in V, so by the maximality of V, there are
S
V11 , V21 , · · · Vnk k such that Sk ∪V11 ∪· · ·∪Vnk k = X. But then ∩nk=1 Sk ∪ ij Vij =.
S
But this implies that V ∪ ij Vij = X, giving a finite subcover of V. This
contradiction shows that X must be compact.
Theorem 6.8 (Tychonoff ’s Theorem) The product of topological spaces is compact if and only if each factor is compact.
Proof:
Since each projection πα is continuous and surjective, compactness of
ΠXα implies the compactness of each Xα .
Conversely, suppose that U is a cover of ΠXα by subbasic open sets. For each
S
α, let Uα = {U : πα−1 [U ] ∈ U}. If Uα is not a cover of Xα , pick xα 6∈ Uα . If no
S
Uα is a cover of the corresponding Xα , then x = (xα ) ∈ ΠXα , but x 6∈ U. But
we assumed that U is a cover of ΠXα , so this means some Uα is an open cover
of Xα . But if U1 , · · · , Un is a finite cover from Uα , then πα−1 [U1 ], · · · , πα−1 [Un ] is
a finite subcover from U. By the Alexander subbase theorem, ΠXα is compact.
In particular, any cube, [0, 1]A is compact and T2 , so is T4 .
We now turn to an important example. Recall that Ω0 is a well-ordered
uncountable set such that whenever x ∈ Ω0 , the set {y ∈ Ω0 : y ≤ x} is
countable. We let ω1 6∈ Ω0 and define Ω = ω0 ∪ {ω1 } with x < ω1 for all x ∈ Ω0 .
Now, if x ∈ Ω0 , we define the successor of x, x+ to be the smallest element of
(x, →). If y = x+ for some x, we say that y is a successor ordinal. Otherwise,
we say that y is a limit ordinal.
We give both Ω and Ω0 the order topology: sub-basic open sets are of the
form (←, x) = {y : y < x} or (x, →) = {y : x < y}. Since both Ω and Ω0 are
linearly ordered, this means that basic open sets are of the form (←, x), (x, →),
or (x, y) = {a : x < z < y}.
In particular, a neighborhood of x contains a set of the form (y, x] = (y, x+).
Theorem 6.9 The space Ω is compact.
67
Proof:
Suppose that {Uα } is an open cover of Ω. Then, there is an α0 such
that ω1 ∈ Uα0 . Since ω1 is not a successor ordinal, there is an x < ω1 such that
(x, ω1 ] ⊆ Uα0 . Let x1 be the smallest such x. If x1 6= 0, then x1 6∈ Uα0 .
Now pick Uα1 such that x1 ∈ Uα1 and the smallest x2 such that (x2 , x1 ] ⊆
Uα1 . If x2 6= 0, then x2 6∈ Uα1 . Continuing in this manner, we either obtain an
infinite decreasing sequence {xn } or we find some xn = 0. The first case violates
the well-ordered property of Ω and the second gives a finite subcover for Ω.
On the other hand, Ω0 is not compact. In fact, it is not even Lindelöf since
the cover {[0, α)} where α ∈ Ω0 is an open cover with no countable subcover.
Lemma 6.10 Suppose that A and B are closed, uncountable subsets of Ω0 .
Then A ∩ B is uncountable.
Proof:
Suppose not. Then there is an x0 ∈ Ω0 such that x ≤ x0 for all
x ∈ A ∩ B. Since A is uncountable and {x : x ≤ x0 } is countable, there is
an a1 ∈ A with x0 < a1 . Now, since B is uncountable and {x : x ≤ a1 } is
countable, there is a b1 ∈ B with a1 < b1 .
Continue in this manner to obtain an ∈ A, bn ∈ B such that x0 < a1 < b1 <
a2 < b2 < · · · . Then y = sup{an } = sup{bn } and y ∈ A ∩ B with x0 < y. This
is a contradiction.
Theorem 6.11 Suppose that f : Ω0 →
and there exists x0 ∈ Ω0 and α0 ∈
R is continuous. Then f is bounded
R such that f (x) = α0 for all x ≥ x0 . In
other words, f is constant on some tail of Ω0 .
Proof: We first show that f is a bounded function. Suppose not. Then, there
is an x1 ∈ Ω1 such that |f (x1 )| > 1. Now, the set [0, x1 ] is compact by the
same proof that shows Ω is compact, so f is bounded on [0, x1 ]. Hence, there
is an x2 > x1 such that |f (x2 )| > 2. Continuing in this manner, we obtain an
increasing sequence {xn } such that |f (xn )| > n. But now, if y = sup{xn }, f
cannot be continuous at y ∈ Ω0 .
Suppose now that f [Ω0 ] ⊆ [−M, M ] ⊆ R.
Claim:
There is an α0 ∈ [−M, M ] such that f −1 [(α0 − ε, α0 + ε)] is
uncountable for every ε > 0.
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CHAPTER 6. COMPACTNESS
Suppose not. Then for each α ∈ [−M, M ], there is an εα > 0 such that
f
−1
[(α−εα , α+εα )] is countable. Let Uα = (α−εα , α+εα ). Then {Uα }α∈[−M,M ]
is an open cover of the compact set [−M, M ], so there is a finite subcover, say
with α1 , · · · , αn ∈ [−M, M ]. This would imply that Ω0 = ∪nk=1 f −1 [Uαn ] is
countbale. This is a contradiction.
We now show that the α0 of the claim is unique. In fact, suppose α1 also
has the property that f −1 [(α1 − ε, α1 + ε)] is uncountable for every ε > 0. If
α0 6= α1 , let ε = |α1 − α0 |/3. Then f −1 [α0 − ε, α0 + ε] and f −1 [α1 − ε, α1 + ε]
are two disjoint, uncountable closed sets in Ω0 in contradiction to the lemma
above.
Now, for each n ∈ N, the closed sets f −1 [α0 − n1 , α0 + n1 ] and f −1 [R \ (α0 −
2
n , α0
+
2
n )]
are disjoint and the first is uncountable. This means the second
is not uncountable, which means it must be countable. But f −1 [R \ {α0 }] =
−1
∪∞
[R \(α0 − n2 , α0 + n2 )], is countable. Choose may now choose any x0 ∈ Ω0
n=1 f
larger than any element of this set to finish the proof.
N∗ = N ∪ {ω0 } with n < ω0 for all n ∈ N given the order
∗
topology. Let T = Ω × N . Then T is a compact T2 space, so is normal. Let
Example:
Let
T ∗ = T \ {(ω1 , ω0 )}. Then T ∗ is a T3 12 space. We show that T ∗ is not normal.
Hence, subspaces of T4 spaces need not be T4 .
∗
In fact, let A = {ω1 } × N and B = Ω × {ω0 }. Then A ∩ T ∗ and B ∩ T ∗
are disjoint closed subsets of T ∗ . Suppose that U and V are open in T ∗ with
A ⊆ U and B ⊆ V . Then, for each n ∈
N, the point (ω1 , n) ∈ U , so there is
an αn ∈ Ω0 such that αn < x implies (n, x) ∈ U . Let α = sup{αn } ∈ Ω0 . Let
β > α in Ω0 . Then (β, ω0 ) ∈ V , so there is an n0 ∈ N such that n > n0 implies
that (β, n) ∈ V . But then, any such (β, n) ∈ U ∩ V , so U and V cannot be
disjoint.
6.1
Local Compactness
Definition 6.12 A topological space X is said to be locally compact if whenever
x ∈ U open, there is an open V such that x ∈ V ⊆ V ⊆ U such that V is
6.1. LOCAL COMPACTNESS
69
compact. In other words,a space is locally compact if the compact neighborhoods
form a neighborhood base at every point.
Notice that a locally compact space is automatically regular. In fact, more
is true.
Proposition 6.13 A locally compact Hausdorff space is completely regular.
Proof:
Suppose that x0 6∈ C closed. Pick an open set V such that x0 ∈ V ⊆
V ⊆ X \ C and such that V is compact. Then V is a compact Hausdorff space,
so is T4 . Also, x0 6∈ V \ V , so there is a continuous function g : V → [0, 1] such
that g(x0 ) = 1 and g = 0 on V \ V .
Now, define f : X → [0, 1] by setting f (x) = g(x) if x ∈ V and f (x) = 0 if
x 6∈ V . Notice that f is then well defined and is continuous on the closed sets
V and X \ V . But then f is continuous on X = V ∪ (X \ V ). Also, f (x0 ) = 1
and f (x) = 0 for x ∈ C.
Notice that we actually get a function that is 0 off the compact set V , i.e. a
function with compact support. This is important in some applications.
Proposition 6.14 Let X be locally compact, K ⊆ U where K is compact and
U is open. Then, there is an open set V with V compact and K ⊆ V ⊆ V ⊆ U .
Proof: For each x ∈ K, there is an open Vx such that x ∈ Vx ⊆ Vx ⊆ U with
Vx compact. But then {Vx } is an open cover of K, so there are x1 , · · · , xn with
S
S
K ⊆ Vxk = V . But then V is open and V = Vxk is compact with V ⊆ U .
Proposition 6.15 A T2 space is locally compact if and only if each point has
a compact neighborhood.
Proof:
One direction is clear. Now suppose that each point has a compact
neighborhood and suppose that x ∈ U open. Also suppose that x ∈ V with
K = V is compact. Then x ∈ U ∩ V and U ∩ V is an open set in K, which
is a compact, T2 space. By regularity of K, there is an open set W such that
x ∈ W ∩K ⊆ ClK (W ∩K) ⊆ U ∩V . But then, x ∈ U ∩V ∩W and ClK (W ∩K) is
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CHAPTER 6. COMPACTNESS
a compact, hence closed subset of X with U ∩ V ∩ W ⊆ ClK (W ∩ K). But then,
U ∩ V ∩ W ⊆ ClK (W ∩ K) ⊆ U ∩ V ⊆ U and U ∩ V ∩ W ⊆ K, so U ∩ V ∩ W
has compact closure in X.
6.2
Compactifications
Definition 6.16 Let X be a topological space. We say that a pair (Y, f ) is a
compactification of X if Y is a compact T2 space, f : X → Y is an embedding
of X to a dense subspace f [X] of Y .
Notice that if X has a compactification, then X is T3 12 .
Example: The pair ([0, 1], i) is a compactification of (0, 1) where i : (0, 1) →
[0, 1] is the inclusion map.
Example: The pair (S 1 , e) is a compactification of (0, 1) where e : (0, 1) →
S 1 is given by e(x) = (cos(2πx), sin(2πx)).
Example: Let f : X → Y be any embedding of X into a compact T2 space.
Then (f [X], f ) is a compactification of X.
Theorem 6.17 Let X be a locally compact T2 space. Define X∞ = X ∪ {∞}
and define U ⊆ X∞ to be open if either U ⊆ X is open in X or ∞ ∈ U and X \U
is compact in X. Then (X∞ , i) is a compactification where i is the inclusion
map.
We call (X∞ , i) the one point compactification of X.
Definition 6.18 Let (Y, f ) and (Z, g) be compactifications of the space X. We
say (Y, f ) ≤ (Z, g) if there is a function h : Z → Y such that f = h ◦ g. Notice
that h is then an extension of the homeomorphism f ◦ g −1 : g[X] → f [X].
By density, this means h is unique if it exists. We say (Y, f ) and (Z, g) are
equivalent compactifications if (Y, f ) ≤ (Z, g) and (Z, g) ≤ (Y, f )
In general, it is common to identify X and f [X] for a compactification (Y, f ).
We call the set Y \ X = Y \ f [X] the remainder of X in Y .
6.2. COMPACTIFICATIONS
71
Lemma 6.19 Suppose that f : X → Y is continuous, where X is T2 . Suppose
that A ⊆ X is dense and f |A is a homeomorphism. Then f [X \ A] ⊆ Y \ f [A].
Proof: Suppose not. Then there are x 6∈ A and a ∈ A such that f (a) = f (x).
Hence, there are disjoint open sets U and V such that x ∈ U and a ∈ V . Now,
A ∩ V is an open set in A, so by the fact that f |A is a homeomorphism, there
is an open set W of Y such that f [A ∩ V ] = W ∩ f [A].
Now, f (x) = f (a) ∈ W ∩ f [A], so x ∈ f −1 [W ]. By density of A, A ∩ U ∩
f −1 [W ] 6= ∅. But, if a0 is in this set, f (a0 ) ∈ W ∩ f [A] = f [A ∩ V ]. Since a0 ∈ U
and U ∩ V = ∅, this contradicts that f is one-to-one on A.
In particular, if h : Z → Y as in the definition of (Y, f ) ≤ (Z, g), then
h[Z \ g[X]] = Y \ f [X]. One inclusion follows from the lemma. The other
follows since h must be onto (Z is compact, so h[Z] is closed in Y and contains
f [X]). Hence, h takes the remainder of X in Z to the remainder of X in Y .
Definition 6.20 Let X be T3 12 and let F denote the collection of all continuous
f : X → [0, 1]. We know that X has the weak topology from the set F, so
e : X → [0, 1]F is an embedding where e(x)f = f (x) for f ∈ F. Then, if we let
βX = e[X], we have that (βX, e) is a compactification of X. This is called the
Stone-Cech compactification of X.
The Stone-Cech compactification has a very useful property: if f : X → [0, 1]
is a continuous map, there is an F : βX → [0, 1] such that F ◦ e = f . In other
words, every continuous function from X into [0, 1] has an extension to βX.
Example: From this, we see that [0, 1] is not equivalent to β(0, 1) since, for
example, f (x) = sin(1/x) has no extension to [0, 1].
More is true about βX:
Theorem 6.21 Suppose X is T3 12 and f : X → Y is continuous where Y is
compact and T2 . Then there is a continuous F : βX → Y such that F ◦ e = f .
Proof:
Let G denote the collection of all continuous g : Y → [0, 1]. There is
an embedding eY : Y → [0, 1]G such that πg ◦ eY = g for all g ∈ G. Since Y is
compact, the image of this map is closed in [0, 1]G .
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CHAPTER 6. COMPACTNESS
For each g ∈ G, the map g ◦ f : X → [0, 1] is continous, so has an extension
Gg : βX → [0, 1] such that Gg ◦e = g◦f . Hence, there is a map G : βX → [0, 1]G
such that πg ◦ G = Gg . Now, for x ∈ X, πg ◦ G ◦ e(x) = Gg ◦ e = g ◦ f (x) =
πg ◦eY ◦f (x). Since this happens for all g ∈ G, we have that G◦e = eY ◦f . Thus,
G[e[X]] ⊆ eY [Y ]. Since e[X] is dense in βX, this shows that G[βX] ⊆ eY [Y ].
But now F = e−1
Y ◦ G : βX → Y satisfies the conditions of the theorem.
As a consequence, we find that certain compactifications are surprisingly
large:
Example: We know that [0, 1][0,1] is separable, so there is a countable dense
subset D ⊆ [0, 1][0,1] . Let f : N → D be any onto map. Then f is continuous so
there is an extension F : β N → [0, 1][0,1] . The image of this map is closed since
β N is compact and dense because it contains D. Hence, F is an onto map. In
particular, card(β N) ≥ card([0, 1][0,1] = 2card([0,1] = card(P([0, 1])).
Another consequence of this result is that whenever (Y, eY ) is a compactification of X, then (βX, e) ≥ (Y, eY ). In other words, (βX, e) is the ’largest’
compactification of X. Furthermore, the only property of βX we used in the
above proof is the fact that all continuous functions from X to [0, 1] can be extended to βX. This means that any other compactification with this property
shares the maximality with βX and so is equivalent to βX.
Example:
We have shown that every continuous function f : Ω0 →
R is
constant on some tail. But if we then define f (ω1 ) to be this constant, we get
and extension of f to Ω. Hence, βΩ0 = Ω.
6.3
Paracompactness
Definition 6.22 Suppose that U = {Uα }α∈I is an open cover of X. Another
open cover V = {Vα }α∈I is said to be a shrinking of U if Vα ⊆ Uα for each
αinI. In particular, the index set of the two covers is the same.
Definition 6.23 Suppse that U is an open cover of X. Another open cover V
is said to be a refinement of U if whenever V ∈ V, there is a U ∈ U such that
V ⊆ U.
6.3. PARACOMPACTNESS
73
To elucidate in the case of indexed covers, the cover V = {Vβ }β∈J is a
refinement of U = {Uα }α∈I if there is a function φ : J → I such that Vβ ⊆ Uφ(β)
for each β ∈ J. We write V ≺ U.
Exercises:
1. Show that a finite union of compact sets is compact.
2. Suppose that X is compact, U ⊆ X is open, and C is a collection of
closed sets such that ∩C ⊆ U . Show there are C1 , · · · , Cn ∈ C such that
∩nk=1 Ck ⊆ U .
3. Suppose that A ⊆ X and B ⊆ Y are compact sets and that A × B ⊆ U ,
which is open in X × Y . Show that there are open sets A ⊆ V and B ⊆ W
such that V × W ⊆ U .
4. Suppose that X is compact. Show that the projection πY : X × Y → Y
is a closed map.
5. Suppose that X is compact, Y is T2 and that f : X → Y is continuous,
one-to-one, and onto. Show that f is a homeomorphism.
6. Suppose that X is compact and A ⊆ X is dense. Show that A is locally
compact if and only if A is open in X.
7. Suppose that X is T2 and that f : X → Y is a closed, onto map with
f −1 (y) compact for each y ∈ Y . Show that Y is T2 .
8. Suppose that X is locally compact, T2 and that f : X → Y is continuous,
closed, open, and onto. Show that Y is locally compact.
9. A space X is said to be countably compact if every countable open cover
has a finite subcover. Show that X is countably compact if and only if every sequence has a cluster point, i.e, that every sequence has a convergent
sub-net.
10. Show that a space is countably compact if and only if whenever C1 ⊇
C2 ⊇ · · · are non-empty closed sets, then ∩∞
n=1 Cn 6= ∅.
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CHAPTER 6. COMPACTNESS
11. Show that Ω0 is countably compact.
12. Show that the continuous image of a countably compact space is countably
compact.
13. A space X is said to be sequentially compact if every sequence has a
convergent subsequence. Show that every sequentially compact space is
countably compact.
14. Show that every first countable countably compact space is sequentially
compact.
15. Show that a metric space is compact if and only if it is sequentially compact.
16. Show that a metric space is compact if and only if it is both complete and
totally bounded.
17. Show that every compact subset of the Sorgenfry line L is countable.
18. Show that equivalence of compactifications is an equivalence relation.
19. Suppose that X is T4 and that A and B are closed disjoint subsets of X.
Show that ClβX A and ClβX B are disjoint.
Chapter 7
Connectedness
7.1
Basic Results
Definition 7.1 We say a topological space X is connected if the only subsets
of X that are both open and closed are the emptyset and X. Otherwise, we say
X is disconnected.
We say that ∅ and X are the trivial open and closed sets in X.
Proposition 7.2 For a topological space X, the following are equivalent:
a) X is disconnected.
b) There are disjoint, non-empty open sets U and V such that X = U ∪ V .
In this case, we call {U, V } a disconnection of X.
c) There are disjoint, non-empty closed sets C and D such that X = C ∪ D.
d) There is a non-constant continuous function f : X → {0, 1}, the two-point
discrete space.
Proof: In fact, if U is both open and closed, then V = X \ U is open and closed
and X = U ∪ V . In this case, defining f (x) = 0 if x ∈ U and f (x) = 1 if x ∈ V
defines a continous function to {0, 1}.
75
76
CHAPTER 7. CONNECTEDNESS
Proposition 7.3 Let X be a topological space and A ⊆ X. Then A is connected
in the subspace topology if and only if whenever U and V are open sets in X
such that A ⊆ U ∪ V and A ∩ U ∩ V = ∅, then A ⊆ U or A ⊆ V .
Proof: Consider the disjoint sets A ∩ U and A ∩ V . These are disjoint sets that
are open in A. Furthermore, every pair of disjoint open sets in A arise in this
way from some U and V with A ∩ U ∩ V = ∅. For connectedness, we need either
A ∩ U = A or A ∩ V = A.
Example:
The closed unit interval [0, 1] is connected. In fact, suppose
that U ⊆ [0, 1] is both open and closed. We may assume 0 ∈ U (otherwise
use [0, 1] \ U . Let A = {x : [0, x) ⊆ U }. Let x0 = sup A. Then [0, x0 ) =
∪x∈A [0, x) ⊆ U , so x0 ∈ A. Notice that x0 > 0 since 0 ∈ U and U is open.
Also, x0 ∈ U since U is closed (otherwise, there would be an ε > 0 such that
(x0 − ε, x0 + ε) ∩ U = ∅, which violates the definition of x0 .). But now, if x0 < 1,
there is a ε > 0 such that (x0 − ε, x0 + ε) ⊆ U , which contradicts x0 = sup A.
But now, [0, 1] = [0, x0 ) ∪ {x0 } ⊆ U . Hence [0, 1] is connected.
Proposition 7.4 Suppose that X is connected and f : X → Y is continuous
and onto. Then Y is connected.
Proof:
f
−1
Suppose not. If U ⊆ Y is a non-trivial open and closed subsets, then
[U ] is a non-trivial (by the fact that f is onto) open and closed subset of X.
In other words, continuous images of connected sets are connected.
The following turns out to be a very useful way to show that sets are connected: show they are unions of connected subsets that overlap.
Theorem 7.5 Suppose {Aα }α∈I are connected subsets of the space X. Suppose
also that ∩α∈I Aα is non-empty. Then ∪α∈I Aα is connected.
Proof:
Let A = ∪α∈I Aα and let x0 ∈ ∩α∈I Aα . Suppose that U and V are
open subsets of X such that A ⊆ U ∪ V and A ∩ U ∩ V = ∅. Without loss of
generality, assume x0 ∈ U . Then, for each α ∈ I, Aα ⊆ U ∪V and Aα ∩U ∩V = ∅.
Since x0 ∈ Aα ∩ U , the connectedness of Aα implies that Aα ⊆ U . But then
A = ∪α∈I Aα ⊆ U . Hence, A is connected.
7.1. BASIC RESULTS
Example: Suppose a < b in
77
R. Then [a, b] is a continuous image of [0, 1],
so is connected. But then, [a, b) = ∪n∈N [a, b − n1 ] is connected. Similarly, (a, b]
and (a, b) are connected. Also, [a, ∞) = ∪n∈N [a, a + n], (−∞, b] and
1
2
2
R are
2
connected. As another example, the set S = {(x, y) ∈ R : x + y = 1} is the
image of [0, 1] under f (t) = (cos 2πt, sin 2πt), so is connected.
Theorem 7.6 (Intermediate Value Theorem) Suppose that X is connected
and f : X →
R is continuous. Suppose that x, y ∈ X and c ∈ R such that
f (x) < c < f (y). Then, there exists a z ∈ X such that f (z) = c.
Proof: Otherwise U = f −1 (−∞, c) and V = f −1 (c, ∞) would give a non-trivial
disconnection of X.
Theorem 7.7 Suppose that A ⊆ X is a connected subset and that A ⊆ B ⊆ A.
Then B is connected. In particular, the closure of a connected set is connected.
Proof:
Suppose that U and V are open in X such that B ⊆ U ∪ V and
B ∩ U ∩ V = ∅. Then A ⊆ U ∪ V and A ∩ U ∩ V = ∅, so by connectedness of A,
we can assume that A ⊆ U . But then A ⊆ X \ V , which is closed, so A ⊆ X \ V .
But then B ⊆ A ⊆ U . Hence B is connected.
Example:
Let X = ({0} × [0, 1]) ∪ {(x, sin( x1 )) : x ∈ (0, 1]}. Then X is
the closure of the set {(x, sin( x1 ) : x ∈ (0, 1]}, which is the continuous image of
2
(0, 1]. Hence, X is a connected set in R . The space X is called the Topologist’s
Sine Curve.
We now turn to products of connected spaces.
Theorem 7.8 Let X and Y be connected spaces. Then X × Y is connected.
Proof: We may assume that both X and Y are non-empty. Let (x0 , y0 ) ∈ X×Y .
Then, the set {x0 } × Y is connected and for each y ∈ Y , the set X × {y} is
connected. Since (x0 , y) ∈ ({x0 }×Y )∩(X ×{y}), so Ay = ({x0 }×Y )∪(X ×{y})
is connected. But now, X × Y = ∪y∈Y Ay and (x0 , y0 ) ∈ Ay for all y, so X × Y
is connected.
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CHAPTER 7. CONNECTEDNESS
Theorem 7.9 Let {Xα }α∈I be any collection of connected spaces. Then Πα∈I Xα
is connected.
Proof:
Let a ∈ X = ΠXα . For each finite set F ⊆ I, the set XF = {x ∈ X :
xα = aα , α 6∈ F } is homeomorphic to the space Πα∈F Xα so it is connected.
S
Also, x ∈ XF for all finite F ⊆ I. Thus, F ⊆I XF is a connected subset of
X. But, this union is dense in X (open sets restrict on at most finitely many
coordinates), so the closure, X, is connected.
7.2
Components and Local Connectedness
Definition 7.10 Let X be a topological space and x ∈ X. The component of x
S
in X is the set {C : x ∈ C, C connected}. A subset of X is called a component
of X if it is the component in X of some x ∈ X.
Notice that the component of x is a connected set since x is an element of
corresponding intersection. Hence, the component is the largest connected set
in X containing x.
Theorem 7.11 Every component of X is closed in X. Distinct components of
X are disjoint sets. Hence, the collection of components of X forms a partition
of X into closed subsets.
Proof: Since components are maximal connected sets and since the closure of a
connected set is connected, each component must be closed. Again, if C and D
are components with C ∩ D 6= ∅, then C ∪ D is connected and contains both C
and D. Hence, C = C ∪ D = D by maximality. This shows that the collection
of components is a partition of X.
Definition 7.12 We say that a space X is totally disconnected if all the components of x are singletons.
Example:
The product X = {0, 1}N of countably many 2-point spaces
is totally disconnected since the projection of any connected subset must be
7.2. COMPONENTS AND LOCAL CONNECTEDNESS
79
a point. Since this space is compact and infinite, it cannot be discrete. In
particular, no singleton in X is open. There is another important description
of this same space.
a+2b
Let [a, b] be any closed interval. Define E(I) = [a, 2a+b
3 ] ∪ [ 3 , b]. So
E(I) is obtained from I by removing the middle third of the interval. Now, if
A = ∪nk=1 Ik is a disjoint union of closed intervals, define I(A) = ∪nk=1 E(Ik ).
Notice that I(A) is a disjoint union of closed intervals and that I(A) ⊆ A. Also
notice that if x is an endpoint of some interval in A, then x is also an endpoint
of some interval in I(A).
Now, define inductively, E0 = [0, 1], En+1 = E(En ). Then En is a disjoint
union of 2n closed intervals, each of length 1/3n . Finally, let E = ∩∞
n=1 En . We
call E the Cantor middle thirds set.
Define a function f : X → E by setting f ((εn )) =
P∞
2εn
n=1 3n .
We leave it as
an e xercise to show that f is a continuous, one-to-one function from X onto E.
Hence, f is a homeomorphism. In particular,
1
4
= f ((0, 1, 0, 1, 0, 1, · · · )) ∈ E.
Theorem 7.13 Let X be a compact Hausdorff space and x ∈ X. Then the
component of x is the intersection of all subsets containing x which are both
open and closed.
Proof:
Let C be the component of x. Suppose that U is an open and closed
set with x ∈ U . Then, C ∩ U is an open and closed subset of C. Since C is
connected, and x ∈ U ∩ C, we must have C ⊆, C ∩ U , i.e, C ⊆ U . In particular,
T
if A = {U : x ∈ U, U open and closed}, then C ⊆ A = K.
We now show that K is connected. Since C is maximal connected, this will
show K ⊆ C, yielding C = K, as claimed. Notice that K is closed since it is
the intersection of closed sets.
So, suppose that K = K1 ∪ K2 with K1 , K2 disjoint closed sets in X. Since
X is T4 , there are disjoint open sets V1 , V2 with Ki ⊆ Vi for i = 1, 2. But now,
T
A ⊆ V1 ∪ V2 , and X is compact, so there are finitely many Uj in A such
that K ⊆ ∩nj=1 Uj ⊆ V1 ∪ V2 . Let U = ∩j Uj so that U is open and closed and
K ⊆ U ⊆ V1 ∪ V2 . Suppose that x ∈ V1 .
80
CHAPTER 7. CONNECTEDNESS
But now, x ∈ U ∩ V1 and U ∩ V1 is open in X. But also, U ∩ V1 = U \ V2
is closed in X. So, x ∈ K ⊆ U ∩ V1 . But then K2 ⊆ K \ V1 = ∅. Thus, K is
connected.
We actually showed a bit more than was claimed. In fact, the proof shows
that if C is a component of X and if C ⊆ U with U open, then there is a set V
which is both open and closed with C ⊆ V ⊆ U . In particular, the closed and
open subsets form a base for the topology of any totally disconnected, compact
Hausdorff space.
Theorem 7.14 Suppose C1 ⊇ C2 ⊇ C3 ⊇ · · · is a decreasing sequence of
T∞
compact, connected, T2 spaces. Then n=1 Cn is connected (and compact).
Proof:
Each of the spaces Cn is closed in C1 , so
T
Cn is also. By the finite
intersection property, the intersection is also non-empty. Now, suppose that we
T
can decompose Cn = K ∪ L where K and L are disjoint, non-empty, closed
sets. Since C1 is a T4 space, there are disjoint open sets U and V with K ⊆ U
T
and L ⊆ V . Then Cn ⊆ U ∪ V .
Now, by compactness, there is a Cn with Cn ⊆ U ∪ V . But, since U ∩ Cn
and V ∩ Cn are non-empty, this contradicts the connectedness of Cn .
In some senses, the opposite extreme from total disonnectedness is local
connectedness:
Definition 7.15 We say a space X is locally connected if whenever x ∈ U with
U open, there is a connected neighborhood V of x such that x ∈ V ⊆ U .
Theorem 7.16 A space X is locally connected if and only if every component
of every open set is open.
Proof: If every component of every open set is open, suppose that x ∈ U and
U is open. Let V be the component of U containing x. Then by assumption
V is open, so it is a neighborhood of x and x ∈ V ⊆ U . Hence X is locally
connected.
Conversely, suppose that X is locally connected and that C is a component
of the open set U . Let x ∈ C. Then x ∈ U , so there is a connected neighborhood
7.3. PATH CONNECTEDNESS
81
V of x with x ∈ V ⊆ U . But now, since C is the component of x and since V is
connected, x ∈ V ⊆ C. Hence, C contains a neighborhood of each of its points,
so C is open.
Example:
It is possible to be connected and not locally connected. Let
X = ([0, 1] × {0}) ∪ ({0} ∪ (({1/n : n ∈ N}) × [0, 1]). Then X is connected. But
the open set U = X \ ([0, 1] × {0}) has {0} × [0, 1] as a component, but this
component is not open. Alternatively, notice that (0, 1) ∈ U , but there is no
connected neighborhood of (0, 1) contained in U .
7.3
Path Connectedness
Definition 7.17 Let X be a topological space and x, y ∈ X. A path from x to
y is a map p : [0, 1] → X such that p(0) = x and p(1) = y.
We say that X is path connected if for every x, y ∈ X there is a path from
x to y. We say that X is locally path connected if each point has a base of path
connected neighborhoods.
Proposition 7.18 Let X be a topological space and define the relation x y if
there is a path from x to y in X. Then
is an equivalence relation on X. The
equivalence classes are called path components of X.
Proof: If x ∈ X, then the constant path p(t) = x for all t ∈ [0, 1] is a path from
x to x, so x x.
Suppose x y and that p : [0, 1] → X is a path from x to y. Define q(t) =
p(1 − t), so q : [0, 1] → X is a path from y to x, so y x.
Finally, assume that x y and y z. Let p be a path from x to y and q a path
from y to z. Define r : [0, 1] → X by

 p(2t)
0 ≤ t ≤ 21
r(t) =
 q(2t − 1) 1 ≤ t ≤ 1
2
Then r is continuous because its restriction is continuous on the closed sets [0, 12 ]
and [ 12 , 1] and r( 12 ) is well defined. Then r is a path from r(0) = x to r(1) = z,
so x z.
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CHAPTER 7. CONNECTEDNESS
Proposition 7.19 A path connected space is connected.
Proof:
Fix any x ∈ X. For each y ∈ X, there is a path py from x to y. But
then, the image p[0, 1] is a connected set containing both x and y, so x and y
are in the same component of X. Hence, X has exactly one component, so is
connected.
Proposition 7.20 A connected, locally path connected space is path connected.
Proof: By local path connectivity, the path components are all open. But the
complement of a path component is the union of all the other path components.
Hence the complement is also open. Connectivity then shows there can be only
one path component.
Example:
The topologist’s sine curve, X, is connected but not path con-
nected. In fact, suppose that p : [0, 1] → X is a path from (0, 0) to (1, sin 1).
Let f = π1 ◦ p, so f (0) = 0 and f (1) = 1. By continuity, there is a t1 ∈ (0, 1)
such that f (t1 ) =
2
π.
Then, there is a t2 ∈ (0, t1 ) such that f (t2 ) =
Continue in this manner to find t1 > t2 > t3 > · · · with f (tn ) =
2
3π .
2
(2n−1)π .
Let x0 = inf{tn : n ≥ 1} = lim tn ∈ [0, 1]. Now let g = π2 ◦ p. Then
= (−1)n−1 . Then continuity of g is violated at x0 .
g(tn ) = sin (2n−1)π
2
Exercises:
1. Show that a space is connected if and only if every continuous function
into the discrete space {0, 1} is constant.
2. Suppose that {Cn }∞
n=1 are connected sets such that Cn ∩ Cn+1 6= ∅ for all
S∞
n ≥ 1. Show that n=1 Cn is connected.
3. Suppose that X is connected and locally connected and C ⊆ X is a connected, closed subspace. Suppose that U is a component of X \ C. Show
that X \ U is connected.
4. Suppose that f : X → Y is continuous, closed and f −1 (y) is connected
for all y ∈ Y . Suppse that C ⊆ Y is connected. Show that f −1 [C] is
connected.
7.3. PATH CONNECTEDNESS
83
5. Suppose that X is a compact T2 space and define xRy when there is a connected C with x, y ∈ C. Show that X/R is a compact totally disconnected
T2 space and that the quotient q : X → X/R satisfies the conditions of
the previous problem.
6. Show that products and continuous images of path connected spaces are
path connected.
7. Show that products and subspaces of totally disconnected spaces are totally disconnected.
8. Show that
R2 \ Q2 is connected.
9. Find two disjoint connected sets A and B such that A ∪ B is connected.
10. Show that a connected, T4 space with at least two points is uncountable.
11. Show that a T3 12 space X is connected if and only if the Stone-Cech compactification βX is connected.
12. Show that β R \ R is not connected.
13. Show that β R \ R has exactly two components.
2
2
14. Show that β R \ R is connected.
84
CHAPTER 7. CONNECTEDNESS
Chapter 8
Homotopy And
Fundamental Groups
8.1
Fundamental Group
Definition 8.1 Let x, y ∈ X be two points in a topological space. Define Px,y
to be the collection of paths p : [0, 1] → X from x to y, i.e, p(0) = x, p(1) = y.
We say that two paths, p, q ∈ Px,y are homotopic if there is a continuous
H : [0, 1] × [0, 1] → X such that H(t, 0) = p(t), H(t, 1) = q(t) for all t ∈ [0, 1]
and H(0, s) = x, H(1, s) = y for all s ∈ [0, 1]. We say that H is a homotopy
from p to q. We write p ' q or H : p ' q if we want to make H explicit.
Definition 8.2 Let p ∈ Px,y and q ∈ Py,z be paths. Define
n
p(2t) 0≤t≤1/2
p ∗ q(t) = q(2t−1)
1/2≤t≤1
Then p ∗ q ∈ Px,z .
Also, for p ∈ Px,y , define p̂(t) = p(1 − t), so p̂ ∈ Py,x .
Finally, if x ∈ X, define the constant path cx (t) = x for all t.
Proposition 8.3 Homotopy of paths in Px,y is an equivalence relation.
Definition 8.4 Define π1 (X; x, y) = Px,y / '. We simplify if x = y and write
π1 (X, x) = π1 (X; x, x).
85
86
CHAPTER 8. HOMOTOPY AND FUNDAMENTAL GROUPS
Proposition 8.5 Suppose p, p0 ∈ Px,y and q, q 0 ∈ Py,z . Suppose also that p ' p0
and q ' q 0 . Then p ∗ q ' p0 ∗ q 0 .
In other words, we can define an operation π1 (X; x, y) × π1 (X; y, z) →
π1 (X; x, z) by [p] ∗ [q] = [p ∗ q]. The proposition says that this operation is
well defined.
Proposition 8.6 For p ∈ Px,y , q ∈ Py,z , r ∈ Pz,w , we have p∗(q ∗r) ' (p∗q)∗r.
Also, for p ∈ Px,y , we have cx ∗ p ' p ' p ∗ cy and cx ' p ∗ p̂ and p̂ ∗ p ' cy .
Theorem 8.7 The set π1 (X, x) with the operation [p] ∗ [q] = [p ∗ q] is a group
with identity [cx ] and inverse [p]−1 = [p̂]. Furthermore, if x and y are in the
same path component of X, then π1 (X, x) is isomorphic to π1 (X, y).
Proposition 8.8 Let f : X → Y be a function.
Then f∗ : π1 (X, x) →
π1 (Y, f (x)) defined by f∗ [p] = [f ◦ p] is well defined and is a homomorphism
of groups.
Proposition 8.9 We have (1X )∗ = 1π1 (X,x) , and (f ◦ g)∗ = f∗ ◦ g∗
8.2
Homotopy
Definition 8.10 Let f, g : X → Y be two continous functions. We say that f
and g are homotopic and write f ' g is there is a continous H : X × [0, 1] → Y
such that H(x, 0) = F (x) and H(x, 1) = g(x) for all x ∈ X. In this case, we
say that H is a homotopy from f to g. We can also write H : f ' g to explicitly
mention the homotopy H.
If, in addition A ⊆ X and f |A = g|A , we say that the homotopy H is relative
to A if H(a, t) = f (a) = g(a) for all 0 ≤ t ≤ 1. In this case, we write H : f ' g
(rel A).
Definition 8.11 A topological pair (X, A) is a topological space X with a subspace A ⊆ X. A map f : (X, A) → (Y, B) is a continous function f : X → Y
with f [A] ⊆ B.
A pointed topological space (X, x) is a topological pair (X, {x}).
8.2. HOMOTOPY
87
Notice that this is consistent with our previous definition of homotopy of
paths if we require the homotopy to be relative to {0, 1} ⊆ [0, 1].
n
Example: Let Y ⊆ R be a convex subset (i.e, if x, y ∈ Y and 0 ≤ t ≤ 1, then
tx + (1 − t)y ∈ Y ). Then any two maps f, g : X → Y are homotopic via the
homotopy H : X × [0, 1] → Y defined by H(x, t) = tg(x) + (1 − t)f (x). Notice
that if f |A = g|A , then H is a homotopy relative to A.
Corollary 8.12 If X ⊆
Rn is a convex set and x ∈ X, then π1 (X, x) is the
trivial group.
Definition 8.13 If X and Y are topological spaces and y ∈ Y , we denote the
constant map cy : X → Y where cy (x) = y for all x ∈ X.
Example:
If X ⊆
Rn is convex, x0 ∈ X and f : X → Y is any map, then
f is homotopic to cy where y = f (x0 ). A homotopy showing this is given by
H(x, t) = f ((1 − t)x + tx0 ).
Proposition 8.14 If X and Y are topological spaces and A ⊆ X, then homotopy relative to A is an equivalence relation on C(X, Y ), the collection of
continuous functions from X to Y .
Proposition 8.15 Suppose that f1 ' f2 : X → Y and g1 ' g2 : Y → Z. Then
g1 ◦ f1 ' g2 ◦ f2 : X → Z.
Proof: Suppose that H : f1 ' f2 . Then g1 ◦ H : X × [0, 1] → Z is a homotopy
g1 ◦ f1 ' g1 ◦ f2 . Also, if G : g1 ' g2 , and we define G1 (x, t) = G(f2 (x), t), then
G1 : g1 ◦ f2 ' g2 ◦ f2 .
Definition 8.16 We say that two spaces X and Y are homotopically equivalent
if there are maps f : X → Y and g : Y → X such that f ◦g ' 1Y and g ◦f ' 1X .
Proposition 8.17 Homotopy equivalence is an equivalence relation on the collection of topological spaces.
equivalent.
Also, homeomorphic spaces are homotopically
88
CHAPTER 8. HOMOTOPY AND FUNDAMENTAL GROUPS
The easiest situation is that of contractible spaces.
Definition 8.18 A space X is said to be contractible if X is homotopy equivalent to a singleton {y}.
Theorem 8.19 For a space X, the following are equivalent:
a) X is contractible
b) The identity map 1X is homotopic to a constant map cx0 : X → X for
some x0 ∈ X.
c) There is an x0 ∈ X and a homotopy H : X × [0, 1] → X such that
H(x, 0) = x and H(x, 1) = x0 for all x ∈ X.
d) Whenever Y is a space and f, g : Y → X are continuous, then f ' g.
Proposition 8.20 Convex subsets of
Rn are contractible.
Exercises
1. Show that the map f : π1 (X × Y, (x, y)) → π1 (X, x) × π1 (Y, y) given by
f [p] = (π1∗ [p], π2∗ [p]) is an isomorphism.
2. Let p ∈ Px,y , q ∈ Py,z , r ∈ Pz,w . Give a homotopy H : p∗(q ∗r) ' (p∗q)∗r.
3. We say that a pair (X, A) with A ⊆ X has the homotopy extension property if whenever f : X → Y and H : A × [0, 1] → Y are maps with
H(a, 0) = f (a) for a ∈ A, then there is an extension G : X×[0, 1] → Y such
that G(x, 0) = f (x) for x ∈ X and G(a, t) = H(a, t) for a ∈ A, t ∈ [0, 1].
Show that the pair (X, A) has the homotopy extension property if and
only if (X × {0}) ∪ (A × [0, 1]) is a retract of X × [0, 1].
4. Show that ([0, 1], {0, 1}) has the homotopy extension property.
5. Suppose that H : [0.1] × [0, 1] → X is a map. Let p(t) = H(t, 0), q(t) =
H(1, t), r(t) = H(0, t), s(t) = H(t, 1). Show that p ∗ q ' r ∗ s as paths in
X.