Download Topology I Lecture Notes

Topology I Lecture Notes
A80029
Version 2.3a, July 5, 2007
by Mahmoud Filali
Edited by V-M Sarenius according to lectures in fall 1998 and 1999.
Further editing in spring 2007 by Iikka Mylly.
Topologically Speaking
Topologically speaking, circles are the same as squares.
Topologically speaking, people are the same as bears.
Your donut is the same as the coffee cup you dunk it in each day.
"Upside-down’s the same as downside-up," a topologist might say.
Topologically speaking, a quarter is the same as a dime.
Topologists can count, but rarely finite amounts.
You can bet they haven’t yet to tell a 6 from a 9.
A topologist can make a t-shirt with a piece of paper and a three-hole punch.
The topological secret is the homeomorphic scrunch!
Topolocical spaces are the places where topologists live.
They like to drive compact manifolds, for when they bump into each other they give.
You can visit your favorite topologist in a land called RP2,
and if the hand that you favor is right you just might become a new left-handed you!
Topologically speaking, peaches are the same as pears.
Topologically speaking, people are the same everywhere.
Topologically faces all look just like the one on you!
Topologically races all share one gender, shape and hue.
Topological spaces will expand your point of view.
’Cause topologically speaking, I’m just the same as you.
It’s true I’m just the same as you.
We’re homeomorphic!
I’m just the same as you!
c
Monty
Harper
Contents
1
1.1
1.2
2
2.1
2.2
2.3
2.4
3
3.1
3.2
3.3
4
4.1
4.2
5
5.1
5.2
6
7
8
9
A
A.1
A.2
A.3
A.4
A.5
A.6
A.7
A.8
A.9
Preliminaries 1
Basic set theory 1
Functions 3
Metric Spaces 6
Distance 6
Interior and closure 9
Sequences 10
Continuous functions 13
Topological spaces 15
Topology 15
Continuous functions 20
Subspaces 21
Operations on topological spaces
Sum of spaces 24
Product spaces 25
Convergence 29
Sequences 29
Nets 31
Separation axioms 35
The T4 -spaces 39
Compact spaces 44
Connected spaces 48
Exercises 56
Exercises for Chapter 1 56
Exercises for Chapter 2 57
Exercises for Chapter 3 60
Exercises for Chapter 4 64
Exercises for Chapter 5 64
Exercises for Chapter 6 66
Exercises for Chapter 7 67
Exercises for Chapter 8 68
Exercises for Chapter 9 69
24
Chapter 1
Preliminaries
This chapter discusses the very basics of set theory and functions. It is intended
as a review of facts the reader is already assumed to be somewhat familiar
with, and should not be seen as a rigorous introduction to the subject. However,
all the rest of the material in these notes should be intelligible to anyone able
to understand the basic results presented here.
1.1
Basic set theory
The discussion in this section will be based on the so called intuitive or naïve
set theory. Although a more exact axiomatic approach is available, given proper
caution the approach taken here will work fine for us.
Definition 1.1. A set, family or collection is a composition of objects, called
the elements or points of the set. If an object a is an element of a given set A,
we write a ∈ A. Otherwise, we write a ∈
/ A.
A straightforward way of describing a set is to list all of its elements, for example the set of symbols in the binary numeral system can be written as {0, 1},
whereas the set of all odd integers could be written as {±1, ±3, ±5, . . . }. It is
easy to realize that this method will be impossible to apply in many cases, even
with sets with finitely many elements. Try listing the set of all the cells in your
body, for instance! To avoid such troubles, we give the following definition.
Definition 1.2. If S is a statement which applies to some of the elements of a
given set A, then {a ∈ A : S(a)} is the set of all elements of A for which S(a) is
true.
One might wonder what was to happen if S from Definition 1.2 was a universally false sentence, like x 6= x, for one. Our next definition answers that
question.
Definition 1.3. Let A be a given set. Then the set
{x ∈ A : x 6= x},
which clearly has no elements, is called the empty set, and denoted by ∅.
Example 1.1. Here we have some examples of how Definition 1.2 can be applied.
1.1 Basic set theory
1)
2)
3)
4)
5)
2
{n ∈ N : 0 < n < 4} = {1, 2, 3}.
{n ∈ Z : n is odd} = {±1, ±3, ±5, . . . } .
{z ∈ C : |z| = 1} = T (the complex unit circle).
{x ∈ R : 0 ≤ x ≤ 1} = [0, 1].
{x ∈ R : 0 < x < 1} = ]0, 1[.
Unfortunately, our naïve definition of sets leads to unexpected difficulties, of
which the next famous result is the prime example.
Russel’s Paradox 1901. First note that a set can easily have elements which
are themselves sets, like in {0, {1, 2}, 5}. So, a set might well contain itself as
one of its elements. In such a situation we say that our set is exceptional. Otherwise we say that it is ordinary. Let now A be the set of all ordinary sets. We
are running into a logical contradiction; this set is both ordinary and exceptional for if A is exceptional then it must be one of its own elements and so it is
ordinary. If it is ordinary then it is contained in A and so it is exceptional.
Remark. Because of Russel’s paradox, we agree from now on that no set shall
contain itself as an element.
Definition 1.4. A set A is said to be a subset of a set B if each element in A
also belongs to B. We denote this by A ⊆ B. The set B is said to be a superset
of A, denoted by B ⊇ A. Two sets A and B are equal if A ⊆ B and B ⊆ A.
Remark. It follows from the definition above, that the empty set ∅ is unique.
Definition 1.5. The union of two sets A and B is the set:
A ∪ B = {a : a ∈ A or a ∈ B}.
The intersection of two sets A and B is the set:
A ∩ B = {a : a ∈ A and a ∈ B}.
The concept of union and intersection can be extended to any collection of sets.
Let {Ai : i ∈ I} = {Ai }i∈I be the collection of sets Ai indexed by elements of a
given set I. The union of this collection is the set
[
Ai = {a : a ∈ Ai for some i ∈ I},
i∈I
and the intersection is the set
\
Ai = {a : a ∈ Ai for all i ∈ I}.
i∈I
The difference of two sets A and B is the set:
A \ B = A − B = {a : a ∈ A and a ∈
/ B}.
Let A and X be sets, with A defined as A = {a ∈ X : S(a)}. Then the complement of the set A in X, denoted by A∁ , is the set of elements x ∈ X which are
not elements of A, i.e., for which S(x) is false. Note that
A∁ = {x ∈ X : x ∈ X and a ∈
/ A} = X \ A.
1.2 Functions
3
The product of two sets A and B is the set of ordered pairs
A × B = {(a, b) : a ∈ A and b ∈ B}.
For any finite collection A1 , . . . , An of sets, the product of the sets is defined as
the set of ordered n-tuplets
A1 × · · · × An = {(x1 , . . . , xn ) : xi ∈ Ai , i = 1, . . . , n}.
If for each i = 1, . . . , n, Ai = A, the product can be abbreviated as An .
The power set of a set A is the set of all subsets of A. The power set is denoted
by P(A) or 2A .
Theorem 1.1. Let X be a given set, let {Ai }i∈I be a family of subsets of X, and
let P(A) be the power set of A. Then
S T
1) X \
Ai =
(X \ Ai ),
i∈I
i∈I
T S
2) X \
Ai =
(X \ Ai ),
i∈I
i∈I
S S
3) A ∩
Ai =
(A ∩ Ai ),
i∈I
i∈I
T T
4) A ∩
Ai =
(A ∩ Ai ),
i∈I
i∈I
T
T 5)
P(Ai ) = P
Ai ,
i∈I
i∈I
S
S 6)
P(Ai ) ⊆ P
Ai .
i∈I
i∈I
The first two identities are called the de Morgan laws.
Proof. See Exercise 5.
1.2
Functions
Definition 1.6. A function (or mapping) consists of three objects: two nonempty sets X and Y and a rule f which assigns to each element x ∈ X a single
element y ∈ Y . We write y = f (x) and say that y is the image of x under f or
the value of f at x. We can also say that f is a function from X to Y , and denote
this by f : X → Y .
The set X is called the domain of the function f and the set Y the range of f .
Definition 1.7. Let A ⊆ X and B ⊆ Y . Then the image of A under f is the set
f (A) ⊆ Y defined by
f (A) = {f (x) ∈ Y : x ∈ A}.
The inverse image of B under f is the subset f −1 (B) ⊆ X with
f −1 (B) = {x ∈ X : f (x) ∈ B}.
Theorem 1.2. Let f : X → Y be a function. Let {Ai }i∈I and {Bj }j∈J be families
of subsets of X and Y respectively. Then
S S
1) f
Ai =
f (Ai ),
i∈I
i∈I
1.2 Functions
2) f
T
i∈I
3) f −1
T
Ai ⊆
f (Ai ),
i∈I
S
j∈J
4) f
−1
4
T
S −1
Bj =
f (Bj ),
j∈J
Bj =
j∈J
−1
T
f −1 (Bj ),
j∈J
f (A) for any A ⊆ X,
6) f f −1 (B) ⊆ B for any B ⊆ Y .
5) A ⊆ f
Proof. See Exercise 6.
Remark. The equality in 2) does not hold in general. For example, let f : R →
R be the constant function f (x) = 1. Let A1 = [0, 1] and A2 = [2, 3]. Then
∅ = f (∅) = f (A1 ∩ A2 ) $ f (A1 ) ∩ f (A2 ) = {1} ∩ {1} = {1}.
Definition 1.8. Let X and Y be non-empty sets, and let f : X → Y be a
function. We say that f is surjective , a surjection or onto when f (X) = Y. We
say that f is injective , an injection or one-to-one when f (x1 ) = f (x2 ) implies
x1 = x2 . When f is both injective and surjective, we say that it is bijective or a
bijection .
Definition 1.9. A set X is said to be countable if it is either finite (in other
words has a finite number of members) or there is a bijection between X and
N. If X is not countable, it is said to be uncountable .
Example 1.2.
1) The set Z of integers is countable. A bijection between Z and N can be
given by
(
2n
if n ≥ 0,
f : Z → N, n 7→
1 − 2n if n < 0.
2) The set Q of rationals is countable. Consider any rational pq . Both p and q
have a unique prime number decomposition, i.e., can be written uniquely
as
mj
p = pn1 1 pn2 2 · · · pni i , q = q1m1 q2m2 · · · qj
for some prime numbers p1 , . . . , pi and q1 , . . . , qj . Now a bijection between
Q and Z is established by letting
g : Q → Z,
p
2m −1
1 2n2
· · · pi2ni q12m1 −1 q22m2 −1 · · · qj j .
p
7→ p2n
q
| 1 2 {z
}|
{z
}
even powers
odd powers
The composition g ◦ f is a bijection between Q and N.
3) The set R of reals is uncountable. We will use a version of Cantor’s diagonal argument to show this. Consider the set B of all real numbers
between 0 and 1 with a decimal expansion consisting only of the digits 0
1.2 Functions
5
and 1. Let h : N → B be an injective mapping such that
1 7→ 0.d11 d12 d13 . . .
2 7→ 0.d21 d22 d23 . . .
3 7→ 0.d31 d32 d33 . . .
..
.
k 7→ 0.dk1 dk2 dk3 . . .
..
.
where dij ∈ {0, 1}. Now define an element x in B by letting
x = 0.c1 c2 c3 . . .
where cn = 0 if dnn = 1 and cn = 1 if dnn = 0, for all n ∈ N. Consequently,
there is no natural number that maps to x, and thus h cannot be surjective. This shows that there exists no bijection between B and N, and so
B is uncountable. This clearly implies that the whole of R is uncountable
as well.
4) The Cantor set. Let C be a set obtained from [0, 1] by removing first the
open middle third ] 13 , 32 [, so we are left with C1 = [0, 31 ] ∪ [ 32 , 1]. Next, remove the open middle thirds ] 19 , 92 [ and ] 97 , 98 [ from [0, 31 ] and [ 23 , 1], respectively. We are left with C2 = [0, 19 ]∪[ 29 , 31 ]∪[ 23 , 97 ]∪[ 98 , 1]. With this process,
∞
we obtain a sequence of closed sets Cn . The Cantor set is C = ∩ Cn . The
n=1
Cantor set is uncountable (see Exercise 30).
Chapter 2
Metric Spaces
The concept of a metric space was first introduced by Maurice Fréchet in his
doctoral dissertation, published in 1906. The fundamental idea of Fréchet’s
work was to generalize the distance properties of the real line, and subsequently extend the concept of convergence to a more abstract setting. The most
important result in this chapter, and arguably one of the key theorems in analysis on metric spaces in general, is Baire’s theorem. It was given by René Baire
in the beginning of 20th century, and has since helped produce significant results such as the Principle of Uniform Boundedness and Open Mapping theorem.
2.1
Distance
Definition 2.1. Let X be a non-empty set. A metric (or distance ) on X is a
function d : X × X → R satisfying, for x, y, z ∈ X, the following conditions.
M1.
M2.
M3.
M4.
d(x, y) ≥ 0.
d(x, y) = 0 if and only if x = y.
d(x, y) = d(y, x).
d(x, y) ≤ d(x, z) + d(z, y).
(symmetry)
(triangular, triangle or △- inequality)
The image d(x, y) is called the distance from x to y, and the pair (X, d) is called
a metric space.
Remark. Often several different metrics can be defined on a given non-empty
set.
Definition 2.2. Let (X, d) be a metric space. Let A and B be subsets of X and
let x ∈ X. Then
d(x, A) = inf{d(x, y) : y ∈ A},
d(A, B) = inf{d(x, y) : x ∈ A and y ∈ B},
d(A) = sup{d(x, y) : x, y ∈ A}.
(distance from point to set)
(distance between two sets)
(diameter of A)
Definition 2.3. Let (X, d) be a metric space, let x ∈ X and r > 0. The set
B(x, r) = {y ∈ X : d(x, y) < r}
is the open ball of centre x and radius r.
2.1 Distance
7
Example 2.1.
1) The real line R with d(x, y) = |x − y| is a metric space:
M1. |x − y| ≥ 0 for all x, y ∈ R.
M2. |x − y| = 0 ⇔ x − y = 0 ⇔ x = y for all x, y ∈ R.
M3. |x − y| = | − (x − y)| = |y − x| for all x, y ∈ R.
M4. |x − y| = |x − z + z − y| ≤ |x − z| + |z − y| for all x, y, z ∈ R.
Now the open ball of centre x and radius r is
B(x, r) = {y ∈ R : d(x, y) < r}
= {y ∈ R : |x − y| < r}
= {y ∈ R : −r < x − y < r}
= {y ∈ R : x − r < y < x + r} = ]x − r, x + r[.
2) More generally, consider the space Rn and let x = (x1 , x2 , . . . , xn ) and
y = (y1 , y2 , . . . , yn ). Define d1 , d2 and d∞ as follows:
d1 (x, y) =
d2 (x, y) =
n
X
|xk − yk |,
k=1
n
X
(xk − yk )2
k=1
12
,
d∞ (x, y) = max{|xk − yk | : k = 1, 2, 3, . . . , n}.
The functions d1 , d2 and d∞ are all metrics on Rn (see Exercise 16).
3) Let X 6= ∅ and define d : X × X → R by
(
0 x = y,
d(x, y) =
1 x 6= y.
Then (X, d) is a metric space, and d is called the discrete metric. The
open balls are given by
B(x, r) = {y ∈ X : d(x, y) < r ≤ 1} = {x}
for r ≤ 1, and
B(x, r) = {y ∈ X : d(x, y) < r} = X
for r > 1.
Definition 2.4. Let (X, d) be a metric space and Y be a non-empty subset of X.
Regard d as a function on Y × Y . Then (Y, d) is also a metric space. It is called
a subspace of (X, d).
Example 2.2. The spaces ([0, 1], | · |), (Q, | · |), (N, | · |) are subspaces of (R, | · |).
The unit circle (T, | · |) and (R, | · |) are subspaces of (C, | · |).
Remark. From now on, if the metric in question is either evident from the
context of the statement or is arbitrary, we may refer to a metric space (X, d)
briefly as metric space X.
2.1 Distance
8
Definition 2.5. Let (X, d) be a metric space. A subset A of X is said to be open,
if for every x ∈ A there exists r > 0 such that B(x, r) ⊆ A. A subset of X is said
to be closed if its complement is open.
Theorem 2.1. Let (X, d) be a metric space. Then each open ball in (X, d) is an
open set.
Proof. Let B(x0 , r) be an open ball in (X, d). Now let x ∈ B(x0 , r) and fix ρ =
r − d(x0 , x). Since x ∈ B(x0 , r), we have d(x0 , x) < r and thus ρ > 0. To show
that B(x, ρ) ⊆ B(x0 , r), take y ∈ B(x, ρ). This means precisely that d(x, y) <
ρ = r − d(x0 , x), and by applying the triangle inequality, we obtain
d(x0 , y) ≤ d(x0 , x) + d(x, y)
< d(x0 , x) + r − d(x0 , x)
= r.
This shows that for all x ∈ B(x0 , r), there exists ρ > 0 such that B(x, ρ) ⊆
B(x0 , r); since x0 ∈ X and r > 0 were arbitrary, we have our desired result.
Theorem 2.2. Let (X, d) be a metric space. Then the following statements are
true.
1) The empty set ∅ and the whole space X are open.
2) Any union of open sets is open.
3) Any finite intersection of open sets is open.
Proof.
1) Clear.
2) Let {Ai }i∈I be an arbitrary family of open sets in X. Let x ∈ ∪ Ai . Then
i∈I
x ∈ Ai0 for some i0 ∈ I. Since Ai0 is open, there exists r > 0 such that
B(x, r) ⊆ Ai0 . Thus B(x, r) ⊆ ∪ Ai , and so ∪ Ai is open.
i∈I
i∈I
3) Let {A1 , A2 , . . . , An } be a finite family of open sets in X and let
n
x ∈ ∩ Ai . Then x ∈ Ai for every i = 1, 2, . . . , n. Since each Ai is open,
i=1
there exists ri > 0 for each i = 1, 2, . . . , n such that B(x, ri ) ⊆ Ai . Let r =
min{r1 , r2 , . . . , rn }. Then B(x, r) ⊆ B(x, ri ) ⊆ Ai for each i = 1, 2, . . . , n.
n
n
i=1
i=1
Accordingly, B(x, r) ⊆ ∩ Ai , and so ∩ Ai is open.
Theorem 2.3. Let (X, d) be a metric space. Then
1) The empty set ∅ and the whole space X are closed.
2) Any intersection of closed sets is closed.
3) Any finite union of closed sets is closed.
Proof. See Exercise 19.
Remark.
1) Infinite intersection of open sets is not, in general, open. Let (R, | · |) be
our metric space, and consider the family {]− n1 , n1 [}∞
n=1 of open sets. Then
∞
∩ ] − n1 , n1 [ = {0}. The singleton {0} is not open, since it cannot contain
n=1
an open interval.
2.2 Interior and closure
9
2) An infinite union of closed sets is not necessarily closed. Take for example ∪ {a} =]0, 1[, which is not closed although each singleton {a} is
a∈]0,1[
closed.
Definition 2.6. Let (X, d) be a metric space. Let x ∈ X and r ≥ 0. Then
B ′ (x, r) = {y ∈ X : d(x, y) ≤ r}
is called the closed ball of centre x and radius r. Note that when r = 0,
B ′ (x, r) = {x}.
Theorem 2.4. Let (X, d) be a metric space. Then each closed ball in (X, d) is a
closed set.
Proof. Let B ′ (x, r) be any closed ball in (X, d). For B ′ (x, r) to be closed we need
to show that X \ B ′ (x, y) is open. Let y ∈ X \ B ′ (x, y), i.e., d(x, y) > r and so
d(x, y) − r > 0. Fix ρ = d(x, y) − r. We check that B(y, ρ) ⊆ X \ B ′ (x, r). Let
z ∈ B ′ (y, ρ). Then d(y, z) < ρ, and so by triangle inequality
d(x, z) ≥d(x, y) − d(y, z)
>d(x, y) + r − d(x, y)
=r,
showing that z ∈
/ B ′ (x, r). In other words, z ∈ X \ B ′ (x, r) and thus X \ B ′ (x, r)
is open, so our result follows.
2.2
Interior and closure
Definition 2.7. Let X be a metric space. Let A be a subset of X. A point x ∈ A
is an interior point of A if it is the centre of some open ball contained in A. In
other words, if there exists r > 0 such that B(x, r) ⊆ A.
◦
We call the set of all such points the interior of A and denote it by Int(A) or A.
Theorem 2.5. Let X be a metric space and A a subset of X. Then
a) Int A is open.
b) A is open if and only if A = Int A.
Proof. See Exercise 25.
Definition 2.8. Let X be a metric space and let A be a subset of X. A point
x ∈ X is a limit point or accumulation point of A if each open ball with centre
x contains at least one point of A different from x. In other words, for all r > 0,
B(x, r) ∩ (A \ {x}) 6= ∅.
Definition 2.9. The closure of A, denoted by A or Cl(A) is the union of A and
the set of all its limit points, i.e.,
A = A ∪ {limit points of A}.
2.3 Sequences
10
Definition 2.10. Let X be a metric space and A ⊆ X. When A = X, we say
that A is dense in X.
If there exists some A countable and dense in X, we say that X is a separable.
Finally, if Int(A) = ∅, we say that A is nowhere dense in X.
Example 2.3.
1) Let N ⊆ (R, | · |). Then N = N. However, Int(N) = ∅, so N is nowhere dense
in R.
2) Int([0, 1]) = ]0, 1[.
3) The space ]0, 1] has 0, which is not in the set, as a limit point. All other
points of ]0, 1] are also limit points, and so ]0, 1] = [0, 1].
4) The subset {1, 12 , 31 , . . . } of (R, | · |) has 0 as the only limit point. Now
{1, 12 , 31 , . . . } = {0, 1, 12 , 31 , . . . } and Int({1, 12 , 31 , . . . }) = ∅, so this set is
nowhere dense in R.
5) Q = R, so Q is dense in R. The set Q is also countable and thus R is
separable.
Theorem 2.6. Let X be a metric space and A a subset of X. Then A is closed if
and only if A = A. In other words, A contains all its limit points.
Proof. Suppose first that A is closed. Clearly A ⊆ A, so we only need to show
that A ⊆ A. Let x ∈ A. Then B(x, r) ∩ A 6= ∅ for all r > 0, and so B(x, r) * X \ A
for all r > 0. Since X \ A is open, the point x cannot be in X \ A, and thus x ∈ A.
Therefore A ⊆ A, as required.
Conversely, suppose that A contains all of its limit points, and proceed to show
that X \A is open. If X \A = ∅, the statement follows immediately. Assume that
X \ A 6= ∅, and let x ∈ X \ A. Then x is not a limit point of A, so B(x, r) ∩ A = ∅
for some r > 0. Therefore B(x, r) ⊆ X \ A for some r > 0, showing that X \ A is
open, as required.
Remark. In general B(x, r) 6= B ′ (x, r). For example, let d be the discrete metric
on some set X which has more than two elements. Then B(x, 1) 6= B ′ (x, 1).
2.3
Sequences
Definition 2.11. Let (X, d) be a metric space and let (xn )∞
n=1 be a sequence of
∞
points in X. We say that (xn )n=1 is convergent in X if there exists a point x in
X, such that for every r > 0, there exists n0 so that n ≥ n0 implies d(xn , x) < r.
Equivalently, (xn )∞
n=1 is convergent, if for all r > 0 there exists n0 such that
n ≥ n0 implies xn ∈ B(x, r).
If such a point x exists, we say that (xn )∞
n=1 converges to x, and denote this by
xn → x or lim xn = x. The point x is called the limit of (xn )∞
n=1 . Note that the
x→∞
limit of a sequence and a limit point of a set are different concepts.
Theorem 2.7. Let X be a metric space and let (xn )∞
n=1 be a sequence in X. Then
(xn )∞
has
exactly
one
limit.
n=1
2.3 Sequences
11
Proof. Suppose that xn → x and xn → y. Then for every r > 0, there exists
n0 ∈ N such that d(xn , x) < 2r whenever n ≥ n0 ; and there exists n1 ∈ N such
that d(xn , y) < 2r whenever n ≥ n1 . Let n2 = max{n0 , n1 }. Then
0 ≤ d(x, y) ≤d(x, xn ) + d(xn , y)
r
r
< + ,
2 2
whenever n ≥ n2 . Thus, for all r > 0, we have 0 ≤ d(x, y) < r. This implies that
d(x, y) = 0, and so x = y.
Theorem 2.8. Let (X, d) be a metric space, A be a nonempty subset of X and
x ∈ X. Then x ∈ A if and only if there is a sequence (xn )∞
n=1 in A such that
xn → x.
Proof. Let x ∈ A. If x ∈ A, then the constant sequence with xn = x for all
n = 1, 2, . . . converges to x. Suppose that x is a limit point of A. Then for each
n ∈ N, B(x, n1 ) ∩ A 6= ∅, so we may take xn ∈ B(x, n1 ) ∩ A. In this way, we
form a sequence (xn )∞
n=1 in A which converges to x. To see this, let r > 0.
Then there exists n0 ∈ N such that 0 < n1 ≤ r for every n ≥ n0 . Accordingly,
d(xn , x) < n1 ≤ r, whenever n ≥ n0 , as required.
Conversely, suppose that (xn )∞
n=1 is a sequence in A converging to some point
x in X. Then for every r > 0, there exists n0 ∈ N such that xn ∈ B(x, r) ∩ A
whenever n ≥ n0 . Thus for every r > 0, B(x, r) ∩ A 6= ∅, giving x ∈ A.
Definition 2.12. A sequence (xn )∞
n=1 is said to be a Cauchy sequence, if for
every r > 0, there exists n0 ∈ N such that n, m ≥ n0 implies d(xn , xm ) < r.
Theorem 2.9. Let X be a metric space and let (xn )∞
n=1 be a convergent sequence
in X. Then (xn )∞
is
a
Cauchy
sequence.
n=1
Proof. Let (xn )∞
n=1 be a sequence in X converging to some point x ∈ X. Then,
for every r > 0, there exists nr ∈ N such that d(x, xn ) < 2r , whenever n ≥ n0 .
Now by the triangle inequality
d(xn , xm ) ≤ d(xn , x) + d(xm , x) <
r
r
+ = r, whenever n, m ≥ n0 .
2 2
Thus (xn )∞
n=1 is a Cauchy sequence.
Example 2.4. In general, a Cauchy sequence need not converge. Let (]0, 1], |·|)
be our metric space and consider the sequence ( n1 )∞
n=1 . This is a Cauchy se1
1
1 1
1
quence because | n − m | ≤ max{ n , m } ≤ n0 < r for all n ≥ n0 , where n0 > 1r .
The sequence ( n1 )∞
n=1 , however, is not convergent in ]0, 1], since its limit 0 is not
in ]0, 1].
Definition 2.13. A metric space X is said to be complete if every Cauchy sequence in X is convergent in X.
Example 2.5.
1) The space (]0, 1], | · |) is not complete (see Example 2.4).
2) The space ([0, 1], | · |) is complete. This is the completion of the space
above.
3) The space (Q, | · |) is not complete.
2.3 Sequences
12
4) The space (R, | · |) is complete (see Exercise 32).
5) The space (Rn , d2 ) is complete (see Exercise 33).
Theorem 2.10. Let X be a complete metric space and let Y be a subspace of X.
Then Y is complete if and only if Y is closed.
Proof. Suppose that Y is complete, and proceed to show that Y = Y . Let y ∈ Y .
Then by Theorems 2.7 and 2.8, there is a sequence (yn )∞
n=1 in Y converging to
∞
some unique y in X. By Theorem 2.9, (yn )n=1 is a Cauchy sequence in Y , and
since Y is complete, (yn )∞
n=1 converges in Y . So y = lim yn ∈ Y . Hence Y = Y ,
n→∞
and so Y is closed.
Conversely, suppose that Y is closed and let (yn )∞
n=1 be a Cauchy sequence in
Y . Then (yn )∞
is
also
a
Cauchy
sequence
in
X,
and so (yn )∞
n=1
n=1 is convergent
to some point y in X since X is complete. But Y is closed, so by Theorem 2.8,
the limit y must be in Y . Thus Y is complete.
Baire’s theorem. Let X be a complete metric space and let {An : n ∈ N} be
a countable family of open subsets of X, each of which is dense in X (in other
∞
words An = X for all n ∈ N). Then ∩ An is also dense in X.
n=1
∞
Proof. We shall show that ∩ An = X. Let x1 be a fixed but arbitrary point in
n=1
∞
X, and let r1 > 0. First we show that B(x1 , r1 ) ∩ ( ∩ An ) 6= ∅.
n=1
Since A1 is dense in X, we see that B(x1 , r1 ) ∩ A1 6= ∅. Pick x2 in B(x1 , r1 ) ∩ A1 .
Since B(x1 , r1 ) ∩ A1 is open, there exists r2 > 0 such that
B ′ (x2 , r2 ) ⊆ B(x2 , 2r2 ) ⊆ B(x1 , r1 ) ∩ A1 .
Since A2 is also dense in X, then x2 ∈ A2 and so we may pick x3 from B(x2 , r2 )∩
A2 . Again, since B(x2 , r2 ) ∩ A2 is open, there exists r3 > 0 such that
B ′ (x3 , r3 ) ⊆ B(x3 , 2x3 ) ⊆ B(x2 , r2 ) ∩ A2 .
∞
Inductively, we form a sequence (xn )∞
n=1 in X and a sequence (rn )n=1 of positive
real numbers such that
B ′ (xn+1 , rn+1 ) ⊆ B(xn , rn ) ∩ An .
Note that rn < 2rn1 . It follows that if n ≥ m, then xn ∈ B(xm , rn ), and so
d(xn , xm ) < rm ≤ 2rm1 . Thus, for all r > 0, there exists m0 such that n, m ≥ m0
implies
1 1
r1
d(xn , xm ) < rm < m ≤ max{ , } < r.
2
n m
Hence (xn )∞
n=1 is a Cauchy sequence, and since X is complete, it converges to
some point x0 in X.
Now since xn ∈ B ′ (xm , rm ) for all n ≥ m and for all m ∈ N, we see that x0 ∈
B ′ (xm , rm ) for all m ∈ N. In particular, x0 ∈ B ′ (x2 , r2 ) ⊆ B(x1 , r1 ). We also have
x0 ∈ B ′ (xn+1 , rn+1 ) ⊆ B(xn , rn ) ∩ An for all n = 1, 2, . . .. Thus x0 ∈ An for all
∞
n ∈ N, and accordingly x0 ∈ B(x1 , r1 ) ∩ ( ∩ An ), as required.
n=1
2.4 Continuous functions
13
Corollary. Let X be a complete metric space and let {Xn : n ∈ N} be a countable
∞
family of subsets of X such that ∪ Xn = X. Then Int(Xn ) 6= ∅ for some n ∈ N.
n=1
In other words, X cannot be covered by a countable union of nowhere dense
subsets.
∞
Proof. Suppose that X = ∪ Xn and Int(Xn ) = ∅ for all n ∈ N. Then (by Exn=1
ercise 25), X \ Xn is dense in X for each n ∈ N. Since Xn is closed, X \ Xn is
open for each n ∈ N. Baire’s Theorem implies that ∩ X \ Xn = X. However,
n∈N
we have
∞
∅ = X \ X = X \ ∪ Xn
n=1
∞
∞
= ∩ (X \ Xn ) ⊇ ∩ (X \ Xn ),
n=1
n=1
implying that ∩ (X \ Xn ) = ∅. Clearly this is a contradiction, and so we must
n∈N
have Int(Xn ) 6= ∅ for some n ∈ N.
Example 2.6. Baire’s theorem can be used to show that there are plenty of
continuous functions that are nowhere differentiable. In fact, in a topological
sense there are more continuous real functions nowhere differentiable on some
interval, say [0, 1], than those differentiable at even a single point. This is established by showing that the set of continuous functions on [0, 1], that are differentiable at some point in [0, 1] is of first category in the space of continuous
real functions on [0, 1], i.e., it is a countable union of nowhere dense subsets.
Let b be an odd integer and 0 < a < 1 such that ab > 1 + 32 π. Then the function
defined on R by
∞
X
an cos(bn πx)
W∞ (x) =
n=0
is an example of a function that is continuous but nowhere differentiable. This
was the first concrete example of such a function, presented by Weierstrass in
1872. One of the simplest examples of this type of functions, found later in 1930
by van der Waerden, can be written as
V∞ (x) =
∞
X
{10n x}
,
10n
n=0
where {a} is the difference between a and the nearest integer.
2.4
Continuous functions
Definition 2.14. Let (X, d1 ) and (Y, d2 ) be metric spaces and let f : X → Y .
We say that f is continuous at some point x0 in X, when for every r > 0, there
exists δ > 0 such that d1 (x0 , x) < δ implies d2 (f (x0 ), f (x)) < r. We say that f is
continuous on X if it is continuous at every point x ∈ X.
Theorem 2.11. Let X, Y, f and x0 be as in Definition 2.14. Then the following
statements are equivalent.
2.4 Continuous functions
14
a) The function f is continuous at x0 .
b) For every r > 0, there exists δ > 0 such that f (B1 (x0 , δ)) ⊆ B2 (f (x0 ), r).
c) For every open set V in Y containing f (x0 ), there exists an open set U in
X containing x0 such that f (U) ⊆ V.
d) Whenever (xn )∞
n=1 is a sequence in X with xn → x0 , we have f (xn ) →
f (x0 ) in Y .
Proof.
a) ⇔ b) Let r > 0 and δ > 0 be such that if d1 (x0 , x) < δ then d2 (f (x0 ), f (x)) <
r. In other words, if x ∈ B1 (x0 , δ), then f (x) ∈ B2 (f (x0 ), r). This says
precisely that f (B1 (x0 , δ)) ⊆ B2 (f (x0 ), r). The converse is obvious.
b) ⇒ c) Let V be an open set in Y containing f (x0 ). Then there exists r > 0 such
that B2 (f (x0 ), r) ⊆ V . By b) there exists δ > 0 such that f (B1 (x0 , δ)) ⊆
B2 (f (x0 ), r). We simply take U = B1 (x0 , δ) to obtain f (U) ⊆ V.
c) ⇒ b) Let r > 0. Then B2 (f (x0 ), r) is an open set in Y containing f (x0 ). By
c), there exists an open set U in X containing x0 such that f (U) ⊆
B2 (f (x0 ), r). Since U is open, there exists δ > 0 such that B1 (x0 , δ) ⊆ U.
Thus,
f (B1 (x0 , δ)) ⊆ f (U) ⊆ B2 (f (x0 ), r).
a) ⇒ d) Let (xn )∞
n=1 be a convergent sequence in X with lim xn = x0 , and let
r > 0. By a), there exists δ > 0 such that
d1 (x0 , x) < δ implies d2 (f (x0 ), f (x)) < r.
Let n0 ∈ N be such that d1 (xn , xn ) < δ for all n ≥ n0 , thus implying that
d2 (f (x0 ), f (xn )) < r whenever n ≥ n0 . Hence, f (xn ) → f (x) in Y .
d) ⇒ a) We argue by contradiction, and suppose that f is not continuous at x0 .
So there exists r > 0, such that for every δ > 0, there exists x ∈ X with
d1 (x0 , x) < δ but d2 (f (x0 ), f (x)) ≥ r. Accordingly, for each n ∈ N, we may
pick xn ∈ X such that d1 (x0 , xn ) < n1 but d2 (f (x0 ), f (xn )) ≥ r. In this
way, we construct a sequence (xn )∞
n=1 in X for which xn → x0 in X, but
f (xn ) 9 f (x0 ) in Y .
Corollary. Let X and Y be metric spaces. A function f : X → Y is continuous
on X if and only if f (xn ) → f (x) in Y whenever xn → x in X.
Proof. See Exercise 36.
Chapter 3
Topological spaces
In this chapter we will abandon the concept of distance, and move on to even
more abstract ways of characterizing openness, continuity and other such properties. This approach, although in a way a joint result of the work of dozens of
the most brilliant mathematical minds of late 19th and early 20th century, is
often said to have its origins in the book Basics of Set Theory by Felix Hausdorff, published in 1914.
3.1
Topology
Definition 3.1. A topological space is pair the (X, T ), where X is a non-empty
set and T is a family of subsets of X satisfying the following conditions.
T1. The empty set ∅ and the whole space X belong to T .
T2. If {Ai }i∈I is a family of sets in T , then ∪ Ai ∈ T .
i∈I
n
T3. If {A1 , A2 , . . . , An } is a finite family of sets in T , then ∩ Ai ∈ T .
i=1
The family T is called a topology on the set X and the elements of T are called
open sets of X.
Remark. As before with metric spaces, where there is little or no possibility of
misinterpretation, we may denote a topological space (X, T ) briefly by topological space X.
Definition 3.2. Let X be a topological space and let A be a subset of X. We
say that A is closed when X \ A is open.
Definition 3.3. If x ∈ X and U ∈ T are such that x ∈ U, we say that U is a
neighbourhood of x.
Remark. A given set A in a topological space X is open if and only if for every
x ∈ A there is a neighbourhood Ux of x such that Ux ⊆ A.
Proof. Suppose that A is open and let x ∈ A. We simply take Ux = A. Conversely, suppose that for every x ∈ A there is a neighbourhood Ux with x ∈
Ux ⊆ A. Then A = ∪ Ux , and A is open by T2.
x∈A
Definition 3.4. A subfamily B ⊆ T is called a base for X, if every A in T is a
union of sets from B.
3.1 Topology
16
Remark. A family B is a base for X if and only if for all x ∈ X and for every
neighbourhood Ux of x, there exists B ∈ B such that x ∈ B ⊆ Ux .
Proof. Let B be a base for X, let x ∈ X and Ux be a neighbourhood of x. Then
since Ux is open, there is a family {Bi : i ∈ I} ⊆ B with Ux = ∪ Bi . So, there
i∈I
exists i ∈ I with x ∈ Bi ⊆ Ux . Conversely, let A ∈ T be arbitrary. Then for every
x ∈ A, there exists some Bx ∈ B with x ∈ Bx ⊆ A. Thus A = ∪ Bx , and so B is
x∈A
a base.
Definition 3.5. A base at a point x ∈ X is a family B(x) of neighbourhoods of
x, such that for every U containing x, there exists B ∈ B(x) with x ∈ B ⊆ U.
Definition 3.6. A family C is a subbase for X if the family of all finite intersections C1 ∩ C2 ∩ · · · ∩ Cn , where each Ci is in C, is a base for X.
Example 3.1.
1) A metric space is a topological space where the topology is defined by
T = {all open sets of (X, d) in the sense of Definition 2.5}.
A base for T can be given by
B ={B(x, r) : x ∈ X, r ∈ R+ },
B ′ ={B(x, r) : x ∈ X, r ∈ Q+ } or
1
B ′′ ={B(x, ) : x ∈ X, n ∈ N}.
n
2) Let X 6= ∅. Then T = P(X) is a topology on X. This is called the discrete
topology, and it is induced by the discrete metric (see Exercise 41).
3) The trivial (or indiscrete ) topology on a set X is T = {∅, X}.
4) Are all topological spaces metric spaces? That is, does there exist a distance d such that T = Td . The following counterexample shows that this
is not true.
Let X be the Sierpinski space. In other words, let X = {a, b} and let
T = {∅, X, {a}}. Then (X, T ) is a topological space. This topology does
not come from any metric.
Proof. Sierpinski topology is clearly a topology. For the second claim suppose that there is d such that Td = T . Then there exists r > 0 such that
B(a, r) ⊆ {a}, since {a} ∈ Td and a ∈ {a}. This means that d(a, x) < r,
implying that x ∈ {a}, i.e., x = a. Hence d(a, b) ≥ r, and B(b, r) = {b} is
an open set in Td = T , which is a contradiction.
If there exists such a distance d on X that T = Td , we say that the
topological space X is metrizable .
5) Let X = R2 . Then {L ⊆ R2 : L is a line} is not a topology (and cannot be
a base for any topology on X) because L1 ∩ L2 = {a}. The set {L ⊆ R2 :
L is a line} is, however, a subbase for the discrete topology.
Theorem 3.1. Let X be a topological space. Then the following statements are
true.
T1’. The empty set ∅ and the whole space X are closed.
3.1 Topology
17
T2’. Any intersection of closed sets is closed.
T3’. Any finite union of closed sets is closed.
Proof. See Exercise 53.
Definition 3.7. Let X be a topological space and let A ⊆ X. Then the closure
of A is the set
A = Cl(A) =
\
{F ⊆ X : F is closed and A ⊆ F }.
Remark. Note that A is closed by Theorem 3.1, and that A is the smallest closed
set containing A.
Definition 3.8. As with metric spaces, we say that a point x in X is a limit
point or accumulation point of a set A ⊆ X, if each neighbourhood of x contains
a point of A \ {x}.
Remark. Let L(A) be the set of all limit points of A. Then
A = A ∪ L(A) = {x ∈ X : for every neighbourhood U of x, U ∩ A 6= ∅}.
Proof. Suppose that x ∈
/ A ∪ L(A). Then there exists a neighbourhood U of x
such that U ∩ A = ∅. This means that A ⊆ X \ U, where X \ U is closed since
U is open. Therefore x ∈
/ A. This shows that A ⊆ A ∪ L(A). Conversely, suppose
that x ∈
/ A. Then x ∈ X \ A. Since A is closed, X \ A is open, and so X \ A is a
/ A ∪ L(A). This shows
neighbourhood of x. Now (X \ A) ∩ A = ∅, and thus x ∈
that A ∪ L(A) ⊆ A.
Theorem 3.2. Let X be a topological space and let A and B be subsets of X.
Then
1)
2)
3)
4)
5)
A ⊆ A.
A ⊆ B ⇒ A ⊆ B.
A = A.
A ∪ B = A ∪ B.
A is closed if and only if A = A.
Proof.
1) Follows immediately from the definition of closure.
2) Let A ⊆ B. Statement 1) says that A ⊆ B ⊆ B. It follows that A ⊆ B,
since A is the smallest closed set containing A.
S
3) From 1) A ⊆ A. A = {F ⊆ X : F closed and A ⊆ F } ⊆ A, since A is
closed and is clearly one of the sets F . Thus A = A.
4) Since A ⊆ A ∪ B, it follows from b) that A ⊆ A ∪ B. Similarly B ⊆ A ∪ B
so A ∪ B ⊆ A ∪ B. Now A ∪ B is closed and A ∪ B ⊆ A ∪ B, so we must
have A ∪ B ⊆ A ∪ B.
5) Let A be closed. Then A = {F ⊆ X : F closed and A ⊆ F } ⊆ A, and so by
1) A = A. Conversely, A = A makes A closed by definition.
3.1 Topology
18
Definition 3.9. Let X be a topological space and A ⊆ X. The interior of A is
the set
[
◦
Int(A) = A = {O ⊆ X : O is open and O ⊆ A}.
◦
Note that A is open and it is the largest open subset of A.
Remark. As with metric spaces, we may also say that
◦
A = {x ∈ A : there exists a neighbourhood U of x such that x ∈ U ⊆ A}.
Proof. Let X be a topological space and A ⊆ X. Then denote
{x ∈ A : there exists a neighbourhood U of x such that x ∈ U ⊆ A} = I(A).
◦
Let x ∈ A. It follows that there exists a set O open in X such that O ⊆ A and
x ∈ O. This says precisely that O is a neighbourhood of x and x ∈ O ⊆ A, so
x ∈ I(A). Conversely, let x ∈ I(A). Now there exists a neighbourhood U of x
◦
with x ∈ U ⊆ A, and so x ∈ A.
Theorem 3.3. Let X be a topological space and A ⊆ X. Then
X \ A = Int(X \ A).
Proof. Let x ∈ X \ A, i.e., x ∈
/ A. Then there exists a neighbourhood U of x such
that U ∩ A = ∅. Equivalently, U ⊆ X \ A, and so x ∈ Int(X \ A).
Theorem 3.4. Let X be a topological space and let A, B ∈ P(X). Then
1)
2)
3)
4)
Int(A) ⊆ A.
A is open if and only if A = Int(A).
Int(Int(A)) = Int(A).
Int(A ∩ B) = Int(A) ∩ Int(B).
Proof.
1. Evident from the definition.
2. A is open ⇒ A ⊆ Int(A) ⇒ A = Int(A) (by 1)). The converse is clear.
3. Int(A) is open, so by 2), Int(Int(A)) = Int(A).
4.
)
A ∩ B ⊆ A ⇒ Int(A ∩ B) ⊆ Int(A)
⇒ Int(A ∩ B) ⊆ Int(A) ∩ Int(B).
A ∩ B ⊆ B ⇒ Int(A ∩ B) ⊆ Int(B)
Now Int(A) ∩ Int(B) is open and contained in A ∩ B. Thus we must have
Int(A) ∩ Int(B) ⊆ Int(A ∩ B), since Int(A ∩ B) is the largest open subset
of A ∩ B.
Remark. Try to do the proof using Theorems 3.2 and 3.3 (see Exercise 59).
Definition 3.10. Let X be a topological space and A ⊆ X. The frontier or the
boundary of A is the set
δ(A) = Fr(A) = A ∩ (X \ A).
3.1 Topology
19
Note that Fr(A) is closed and that
Fr(X \ A) = X \ A ∩ X \ (X \ A) = X \ A ∩ A = Fr(A).
Example 3.2.
a) Fr(]a, b[) = [a, b] ∩ (] − ∞, a[ ∪ ]b, ∞[) = {a, b}.
b) Fr(X) = X ∩ X \ X = X ∩ ∅ = ∅.
c) Fr(Q) = Q ∩ R \ Q = R ∩ R = R.
d) The frontier of R in R is the set FrR (R) = R ∩ R \ R = ∅.
e) The frontier of R in R2 is the set FrR2 (R) = R ∩ R2 \ R = R ∩ R2 = R.
As items d) and e) above show, the frontier of a set depends on the space where
the set is considered.
Theorem 3.5. Let X be a topological space and let A and B be subsets of X.
Then
1) A = A ∪ Fr(A) = Int(A) ∪ Fr(A).
2) Int(A) = A \ Fr(A).
3) Int(A) ∪ Fr(A) ∪ Int(X \ A) = X.
4) Fr(A ∪ B) ⊆ Fr(A) ∪ Fr(B).
5) Fr(A ∩ B) ⊆ Fr(A) ∪ Fr(B).
6) Fr(A) = A \ A when A is open.
7) Fr(A) = ∅ if and only if A is closed and open (clopen).
Proof.
1) First note that
A ∪ Fr(A) = A ∪ (A ∩ (X \ A) = (A ∪ A) ∩ (A ∪ X \ A) = A ∩ X = A.
Then for the other equality consider
Int(A) ∪ Fr(A) = Int(A) ∪ (A ∩ X \ A)
\
= (Int(A) ∪ A) (Int(A) ∪ X \ A)
= A ∩ [Int(A) ∪ (X \ Int(A))]
= A ∩ X = A.
2) Now A \ Fr(A) = A \ (A ∩ (X \ A)). By de Morgan’s laws and Theorem
3.3, A \ (A ∩ (X \ A)) = (A \ A) ∪ A \ (X \ A) = ∅ ∪ (A ∩ X \ X \ A) =
A ∩ Int(A) = Int(A).
3) Using 1) and Theorem 3.3, we obtain
Int(A) ∪ Fr(A) ∪ Int(X \ A) = A ∪ Int(X \ A) = A ∪ (X \ A) = X.
4) With the help of Theorem 3.2 and by meticulous use of Definition 3.10,
3.2 Continuous functions
20
we can write
Fr(A ∪ B) = A ∪ B ∩ X \ A ∪ B
= (A ∪ B) ∩ (X \ A) ∩ (X \ B)
⊆ (A ∪ B) ∩ (X \ A ∩ X \ B)
= A ∩ (X \ A ∩ X \ B) ∪ B ∩ (X \ A ∩ X \ B)
⊆ (A ∩ X \ A) ∪ (B ∩ X \ B)
= Fr(A) ∪ Fr(B).
5) Again using the properties of closure, we have
Fr(A ∩ B) = (A ∩ B) ∩ (X \ A ∩ B)
⊆ (A ∩ B) ∩ (X \ A ∪ X \ B)
= (A ∩ B ∩ X \ A) ∪ (A ∩ B ∩ X \ B)
⊆ (A ∩ X \ A) ∪ (B ∩ X \ B)
= Fr(A) ∪ Fr(B).
6) Let A be open, implying that X\A is closed. In other words X \ A = X\A,
giving Fr(A) = A ∩ (X \ A) = A ∩ (X \ A) = A \ A.
7) Let A be open and closed. Then Fr(A) = A ∩ X \ A = A ∩ (X \ Int(A)) =
A ∩ (X \ A) = ∅. Conversely, let Fr(A) = A ∩ X \ A = ∅. Then by Theorem
3.3, A ⊆ A ⊆ X \ (X \ A) = Int(A) ⊆ A ⊆ A, giving A = Int(A) = A, and
so A is both open and closed.
3.2
Continuous functions
Definition 3.11. Let X and Y be topological spaces and f : X → Y be a
function (mapping). We say that f is continuous at a point x ∈ X if for every
neighbourhood V of f (x) in Y , there exists a neighbourhood U of x in X such
that f (U) ⊆ V. We say that f is continuous on X when f is continuous at each
point in X.
Theorem 3.6. Let X and Y be topological spaces and f be a function from X
to Y . Then the following statements are equivalent.
a)
b)
c)
d)
The function f is continuous on X.
For each open set V in Y , f −1 (V) is open in X.
For each closed set F in Y , f −1 (F ) is closed in X.
f (A) ⊆ f (A), for every A ⊆ X, i.e., f (ClX (A)) ⊆ ClY (f (A)).
Proof.
a) ⇒ b) Let V be on open subset of Y and let x ∈ f −1 (V). Then f (x) ∈ V , and
so there exists a neighbourhood U of x such that f (U) ⊆ V, i.e., x ∈ U ⊆
f −1 (V). Thus f −1 (V) is open.
3.3 Subspaces
21
b) ⇒ c) Let F be closed in Y , implying that Y \ F is open in Y . By b), f −1 (Y \ F )
is open in X, giving that X \ f −1 (F ) is open in X. Hence f −1 (F ) is closed
in X.
c) ⇒ d) Let A ⊆ X. Then A ⊆ f −1 (f (A)) ⊆ f −1 (f (A)). Now by taking closure, we
obtain
c)
A ⊆ f −1 (f (A)) = f −1 (f (A)),
showing that f (A) ⊆ f (A).
d) ⇒ a) Let x ∈ X. We want to show that if d) holds, then f is continuous at x.
Let V be an open set in Y with f (x) ∈ V, and let U = Int(f −1 (V)). Then
U is open in X and f (U) = f (Int(f −1 (V))) ⊆ f (f −1 (V)) = V. So f (U) ⊆ V.
We only need to check that x ∈ U. Suppose that x ∈
/ U. Then
x ∈ X \ U = X \ Int(f −1 (V)) = X \ f −1 (V).
Hence f (x) ∈ f (X \ f −1 (V)), which by d) is contained in f (X \ f −1 (V)) =
f (X) \ f (f −1 (V)), meaning that f (x) ∈ Y \ V = Y \ V. This shows that
f (x) ∈
/ V, which is a contradiction. Thus x ∈ U.
Definition 3.12. Let X and Y be topological spaces. We say that f : X → Y is
a homeomorphism when
i) f is bijective,
ii) f is continuous and
iii) f −1 : Y → X is continuous.
If there exists such a homeomorphism, we say that X and Y are homeomorphic.
Definition 3.13. Let X and Y be topological spaces. We say that f : X → Y is
an embedding of X into Y , when
i) f is injective,
ii) f is continuous and
iii) f −1 : Y → X is continuous.
When such an embedding exists, we say that X is embedded into Y . We will
denote this by X ֒→ Y .
Example 3.3.
1) Let a < b be in R. Then ]a, b[ and ]0, 1[ are homeomorphic. The homeomorphism f : ]a, b[ → ]0, 1[ is given by f (x) = x−a
b−a .
2) The sets ]1, +∞[ and ]0, 1[ are homeomorphic, the homeomorphism being
f (x) = x1 .
3) The real line R and ] − π2 , π2 [ are homeomorphic, where the homeomorphism f : R → ] − π2 , π2 [ is given by f (x) = arctan(x).
3.3
Subspaces
Definition 3.14. Let (X, T ) be a topological space and let Y ⊆ X. Then the
family
T ′ = {A ∩ Y : A ∈ T }
3.3 Subspaces
22
is a topology on Y (see Exercise 58). It is called the relative or the induced
topology on Y , and (Y, T ′ ) a subspace of (X, T ).
Example 3.4.
1) The space R with its usual topology is a subspace of the Euclidean space
R2 .
2) Usually Q is regarded as a subspace of R.
3) A subspace Z of a subspace Y of X is a subspace of X (see Exercise 71).
Theorem 3.7. Let (X, T ) be a topological space, (Y, T ′ ) be a subspace of X and
A ⊆ Y . Then
a) The set A is closed in Y if and only if A = F ∩ Y , where F is closed in X.
b) ClY (A) = ClX (A) ∩ Y .
c) If B is a base for (X, T ) then B ′ = {B ∩ Y : B ∈ B} is a base for (Y, T ′ ).
Proof.
a) Let A be closed in Y , and so Y \ A is open in Y . Then Y \ A = U ∩ Y for
some open subset U of X. It follows that
A =Y \ (Y \ A) = Y \ (U ∩ Y ) = (Y \ U) ∪ (Y \ Y )
=Y \ U = (X \ U) ∩ Y.
So A = (X \ U) ∩ Y , where X \ U is closed in X.
Conversely, suppose that A = F ∩ Y with F closed in X. Then
Y \ A =Y \ (F ∩ Y ) = (Y \ F ) ∪ (Y \ Y )
=Y \ F = (X \ F ) ∩ Y,
where X \ F is open in X, and so Y \ A is open in Y . Thus A is closed in
Y.
b) From the definition of closure, we obtain
\
ClY (A) = {K ⊆ Y : K closed in Y and A ⊆ K}
\
a)
= {F ∩ Y : F closed in X and A ⊆ F }
\
=
{F : F closed in X and A ⊆ F } ∩ Y
= ClX (A) ∩ Y.
c) We need to show that for any open set U ′ in Y and for every x ∈ U ′ , there
exists B ′ ∈ B ′ such that x ∈ B ′ ⊆ U ′ . By a), U ′ = U ∩ Y for some open
set U in X. Since B is a base for X, we pick B ∈ B with x ∈ B ⊆ U. So
x ∈ B ∩ Y ⊆ U ∩ Y = U ′ , as required.
Remark. There is no mention about interior or frontier in Theorem 3.7. The
reason is as follows:
Let X = R2 , Y = R and A = R. Then IntY (A) = IntR (R) = R, whereas IntX (A) =
IntR2 (R) = ∅. So IntY (A) 6= IntX (A) ∩ Y .
3.3 Subspaces
23
Similarly for frontiers, we have FrY (A) = FrR (R) = ClR (R) ∩ ClR (R \ R) = ∅,
whereas
FrX (A) = FrR2 (R) = ClR2 (R) ∩ ClR2 (R2 \ R) = R ∩ R = R,
so FrY (A) 6= FrX (A) ∩ Y .
However, we still have
FrY (A) ⊆ FrX (A) ∩ Y,
IntY (A) ⊇ IntX (A) ∩ Y
(see Exercise 58).
Chapter 4
Operations on topological spaces
In Chapter 3 we saw how a subset of a given topological space naturally inherits a topology from the whole space, in a sense through the operation of set
intersection. In this chapter we show that there are natural topologies corresponding to disjoint unions and cartesian products of sets as well. The latter,
called product or Tychonoff topology, is vital in many important applications,
and gives rise to one of the most universally applied theorems in general topology, the Tychonoff theorem (introduced in Chapter 8).
4.1
Sum of spaces
Let {Xs }s∈S be a family of disjoint topological spaces, that is, Xs ∩ Xt = ∅ for
S
all s 6= t. Let X =
Xs and T = {A ⊆ X : A ∩ Xs is open in Xs ∀s ∈ S}. Then
s∈S
T is a topology on X (see Exercise 76).
Definition 4.1. The set X with topology T (from above) is called the sum of
L
the spaces Xs . We write (X, T ) =
Xs .
s∈S
L
Theorem 4.1. A set F ⊆
Xs is closed if and only if F ∩ Xs is closed in Xs for
s∈S
all s ∈ S.
L
Proof. Let F be a closed set in
Xs . This implies that the complement X \ F
s∈S
L
L
is open in
Xs . Now the set X \ F is open in
Xs if and only if (X \ F ) ∩ Xs
s∈S
s∈S
is open in Xs for all s ∈ S. This is again equivalent with
[X \ ((X \ F ) ∩ Xs )] ∩ Xs =(F ∩ Xs ) ∪ [(X \ Xs ) ∩ Xs ]
=(F ∩ Xs ) ∪ ∅ = F ∩ Xs
being closed in Xs for all s ∈ S, as required.
Remark.
a) Each set Xs is open and closed in
L
Xs , since Xs = Xs ∩ Xs is open in
s∈S
Xs and Xs = Xs ∩ Xs is closed in Xs for all s ∈ S.
L
b) Each Xs is a subspace of
Xs , as is evident from the definition of the
s∈S
topology.
4.2 Product spaces
25
S
Theorem 4.2. Let X be a topological space and X =
Xs , where each Xs is
s∈S
L
open in X and Xs ∩ Xt = ∅ when s =
6 t. Then X =
Xs , where each Xs is
s∈S
equipped with the relative topology from X.
L
Proof. Since X and
Xs have the same points, we only need to prove that
s∈S
they have the same open sets. Let U be open in X, implying that U ∩ Xs is open
L
in Xs for all s ∈ S. This says exactly that U is open in
Xs .
s∈S
Conversely, let U be open in
L
Xs . This implies that U ∩ Xs is open in Xs for
s∈S
all s ∈ S. Now each Xs is equipped with the subspace topology coming from X,
and so we must have U ∩ Xs = V ∩ Xs for some V open in X. Since the sets
Xs are open in X for all s ∈ S as well, the intersection U ∩ Xs is open in X.
Ultimately
∪ (U ∩ Xs ) = U ∩ ∪ Xs = U ∩ X = U
s∈S
s∈S
is open in X, being a union of open subsets of X.
Example 4.1.
1. The real line with its usual topology cannot be written as a sum of two
topological spaces.
2. Let E be the Sorgenfrey line, i.e., E = (R, T ) where B = {[a, r[: a ∈
R and r ∈ Q} is a base for T . Then let a ∈ E be any point, r ∈ Q such
that a < r and X1 = [a, r[ . Now E = X1 ⊕ (E \ X1 ).
4.2
Product spaces
Definition 4.2. Let {Xα }α∈I be a family of non-empty sets. The cartesian product of the sets Xα is
Y
Xα = {x : I → ∪ Xα such that x(α) ∈ Xα for all α ∈ I}
α∈I
α∈I
Usually we write
Q
Xα instead of
Q
Xα and xα for x(α).
α∈I
We also say that xα is the α:th coordinate of x.
Q
The function (map) πβ : Xα → Xβ , defined by πβ (x) = xβ , is the β:th projection.
Definition 4.3. Let X be a topological space with topologies T1 and T2 . We say
that T1 is weaker than T2 or T2 is stronger than T1 when T1 ⊆ T2 .
Example 4.2.
1) When I = {1, 2, 3, . . . , n}, it is customary to write
Y
α∈I
Xα =
n
Y
α=1
Xα = {(x1 , x2 , . . . , xn ) : xk ∈ Xk , k = 1, 2, 3, . . . , n}.
4.2 Product spaces
26
2) Also in the case I = N, we may write
Y
Xα =
α∈N
∞
Y
Xα = {(x1 , x2 , . . .) : xk ∈ Xk for every k ∈ N}.
α=1
3) If Xα = X for all α ∈ I, then
Q
Xα is the set of all functions from I to
Q
X. The notation can be simplified by writing
Xα = X I .
α∈I
α∈I
Warning. The statement that an arbitrary product space
Q
Xα is non-empty
α∈I
is equivalent to the Axiom of choice.
Definition 4.4. The Tychonoff topology or the product topology on
Q
obtained by taking as a base the sets of the form
Aα , where
Q
Xα is
α∈I
α∈I
a) Aα is open in Xα for each α ∈ I,
b) Aα = Xα for all but finitely many α ∈ I.
Remark. Condition b) says that Aα = Xα except for α1 , α2 , . . . , αn , so we can
write
Y
Y
\ Y
\ Y 2\
Aα =
···
A1α
Aα
Anα ,
α∈I
where
Akα
α∈I
α∈I
α∈I
= Xα for all but αk (k = 1, 2, . . . , n). In other words,
Y
Aα = πα−1
(Aα1 ) ∩ πα−1
(Aα2 ) ∩ · · · ∩ πα−1
(Aαn ).
1
2
n
α∈I
This says that the family C = πα−1 (Aα ) : Aα open in Xα is a subbase for the
Tychonoff topology.
Example 4.3.
1. Let S 1 be the unit circle in R2 . Then S 1 × [0, 1] is a cylinder and S 1 × S 1
is a torus.
Q
2. For each α ∈ I, let Xα be discrete. Then
Xα is discrete if and only if
α∈I
I is finite.
Q
3. Let X = RR =
R be the set of all real-valued functions on R, i.e.,
α∈R
X = {f : R → R}.
Let f ∈ X, and proceed to inspect what a neighbourhood of f looks like.
Q
Let U be a neighbourhood of f . Then f ∈ U =
Aα , where Aα = R for
α∈R
all α ∈ I, but for some α1 , α2 , . . . , αn . Then for each k = 1, 2, . . . , n, there
exists rk > 0 such that f (αk ) ∈ Aαk =] − rk + f (αk ), rk + f (αk )[ . We can
give the neighbourhood of f in RR as
U(f ; α1 , α2 , . . . , αn ; r1 , r2 , . . . , rn ) =
Y
α∈R
where Vα = R except for αk , k = 1, 2, . . . , n, for which
Vαk =] − rk + f (αk ), rk + f (αk )[.
Vα ,
4.2 Product spaces
27
Perhaps a more illustrative way of describing U is to give it as the set
{g ∈ RR : |g(αk ) − f (αk )| < rk , k = 1, 2, . . . , n}.
A simpler neighbourhood is obtained by letting r = min{r1 , r2 , . . . , rn }
and F = {α1 , α2 , . . . , αn }. Then
] − r + f (α), r + f (α)[ ⊆ ] − rk + f (αk ), rk + f (αk )[
for each k = 1, . . . , n, and so f ∈
Q
Wα ⊆
α∈R
Q
α∈R
Vα , where Wα = R for all
but α ∈ F . For α ∈ F , Wα =] − r + f (α), r + f (α)[. Again, this neighbourhood can be written simply as
Y
Wα = W(f ; F, r) = {g ∈ RR : |g(α) − f (α)| < r, α ∈ F }
α∈R
. . .
f
. . .
α1
α2
α3
αn
Figure 4.1: The neighbourhood of f in Example 4.3 part 3.
Q
Theorem 4.3. Each projection πβ :
Xα → Xβ is continuous and open, but
need not to be closed. Furthermore, the Tychonoff topology is the weakest topolQ
ogy on Xα for which each projection is continuous.
Q
Proof. Let Aβ ⊆ Xβ be open. Then πβ−1 (Aβ ) = Uα , where Uβ = Aβ and Uα =
Q
Xα for all α 6= β, is open in Xα . This shows that πβ is continuous. Let πγ−1 (Aγ )
be an element in the subbase of the Tychonoff topology (in other words, Aγ is
Q
open in Xγ ). Then πγ−1 (Aγ ) = Uα , where Uγ = Aγ and Uα = Xα for all α 6= γ.
Q
The projection πβ (πγ−1 (Aγ )) = πβ ( Uα ) = Aγ if β = γ or Xβ if β 6= γ. In both
Q
cases πβ ( Uα ) is open, and so πβ is an open function.
Proceed to show that the Tychonoff topology T is the weakest topology for which
Q
each πβ is continuous. Let T ′ be another topology on Xα with this property.
We need to show that T ⊆ T ′ . Let πα−1 (Aα ) be an arbitrary element from the
subbase of T . Then Aα is open in Xα , and, since πα is continuous with respect
Q
to T ′ , πα−1 (Aα ) is open in ( Xα , T ′ ). This shows that each member of T is also
in T ′ , i.e., T ⊆ T ′ .
4.2 Product spaces
28
Theorem 4.4. Let X be a topological space and let
Tychonoff topology. Then
Q
Xα
f :X
πα ◦f
Q
Xα be equipped with the
πα
Xα
is continuous if and only if πα ◦ f is continuous for each α ∈ I.
Proof. Necessity is immediate, since composition of continuous functions is
continuous (see Exercise 62).
Conversely let πα−1 (Aα ) be an arbitrary element from the subbase of the Tychonoff topology. Then f −1 (πα−1 (Aα )) = (πα ◦ f )−1 (Aα ) is open in X since Aα is
open in Xα and πα ◦ f is continuous by assumption. It follows that f is continuous.
Remark.
1) Let X be a set and {Xα }α∈I be a family of topological spaces, and let
fα : X → Xα . The weak topology induced on X by the family {fα }α∈I
of functions is the smallest topology on X making each fα continuous. A
subbase for this topology is
{fα−1 (Aα ) : Aα is open in Xα , α ∈ I}.
Theorem 4.2 says that the Tychonoff topology is the weak topology on
Q
Xα induced by {πα }α∈I .
2) Dually to the weak topology, we also have the notion of strong topology
(or the largest topology ) induced on Y by the family of functions fα :
Yα → Y , where Yα is a topological space for each α. This is called the
quotient topology . Details on the quotient topology can be found in most
books on general topology. Let f : X → Y , f continuous. Is the quotient
space Y open or closed?
3) Another way of constructing topological spaces is with the so-called inverse systems and their limits.
Chapter 5
Convergence
In Chapter 2 we saw how useful sequences can be in describing the topology
of a metric space, for example by giving characterizations of closedness and
completeness in terms of convergence. This chapter examines convergence in a
more general setting, and will, to a certain extent, show where sequences are
sufficient to describe the topology of a space. Later, however, we shall see that
it takes a more general notion of convergence to describe some topologies, and
new concepts are introduced to address this problem.
5.1
Sequences
Definition 5.1. A sequence (xn )∞
n=1 = (xn )n∈N in a topological space X converges to x in X, when for every neighbourhood U of x, there exists n0 ∈ N such
that xn ∈ U, whenever n ≥ n0 . We say that (xn )∞
n=1 is eventually in U and write
xn → x or lim xn = x.
n→∞
Remark. Let (X, d) be a metric space. Then Definition 2.11 on page 10 and
Definition 5.1 coincide.
Definition 5.2. A topological space X is first countable when each of its points
has a countable base.
Example 5.1. Any metric space X is first countable, since for each x ∈ X,
B(x) = {B(x, r) : r ∈ Q} is a countable base.
Theorem 5.1. Let X be a first countable space and let E ⊆ X. Then
a) x ∈ E if and only if there exists a sequence (xn )∞
n=1 in E such that xn → x.
b) E is open in X if and only if for every x in E and every sequence (xn )∞
n=1
in E with xn → x, there exists n0 ∈ N such that xn ∈ E whenever n ≥ n0 .
c) F is closed in X if and only if whenever (xn )∞
n=1 ⊆ F and xn → x, then
x ∈ F.
d) If Y is also a first countable space, then f : X → Y is continuous if and
only if f (xn ) → f (x) in Y whenever xn → x in X.
Proof. See Exercise 82.
5.1 Sequences
30
Remark. Theorem 5.1 means that in first countable spaces, in particular metric
spaces, the topology is fully described by the convergence of sequences. However, the following examples show that this is not true in general.
Example 5.2. Let X = (N × N) ∪ {(0, 0)}. Let N × N carry the discrete topology. The neighbourhoods of (0, 0) consist of (0, 0) and all but a finite number of
points in all but a finite number of columns Nk = {k} × N, for example, (X \ N1 )
is a neighbourhood of (0, 0). Denote by Nk = N × {k} (the rows). Then N1 × N1
is also a neighbourhood of x.
Show that the above system is a topology for X = (N×N)∪{(0, 0)} (see Exercise
86). Then proceed to show that (0, 0) ∈ N × N, but there is no sequence in N × N
which can converge to (0, 0).
Proof. Now (0, 0) ∈ N × N because every neighbourhood of (0, 0) meets N × N by
definition. Let (an )∞
n=1 be a sequence in N × N with an → (0, 0). Let U1 = X \ N1 .
Then U1 is a neighbourhood of (0, 0) and so (an )∞
n=1 is eventually in U1 . Let
U2 = X \ N2 . Then U2 is a neighbourhood of (0, 0) and so (an )∞
n=1 is eventually
in U2 . We continue this process to obtain for each k ∈ N, Uk = X \ Nk . Then
Uk is a neighbourhood of (0, 0) and so (an )∞
n=1 is eventually in Uk . Thus there
are at most finitely many points from (an )∞
n=1 in Nk . Remove these points from
∞
Nk to obtain a set N′k . Let U = {(0, 0)} ∪ ( ∪ N′k ). Then U is a neighbourhood
k=1
∞
of (0, 0) that contains no point from (an )∞
n=1 . Hence (an )n=1 cannot converge to
(0, 0).
Q
Example 5.3. Let X = RR =
R = {f : R → R} with the Tychonoff topology.
α∈R
Let E = {f ∈ RR : f (x) = 0 or 1, and f (x) = 0 only for finitely many x ∈ R},
let g ∈ RR be the trivial function, i.e., g(x) = 0 for all x ∈ R, and U(g, F, r) be a
neighbourhood of g. Then r > 0,
F = {x1 , x2 , . . . , xn } ⊆ R
and
U(g, F, r) = {h ∈ RR : |h(y) − g(y)| < r for all y ∈ F }
= {h ∈ RR : |h(y)| < r for all y ∈ F }.
Now U(g, F, r) ∩ E 6= ∅ because it contains the function which is 0 on F and 1
elsewhere, and so g ∈ E. On the other hand, let (fn ) ∈ E. Then, for every n ∈ N,
there exists An , a finite subset of R, such that fn (x) = 0 for all x ∈ An .
∞
Let f = lim fn , and let x ∈
/ ∪ An . Then for r =
n→∞
n ≥ n0 implies |fn (x) − f (x)|
n=1
< 21 ,
1
2,
there exists n0 such that
i.e., for n ≥ n0 we have |1 − f (x)| < 21 . This in
∞
∞
n=1
n=1
turn shows that f (x) ≥ 12 , and thus f is non-zero outside ∪ An . Since ∪ An is
countable, we see that f can be zero at most on a countable set. Now g = 0 on
all of R, which of course is uncountable, and so g cannot be the limit of (fn )∞
n=1 .
5.2 Nets
31
Example 5.4. Let Ω be the set of ordinals ≤ ω1 , where ω1 is the first uncountable ordinal and let Ω0 = Ω \ {ω1 }.
Equip Ω with the order topology. Now the topology on Ω has a subbase that
consists of all sets
[1, α[ = {γ ∈ Ω : 1 ≤ γ < α, α ∈ Ω}
together with the sets
]β, ω1 ] = {γ ∈ Ω : β < γ ≤ ω1 , β ∈ Ω}.
Again, sequences fail to describe the topology on Ω.
5.2
Nets
A more generally applicable form of convergence is established via the so called
nets. A third important type of convergence is obtained with objects known as
filters (see Exercise 87).
Definition 5.3. A non-empty set Λ is said to be directed by a relation 6 if
a) λ 6 λ for every λ ∈ Λ,
(reflexivity)
b) λ1 6 λ2 and λ2 6 λ3 ⇒ λ1 6 λ3 ,
(transitivity)
c) λ1 , λ2 ∈ Λ ⇒ there exists λ3 ∈ Λ such that λ1 6 λ3 and λ2 6 λ3 .
The pair (Λ, 6) is called a directed set.
Remark. If possible without risk of misinterpretation, from now on we may
denote a directed set (Λ, 6) by Λ.
Example 5.5.
1) Let X 6= ∅. Then (P(X), ⊇) and (P(X), ⊆) are directed sets.
2) Let (X, T ) be a topological space and let F be the family of all closed sets
in X. Then (T, ⊇), (T, ⊆), (F, ⊇) and (F, ⊆) are all directed sets.
3) Consider [a, b], and remember that a partition of [a, b] is a finite set
{x0 , x1 , x2 , . . . , xn } ⊆ [a, b]
such that a = x0 < x1 < x2 < · · · < xn = b. Let P ([a, b]) be the set of all
the partitions of [a, b]. Then (P ([a, b]), ⊇) is a directed set.
4) The set N is a directed set with the usual relation ≥. We may regard a
sequence (xn )∞
n=1 in X as a function defined on N by
x : N0 → X
n 7→ x(n) = xn
This is generalized by allowing the domain to be any directed set.
Definition 5.4. A net (or generalized sequence ) in a set X is a function x :
Λ → X, where Λ is a directed set. We usually use xλ for x(λ), and speak of the
net (xλ )λ∈Λ , or simply (xλ ).
5.2 Nets
32
Definition 5.5. A subnet of a net (xλ )λ∈Λ is the composition x ◦ ϕ, where ϕ :
M → Λ, M is a directed set and ϕ is increasing and cofinal, i.e.,
1) ϕ(µ1 ) 6 ϕ(µ2 ) in Λ whenever µ1 6 µ2 in M ,
2) for every λ ∈ Λ there exists µ ∈ M such that λ 6 ϕ(µ).
(increasing)
(cofinal)
Remark. A subsequence of a sequence is a subnet, but a subnet of a sequence
does not need to be a sequence (see Exercise 90).
Definition 5.6. Let X be a topological space and let (xλ ) be a net in X. We
say that (xλ ) converges to x ∈ X if for each neighbourhood U of x, there exists
λ0 ∈ Λ such that xλ ∈ U whenever λ > λ0 . This can be expressed also by saying
that (xλ ) is residually or eventually in every neighbourhood of x.
A point x ∈ X is a cluster point of a net (xλ ) when for each neighbourhood U
of x and for every λ0 ∈ Λ, there exists λ > λ0 such that xλ ∈ U. Another way of
phrasing this is to say that (xλ ) is cofinally or frequently in every neighbourhood of x.
Example 5.6. Let X be a topological space, let x ∈ X and let Λ be a fixed base
for x. Then (Λ, ⊇) is a directed set.
Proof. Clearly set inclusion is reflexive and transitive, so we only need to check
the third condition. Let U1 , U2 ∈ Λ. Then U1 ∩U2 is open and x ∈ U1 ∩U2 , implying
that there exists U3 ∈ Λ such that U3 ⊆ U1 ∩ U2 (because Λ is a base). Now
U1 ⊇ U3 and U2 ⊇ U3 as required.
For each U ∈ Λ pick some xU ∈ U and form a net (xU )U ∈Λ in X. We proceed
to show that xU → x. Let V be any neighbourhood of x. Since Λ is a base for
x, there exists U0 ∈ Λ with x ∈ U0 ⊆ V. Then, for every U > U0 we have
xU ∈ U ⊆ U0 ⊆ V, showing that (xU ) is residually in every neighbourhood of x.
Theorem 5.2. Let X be a topological space and let (xλ ) be a net in X. Then
(xλ ) has a cluster point x if and only if it has a subnet which converges to x.
Proof. Let x be a cluster point of (xλ ). Then for every neighbourhood U of x and
for every λ0 ∈ Λ, there exists λ > λ0 such that xλ ∈ U. Define M = {(λ, U) :
U is a neighbourhood of x and λ ∈ Λ such that xλ ∈ U }. Also define a relation
6 on M by (λ1 , U1 ) 6 (λ2 , U2 ) if and only if λ1 6 λ2 and U1 ⊇ U2 . Then (M, 6) is
a directed set.
Reflexivity and transitivity follow immediately. For the third condition, let
(λ1 , U1 ) and (λ2 , U2 ) be in M . Then there exists λ′ ∈ Λ such that λ1 6 λ′ and
λ2 6 λ′ . Let U3 = U1 ∩ U2 , and note that U1 ⊇ U3 and U2 ⊇ U3 , and that U3
also is a neighbourhood of x. Now since x is a cluster point of (xλ ), there exists λ3 > λ′ such that xλ3 ∈ U3 . This shows that (λ3 , U3 ) ∈ M , and that both
(λ1 , U1 ) 6 (λ3 , U3 ) and (λ2 , U2 ) 6 (λ3 , U3 ). Thus M indeed is directed by 6.
Define now ϕ : M → Λ by ϕ((λ, U)) = λ. Then ϕ is increasing because (λ1 , U1 ) 6
(λ2 , U2 ) ⇔ λ1 6 λ2 , implying ϕ((λ1 , U1 )) 6 ϕ((λ2 , U2 )). The mapping ϕ is also
cofinal, for let λ0 ∈ Λ and let U be a neighbourhood of x. Then xλ ∈ U for some
λ > λ0 , and thus (λ, U) ∈ M .
5.2 Nets
33
We have constructed a subnet (xϕ((λ,U )) ) of (xλ ). Next we show that (xϕ((λ,U )) )
converges to x.
Let U0 be some neighbourhood of x. Since (xλ ) is cofinally in every neighbourhood of x, we may pick some λ0 ∈ Λ such that xλ0 ∈ U0 . Then note that
(λ0 , U0 ) ∈ M , and let (λ, U) ∈ M be such that (λ0 , U0 ) 6 (λ, U). Now xλ ∈ U
and U0 ⊇ U, and so we have
xϕ((λ,U )) = xλ ∈ U ⊆ U0 ,
for every (λ, U) > (λ0 , U0 ). This means precisely that xϕ((λ,U )) → x, as required.
Conversely, let (xϕ(µ)µ∈M ) be a subnet of (xλ )λ∈Λ and suppose that xϕ(µ) → x.
⋆
Then for every U neighbourhood of x, there exists µU ∈ M such that µ > µU ⇒
xϕ(µ) ∈ U. Let λ0 ∈ Λ be given. Since ϕ is cofinal, there is µ0 ∈ M such that
ϕ(µ0 ) > λ0 . Pick µ ∈ M such that µ > µ0 and µ > µU , then λ = ϕ(µ) > ϕ(µ0 ) >
λ0 and xλ = xϕ(µ) ∈ U (by ⋆). Then x is a cluster point of (xλ ).
Theorem 5.3. Let E be a subset of a topological space X. Then x ∈ E if and
only if there exists a net (xλ )λ∈Λ in E such that xλ → x.
Proof. Let x ∈ E, and let Λ be a neighbourhood base for x. Then U ∩ E 6= ∅ for
every U ∈ Λ. As in Example 5.6, pick a point xU ∈ U ∩ E for each U ∈ Λ, and
form the net (xU )U ∈Λ (remember that the direction in Λ is given by ⊇). Then
the net (xU )U ∈Λ is in E and converges to x.
Conversely let (xλ ) be any net in E which converges to x. By definition, this
means that for every U neighbourhood of x, there exists λ0 ∈ Λ such that xλ ∈ U
for every λ > λ0 . Since (xλ ) is a net in E, for every neighbourhood U of x we
have U ∩ E 6= ∅, showing that x ∈ E.
Theorem 5.4. Let X and Y be topological spaces. Then f : X → Y is continuous
at point x ∈ X if and only if f (xλ ) → f (x) in Y whenever xλ → x in X.
Proof. Suppose that f is continuous at some x in X and let xλ → x in X. Let
V be a neighbourhood of f (x) in Y . Then f −1 (V) is a neighbourhood of x in X,
and so there exists λ0 such that xλ ∈ f −1 (V) for every λ > λ0 . In other words,
λ > λ0 ⇒ f (xλ ) ∈ V, which implies that f (xλ ) → f (x).
Conversely suppose that f is not continuous at some x ∈ X. Then there exists V
neighbourhood of f (x) in Y such that f (U) * V whenever U is a neighbourhood
of x in X. As in Example 5.6, we may form a net converging to x by picking
xU ∈ U from every U neighbourhood of x. Still, we have f (xU ) ∈
/ V for every U
neighbourhood of x, implying precisely that f (xU ) 9 f (x).
Q
Theorem 5.5. A net (xλ )λ∈Λ in
Xα converges to x if and only if πα (xλ ) →
α∈I
πα (x) in each factor space Xα .
Q
Q
Proof. Since πα : Xα → Xα is continuous, xλ → x in Xα trivially implies
πα (xλ ) → πα (x) for each α ∈ I.
Q
Q
Let U be a neighbourhood of x in Xα . We may assume that U = Aα , where
Aα = Xα for every α ∈ I, except for some α1 , α2 , . . . , αn . We have open sets
5.2 Nets
34
Aα1 , Aα2 , . . . , Aαn in Xα1 , Xα2 , . . . , Xαn , respectively. Using them, we may write
(Aαn ). Then for each i = 1, 2, . . . , n, there
(Aα2 ) ∩ · · · ∩ πα−1
(Aα1 ) ∩ πα−1
U = πα−1
n
2
1
exists λi ∈ Λ such that παi (xλ ) ∈ Aαi for every λ > λi (since παi (xλ ) → παi (x)
and παi (x) ∈ Aαi ).
Let λ0 > λ1 , λ2 , . . . , λn . Then λ > λ0 implies παi (xλ ) ∈ Aαi for every i =
n
(Aαi ) = U. This means precisely that xλ → x
1, 2, . . . , n, and so xλ ∈ ∩ πα−1
i
i=1
Q
in
Xα , as required.
α∈I
Remark. Consider Y X =
Q
Y = {f : X → Y }. Then by Theorem 5.5, fλ → f
x∈X
in Y X if and only if πx (fλ ) → πx (f ) in Y for each x ∈ X. Now πx (fλ ) = fλ (x)
and πx (f ) = f (x) ∀x ∈ X, so Theorem 5.5 says that fλ → f in Y X if and only if
fλ (x) → f (x) for each x ∈ X.
In other words, the convergence in Y X with the Tychonoff topology and the
pointwise convergence in Y X are the same.
Chapter 6
Separation axioms
The concept of separation in topological spaces was introduced by Alexandroff
and Hopf in 1927.
Definition 6.1. Let X be a topological space. We say that X is a
1) T0 -space, when for every pair of distinct elements x and y in X, there
exists an open set U containing one, but not the other. In other words
x ∈ U and y ∈
/ U or y ∈ U and x ∈
/ U.
2) T1 -space, when for every pair of distinct elements x and y in X, there
exists a neighbourhood U of x and a neighbourhood V of y such that
y∈
/ U and x ∈
/ V.
3) T2 -space or a Hausdorff space, when for every pair of distinct elements
x and y in X, there exist disjoint open sets U and V such that x ∈ U and
y ∈ V.
/ A,
4) regular space, when for every closed set A and every element x ∈
there exist two disjoint open sets U and V such that x ∈ U and A ⊆ V.
5) T3 -space, when X is a T1 -space and regular.
6) completely regular space, when for every closed set A and every element
x∈
/ A, there exists a continuous function f : X → [0, 1] such that f (A) =
{1} and f (x) = 0.
7) T3 21 -space, or a Tychonoff space when X is a T1 -space and completely
regular.
Remark. The abbreviation T comes from the German word Trennung (separation).
Remark. Clearly we have T2 ⇒ T1 ⇒ T0 .
Example 6.1.
1) Let X be a set with more than one element, and give it the trivial topology T = {∅, X}. Then (X, T ) is not even a T0 -space.
2) Let X = {a, b} and T = {∅, {a}, X}. Then X is a T0 -space, but not a
T1 -space.
3) Let X be an infinite set with the cofinite topology. Then X is a T1 -space,
since if x 6= y, let U = X \{y} and V = X \{x}. Then U is a neighbourhood
of x which does not contain y and V is a neighbourhood of y which does
not contain x. The space X is not however Hausdorff.
4) Every metric space is a Hausdorff space (see Exercise 102).
36
Theorem 6.1. Let X be a topological space. Then X is T1 -space if and only if
{x} is closed for every x ∈ X.
Proof. Suppose that X is a T1 -space and let x ∈ X. We show that X \ {x} is
open. Let y ∈ X \ {x}. Then y 6= x, and so there exists V neighbourhood of y
which does not contain x. In other words, y ∈ V ⊆ X \ {x}, showing that X \ {x}
is open.
Conversely, suppose that each singleton is closed in X and let x and y be two
distinct points in X. Then {x} and {y} are closed, and so X \ {x} and X \ {y}
are the required open sets.
Remark. It is clear that T3 ⇒ T2 . It is also easy to see that T3 21 ⇒ T3 (see
Exercise 108).
Theorem 6.2. Let X be a topological space. Then the following statements are
equivalent.
a) X is a Hausdorff space.
b) Limits in X are unique (in other words there is no net or filter that converges to more than one point in X).
c) The diagonal ∆ = {(x, x) : x ∈ X} is closed in X × X.
Proof.
a) ⇒ b) Let (xα ) be a net in X such that xα → x and xα → y. Suppose that x 6= y.
Then for every neighbourhood U and V of x and y, respectively, there
exists α0 such that α ≥ α0 implies xα ∈ U ∩ V. Thus U ∩ V =
6 ∅, and X is
not a Hausdorff space.
b) ⇒ c) Suppose that ∆ is not closed in X ×X. Then by Theorem 5.3, there exists
a net (xλ , xλ ) in ∆ which converges to (x, y) with x 6= y. So xλ → x and
xλ → y, contradicting b).
c) ⇒ a) Suppose that ∆ is closed and let x 6= y. Then (x, y) ∈
/ ∆. In other words,
(x, y) ∈ (X × X) \ ∆, which is open. Now there exist open sets U and V in
X such that (x, y) ∈ U × V ⊆ (X × X) \ ∆. This means that x ∈ U, y ∈ V
and U ∩ V = ∅, showing that X indeed is Hausdorff.
Theorem 6.3. Let X be a topological space. Then X is regular if and only if for
every x in X and for every neighbourhood U of x, there exists a neighbourhood
V of x such that x ∈ V ⊆ V ⊆ U.
Proof. Suppose that X is regular and let U be a neighbourhood of some x in
X. Then X \ U is closed and x ∈
/ X \ U. So, there exist disjoint neighbourhoods V and W of x and X \ U, respectively. Now X \ W is closed, and so
x ∈ V ⊆ V ⊆ X \ W = X \ W ⊆ U.
Conversely, let F be a closed set not containing x. Then x ∈ X \ F and X \ F
is a neighbourhood of x. By assumption, there exists an open set V such that
x ∈ V ⊆ V ⊆ X \ F , and thus V and X \ V are disjoint neighbourhoods of x and
F as required.
Theorem 6.4. Let i = 0, 1, 2, 3 or 3 12 . Then
37
a)
b)
c)
d)
Every subspace of a Ti -space is a Ti -space.
Every subspace of a regular space is regular.
Every subspace of a completely regular space is completely regular.
The product of Ti -spaces (or regular or completely regular spaces) is a
Ti -space (or a regular or a completely regular space).
Proof. We will begin with the subspace arguments, and with i = 2. The cases
i = 0, 1 are left as exercises (see Exercise 95). Suppose that (X, T ) is a T2 -space
and let Y be a subspace of X. Then x 6= y in Y implies x 6= y in X. Since X is
T2 , there exist U, V ∈ T such that x ∈ U, y ∈ V and U ∩ V = ∅. Now U ∩ Y and
V ∩ Y , respectively, are neighbourhoods of x and y in Y and
(U ∩ Y ) ∩ (V ∩ Y ) = ∅,
showing that Y is a T2 -space.
Next, let X be a regular space and Y a subspace of X, A be closed in Y and
x ∈ Y \ A. Then A = F ∩ Y , where F is closed in X. So x ∈
/ F , and we can
separate F and x by open sets U and V in X. Thus U ∩ Y and V ∩ Y are the open
sets in Y separating x and A, and so Y is regular.
Now if X is a T3 -space, it is both T1 and regular. Hence by above, a subspace Y
of X is T1 and regular, and subsequently a T3 -space.
Then suppose that X is completely regular, let A be a closed subset of Y subspace of X and x ∈ Y \ A. Write A = F ∩ Y , where F is closed in X. Then x ∈
/ F,
and there exists a continuous function f : X → [0, 1] such that f (x) = 1 and
f (F ) = {0}. Then f|Y : Y → [0, 1] is continuous with f|Y (x) = 1 and f|Y (A) = 0,
showing that Y is completely regular.
Finally, if X is a T3 21 -space, it’s both T1 and completely regular. Again by above,
a subspace Y of X is T1 and completely regular, i.e., a T3 21 -space.
For d), consider the following argument.
Let {Xα }α∈I be a family of Ti -spaces. We start with i = 2, and again leave the
cases i = 0, 1 as exercises (see Exercise 97). Let x and y be a pair of distinct
Q
points in
Xα . Then for some α ∈ I, xα 6= yα in Xα , and so there exists Uα
α∈I
a neighbourhood of xα in Xα and Vα a neighbourhood of yα in Xα such that
Uα ∩ Vα = ∅. Then πα−1 (Uα ) and πα−1 (Vα ) are the required neighbourhoods of x
Q
and y in
Xα . Thus X is a T2 -space.
α∈I
Now let Xα be a completely regular space for each α ∈ I, A be a closed set in
Q
Xα = X and x ∈ X \ A. Then X \ A is open in X, and so there exists some
α∈I
neighbourhood πα−1
(U1 ) ∩ · · · ∩ πα−1
(Un ) of x, that has empty intersection with
1
n
A. The factor spaces Xαi are completely regular for each i = 1, . . . , n, and so for
each i = 1, . . . , n, there exists a continuous fi : Xαi → [0, 1] such that fi (xαi ) = 1
Q
and fi (Xαi \ Ui ) = {0}. Define g :
→ [0, 1] by
α∈I
g(y) = min{fi (yαi ) : i = 1, . . . , n}.
38
Then g is continuous, g(x) = 1 and g(X \ A) = 0. Thus
Q
Xα is completely
α∈I
regular.
If Xα is a T3 21 -space for each α ∈ I, it follows that Xα is a T1 -space and comQ
pletely regular for each x ∈ I. By previous results,
Xα is both T1 and comα∈I
Q
pletely regular, and so
Xα is a T3 12 -space.
α∈I
Next let Xα be a regular space for each α ∈ I, let x ∈
Q
Xα and
α∈I
U = πα−1
(U1 ) ∩ πα−1
(U2 ) ∩ · · · ∩ πα−1
(Un )
1
2
n
be a neighbourhood of x. Since each Xα is regular, there exists Vi a neighbourhood of xαi = παi (x) in Xαi such that xαi ∈ Vi ⊆ Vi ⊆ Ui ∀i = 1, 2, . . . , n. Then
x ∈ πα−1
(V1 ) ∩ πα−1
(V2 ) ∩ · · · ∩ πα−1
(Vn ) ⊆ πα−1
(V1 ) ∩ πα−1
(V2 ) ∩ · · · ∩ πα−1
(Vn )
2
n
2
n
| 1
{z
} | 1
{z
}
V
⊆
πα−1
(U1 )
1
|
∩
V
πα−1
(U2 )
2
∩ ··· ∩
{z
πα−1
(Un ),
n
}
U
that is, x ∈ V ⊆ V ⊆ U. From Theorem 6.3 we deduce that
Q
Xα is regular.
α∈I
Again Xα is a T3 -space for all α ∈ I implies that Xα is a T1 -space and regular,
Q
immediately giving that
Xα is a T3 -space.
α∈I
Note. In fact we have equivalence, that is
Q
Xα being a Ti -space implies that
α∈I
each Xα is a Ti -space as well (or regular or completely regular).
Remark.
1) T3 21 ⇒ T3 ⇒ T2 ⇒ T1 ⇒ T0 , but T0 ; T1 ; T2 ; T3 ; T3 21 . To show
that T2 ; T3 , take X = R with the topology
T = {neighbourhoods U of x 6= 0 are the usual ones,
1 1
neighbourhoods of 0 are the form U \ {1, , , . . .}}.
2 3
Let A = {1, 21 , 31 , . . .}. Then A is closed, 0 ∈
/ A, but A and 0 cannot be
separated by open sets. Similar counterexamples will be discussed in
Exercise 108.
2) With the help of quotient topology, one can produce examples for the
following purposes:
•
The continuous, closed image of a T2 -space is not a T2 -space.
•
The continuous, open image of a T2 -space is not a T2 -space
•
The continuous, closed image of a T3 -space is not a T3 -space.
•
The continuous, open image of a T3 -space is not regular.
Chapter 7
The T4 -spaces
The separation axioms treated in Chapter 6 behave rather well when one considers the products or subspaces of spaces carrying a certain separation property. Such is not the case with the form of separation introduced in this chapter.
This subject deserves a chapter of its own not only due to ill behavior, but also
because of some exceptional properties and strong theorems associated with it.
Definition 7.1. Let X be a topological space. We say that X is normal if for
each pair of disjoint closed subsets A and B of X, there exist open sets U and V
in X such that A ⊆ U and B ⊆ V and U ∩ V = ∅.
Definition 7.2. A normal T1 -space is called a T4 -space.
Note. In books Ti -spaces might have different definitions, depending on the
notation.
Example 7.1.
1) Every metric space (X, d) is T4 .
Proof. Clearly every metric space is T1 , so it suffices to show that X is
normal.
Let A, B ⊆ X be closed and disjoint. For every x ∈ A, pick rx > 0 such
that B(x, rx )∩B = ∅. For every y ∈ B, pick ρy > 0 such that B(y, ρy )∩A =
ρ
∅. Let U = ∪ B(x, r3x ) and V = ∪ B(y, 3y ). Then U and V are open,
x∈A
y∈B
A ⊆ U and B ⊆ V. Assume now that V ∩ U =
6 ∅. Then there exists some
ρ
z ∈ U ∩ V, implying that z ∈ B(x, r3x ) and z ∈ B(y, 3y ) for some x ∈ A and
ρ
y ∈ B. In other words, d(x, z) < r3x and d(y, z) < 3y . Now by the triangle
inequality,
d(x, y) ≤d(x, z) + d(z, y)
rx
ρy
< +
3
3
< max{rx , ρy }.
We can assume that max{rx , ρy } = rx , giving y ∈ B(x, rx ). This is a
contradiction, and so we must have U ∩ V = ∅, showing that X is normal.
40
2) A normal space is not necessarily regular. For example, equip R with the
topology T = {]a, ∞[: a ∈ R}. Now (X, T ) is normal, but not regular (see
Exercise 109).
3) T4 implies T3 , but normal does not necessarily imply T3 .
4) The Moore plane Γ is not normal (see Exercise 115).
5) Let E be the Sorgenfrey line (see page 25), then E is normal, but E × E
is not normal.
Theorem 7.1.
a) Closed subspaces of normal (T4 -) spaces are normal (T4 ).
b) The closed continuous image of a normal (T4 -) space is normal (T4 ).
Proof.
a) Let Y be a closed subspace of a normal space X and let A and B be closed
subsets of Y . Then A = F ∩ Y and B = F ′ ∩ Y , where F and F ′ are closed
in X. Furthermore, F ∩ Y and F ′ ∩ Y are closed and disjoint in X, and
so there are open disjoint subsets U and V in X such that F ∩ Y ⊆ U
and F ′ ∩ Y ⊆ V, accordingly. Now U ∩ Y and V ∩ Y are the separating
open sets in Y . In case of X being T4 , we use Theorem 6.4 to see that any
subspace of a T1 -space is T1 as well, and thus any closed subspace of a
T4 -space is T4 .
b) Let f : X → Y be a continuous and closed function from a normal space
X onto Y . Let A and B be closed and disjoint subsets of Y . Then f −1 (A)
and f −1 (B) are closed and disjoint subsets of X because f is continuous.
Since X is normal, there are open disjoint subsets U and V in X with
f −1 (A) ⊆ U and f −1 (B) ⊆ V. Now X \ U and X \ V are closed in X,
and so are their images f (X \ U) and f (X \ V) in Y since f is closed. It
follows that Y \ f (X \ U) and Y \ f (X \ V) are open, disjoint subsets of
Y containing A and B respectively, so Y is normal. In case of X being
T4 , we only need to use the fact that the closed continuous image of a
T1 -space is again T1 (see Exercise 94).
Remark. Arbitrary subspaces of a normal space X are not necessarily normal.
Let Γ be the Moore plane. Then Γ is a T3 21 -space and so it can be seen as a
Q
subspace of some cube
[0, 1]. It can be shown that any cube is normal, but
α∈A
in Exercise 115 we show that Γ is not normal.
Urysohn’s lemma. X is a normal space if and only if whenever A and B are
disjoint closed subsets of X, there exists a continuous function f : X → [0, 1]
such that f (A) = {0} and f (B) = {1}.
Proof. Suppose that X is normal and let A and B be closed and disjoint subsets
of X. Then there exist disjoint open sets U and V with A ⊆ U and B ⊆ V. Now
for all x ∈ B we have some V neighbourhood of x with V ∩ U = ∅, implying
x∈
/ U, and so B ∩ U = ∅. Thus there exists a neighbourhood U 21 of A such that
A ⊆ U 21 ⊆ U 21 ⊆ X \ B. Now A and X \ U 21 are closed and disjoint, and so are
41
U 12 and B. As above, there are open sets U 14 and U 43 such that A ⊆ U 41 ⊆ U 41 ⊆
X \ (X \ U 21 ) = U 21 ⊆ U 21 ⊆ U 34 ⊆ U 34 ⊆ X \ B.
By induction we have for each rational number r of the form r = 2kn , n > 0
where k = 1, 2, . . . , 2n − 1, an open set. Ur is such that r1 ≤ r2 ⇒ A ⊆ Ur1 ⊆
Ur1 ⊆ Ur2 ⊆ Ur2 ⊆ X \ B.
Define now f : X → [0, 1] by

1
f (x) =
inf{r : x ∈ U }
r
if x ∈
/ ∪Ur ,
r
otherwise.
Then f (A) = 0 since A ⊆ U 21n for all n ∈ N, i.e. f (x) ≤ 21n ∀n ∈ N for every
x ∈ A. Furthermore, f (B) = 1 because Ur ⊆ X \ B ∀r, and so B ∩ Ur = ∅.
It remains to show that f is continuous. Let f (x0 ) = r0 . Then we have the
following three cases.
Case 1. r0 6= 0 and r0 6= 1: Take r1 and r2 such that r0 − ε < r1 < r0 < r2 < r0 + ε.
Then x0 ∈ Ur2 ∩ (X \ Ur1 ), and Ur2 ∩ (X \ Ur1 ) is a neighbourhood of x0 .
Now let y ∈ Ur2 ∩ (X \ Ur1 ). We have both f (y) ≤ r2 and f (y) ≥ r1 , i.e.
f (y) ∈ [r1 , r2 ] ⊂]r0 − ε, r0 + ε[.
Case 2. r0 = 0: Let 0 < r2 < ε. Then Ur2 is a neighbourhood of x0 , and clearly
f (Ur2 ) ⊆ [0, r2 ] ⊆ [0, ε[.
Case 3. r0 = 1: Let 1 − ε < r1 < 1. Then X \ Ur1 is a neighbourhood of x0 and
f (X \ Ur1 ) ⊆ [r1 , 1] ⊆ ]1 − ε, 1].
It follows from above that f is continuous.
Conversely let A and B be closed and disjoint in X. Suppose that there is a
continuous function f : X → [0, 1] such that f (A) = 0 and f (B) = 1. Then
f −1 ([0, 21 [) and f −1 (] 12 , 1]) are open and disjoint with A ⊆ f −1 ([0, 12 [) and B ⊆
f −1 (] 21 , 1]), so X is normal.
Corollary. Let X be a normal space, let A and B be disjoint and closed and let
a < b. Then there exists a continuous function f : X → [a, b] such that f (A) = {a}
and f (B) = {b}.
Proof. See Exercise 112.
Tietze’s extension theorem. A topological space X is normal if and only if
whenever A is a closed subset of X and f : A → [−1, 1] is continuous, there exists
a continuous extension of f to all X, i.e., a continuous function f˜ : X → [−1, 1]
˜ A= f .
such that f|
Proof. Let A1 = {x ∈ A : f (x) ≥ 13 } and B1 = {x ∈ A : f (x) ≤ − 31 }. Then A1
and B1 are closed and disjoint in X, and by the corollary above we may pick a
continuous function f1 : X → [− 13 , 31 ] such that f1 (A1 ) = 31 and f1 (B1 ) = − 13 .
Note that |f (x)−f1 (x)| ≤ 32 for all x ∈ A, and thus f −f1 is a continuous function
42
from A onto [− 32 , 32 ]. Repeat the process with f − f1 , dividing [− 32 , 32 ] into thirds
and let A2 = {x ∈ A : (f − f1 )(x) ≥ 29 } and B2 = {x ∈ A : (f − f1 )(x) ≤ − 92 }.
Again by the corollary of Urysohn’s lemma, there exists a continuous function
f2 : X → [− 92 , 92 ] such that f2 (A2 ) = 29 and f2 (B2 ) = − 92 . Note again that
2
|(f − f1 )(x) − f2 (x)| ≤ 32 for all x ∈ A. With this process we obtain a sequence
1 2 n−1
for all x ∈ X,
(fn )∞
n=1 of continuous functions on X such that |fn (x)| ≤ 3 3
N
P
n
fn (x)| ≤ 23 for all x ∈ A.
and |f (x) −
n=1
∞
P
Define now f˜ : X → [−1, 1] by f˜(x) =
fn (x). Then from the inequality
n=1
above, it follows that f˜(x) = f (x) for all x ∈ A, and we only need to show that
f˜ is continuous.
∞
P
2 n
Let x ∈ X and ε > 0. Now pick N ∈ N such that
< 2ε . Since fn is
3
n=N +1
continuous for n = 1, 2, . . . , N , we can pick Un neighbourhood of x such that y ∈
ε
. Then U = U1 ∩ U2 ∩ · · · ∩ UN is a neighbourhood
Un implies |fn (y) − fn (x)| ≤ 2N
of x. Let y ∈ U, and note that
∞
∞
X
△-ineq. X
˜
˜
|f (y) − f (x)| = fn (y) − fn (x) ≤
|fn (y) − fn (x)|
n=1
≤
N
X
n=1
|fn (y) − fn (x)| +
n=1
<N·
∞
X
∞
∞
X
X
ε
|fn (x)|
|fn (y)| +
+
2N
n=N +1
≤
ε
+
2
|fn (y) − fn (x)|
n=N +1
∞
X
n=N +1
∞
X
ε
= +
2
n=N +1
1 2 n−1
+
3 3
n=N +1
∞
X
n=N +1
2 n
ε ε
< + = ε,
3
2 2
1 2 n−1
3 3
showing that f˜ is continuous.
Conversely, suppose that each continuous function on some closed non-empty
subset of X can be continuously extented to all X and let A and B be closed and
disjoint subsets of X. Then A ∪ B is closed and the function f : A ∪ B → [0, 1]
defined by f (A) = {0} and f (B) = {1} is continuous on A ∪ B, and so we
can extend f to f˜ defined on all of X. It follows that f˜(A) = f (A) = {0} and
f˜(B) = f (B) = {1}, and so X is normal by Urysohn’s lemma.
Corollary. A topological space X is normal if and only if whenever A is a closed
subset of X and f : A → R is continuous, there is a continuous extension of f to
all X.
Proof. See Exercise 118.
Remark. Even in recent years, new shorter and simpler proofs for Tietze’s extension theorem continue to appear. Sometimes they don’t use the uniform convergence. See for example: Erich Ossa, A simple proof of the Tietze-Urysohn
43
extension theorem in Arch. Math. 71 (1998) 331–332 or Mark Mandelkern in
Arch. Math. 60 (1993) 364–366.
Chapter 8
Compact spaces
Closed and bounded intervals on the real line are the domain of many important results in classical analysis. This is due to the Heine-Borel property they
possess, i.e., every cover of a closed bounded interval by open sets has a finite
subcover. It turns out that sets satisfying an analogous property in a topological space have some formidable qualities.
Definition 8.1. A topological space X is compact if each open cover of X has
a finite subcover, i.e., if X = ∪ Ui where Ui is open for all i ∈ I, there exist
i∈I
n
finitely many Ui1 , Ui2 , . . . , Uin such that X = ∪ Uik .
k=1
Definition 8.2. A space X is said to be locally compact when each of its points
has a neighbourhood with compact closure.
Example 8.1.
1) R is not compact, for {] − n, n[ : n = 1, 2, 3, . . . } is an open cover of R,
N
but it has no finite subcover. Otherwise we would have R = ∪ ] − n, n[=
n=1
] − N, N [ for some N ∈ N.
2) N is not compact. In fact a discrete space is compact if and only if it is
finite (see Exercise 119).
3) ] − 1, 1[ is not compact (see Exercise 119).
4) [−1, 1] is compact.
5) The Cantor set is compact.
6) The ordinal space Ω is compact.
Definition 8.3. A family C of subsets of X is said to have the finite intersection
property (FIP) if every finite subfamily of C has a non-empty intersection, i.e.,
n
for all n ∈ N and for every C1 , C2 , . . . , Cn ∈ C, we have ∩ Ck 6= ∅.
k=1
Definition 8.4. A net (xλ ) is said to be an ultranet if for every U ⊆ X, (xλ ) is
either eventually in U or X \ U.
Theorem 8.1. Let X be a topological space. Then the following statements are
equivalent.
a) X is compact.
b) Every family of closed subsets of X with the
section.
c) Each net in X has a cluster point.
FIP
has a non-empty inter-
45
d) Each net in X has a convergent subnet.
e) Each ultranet in X converges.
Proof.
a) ⇒ b) Let C = {Ci }i∈I be a family of closed sets in X. Then {X \ Ci }i∈I is a
family of open sets in X. Now if ∩ Ci = ∅, we have that
i∈I
X = X \ ∅ = X \ ∩ Ci = ∪ (X \ Ci ),
i∈I
i∈I
and so {X \ Ci }i∈I is an open cover of X. Since X is compact, we have
some finite subcover {X \ Cik }nk=1 . This gives us
n
n
k=1
k=1
X = ∪ (X \ Cik ) = X \ ∩ Cik ,
n
showing that ∩ Cik = ∅. Hence C does not have the
k=1
FIP.
b) ⇒ a) Let {Ui }i∈I be an open cover of X. This gives us the empty set as
∅ = X \ ∪ Ui = ∩ (X \ Ui ),
i∈I
i∈I
and from b) we have that {X \ U1 } does not have the FIP . This is to say,
n
n
k=1
k=1
that for some i1 , i2 , . . . , in , ∩ (X \ Uik ) = ∅, giving that ∅ = X \ ∪ Uik ,
n
implying X = ∪ Uik as required.
k=1
b) ⇒ c) Let (xλ ) be a net in X and let Tλ0 = {xλ : λ > λ0 }. Then {Tλ0 }λ0 ∈Λ is a
family of closed sets with FIP , and by b) ∩ Tλ0 6= ∅. This in turn means
λ0 ∈Λ
that (xλ ) has a cluster point (see Exercise 91).
c) ⇔ d) Follows from theorem 5.2 on page 32.
c) ⇒ b) Let C = {Ci }i∈I be a family of closed sets with the FIP. Let Λ be the
family of all finite subfamilies of C. We define the relation 6 on Λ by
λ = {C1 , C2 , . . . , Cn } 6 λ′ = {C1′ , C2′ , . . . , Cn′ } ⇔
C1 ∩ C2 ∩ · · · ∩ Cn ⊇ C1′ ∩ C2′ ∩ · · · ∩ Cn′ .
Then (Λ, 6) is a directed set. Pick xλ from each λ = C1 ∩ C2 ∩ · · · ∩ Cn ,
forming a function x : Λ → X, λ = {C1 , C2 , . . . , Cn } 7→ xλ , where xλ ∈
C1 ∩ C2 ∩ · · · ∩ Cn . We now have a net (xλ ) in X, and so by assumption
(xλ ) has a cluster point x. It remains to show that x ∈ ∩ Ci .
i∈I
Let C0 be any element in C and let U be a neighbourhood of x. Then there
exists λ = {C1 , C2 , . . . , Cn } > λ0 = {C0 } such that xλ ∈ U (x is a cluster
point of xλ ). Since xλ ∈ C1 ∩ C2 ∩ · · · ∩ Cn ⊆ C0 , xλ is also in C0 . Thus
xλ ∈ C0 ∩ U, i.e., C0 ∩ U =
6 ∅. It follows that x ∈ C0 = C0 , and since C0 is
arbitrary, x ∈ ∩ Ci .
i∈I
d) ⇒ e) An ultranet is a net and so it has a convergent subnet by d). Now by
Exercise 89 the ultranet converges.
e) ⇒ d) Follows from the fact that each net has an ultranet as a subnet, see
Exercise 89.
46
Theorem 8.2. A closed subspace Y of a compact space X is compact.
Proof. Let {Vi }i∈I be an open cover of Y . Then for each i ∈ I, let Ui be an open
set in X such that Vi = Ui ∩ Y . Now {Ui }i∈I ∪ {X \ Y } is an open cover of X.
Since X is compact, we have a finite subcover, i.e., there exist Ui1 , Ui2 , . . . , Uin
n
n
k=1
k=1
in {Ui }i∈I such that ∪ Uik ∪ (X \ Y ) = X. Clearly Y ⊆ ∪ Uik , and so
n
n
n
k=1
k=1
k=1
Y = Y ∩ ∪ Uik = ∪ (Uik ∩ Y ) = ∪ Vik .
Thus {Vik }nk=1 is a finite subcover of {Vi }i∈I , and subsequently Y is compact.
Theorem 8.3. A compact subset Y of a Hausdorff space X is closed.
Proof. Let x ∈ Y , and pick a net (xλ ) in Y with lim xλ = x. Because Y is
compact, (xλ ) has a subnet (xλρ ) with lim xλρ = y in Y . Now X is Hausdorff,
and so x = y. Thus Y = Y and Y is closed.
Theorem 8.4. The continuous image f (X) of a compact space X is compact.
Proof. Let f : X → Y be continuous and let X be compact. Let {Vi }i∈I be an
open cover of f (X). Then {f −1 (Vi )}i∈I is an open cover of X. Since X is compact,
n
X = ∪ f −1 (Vik ) for some i1 , . . . , in ∈ I. Now it follows that
k=1
n
n
k=1
k=1
f (X) = f ( ∪ f −1 (Vik )) = ∪ Vik ,
showing that f (X) indeed is compact.
Remark. Using nets, one can give shorter proofs for Theorems 8.2 and 8.4.
Q
Tychonoff theorem. The non-empty product X =
Xα is compact if and
α∈I
only if each factor Xα is compact.
Proof. Suppose that X is compact. Since the projections are continuous, it follows from Theorem 8.4 that the factors Xα = πα (X) are compact for each α ∈ I.
Conversely, let (xλ )λ∈Λ be an ultranet in X. Then for each α ∈ I, (πα (xλ ))λ∈Λ
is an ultranet in Xα (see Exercise 89). Now (πα (xλ ))λ∈Λ is convergent in Xα
by Theorem 8.1 e). Theorem 5.5, on the other hand, states that (xλ )λ∈Λ is then
convergent in X, and Theorem 8.1 e) again gives that X is compact.
Remark. Try to prove the Tychonoff theorem by using the definition or statement b) in Theorem 8.1.
Definition 8.5. A topological space X is said to be a Lindelöf space if every
open cover of X has a countable subcover.
Theorem 8.5. A compact Hausdorff space is a T4 -space.
47
Proof. Let X be a compact Hausdorff space. Since X is clearly Lindelöf and
since every regular Lindelöf space is normal (see Exercise 116), it is enough to
show that X is regular.
Let A be a closed subset of X and x ∈
/ A. For each a ∈ A, let Ua and Va be
disjoint neighbourhoods of x and a, respectively. In such a way we cover A
with {Va }a∈A . Since A is closed in X and X is compact, Theorem 8.2 gives
that A is also compact. So, there is a finite subcover Va1 , Va2 , . . . , Van , for which
n
n
k=1
k=1
A ⊆ ∪ Vak = V. Let U = ∩ Uak . Then V is a neighbourhood of A, U is a
neighbourhood of x and we have
n
n
k=1
n
k=1
n
n
k=1
k=1
V ∩ U = ( ∪ Vak ) ∩ ( ∩ Uak ) = ∪ (Vak ∩ ( ∩ Uak )
n
⊆ ∪ (Vak ∩ Uak ) = ∪ ∅ = ∅,
k=1
k=1
showing that X is regular and hence normal. Since X is Hausdorff, it also is T1
and subsequently T4 .
Definition 8.6. A topological space X is countably compact if every countable
open cover has a finite subcover.
Theorem 8.6. A continuous function f : X → R is bounded when X is countably compact.
Proof. Since f is continuous, f −1 (] − n, n[) is open in X for all n = 1, 2, . . . .
∞
Clearly X = ∪ f −1 (]−n, n[), i.e., {f −1 (]−n, n[)}∞
n=1 is a countable open cover of
n=1
N
X. Since X is countably compact, there exists N ∈ N such that X = ∪ f −1 (] −
n=1
n, n[) = f −1 (] − N, N [). This simply means that |f (x)| ≤ N for all x ∈ X.
Theorem 8.7. If f is a continuous bijection from a compact space X to a Hausdorff space Y , then f is a homeomorphism.
Proof. We only need to check that f is closed. Let F be closed in X. Then F is
compact in X by Theorem 8.2, and further, f (F ) is compact in Y by Theorem
8.4. Finally, f (F ) is closed in Y by Theorem 8.3.
We end this chapter with the following results on the compactness of metric
spaces. The proofs are left for the reader (see Exercises 127–129).
Theorem 8.8. A metric space X is compact if and only if every sequence in X
has a convergent subsequence.
Definition 8.7. For ε > 0 a subset A of X is an ε-net if it is finite and X =
∪ B(a, ε). We say that X is totally bounded if it has an ε-net for each ε > 0.
a∈A
Theorem 8.9. A metric space is compact if and only if it is complete and totally
bounded.
Corollary. A closed subspace of a complete metric space is compact if and only
if it is totally bounded.
Chapter 9
Connected spaces
In this final chapter we will explore the highly visual concept of connectedness in topological spaces. One of the main results addressed here is that the
continuous image of a connected space is again connected, which yields a generalization for the classical Intermediate Value Theorem.
Definition 9.1. A topological space X is disconnected if X = H ∪ K, where H
and K are disjoint open non-empty sets in X. When there is no such disconnection, we say that X is connected.
Remark. Note that we can replace “open” by “closed” in this definition, since
H = X \ K and K = X \ H.
Example 9.1.
1) Any discrete space with more than one element is disconnected.
2) If {x} is open in X, we say that x is isolated in X. If X is a T1 -space with
an isolated point, then X is disconnected (remember that singletons are
closed in T1 -spaces). In particular, Ω and Ω0 are disconnected, but the
long line is connected.
3) (Q, | · |) is disconnected, since (] − ∞, x[ ∩Q) and (]x, ∞[ ∩Q) are open in
Q and disjoint, and their union is empty for any irrational number x.
4) The Sorgenfrey line (see page 25) is disconnected because [x, r[ is both
open and closed.
5) (R, | · |) is connected.
6) ([0, 1], | · |) is connected.
Proof. Suppose that [0, 1] = H ∪ K, where H and K are open and disjoint
in [0, 1]. Suppose that 1 ∈ H, let c = sup K, and note that then c 6= 1.
Now c is either in H or K, so there exists a neighbourhood of c which
contained in H or K, but this is not possible because any neighbourhood
of c intersects K and H.
Remark. A subspace of a connected space is not always connected!
Theorem 9.1. A subspace Y of a topological space X is connected if and only if
there are no non-empty sets H and K in X such that
i) Y = H ∪ K,
ii) H ∩ ClX (K) = ClX (H) ∩ K = ∅.
Note that X is not assumed to be connected.
49
Proof. Suppose that Y is disconnected and let Y = H ∪ K, where H and K are
non-empty, disjoint open sets in Y . Then H and K are also closed in Y , and so
ClY (H) = H and ClY (K) = K. It follows that
H ∩ ClX (K) =(H ∩ Y ) ∩ ClX (K) = H ∩ (Y ∩ ClX (K))
=H ∩ ClY (K) = H ∩ K
=∅.
Similarly, K ∩ ClX (H) = ∅.
Conversely, suppose that Y = H ∪ K, where H ∩ ClX (K) = K ∩ ClX (H) = ∅.
Then
ClY (H) =Y ∩ ClX (H) = (H ∪ K) ∩ ClX (H)
=(H ∩ ClX (H)) ∪ (K ∩ ClX (H)) = H.
So H is closed in Y . Similarly ClX (K) = K, and K is closed in Y . It follows that
Y is the union of non-empty closed disjoint sets of Y , i.e., Y is disconnected.
Corollary. If Y ⊆ H ∪ K, where H and K are such that H ∩ ClX (K) = K ∩
ClX (H) = ∅ and Y is connected, then Y ⊆ H or Y ⊆ K.
Proof. Let H ∩ ClX (K) = K ∩ ClX (H) = ∅. Then
(H ∩ Y ) ∩ (ClX (K ∩ Y )) = (K ∩ Y ) ∩ (ClX (H ∩ Y )) = ∅,
and thus Y ⊆ H ∪ K, meaning that Y = (H ∩ Y ) ∪ (K ∩ Y ). By Theorem 9.1 we
have H∩ = ∅ or K ∩ Y = ∅, showing that Y ⊆ H or Y ⊆ K.
Theorem 9.2. Let X be a topological space. Then
a) If X = ∪ Xα , where each Xα is connected and ∩ Xα 6= ∅, then X is
α∈A
α∈A
connected.
b) If each pair of points x and y in X lie in some connected subset Exy of X,
then X is connected.
∞
c) If X = ∪ Xn , where each Xn is connected and Xn ∩ Xn+1 6= ∅ for each
n=1
n ≥ 1, then X is connected.
Proof.
a) Let X = H ∪ K, where H ∩ K = H ∩ K = ∅. Since Xα is connected, we
have Xα ⊆ H or Xα ⊆ K by the corollary above. Now the spaces Xα are
not disjoint while H ∩ K = ∅, and so we must have Xα ⊆ H for all α ∈ A
or Xα ⊆ K for all α ∈ A. Then X ⊆ H or X ⊆ K, and either H or K is ∅.
Thus X is connected.
b) Fix y ∈ X. Then the fact that X = ∪ Exy is connected follows from a).
x∈X
c) Let X1 and X2 connected. Then by a), X1 ∪ X2 is connected. Again by a),
(X1 ∪ X2 ) ∪ X3 is connected. Furthermore, by a), An = X1 ∪ X2 ∪ · · · ∪ Xn
∞
∞
n=1
n=1
is connected for each n ≥ 1, and so inductively X = ∪ Xn = ∪ An is
connected.
50
∞
Example 9.2. R = ∪ [−n, n] is connected by Theorem 9.2, since [−n, n] is
n=1
homeomorphic to [0, 1] and [0, 1] is connected as seen earlier. The homeomorphism can be given as
f : [0, 1] → [−n, n]
x 7→ n(2x − 1)
Definition 9.2. Let X be a topological space and let x ∈ X. The largest connected subset Cx of X which contains x is called the component of x.
Remark.
1) The component Cx exists because it is just the union of all connected
subsets of X which contain x.
2) If x 6= y, then Cx = Cy or Cx ∩ Cy = ∅. Suppose that Cx ∩ Cy 6= ∅. This
means that Cx ∪ Cy is connected and so Cx = Cy , since Cx and Cy are the
largest connected sets containing x and y, respectively. This shows that
the components form a partition of X into maximal connected subsets.
Theorem 9.3. Let X be a topological space. Then
a) The closure of a connected subset of X is connected.
b) The components are closed in X.
c) If A ⊆ X is connected and f : X → Y is continuous, f (A) ⊆ Y is connected.
Proof. See Exercises 133–135.
Example 9.3.
1) Let X be connected and x ∈ X. Then Cx = X.
2) The components of Q are the points.
3) The components of the Cantor set are the points.
Q
Theorem 9.4. A non-empty product space X =
Xα is connected if and only
α∈A
if each factor Xα is connected.
Proof. The necessity follows from the continuity of the projection πα : X → Xα ,
and from the fact that the continuous image of a connected space is connected
by Theorem 9.3.
Q
Conversely suppose that each Xα is connected. Pick a ∈
Xα , denote by E
α∈A
the component of a. Then E is connected by theorem 9.2 b).
We show that E is dense in X, i.e., E = X. Let
(Un )
(U2 ) ∩ · · · ∩ πα−1
(U1 ) ∩ πα−1
U = πα−1
n
2
1
be an open set in X. For each i = 1, . . . , n pick bαi ∈ Ui and define the sets
51
E1 = {x ∈
E2 = {x ∈
..
.
En = {x ∈
Y
Y
Y
Xα : xα1 is arbitrary, xα = aα otherwise },
Xα : xα1 = bα1 , xα2 is arbitrary, xα = aα otherwise },
Xα : xαi = bαi , for i = 1, 2, 3, . . . , n − 1, xαn is arbitrary,
xα = aα otherwise }.
Then Ek is homeomorphic to Xαk and so Ek is connected. Also, Ek ∩ Ek + 1 6= ∅
n
for k = 1, . . . , n − 1, implying that ∪ Ek is connected as well. Now a ∈ F , and
k=1
so we must have F ⊆ E. Furthermore, since b ∈ F ∩ U, we obtain E ∩ U =
6 ∅ as
required.
Now X = E is connected, coming from either the fact that the closure of a
connected set is connected or from the fact that E is the component of a and
thus closed (see Exercises 133–134).
Definition 9.3. A topological space X is totally disconnected if and only if
the components in X are the singletons. Equivalently we may define that X is
totally disconnected if and only if the only nonempty connected subsets of X
are the one-point sets.
Index
accumulation point
in metric space, 9
in topological space, 17
in topological space, 65
Baire’s theorem, 12
base
at a point, 16
of a topology, 15
be eventually in, 29
bijection, 4
bijective, 4
boundary
in topological space, 18
Cantor set, 5
Cantor’s Teepee, 70
Cantor’s Leaky Tent, 70
cartesian product, 25
Cauchy sequence, 11
closed ball, 9
closed set
of a metric space, 8
of a topological space, 15
closure
in metric space, 9
in topological space, 17
cluster point, 32
cofinally, 32
cofinite topology, 61
compact space, 44
complement, 2
complete metric space, 11
completely regular space, 35
completion, 11
component, 50
connected space, 48
continuous function
in metric space, 13
in topological space, 20
converge
net, 32
sequence, 10
convergent net, 32
convergent sequence, 10
countable set, 4
countably compact space, 47
cylinder, 26
de Morgan laws, 3
dense, 10
diameter, 6
difference, 2
directed set, 31
disconnected space, 48
discrete metric, 7
discrete topology, 16
distance, 6
domain, 3
embedded into, 21
embedding, 21
ε-net, 47
Euclidean metric, 57
eventually, 29
filter, 31, 65
finite intersection property, 44
first category, 13
first countable, 29
frontier
in topological space, 18
function, 3
generalized sequence, 31
Hausdorff space, 35
Hilbert space, 68
homeomorphic, 21
homeomorphism, 21
image, 3
INDEX
indiscrete topology, 16
induced topology, 22
injection, 4
injective, 4
interior
in metric space, 9
in topological space, 18
interior point, 9
intersection, 2
isolated point, 48
largest topology, 28
limit, 10
limit point
in metric space, 9
in topological space, 17
Lindelöf space, 46
locally compact, 44
mapping, 3
metric, 6
metric space, 6
metrizable, 16
Moore plane, 63
n-dimensional Euclidean space, 57
natural topology, 61
neighbourhood, 15
net, 31
normal space, 39
nowhere dense, 10
53
radial metric, 58
range, 3
regular space, 35
relative topology, 22
residually, 32
river distance, 57
Russel’s paradox, 2
scattered line, 63
separable
metric space, 10
Sierpinski space, 16
Sierpinski’s metric space, 59
slotted plane, 63
Sorgenfrey line, 25, 65
strong topology, 28
subbase, 16
subnet, 32
subset, 2
subspace
of a metric space, 7
of a topological space, 22
sum of spaces, 24
superset, 2
surjection, 4
surjective, 4
perfect set, 59
post office metric, 58
power set, 3
product, 3
product of spaces, 25
product topology, 26
T0 -space, 35
T1 -space, 35
T2 -space, 35
T3 -space, 35
T4 -space, 39
T3 21 -space, 35
Tietze’s extension theorem, 41
topologist’s sine curve, 69
topology, 15
torus, 26
totally disconnected, 51
totally bounded space, 47
triangle inequality, 6
△-inequality, 6
triangular inequality, 6
trivial topology, 16
Tychonoff space, 35
Tychonoff theorem, 46
Tychonoff topology, 26
quotient topology, 28
ultranet, 44
one-point compactification, 69
one-to-one, 4
onto, 4
open ball, 6
open set
of a metric space, 8
of a topological space, 15
overlapping interval topology, 61
INDEX
uncountable, 4
union, 2
Urysohn’s lemma, 40
van der Waerden function, 13
weak topology, 25
Weierstrass function, 13
54
Bibliography
[1] R. Engelking. Outline of General Topology. North-Holland Publishing
Company, Amsterdam, 1968.
[2] P.R. Halmos. Naive Set Theory. Van Nostrand Reinhold, New York, 1960.
[3] J. L. Kelley. General Topology. Van Nostrand, Princeton, 1955.
[4] S. Lipschutz. General topology. In Schaum’s Outline of Theory and Problems. Schaum Publishing Co, New York, 1965.
[5] A. S. Lynn and J. A. Seebach, Jr. Counterexamples in Topology. Dover
Publications, inc., New York, 1995.
[6] C.G.C Pitts. Introduction to Metric Spaces. Oliver & Boyd, Edinburgh,
1972.
[7] S. Willard. General Topology. Addison-Wesley, Readin Mass., 1968.
Appendix A
Exercises
A.1 Exercises for Chapter 1
1. Let A, B and C be any sets. Show that
a)
b)
c)
d)
if A ⊆ B and B ⊆ C then A ⊆ C.
B \ A = B ∩ A∁ .
A ∩ B ⊆ A ⊆ A ∪ B.
A ⊆ B if and only if A ∩ B = A.
2. Let A = {a, b, c, {a, b, c}} and B = {a, b, {a, b}} find
a)
b)
c)
d)
A ∪ B,
A ∩ B,
A \ B,
B \ A.
3. Let A, B and C be sets like in the pictures below. Shade the sets
A ∩ (B ∪ C) and C \ (A ∩ B).
A
B
A B
C
C
4. State whether or not each of the diagrams below defines a function from
A = {a, b, c} into B = {x, y, z}.
a) a
x
b) a
x
c) a
x
b
y
b
y
b
y
c
z
c
z
c
z
A.2 Exercises for Chapter 2
57
5. Prove Theorem 1.1 on page 3.
6. Prove what is left to prove in Theorem 1.2 on page 3.
7. (Schroeder-Bernstein theorem) Let X and Y be non-empty subsets. Show
that if there is a one-to-one function f : X → Y and also a one-to-one function
g : Y → X, then there exists a bijection between X and Y .
8. Show that the following sets are countable:
a) {2, 4, 6, 8, . . .}
b) {0, 2, −2, 4, −4, 6, −6, . . .}
c) {1, 4, 9, 16, 25, . . .} (observed already by Galileo Galilei in the the early
17th century).
9. Show that any countable union of countable sets is countable.
10. Show that if X and Y are countable, then X × Y is also countable. Then
use this result to show that Z and Q are countable.
11. In the lectures the uncountability of R has been shown by using Cantor’s
diagonal argument. Find an alternative way of showing that R is uncountable.
A.2 Exercises for Chapter 2
12. Let X =]0, ∞[ and d(x, y) = x1 − y1 for all x, y ∈ X. Show that (X, d) is a
metric space.
13. Let X = R and d(x, y) = max{x − y, 0}. Why is d(x, y) not a metric on X?
What about d1 (x, y) = min{|x − y|, 1}?
14. Let X = R × R, and define for x = (x1 , x2 ) and y = (y1 , y2 )
(
|x2 − y2 |
if x1 = y1
d(x, y) =
|x2 | + |y2 | + |x1 − y1 | if x1 6= y1
a) Show that d is a distance on X. This is called the river distance.
b) Draw the open ball B(a, 1) when a = (0, 0), a = (0, 1) and a = (1, 0). Also
draw the open ball B((0, 1), 2).
15. Give an example of two closed subsets A and B of the real line R such that
d(A, B) = 0, but A ∩ B = ∅.
16. Let d1 , d2 , and d∞ be defined on Rn × Rn , respectively, by:
d1 (x, y) =
n
X
|xi − yi |,
i=1
n
X
d2 (x, y) =
i=1
(xi − yi )2
1/2
,
d∞ (x, y) = max{|x1 − y1 |, |x2 − y2 |, . . . , |xn − yn |},
for all x = (x1 , x2 , . . . , xn ) and y = (y1 , y2 , . . . , yn ) in Rn .
a) Show that d1 , d2 and d∞ are metrics on Rn . The metric d2 is called
the Euclidean metric, and (Rn , d2 ) is called the n-dimensional Euclidean
space.
A.2 Exercises for Chapter 2
58
b) Let n = 2 and x = (0, 0). Draw the open ball B(x, 1) with d1 , d2 and d∞ .
17. Let (X, d) be the Euclidean plane with the ordinary metric d. Let 0 = (0, 0)
be the origin of this plane. Define a new metric d∗ by:
(
d(0, x) + d(0, y) if x 6= y
∗
d (x, y) =
0
if x = y
a) Show that d∗ is a metric. This is called the post office metric.
b) What can we say about open balls B(0, r) and B(x, r), x 6= 0?
18. Let (X, d) be the Euclidean plane with the ordinary metric d. Let 0 = (0, 0)
be the origin of this plane. Define a new metric d∗ by:


0
if x = y




d(x, y)
if x 6= y and the line through
d∗ (x, y) =

x and y passes through the origin




d(x, 0) + d(y, 0) otherwise
a) Show that d∗ is a metric. This is called the radial metric.
b) What do the open balls look like in this metric?
c) Show that (R2 , d∗ ) is not separable.
19. Prove Theorem 2.3 in page 8. Hint: use Theorem 2.2 and the de Morgan
laws.
20. Let X be a non-empty set, let B(X) be the set of all bounded real-valued
functions on X, and define D(f, g) = sup |f (x) − g(x)| for all f and g in B(X).
Show that (B(X), D) is a metric space. Show that (B(X), D) is complete.
21. Let X = [0, 1], and let C([0, 1]) be the set of all continuous (bounded) functions on X. Let d be defined by
Z 1
|f (x) − g(x)|dx for all f, g ∈ C([0, 1]).
d(f, g) =
0
Show that (C([0, 1]), d) is a metric space.
d(x,y)
is
22. Let (X, d) be a metric space. Show that d1 defined by d1 (x, y) = 1+d(x,y)
also a metric on X. Observe that X is a bounded set in (X, d1 ), in other words
d1 (X) < ∞.
23. Let (X, d) be a metric space, and define for x and y in X,
ρ(x, y) = min{d(x, y), 1}.
a) Show that ρ is also a distance on X.
b) Let A ⊆ X. Show that A is open in (X, d) if and only if A is open in (X, ρ).
24. Let (X, d) be a metric space, let x be a point of X, and let r > 0. Recall that
B(x, r) = {y ∈ X : d(x, y) < r}, and B ′ (x, r) = {y ∈ X : d(x, y) ≤ r}. One is
inclined to believe that B(x, r) = B ′ (x, r). Give an example to show that this is
not true.
25. Let (X, d) be a metric space, and let A be a non-empty subset of X. Prove
the following facts:
A.2 Exercises for Chapter 2
a)
b)
c)
d)
e)
59
Int A is open.
A is open if and only if A = Int A.
X \ A = Int(X \ A).
A = {x ∈ X : d(x, A) = 0}.
A is nowhere dense in X if and only if X \ A is dense in X, in other words
Int A = ∅ if and only if X \ A = X.
26. Give an example of
a)
b)
c)
d)
e)
an infinite family of closed sets whose union is not closed.
a set which is both open and closed.
a set which is neither open nor closed.
a set which contains a point which is not a limit point of the set.
a set which contains no point which is not a limit point of the set.
27. Describe the closure and the interior of each of the following subsets of the
real line:
a) N.
e) [0, 1].
b) Q.
f) [0, 1[∪{2, 3}.
c) R \ Q.
g) ]0, ∞[.
d) ]0, 1[.
h) ] − 1, 0[∪]0, 1[.
28. Let A and B be subsets of a metric space X. Show the following:
a)
b)
c)
d)
Int(A) ∪ Int(B) ⊆ Int(A ∪ B).
Int(A) ∩ Int(B) = Int(A ∩ B).
A ∪ B = A ∪ B.
A ∩ B ⊆ A ∩ B.
Give an example of two subsets A and B of the real line for which the inclusion
in statement a) is strict. How about statement d)?
29. Let d be a metric on X. Show that for any subsets A, B ⊆ X:
a) d(A ∪ B) ≤ d(A) + d(B) + d(A, B)
b) d(A) = d(A).
Remember that d(A) is the diameter of A (see Definition 2.2 on page 6).
30. Let C be the Cantor set (see Example 1.2 on page 4).
a) Show that C is closed.
b) Prove that the left-hand side end points of the intervals whose union is
Cn are of the form α31 + α322 + . . . + α3nn , where each αk is 0 or 2.
c) Find how many intervals C25 is the union of.
d) Calculate the sum of the lengths of all the intevals removed.
e) Show that C is uncountable.
f) Show that C is nowhere dense in [0, 1].
g) Show that C is a perfect set, in other words it is closed and every point
in C is a limit point of C.
31. Let X = {xi : i = 1, 2, 3, . . .} be a countable set. Let d(xi , xj ) = 1 +
i 6= j and d(xi , xi ) = 0.
1
i+j
for
a) Show that (X, d) is a metric space. This is called Sierpinski’s metric
space.
b) Show that (X, d) is complete.
A.3 Exercises for Chapter 3
60
32. Show that (R, | · |) is complete.
33. Show that the n-dimensional Euclidean space (Rn , d2 ) is complete. (See
Exercise 16 for the definition of d2 .)
34. Let d be a metric on X and let A be any arbitrary subset of X. Show that
the function f : X → R defined by f (x) = d(x, A) is continuous.
35. We say that f : X → Y is an open function if f (O) is open in Y for every
open set O in X. Give an example of a function that
a)
b)
c)
d)
is continuous but not open.
open but not continuous.
open and continuous.
neither open nor continuous.
36. Let X and Y be metric spaces. Show that a function f : X → Y is continuous on X if and only if f (xn ) → f (x) in Y whenever xn → x in X.
A.3 Exercises for Chapter 3
37. Let X = {a, b, c}. Show that τ = {∅, X, {a}, {a, b}, {a, c}} is a topology on X.
What are the open sets and the closed sets in this case?
38. List all possible topologies for the set X = {a, b}.
39. Let X = {♣, ♦, ♥, ♠}.
a) Say which of the following is a topology on X
T1 ={∅, X, {♣}, {♣, ♦}, {♣, ♥}, {♣, ♦, ♥}},
T2 ={∅, X, {♣}, {♦}, {♣, ♥}, {♣, ♦, ♠}},
T3 ={∅, X, {♣, ♦}, {♥, ♠}},
T4 ={∅, X}.
b) Find the open sets and the closed sets when Ti is a topology.
40. Let X be a non-empty set, and τ be a non-empty family of subsets of X.
Say which of the following is a topology.
a)
b)
c)
d)
e)
τ
τ
τ
τ
τ
is the family of all subsets of X.
is the family of all finite subsets of X.
is the family of all infinite subsets of X.
= τ1 ∪ τ2 , where τ1 and τ2 are topologies on X.
= τ1 ∩ τ2 , where τ1 and τ2 are topologies on X.
41. Show that the discrete topology T = P(X) on X is metrizable. In other
words, find a metric for which T = Td .
42. Let X = {a, b, c, d, e} be equipped with the topology
T = {X, ∅, {a}, {a, b}, {a, c, d}, {a, b, c, d}, {a, b, e}}.
Let A = {a, b, c}.
a) Find Int(A).
A.3 Exercises for Chapter 3
61
b) Find Int(A∁ ).
c) Find Fr(A).
43. Show that if A ∩ B = ∅, then Fr(A ∪ B) = Fr(A) ∪ Fr(B).
44. Let R be the set of real numbers and τ the family of all subsets E of R with
the property that for every x ∈ E there is r > 0 such that ]x − r, x + r[⊆ E.
a) Show that (R, τ ) is a topological space (τ is called the natural topology
on R).
b) Show that the family B of all open intervals with rational end-points is
a base for (R, τ ).
45. Show that Fr(Int(A)) ⊆ Fr(A). Construct an example in which equality does
not hold.
46. 1. Let B be a family of subsets of some nonempty set X. Show that B is a
base for a topology on X if and only if the following two conditions hold.
i) X = ∪ B.
B∈B
ii) Whenever B1 and B2 ∈ B with x ∈ B1 ∩ B2 , there is B3 ∈ B with x ∈
B3 ⊆ B1 ∩ B2 .
47. Consider the discrete topology T (see Example 3.1 on page 16) on X =
{a, b, c, d, e}. Find a subbase C for T which does not contain any singleton sets.
48. Let (X, d) be a metric space, and let τ be the family of all subsets E of X
with the property that for every x ∈ E, there exists r > 0 such that B(x, r) ⊆ E.
a) Show that (X, τ ) is a topological space.
b) Find a base for (X, τ ).
49. Let X = [−1, 1] and generate a topology T from the sets of the form [−1, b[,
b > 0 and ]a, 1], a < 0. Then all sets of the form ]a, b[ are open. Show that (X, T )
is a topological space. T is called the overlapping interval topology.
50. Show that if C is a subbase for a topology T on X, then C \ {X, ∅} is also a
subbase for T .
51. Let X be an infinite set, and τ be the family of all subsets of X whose
complements are finite. Show that τ is a topology on X (this is the cofinite
topology ) .
Is τ still a topology on X if “finite” is replaced by infinite or countable?
52. Let X be an infinite set. Fix x0 in X, and let
τ = {A ⊂ X : x0 ∈
/ A} ∪ {B ⊂ X : X \ B is finite }.
a) Show that (X, τ ) is a topological space.
b) Show that all one-point subsets of X, except {x0 } are open and closed.
What about {x0 }?
c) Show that B = {{x} : x 6= x0 } ∪ {X \ F : F is finite } is a base for (X, τ ).
d) Show that C = {{x} : x 6= x0 } ∪ {X \ {x} : x ∈ X} is a subbase for (X, τ ).
e) Let A be a subset of X. Show that
(
A,
if A is finite
A=
A ∪ {x0 }, if A is infinite.
A.3 Exercises for Chapter 3
f) Show that
Int(A) =
62
(
A,
if X \ A is finite
A \ {x0 },
if X \ A is infinite.
53. Prove Theorem 3.1 from page 16. Hint: use the de Morgan laws.
54. Show that Q is dense in R, and that Z is nowhere dense in R.
55. Let X = {a, b, c, d, e} be equipped with the topology
T = {X, ∅, {a}, {a, b}, {a, c, d}, {a, b, c, d}, {a, b, e}}.
a) List the closed subsets of X.
b) Determine the closure of the sets {a}, {b} and {c, e}.
c) Which sets in b) are dense in X?
56. Let A be a dense subspace of a topological space X. Show that for every
open subset E of X, we have E = E ∩ A.
57. Let X be a topological space, and let A and B be subsets of X. Show that
A∩B ⊆ A ∩ B, and with an example show that the inclusion cannot be replaced
by an equality.
58. Let (X, τ ) be a topological space and Y ⊂ X. Show that τ ′ = {U ∩Y : U ∈ τ }
is a topology on Y . Recall that this is called the relative topology for Y , and Y
is a subspace of X.
59. Prove Theorem 3.4 on page 18 using Theorems 3.2 and 3.3.
60. Let X be any set, and let A be an arbitrary subset of X. Which of the
following is a topology on X:
T1 = {∅, X, A, X \ A},
T2 consists of ∅ and every superset of A,
T3 consists of ∅ and every set which does not meet A.
61. Let A1 , A2 , . . . be a sequence of subsets of a topological space X.
a) Show that
∞
∞
i=1
i=1
∪ Ai = ( ∪ Ai )
[
∞ ∞
( ∩ ∪ Ai+j ).
i=1j=0
b) Show by an example that the equality above is not correct when the
second term is omitted.
62. Let X, Y and Z be topological spaces, f : X → Y and g : Y → Z be
continuous mappings. Show that g ◦ f : X → Z is also continuous.
63. Let X have the discrete topology and let Y be an arbitrary topological
space. Show that every function f : X → Y is continuous.
64. Let X = {a, b, c}.
a) Say which of the following is a topology on X:
T1 = {∅, X, {a}, {a, b}, {a, c}},
T2 = {∅, X, {a}, {b}, {a, b}, {b, c}},
T3 = {∅, X, {a}, {b}, {a, b}},
T4 = {∅, X, {a}, {b}, {a, c}}.
b) Let X = {a, b, c}. Define f : X → X by f (a) = a, f (b) = c, f (c) = b. When
Ti is a topology on X, find out whether f is continuous with respect to
Ti .
A.3 Exercises for Chapter 3
63
c) What about if f (a) = f (b) = c and f (c) = a?
65. Let f : A → B and g : C → D be continuous functions. Show that f × g :
A × C → B × D with (a, c) 7→ (f (a), g(c)) is also continuous.
66. Let f : A → A1 and g : A → A2 be continuous functions. Show that (f, g) :
A → A1 × A2 with a 7→ (f (a), g(a)) is also continuous.
67. Let f : X → Y be a mapping from one topological space into another. Show
that the following statements are equivalent:
a)
b)
c)
d)
f is a homeomorphism.
f (E) is open in Y if and only if E is closed in X.
f (F ) is closed in Y if and only if F is closed in X.
f (ClX (A)) = ClY (f (A)) for every subset A of X.
68. Let X have the discrete topology and let Y be an arbitrary topological
space. Show that every function f : X → Y is continuous.
69. Let f : X → Y be mapping from one topological space into another. Show
that the following statements are equivalent:
a) f is continuous,
b) f −1 (F ) is closed in X whenever F is closed in Y ,
c) f (A) ⊆ f (A) for every subset A of X.
70. Show that f : X → Y is continuous if and only if f −1 (Int(A)) ⊆ Int(f −1 (A))
for every A ⊆ X.
71. Let (X, T ) be a topological space and Y ⊆ X.
a) Show that a subspace Z of a subspace Y of X is a subspace of X.
b) Show that FrY (A) ⊆ FrX (A)∩Y and IntX (A)∩Y ⊆ IntY (A) for any A ⊆ Y .
72. The slotted plane is defined as follows. At each point z in the plane R2 , the
basic neighbourhoods are the sets {z} ∪ A, where A is a disk about z with a
finite number of lines through z removed.
a) Show that the slotted plane is a topological space.
b) What is the topology of a straight line as a subspace of the slotted plane?
73. Let T be the usual topology on R. Define TS = {U ∪V : U ∈ T and V ⊆ R\Q}.
The space S = (R, TS ) is called the scattered line .
a)
b)
c)
d)
Show that the scattered line is a topological space.
Describe the base of S.
Discuss the subbases of S.
Is the scattered line metrizable?
74. Let Γ be the upper closed half plane. For each point in the open half plane,
basic neighbourhoods will be the usual open disks contained in Γ. At the points
z on the x-axis, the basic neighbourhoods will be the sets {z} ∪ A, where A is an
open disk in Γ tangent to the the x-axis at z. The space Γ is called the Moore
plane.
a) Show that the Moore plane is a topological space.
b) Show that Γ is separable.
c) Show that the x-axis, as a subspace of Γ is not separable.
75. Let X be a topological space and Y be a subspace of X. Let f be a continuous function on X. Show that f|Y is a continuous function on Y .
A.4 Exercises for Chapter 4
64
A.4 Exercises for Chapter 4
76. Consider the following problems on sums of spaces.
a) Give examples of topological spaces which can be seen as sums of other
topological spaces.
b) Show that R with its usual topology cannot be seen as the sum of any
two topological spaces.
Q
77. Let {Xα : α ∈ I} be a family of topological spaces, and let Xα be their
Q
Cartesian product. Show that τ = { Uα : Uα open in Xα } is a topology on
Q
Xα . It is called the box topology.
78. Verify that the Tychonoff topology (see Definition 4.4 on page 26) is indeed a
Q
Q
topology on Xα , and that C = { −1 (Uα ) : α ∈ I, Uα open in Xα } is a subbase
α
for this topology.
79. Use the Euclidean plane to show that a projection πβ :
not send closed sets to closed sets.
Q
Xα → Xβ need
80. Let {Xα : α ∈ I} be a family of topological spaces and let X =
Q
Xα be
α∈I
their Cartesian product. Let B be the family of subsets B of X which are of the
Q
form
Uα , where each Uα is open in Xα , and Uα = Xα for all but finitely many
α∈I
α ∈ I.
a) Show that B is a base for the Tychonoff topology on X =
Q
Xα . 26.
α∈I
b) Let Aα ⊆ Xα for each α ∈ I. Show that
Q
α∈I
Aα =
Q
Aα .
α∈I
c) Suppose that each factor space Xα , α ∈ I is discrete and with more than
two points. Show that X is discrete if and only if I is finite.
Q
81. Let {Xα : α ∈ I} be a family of topological spaces and let X =
Xα be
α∈I
their Cartesian product. Let B be the family of subsets B of X which are of the
Q
form
Uα , where each Uα is open in Xα .
α∈I
a) Show that B is a base for the box topology on X =
Q
Xα .
α∈I
b) What do neighbourhoods of f ∈ RR look like in the box topology?
A.5 Exercises for Chapter 5
82. Let X be a first countable space. Show that the following statements are
true.
a) Let E ⊆ X. Then x ∈ E if and only if there exists a sequence (xn )∞
n=1 in
E such that xn → x.
b) E is open in X if and only if for every x in E and every sequence (xn )∞
n=1
in E, with xn → x, there exists n0 ∈ N such that xn ∈ E whenever n ≥ n0 .
c) F is closed in X if and only if whenever (xn )∞
n=1 ⊆ F and xn → x then
x ∈ F.
d) If Y is also a first countable space, then f : X → Y is continuous if and
only if f (xn ) → f (x) in Y whenever xn → x in X.
A.5 Exercises for Chapter 5
65
83. Show that the family B = {[a, r[ , a ∈ R and r ∈ Q} is a base for a topology
on R. Then R equipped with this topology is called the Sorgenfrey line.
84. Let X1 be an uncountable set with the cofinite topology.
Let X2 be an uncountable set with the cocountable topology.
Let X3 = R in which the open sets are of the form ]a, ∞[, a ∈ R.
Let X4 be the Sorgenfrey line.
Let X5 be a discrete space.
Let X6 be the trivial space.
Let X7 be the scattered line (see Exercise 73).
For all spaces above answer the following questions:
a) Which sequences converge to which points?
b) Is Xn first countable?
85. Let X be a topological space with overlapping interval topology (see Exercise 49).
a) Show that X is first countable.
b) What is the limit point of the sequence 0, 12 , 0, 21 , 0, 21 , . . .?
c) A given point p in a topological space X is an accumulation point of
a given sequence (xn ) if every open set containing p contains infinitely
many terms of the sequence. What are the accumulation points of the
sequence in b)?
86. Let X = (N × N) ∪ {(0, 0)}. Let N × N carry the discrete topology. The
neighbourhoods of (0, 0) consist of (0, 0) and all but a finite number of points
in all but a finite number of columns Nk = {k} × N, for example, (X \ N1 ) is a
neighbourhood of (0, 0). Denote by Nk = N × {k} (the rows). Then N1 × N1 is also
a neighbourhood of x.
Show that the above system is a topology for X = (N × N) ∪ {(0, 0)}.
87. Let X be a nonempty set, and let F ⊆ P(X) such that
F1. A 6= ∅ for all A ∈ F,
F2. A ∈ F and B ∈ F ⇒ A ∩ B ∈ F,
F3. A ∈ F and A ⊆ B ⇒ B ∈ F.
Such an F is called a filter in X.
Show that (F, ⊆) and (F, ⊇) are directed sets.
88. Let E = {f ∈ RR : f (x) = 0 or 1 and f (x) = 0 only finitely often}, and let g
be the trivial function in RR . Then as seen in the lectures, g ∈ E. Find a net in
E which converges to g.
89. A net (xλ ) in a set X is an ultranet if for each subset E of X, (xλ ) is either
eventually in E or X \ E.
a) Show that if (xλ ) is an ultranet in X and f : X → Y , then f (xλ ) is an
ultranet in Y .
b) Show that every subnet of an ultranet is an ultranet.
c) Show that if an ultranet has x as a cluster point, then it converges to x.
A.6 Exercises for Chapter 6
66
90. Exhibit a sequence (xn ) on a set X and a subnet of (xn ) which is not a
sequence.
91. If (xλ ) is a net and Tλ0 = {xλ : λ > λ0 }, show that x is a cluster point of
(xλ ) if and only if x ∈ Tλ0 for every λ0 ∈ Λ.
Q
92. Show that if (xλ ) is a net in Xα with a cluster point x, then (πα (xλ ))λ∈Λ
has πα (x) as a cluster point for each α. With an example show that the converse
is false.
A.6 Exercises for Chapter 6
93. Let X be a topological space with overlapping interval topology (see Exercise 49). Show that X is T0 , but not T1 .
94. Show, by a counterexample, that the image of a T1 -space under a continuous map need not be T1 . Then proceed to show that if we instead have a closed
and continuous map, the image of a T1 space will be T1 .
95. Consider the following problems on separation.
a) Show that a subspace of a T0 -space is a T0 -space.
b) Show that a subspace of a T1 -space is a T1 -space.
96. Let X and Y be topological spaces. Let i ∈ {0, 1, 2, 3, 3 12 }. Show that X × Y
is a Ti -space, whenever X and Y are Ti -spaces.
97.
a) Show that the product of T0 -spaces is a T0 -space.
b) Show that the product of T1 -spaces is a T1 -space.
Note. In exercices 98–101, X and Y are topological spaces and f, g : X → Y .
98. If f is continous and Y is Hausdorff, then
{(x1 , x2 ) ∈ X × X : f (x1 ) = f (x2 )}
is closed in X × X.
99. If f is open and onto and the set
{(x1 , x2 ) ∈ X × X : f (x1 ) = f (x2 )}
is closed in X × X, then Y is a Hausdorff space.
100. If f and g are continuous and Y is Hausdorff, then the set
{x ∈ X : f (x) = g(x)}
is closed in X.
101. If f and g are continuous and Y is Hausdorff and f (x) = g(x) on a dense
subset of X, then f = g.
102. Show that every metric space is Hausdorff.
103. Find out whether the topological spaces you have seen so far in this course
are Hausdorff. (See for example Exercises 72–74 and 84.)
104. Give an example of a regular space that is not T1 .
A.7 Exercises for Chapter 7
67
105. Let X be nonempty and A ⊆ X. Show that
T = {B ∈ P(X) : A ⊆ B}
is a topology for X. Under what conditions is T regular or completely regular?
106. Show that the scattered line (see Exercise 73) is regular. What about completely regularity?
107. Let (X, T ) be a topological space and A ⊆ X. Let
TA = {U ∪ (V ∩ A) : U, V ∈ T }.
a) Show that TA is also a topology for X.
b) Show that if T is regular or completely regular, and A is closed, the TA
has the same property.
A.7 Exercises for Chapter 7
108. It has been seen on the lectures that T4 ⇒ T3 21 ⇒ T3 ⇒ T2 ⇒ T1 ⇒
T0 . Show by counterexamples that none of the reverse implications is true in
general.
109. Let R be equipped with the topology T = {]a, ∞[ : a ∈ R}, and call such a
space X.
a) Show that X is normal.
b) Show that X is not regular.
110. Give an example of a space that is normal, but not a T0 -space.
111. Let X be a topological space with overlapping interval topology (see Exercise 49). Find out if X is normal.
112. Let X be a normal space, and let A and B be disjoint closed subsets of X.
Let a and b be two real numbers such that a < b.
a) Show that there exists a continuous function f : X → [a, b] such that
f (A) = a and f (B) = b.
b) Let U be a neighbourhood of A, show that there exists a continuous function f : X → [0, 1] such that f (A) = 1 and f (X \ U) = 0.
113. Let X be a normal space, A a closed subset of X, and f : A → R a continuous function. Show that f has a continuous extension to all of X.
114. Show that the Moore plane Γ (see Excercise 74) is a Tychonoff space (see
Definition 6.1 on page 35.)
115. Show that Γ is not normal.
116. Show that:
a) The continuous image of a Lindelöf space is Lindelöf (see Definition 8.5
on page 46).
b) A closed subspace of a Lindelöf space is a Lindelöf.
c) An arbitary subspace of a Lindelöf space do need to be Lindelöf.
d) A regular Lindelöf space is normal.
A.8 Exercises for Chapter 8
68
117. Let E be the Sorgenfrey line (see Example 2 on page 25.)
a) Show that E is Lindelöf and regular, and so E is a T4 space.
b) Show that E × E is not normal.
c) Show that E × E is not Lindelöf.
118. Show that a topological space X is normal if and only if whenever A is a
closed subset of X and f : A → R is continuous, there is a continuous extension
of f to all X.
A.8 Exercises for Chapter 8
119. Find out whether the following topological spaces are Hausdorff or compact.
a)
b)
c)
d)
The Cantor set with the subspace topology from R (see Example 1.2).
] − 1, 1[ with the subspace topology from R.
An infinite set with the cofinite topology.
The set m of all bounded sequences (xn )n∈N of real numbers with the
metric
d((xn ) − (yn )) = sup{|xn − yn | : n ∈ N}.
e) A discrete space.
120. Which subsets of the following spaces are compact?
a)
b)
c)
d)
Rn .
The Sorgenfrey line.
The slotted plane (see Exercise 72).
The Moore plane (see Exercise 74).
121. Show that a finite union of compact subsets of X is compact.
122. Let X be a topological space with indescrete topology (see Example 3.1
on page 16), and have at least two points. Show that every subset A of X is
compact.
123. Let X be a topological space with overlapping interval topology (see Exercise 49). Show that X is compact.
124. Show that any intersection of compact subsets of a Hausdorff space is
compact.
125. Which of the following spaces are locally compact?
a)
b)
c)
d)
e)
f)
R.
Q.
The Moore plane.
The Sorgenfrey line.
The slotted plane.
The Hilbert space ℓ2 (recall that ℓ2 is the metric space which is made
∞
P
x2n < ∞ and
with all the real sequences x = (xn )∞
n=1 which satisfy
n=1
the distance
v
u∞
uX
d(x, y) = t
(xn − yn )2 .
n=1
A.9 Exercises for Chapter 9
69
126. Let X be a locally compact Hausdorff space. Let ∞ be an object not in X.
Form the set X∞ = X ∪ {∞}, and consider the family T∞ which consists of the
following subsets of X∞ :
(1) The open subsets of X.
(2) The complements in X∞ of all compact subsets of X.
(3) The whole set X∞ .
a) Show that T∞ is a topology on X∞ , and that the given topology on X
equals its relative topology as a subspace of X∞ .
b) Show that, with this topology, X∞ is a compact Hausdorff space. The
compact Hausdorff space X∞ is called the one-point compactification of
X.
c) Deduce that X is a completely regular space.
d) If K is a compact subspace of X and U a neighborhood of K in X, show
that there exists a continuous function f : X → [0, 1] such that f (K) = 1
and f (U ′ ) = 0.
127. Show that a metric space X is compact if and only if every sequence in X
has a convergent subsequence.
128. Show that a metric space is compact if and only if it is complete and totally
bounded.
129. A closed subspace of a complete metric space is compact if and only if it is
totally bounded.
130. Give an example of a T0 -space and a T1 -space in which the compact
subsets are not necessarily closed. This also shows why Hausdorff-property
is needed in Theorem 8.3. Hint: Use R as the set in both spaces.
A.9 Exercises for Chapter 9
131. Show that a subspace of R is connected if and only if it is an interval.
132. Which of the following spaces are connected:
a)
b)
c)
d)
e)
f)
g)
R,
Q,
The Sorgenfrey line,
The slotted plane,
An infinite set with the cofinite topology,
A countable subset of R,
n
o [n
o
1
V = (x, 0) : x ≤ 0
(x, sin ) : x > 0
x
with the relative topology in R2 (this is called the topologist’s sine curve ).
133. Show that the closure of a connected subset of a topological space X is
connected.
134. Show that the component of a point is closed.
135. Show that the continuous image of a connected space is connected.
A.9 Exercises for Chapter 9
70
136. (Knaster and Kuratowski, 1921) Let C be the Cantor set. Let E be the set
of endpoints of the intervals deleted from [0, 1] to obtain C, and let F = C \ E.
Let p be the point ( 12 , 21 ) in the plane and for each c ∈ C let L(c) be the segment
joining p to c. Define
KE = ∪ {(x, y) ∈ L(c) : y ∈ Q} and
c∈E
KF = ∪ {(x, y) ∈ L(c) : y ∈ R \ Q},
c∈F
and let K = KE ∪ KF .
a) Show that K is connected.
b) Show that K \ {p} is totally disconnected.
The set K is known as Cantor’s Teepee and K \ {p} as Cantor’s Leaky Tent.