Introduction to Topology - Ryan Lok-Wing Pang

Introduction to Topology

Ryan Lok Wing [email protected]

May 26, 2014

Contents

1 Preliminaries 5

1.1 Basic Set Theory . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 Useful Inequalities . . . . . . . . . . . . . . . . . . . . . . . . 6

2 Metric Spaces 7

2.1 Definition and Examples of Metric Spaces . . . . . . . . . . . 7

2.2 Topological Properties in Metric Spaces . . . . . . . . . . . . . 8

2.2.1 Open Subset . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2.2 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.2.3 Limit Point . . . . . . . . . . . . . . . . . . . . . . . . 12

2.2.4 Closed Subset . . . . . . . . . . . . . . . . . . . . . . . 13

2.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3 Topological Spaces 17

3.1 Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.2 Basis of a Topology . . . . . . . . . . . . . . . . . . . . . . . . 18

3.3 Comparing Topologies . . . . . . . . . . . . . . . . . . . . . . 18

3.4 Topology as a Language . . . . . . . . . . . . . . . . . . . . . 19

3.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4 Continuity and Homeomorphism 23

4.1 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . 23

4.2 Homeomorphism . . . . . . . . . . . . . . . . . . . . . . . . . 24

4.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

5 Constructing Topologies 27

5.1 Subspace Topology . . . . . . . . . . . . . . . . . . . . . . . . 27

5.2 Product Topology . . . . . . . . . . . . . . . . . . . . . . . . . 29

5.3 Quotient Topology . . . . . . . . . . . . . . . . . . . . . . . . 30

5.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3

4 CONTENTS

6 Hausdorff Spaces 336.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

7 Connected Spaces 357.1 Path Connectedness . . . . . . . . . . . . . . . . . . . . . . . 387.2 Connected Components . . . . . . . . . . . . . . . . . . . . . 407.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

8 Compact Spaces 438.1 Compact Hausdorff Spaces . . . . . . . . . . . . . . . . . . . . 458.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

9 Metric Space Revisited 499.1 Banach Fixed Point Theorem . . . . . . . . . . . . . . . . . . 499.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

10 Separation Axioms 53

11 Graph Theory 5511.1 Seven Bridges of Konigsberg . . . . . . . . . . . . . . . . . . . 55

Chapter 1

Preliminaries

1.1 Basic Set Theory

Theorem 1.1 (De Morgans Law).(

S

)c=

Sc

(

S

)c=

Sc

Theorem 1.2.

U f1(f(U)), f(f1(X)) X

Theorem 1.3.

f

(

S

)

f(S), f1(

S

)=

f1(S)

f

(

S

)=

f(S), f1(

S

)=

f1(S)

Theorem 1.4. If X A, then f(A) f(X) f(AX)

If Y B, then f1(B Y ) = f1(B) f1(Y ).

5

6 CHAPTER 1. PRELIMINARIES

1.2 Useful Inequalities

Theorem 1.5 (Youngs Inequality). For p > 1, let 1p

+ 1q

= 1. If , 0,then

p

p+q

q.

Theorem 1.6 (Holders Inequality). For p > 1, let 1p+1q

= 1. If 1, , n, 1, , n C, then

ni=1

|ii| (

ni=1

|i|p)1/p( n

i=1

|i|q)1/q

Theorem 1.7 (Minkowskis Inequality). For p 1, if x1, , xn, y1, , yn C, then (

ni=1

|xi + yi|p)1/p

(

ni=1

|xi|p)1/p

+

(ni=1

|yi|p)1/p

Similarly, we have the integral versions of these inequalities.

Theorem 1.8 (Holders Inequality). For p > 1, let 1p

+ 1q

= 1. If |f |p and|g|p are Lebesgue integrable on a Lebesgue measurable set X, then

X

|fg| (

X

|f |p)1/p(

X

|g|q)1/q

Theorem 1.9 (Minkowskis Inequality). For p 1, If |f |p and |g|p areLebesgue integrable on a Lebesgue measurable set X, then(

X

|f + g|p)1/p

(

X

|f |p)1/p

+

(X

|g|p)1/p

Chapter 2

Metric Spaces

2.1 Definition and Examples of Metric Spaces

Defintion 2.1 (Metric Space). A Metric space is a nonempty set X with afunction d : X X R such that for every x, y X,

(1) d(x, y) 0 and equality holds if and only if x = y. (Positivity)

(2) d(x, y) = d(y, x). (Symmetry)

(3) d(x, z) d(x, y) + d(y, z). (Triangle Inequality)

Such a function d on X is said to be a metric on X.

Example 2.2. The following are some commonly used metrics on Rn:

d2(x, y) =

(ni=1

(xi yi)2)1/2

(Euclidean metric)

d1(x, y) =ni=1

|xi yi| (Taxicab metric)

d(x, y) = max{|x1 y1|, , |xn yn|} (L-metric)More generally, for p 1, we have

dp(x, y) =

(ni=1

|xi yi|p)1/p

(Lp-metric)

Proof. Exercises.

7

8 CHAPTER 2. METRIC SPACES

Example 2.3. On the set C[0, 1] of all continuous functions on [0, 1], wehave the following metrics similar to the Euclidean space:

d1(f, g) =

10

|f(t)g(t)|dt (L1-metric, Geometrically, it is the area between f and g.)

d(f, g) = max0t1

{|f(t) g(t)|} (L-metric)

dp(f, g) =

( 10

|f(t) g(t)|pdt)1/p

(Lp-metric)

Example 2.4. On any set X, the function

d(x, y) =

{0 if x = y1 if x 6= y.

is a metric, called the discrete metric.

2.2 Topological Properties in Metric Spaces

Defintion 2.5. Let (X, d) be a metric space. All elements and sets men-tioned below are understood to be elements and subsets of X.

1. The open ball of x0 is a set Bd(x0, r) = {x X | d(x0, x) < }. Thenumber is called the radius of Bd(x0, ).

2. x0 is an interior point of A X iff there exists r > 0 such thatBd(x0, ) A. (The set of all interior point of A is denoted A) (Remark:interior point of A are in A.)

3. A is open iff A = A. i.e. for any a A, we have B(a, ) A forsome > 0.

4. x X is a limit point of the set A iff for all > 0, there exists a Asuch that a (Bd(x, ) \ {x}) A. i.e. For any > 0, there exists a point asuch that

(1) a A, (2) a 6= x, (3) d(a, x) < .

(The set of all limit point of A is denoted A.) (Remark: limit point of Amay or may not be in A.)

2.2. TOPOLOGICAL PROPERTIES IN METRIC SPACES 9

5. The closure of A is the set A = A A.

6. A is closed iff A = A, or equivalently A A.

7. A is dense in X iff A = X.

Theorem 2.6. A is closed in X iff X A is open in X. In particular, and X are both open and closed.

Lemma 2.7 (Ball-in-Ball lemma). Let x Bd(x0, r0). Then Bd(x, r) Bd(x0, r0) Bd(x,R) for 0 < r r0 d(x, x0) and R r0 + d(x, x0).Proof. Exercise.

2.2.1 Open Subset

Theorem 2.8 (Structure Theorem for Open Sets). A is open in X iff

A =

Bd(x, r)

for some x X and r > 0.Proof. If A =

Bd(x, r), then every x S is in some Bd(x, r). By

the ball-in-ball lemma, Bd(x, r) B(x, r) A, where r = r d(x, x).Hence A is open in X.

Conversely, if A is open in X, then every x A is an interior point of A.i.e. there exists rx > 0 such that Bd(x, rx) A. Hence

A =xA{x}

xA

Bd(x, rx) A.

Therefore, A =Bd(x, r).

The following result is the most important properties of open sets. Theproperties will become the axioms for the concept of topology.

Theorem 2.9. The open subsets of a metrix space X satisfy the following:

(1) and X are open.

(2) Arbitrary Unions of open subsets are open.

(3) Finite intersections of open subsets are open.


To make use of the above theorem, it is helpful to bear in mind thatthere exists goes with unions and for all goes with intersections. Forexample, the set of all real numbers such that there exists a positive integern with |x n| < 10 is the union over all n of the set of x with |x n| < 10.Perhaps writing this symbolically makes it clearer:

{x : n N such that |x n| < 10} =nN{x : |x n| < 10}

This often makes it possible to show that a set is open by showing that it isa union of sets that are more obviously open.

Example 2.10. The subset

U = {f C[0, 1] : f(t) > 0 0 t 1}

is L-open but not L1-open.

Proof. For any positive function f , we wish to find some > 0 such that allthe function in BL(f, ) are still positive. Now, the continuous function fhas positive lower bound > 0 on the closed and bounded interval [0, 1] (byExtreme Value Theorem). Suppose g BL(f, ), then

|g(t) f(t)| max0t1|g(t) f(t)| < . 0 t 1.

This implies g(t) > f(t) 0. Hence g U.

Now define the function f by

f(t) =

{t/ if 0 t < ,1 if t 1.

Then we have g(t) = 1 U , dL1(f, g) < and f(0) = 0, so that f / U. HenceU is not L1-open.

Example 2.11. Suppose d and dare two metrics on X satisfying d(x, y) cd(x, y) for some c R+. Prove that d-open subsets are d-open.

Proof. Suppose U is d-open, i.e. for all x U , there exists > 0 such thatd(x, y) < y U . Let = /c, then d(x, y) < implies d(x, y) cd(x, y) < c/c = . Hence y U. i.e. U is d-open.


2.2.2 Continuity

Defintion 2.12 (Continuity). A map f : X Y between metric spaces iscontinuous if for any a X and > 0, there exists > 0 such that for allx X,

dX(x, a) < dY (f(x), f(a)) < .Example 2.13. Let X be a discrete metric space and let Y be any metricspace. Then any map f : X Y is continuous. To see this, for anygiven a and , take = 1, then d(x, a) < = 1 implies x = a, henced(f(x), f(a)) = 0 < .

Example 2.14. The integration map

I : C[0, 1]L RL1defined by

I(f(t)) =

10

f(t)dt

is continuous. By taking = , we have

d(f, g) < |I(f) I(g)| 10

|f(t) g(t)|dt < = .

Example 2.15. The evaluation map

E(f) : C[0, 1]L RL1 ,f(t) 7 f(0)

is continuous.

Proof. We have

d1(E(f), E(g)) = |f(0) g(0)| max0t1|f(t) g(t)| = d(f, g).Hence by choosing = , we are done.

Theorem 2.16 (Topological Continuity Theorem). Suppose f : X Y isa map between metric spaces, then TFAE:

(1) f is continuous.

(2) The preimage f1(U) of any open subset U Y is open in X.

(3) The preimage f1(V ) of any closed subset V Y is closed in X


Example 2.17. Example 2.15 shows that the evaluation map E(f) is con-tinuous Therefore the subset

E1((1,)) = {f C[0, 1] : f(0) > 1}is L-open.

Example 2.18. Let X be the set of all real numbers such that there existsa rational number p/q such that |e p/q| < 1/2q. Prove that X is open.Proof. Express X as

X =

p/qQ{ R : e (p/q 1/2q, p/q + 1/2q)} =

p/qQ

f1(Ip/q)

where f(x) = ex which is continuous and Ip/q = (p/q1/2q, p/q+1/2q). Nowsince Ip/q is open, so is f

1(Ip/q), therefore so isp/qQ f

1(Ip/q), done.

Example 2.19. Prove that GLn(R)is open (in any metric).

Proof. Since det : Mn(R) R is continuous and R \ {0} is open, henceGLn(R) = det1(R \ {0}) is open.

2.2.3 Limit Point

Here are some basic facts about limit points:

Theorem 2.20.(1) A B A B(2) (A B) = A B(3) (A B) A B

(4) A A

Theorem 2.21. If x is a limit point of a set A, then every open ball of xcontains infinitely many elements of A.

Proof. We proceed by contradiction. Suppose there is open ball B of xwhich contains only a finite number of elements of A. Let a1, ..., an be thoseelements of B A, which are distinct from x, and put

= min {d(x, a1), ..., d(x, an)}The minimum of a finite set of positive numbers is clearly positive, so that > 0. Then the neighborhood Bd(x, )) contains no elements a of A suchthat a 6= x, a contradiction, so we are done.


Corollary 2.22. A finite set has no limit points.

Corollary 2.23. A finite set A is closed. In particular, a singleton set {x}is closed

Proof. A has no limit points by corollary 2.22, hence A = A, done.Example 2.24. By taking = 1, it is easy to see that any subset of a discretemetric space has no limit point.

Example 2.25. Is the zero function 0 a limit point of the subset A = {f :f(1) = 1} C[0, 1] w.r.t. L-metric, L1-metric?Proof. For the L-metric, the problem is the existence of a function f sat-isfying f(1) = 1, f 6= 0, and |f(t)| < 0 t 1. By choosing = 1, we seethat such a function do not exist.

For the L1-metric, the problem is the existence of a function f satisfyingf(1) = 1, f 6= 0, and 1

0|f(t)|dt < . Now for all > 0, there is a N > 1/

such that f(t) = tN satisfies the three conditions. Therefore 0 is an L1-limitpoint of A.

Theorem 2.26. Let A be a subset of a metric space X, prove that x Aiff there are distinct a1, a2, ... A such that limn an = xProof. If x A, then there is a1 (Bd(x, 1) {x}) A. Define a sequence{an} inductively satisfying

an (Bd(x, n) {x}) A, n = min{d(x, an1), 1n}.

for n 2, then clearly a1, a2, are distinct with limn an = x. Theconverse is obvious.

2.2.4 Closed Subset

Theorem 2.27. A finite set A is closed.

Proof. A has no limit points by corollary 2.22, hence A = A, done.Theorem 2.28. A subset A X is closed iff its complement XA is open.Theorem 2.29. The closed subsets of a metric space X satisfy the following:

(1) and X are closed.

(2) Arbitrary intersections of closed sets are closed.

(3) Finite union of closed sets are closed.


Theorem 2.30 (Topological Continuity Theorem). Suppose f : X Y isa map between metric spaces, then TFAE.

(1) f is continuous.

(2) The preimage f1(U) of any open subset U Y is open in X.

(3) The preimage f1(V ) of any closed subset V Y is closed in XExample 2.31. The solution set of a pair of hyperbola H = {(x, y) R2 :x2 y2 = 1} is closed in the Euclidean metric.Proof. Define f : R2 R by f(x, y) = x2 y2 1. Since f is continuousand {0} is closed, the preimage f1({0}) = H is closed.

In fact, we can generalize it to the following

Example 2.32. Let f : R R be a continuous function, prove that thegraph of f , {(x, y) : y = f(x)} is closed.Proof. The map F (x, y) = y f(x) is continuous, and {0} is closed, hence{(x, y) R2 : y = f(x)} = {(x, y) : y f(x) = 0} = F1({0}) is closed.

2.3 Problems

Problem 2.1. Suppose d(1, 2) = 2 and d(1, 3) = 3. What number can d(2, 3)be so in order for X = {1, 2, 3} to be a metric space?Problem 2.2. Prove that any subset in a finite metric space is open.

Problem 2.3. Prove lemma 2.7.

Problem 2.4. Prove that the map

M : C[0, 1]L RL1f(t) 7 max

t[0,1]f(t)

is continuous.

Problem 2.5. Prove that the following subsets are closed under the Eu-clidean metric.

2.3. PROBLEMS 15

(a) S = {(a, b, c) R3 : 1 x3 y3 + 2z3 xy 2yz + 3xz 3}.

(b) S = {(x, y, z, w) R4 : x4 + y4 = z4 + w4, x3 + z3 = y3 + w3}.

(c) All n n orthogonal matrices.

(d) All n n matrices with all eigenvalues equal to 1.

Chapter 3

Topological Spaces

3.1 Topology

Defintion 3.1 (Topology). A topology on a set X is a collection T of subsetsof X such that

1. , X T .2. If A T , then A T .3. If U, V T , then U V T .

A set X with a topology T is called a topological space. An element of Tis called an open set.

Example 3.2. Let X be any set, then the collection of all subsets of X isa topology on X, called the discrete topology. The collection consisting of Xand only is also a topology on X, called the trivial topology.Example 3.3. Let X be a set and T = {X F : F X, |F | < } {}. Then T is a topology on X, called the finite complement topology, or thecofinite topology.

Example 3.4. Prove that A is open iff for any x A, there is an opensubset U , such that x U A.Proof. Suppose A is open, then for any x A, we may choose U = A.Conversely, suppose for each x A, there exists an open subset Ux such thatx Ux A, then we have

A =xA{x}

xA

Ux A,

This implies A =xA Ux, a union of open subsets, hence open.

17

18 CHAPTER 3. TOPOLOGICAL SPACES

3.2 Basis of a Topology

Usually it is very hard to describe the entire collection T , in most cases, onespecifies a smaller collection of subsets of X instead and defines the topologyin terms of that.

Defintion 3.5 (Basis). Let X be a set. A basis of a topology on X is a col-lection B of subsets in X such that any of the following conditions is satisfied.

1. (a) X =BB B. i.e. x X x B for some B B, and

(b) If B1 and B2 in B overlap at x, then there is a B3 B such thatx B3 B1 B2.

2. (a) X =BB B. i.e. x X x B for some B B, and

(b) B1, B2 B B1 B2 B or B1 B2 = .

3. For each open set U of X and each x U , there is an element B Bsuch that x B U.

4. T = BB B.Example 3.6. For any topological space X, prove that B = {X F : F X, |F |

3.4. TOPOLOGY AS A LANGUAGE 19

In other words, T is coarser than T if T have more open sets than thatof T . Note that two topologies on X need not be comparable.Theorem 3.11. Let B,B be bases for T , T on X. Then TFAE:

1. T is finer than T .

2. For each x X and each base element B B containing x, there is abase element B B such that x B B.Example 3.12. The discrete topology is the finest topology and the trivialtopology is the coarsest topology.

Example 3.13. The lower limit topology Rl is strictly finer than the standardtopology R.

Proof. Suppose x (a, b), then x [x, b) (a, b). On the other hand, givenx [x, d), there is no open interval (a, b) satisfying x (a, b) [x, d).Example 3.14. By Example 2.11. If two metrics satisfy d(x, y) cd(x, y)for some c R+, then d induces coarser topology than d. In particular,if two metrics are equivalent: i.e. there exists c1, c2 > 0 such that for allx, y X

c1d(x, y) d(x, y) c2d(x, y),then the two metrics induce the same topology.

Example 3.15. The topology induced by L1 on C[0, 1] is coarser than thetopology induced by L. i.e. L1-open subset are L-open.

Proof. Follows immediately from the inequality d1 d.Remark. Note that the converse is not true.

3.4 Topology as a Language

Defintion 3.16. Closed Set A is closed if X A is open.Theorem 3.17. 1. X, are closed.

2. Arbitrary intersections of closed sets are closed.

3. Finite unions of closed sets are closed.


Proof. Exercises.

Defintion 3.18 (Closure). The closure A of a set A X is the smallestclosed set containing A. This exists because the intersection of closed sets isclosed. i.e.

A =CA

C,

where the intersection is over all closed sets C.

Defintion 3.19 (Interior). The interior of A, denoted int(A), is the largestopen set contained in A. This exists because the union of open sets is open.i.e.

int(A) =UA

U,

where the union is over all open sets U .

Defintion 3.20 (Boundary). The boundary of A is their difference: A =A intA.

Defintion 3.21 (Density). We say A X is dense if A = X.

An essential notion in calculus is the idea of limit, but it is usually ex-pressed in terms of sequences. How can it be expressed just using open sets?

Defintion 3.22 (Limit Points). Let A be a subset of a topological space X.A point x X is a limit point of A if for any open neighborhood U of x,there exists a such that

1. a A.

2. a 6= x.

3. a U.The three conditions together measns

(A {x}) U 6= .

The set of limit points of A is denoted A.

Example 3.23. Find the limit points of (0, 1) Rl.

3.5. PROBLEMS 21

Proof. For any x Rl, the open neighborhood of x is of the form U = [a, b).

Case 1: 0 x < 1, then ((0, 1) {x}) U = (x,min{1, b}) 6= .

Case 2 : x < 0, then x U = [x, 0) and U contains no point in (0, 1).

Case 3: x 1, then x U = [x, x+ 1) and U contains no point in (0, 1).Hence (0, 1) = [0, 1).

Example 3.24. Prove that if x is a limit point of A, then x is also a limitpoint of A {x}.Proof. For any open neighborhood U of x, we have (A {x}) U 6= byassumption. Hence ((A {x}) {x}) U 6= .Theorem 3.25. Let A be the set of limit points of A, then A = A A.Proof.

Corollary 3.26. A topological space X is closed iff X = X.

Example 3.27 (HKUST Topology Final 13). Let X be a topological space

and A be any subset. Suppose U is an open subset, prove that A U = A U .

Proof. A U A U , hence A U A U. On the other hand, for anyx inA U , x A. If x A, then x A U , which implies x A U. Ifx A A, then x A A. For any open neighborhood V of x, we haveV A 6= . Since UcapV is also an open neighborhood of x, U V A 6= .Hence x (U A) U A. Hence A U U A, so A U A U andthe result follows.

3.5 Problems

Problem 3.1. Prove that if any single point is open, then the topology isdiscrete.

Problem 3.2. Prove theorem 7.11.

Problem 3.3. Prove that any open subset has no limit point in the discretetopology.

Problem 3.4. Prove that in the lower limit topology, x is a limit point of Aiff there is a strictly decreasing sequence in A converging to x.


Problem 3.5. Prove that in the finite complement topology, A = X or according to whether A is infinite or not.

Problem 3.6. Prove that if x is a limit point of A (X, T ), then x is stilla limit point of A in a coarser topology. What about in a finer topology?

Chapter 4

Continuity andHomeomorphism

4.1 Continuous Functions

Defintion 4.1 (Continuous Functions). A function f : X Y is continuousif f1(U) is open for all open sets U Y.

Theorem 4.2. The composition of two continuous functions is continuous.

Proof. Let f : X Y, g : Y Z be continuous, then for all open setsU Z, g1(U) is open in Y since g is continuous. Hence f1(g1(U)) isopen in X since f is continuous. Therefore (g f)1(U) = f1(g1(U)) isopen in X for all open sets U in Z.

Lemma 4.3 (Pasting Lemma). Let X = A B, where A,B are closed (oropen) in X. Let f : A Y and g : B Y be continuous,. If f(x) = g(x)for every x A B, then the function h : X Y defined by

h(x) =

{f(x) if x Ag(x) if x B.

is continuous.

Example 4.4. Every function from a discrete topological space is con- tinu-ous. Analogously, every function to a trivial topological space is continuous.

Example 4.5. Consider the identity map

id : XT XT ,

23

24 CHAPTER 4. CONTINUITY AND HOMEOMORPHISM

between two topologies on X. The identity map is continuous iff T is coarserthan T . The inverse map

id1 : XT XTis continuous iff T is coarser than T . Hence id is a homeomorphism iffT = T .Example 4.6. By Example 3.15, we see that the identity map id : C[0, 1]L C[0, 1]L1 is continuous but the inverse id : C[0, 1]L1 C[0, 1]L is not con-tinuous.

Example 4.7. Prove that f : X Y is continuous iff f1(A) f1(A) forany subset A Y .Proof. Suppose f is continuous. Since A A, we have f1(A) f1(A).Since the closure A is closed, and f is continuous, f1(A) is also closed. Bydefinition, the closure f1(A) is the smallest closed set containing f1(A),hence f1(A) f1(A).

Conversely, if f1(A) f1(A), then by taking A to be a closed subset,we have A = A and f1(A) f1(A). On the other hand, it is always truethat f1(A) f1(A). Hence f1(A) = f1(A). for any closed set A. i.e.f1(A) is closed for any closed A. Hence f is continuous.

4.2 Homeomorphism

Defintion 4.8 (Homeomorphism). A bijection between topological spacesmf : X Y , is a homeomorphism if both f and f1 are continuous. This isthe same as saying that U is open iff f(U) is open.

Remark. It is tempting to think that a bijective continuous map isalways a homeomorphism. This is NOT true. We will later see that acontinuous bijective map between compact Hausdorff topological spaces isalways a homeomorphism.

Example 4.9. R is homeomorphic to the open interval (0, 1).

Proof. First note that all bounded interval (a, b) are related by linear maps(for example: (pi/2, pi/2) is homeomorphic to (0, 1) via x 7 x/pi + 1/2.),and then note that tan : (pi/2, pi/2) R is a homeomorphism.

Let C(X, Y ) denote the set of all continuous map f : X Y . We letC(X) = C(X,R) denote the algebra of real-valued continuous functions on

4.3. PROBLEMS 25

X. It is easy to show that if f and g are continuous, then so are the sum andproduct f + g, fg. Thus C(X) is an algebra. It always contains the constantfunctions R.

Example 4.10. In the discrete topology, all sets are open, and hence allfunctions are continuous, so C(X) = RX . In the trivial topology, only Xand are open, so C(X) = R.Example 4.11. Prove that the only continuous map Rf R are the con-stants.

Proof. Suppose f is a non-constant continuous map, then there exists a, b Rf such that f(a) 6= f(b). Since R is Hausdorff, there exists disjoint openneighborhood Ua, Ub such that f(a) Ua, f(b) Ub and Ua Ub = . Thisimplies a f1(Ua), b f1(Ub) are disjoint open neighborhood of a and brespectively, which is a contradiction because if RF1,RF2 are two opensubsets in R, then (R F1) (R F2) = R (F1 F2) 6= .Example 4.12. Prove that [0, 1] and (0, 1) are not homeomorphic.

Proof. For any continuous map f : [0, 1] (0, 1), f achieves its maximum,say f(x0). Then f(x) f(x0) for all x [0, 1] and f(x0) (0, 1). Hencef([0, 1]) (0, f(x0)] (0, 1), Thus f cannot be surjective.

4.3 Problems

Problem 4.1. Define f : R R by

f(x) =

{ |x| if x Q|x| if x / Q.

Prove that f is continuous at x = 0 but not continuous at other points.

26 CHAPTER 4. CONTINUITY AND HOMEOMORPHISM

Chapter 5

Constructing Topologies

5.1 Subspace Topology

Defintion 5.1 (Subspace Topology). If Y X and X is a topological space,the subspace topology on Y is defined by

TY = {U Y : U TX}

Theorem 5.2 (Universal Property of Subspace Topology). The subspacetopology is the coarsest topology such that the inclusion map Y X iscontinuous. Also, f : Z Y is continuous iff i f : Z X is continuous,where i : Y X is the inclusion map. The diagram is commutative.

Y i - X

Z

f

6

i f

-

Proof. Suppose the inclusion map Y X is continuous, then for U TX , i1(U) = U Y. Since the subspace topology on Y consists of exactlysuch subsets, it is the coarsest topology possible.

if is continuous iff U TX (if)1(U) = f1(i1(U)) = f1(Y U) TZ iff f1(U) TZ iff f is continuous.

Example 5.3 (HKUST Topology Final 13). Consider a map f : X Y .(a) Suppose that X =

U is a union of open subsets. Show that if all

the restrictions f = f |U : U Y are continuous, then f is continuous.

27

28 CHAPTER 5. CONSTRUCTING TOPOLOGIES

(b) Suppose that X =ni=1Ci is a union of finitely manu closed subsets.

Show that if all the restrictions fi = f |Ci :: Ci Y are continuous, then fis continuous.

(c) Suppose that X = U is a locally finite union of closed subsets, i.e.

for any x X, there is an open Ux such that x Ux and Ux C = for allbut finitely many . Show that if all the restrictions f = f |C : C Y arecontinuous, then f is continuous.

Proof. (a) For U TY , f1 (U) = f1(U) U U TX . Hence

f1 (U) =

f1(U) U = f1(U) X = f1(U) TX .

(b) For V Y , f1i (V ) = f1(V ) Ci Ci is closed in X since Ci isclosed in X. Therefore f1i (V ) X is closed. Hence f1(V ) =

ni=1 f

1i (V )

is a finite union of closed sets, which is closed and hence f is continuous.

(c) For any x X, there exists Ux TX such that Ux C = for allbut finitely many . Now

Ux = Ux X = Ux

C =

(Ux C)

since that is a finite union (as Ux C = for all but finitely many ). NowUx C is closed and

f |UxC : Ux C C Y

are continuous w.r.t. subspace topology by (b). Hence

f |Ux : Ux Y

are continuous. But X =xX Ux, hence f : X Y is continuous by

(a).

Defintion 5.4 (Embedding). An injective map f : X Y is an embeddingif the induced invertible map : X f(X) is a homeomorphism, wheref(X) has the subspace topology.

Remark. Note that is invertible because it is both injective (as f is)and surjective. Moreover, is continuous because f is continuous. Thereforethe key for f being an embedding is the continuity of 1.

5.2. PRODUCT TOPOLOGY 29

5.2 Product Topology

Defintion 5.5 (Product Topology). Let X and Y be topological spaces. Thenthe product topology on X Y is the topology induced by the basis BXY ={B1 B2|B1 BX , B2 BY }.Example 5.6. Prove that the projection piX : X Y X maps open setsto open sets, but does not necessarily map closed sets to closed sets.

Proof. Suppose W X Y is open in the product topology, we will showthat piX(W ) TX . Let (x, y) W , then there exists U V BXY suchthat (x, y) U V W. So that x U piX(W ).

Next, the projection R2 R to the first coordinate takes the closedsubset {(t, t1) R2 : t > 0} to the open (and not closed) subset (0,).Theorem 5.7 (Universal Property of Product Topology). Let X and Y betopological spaces. Then the product topology is the coarsest topology suchthat the projection maps piX : X Y X and piY : X Y Y arecontinuous. It is also the unique topology (up to homeomorphism) so thath : Z X Y defined by h(x) = (f(x), g(x)) is continuous iff f = piX hand g = piY h are continuous. The diagram is commutative.

X piX

X Y piY - Y

Z

h

6

g

-

f

Proof. Exercises.

Example 5.8. Prove that a map f : X Y is continuous iff its graph : X X Y, x 7 (x, f(x)) is continuous.Proof. By the Universal Property of Product Topology, is continuous iffboth Id and f are continuous. Since Id is continuous, we are done.

Example 5.9 (HKUST Topology Final 13). For any injective map f : X Y , prove that its graph is an embedding.

Proof. The induced map : X (X) is continuous by example 5.8 and isclearly bijective. Hence it suffices to show that 1 is continuous. Consider


the commutative diagram

X - (X)

i- X Y

X

1

?

piX

we have 1 = piX i. Since piX , i are continuous, the result follows.

5.3 Quotient Topology

Defintion 5.10 (Quotient Topology). Let X be a topological space and f :X Y be a surjective map, then the quotient topology on Y is

TY = {U Y : f1(U) TX}.Theorem 5.11 (Universal Property of Quotient Topology). Let f : X Ybe a surjective map. Then the quotient topology on Y is the finest topologysuch that the quotient map f is continuous. Moreover, a map g : Y Z iscontinuous iff g f : X Z is continuous. The diagram is commutative.

Z

Xf--

g f

-

Y

g

6

Proof. Suppose T Y is another topology on Y , then for all U T Y such thatf1(U) TX , we have U TY by defintion of quotient topology, henceT Y TY .

g f is continuous iff U TZ (g f)1(U) = f1(g1(U)) TX iffg1(U) TY iff g is continuous.

Let X be a topologcial space, f : X Y is a surjective map, then we candefine an equivalence relation on X as x1 x2 if f(x1) = f(x2). It is easyto check that is indeed an equivalence relation. For each x X, dnote theequivalence class of x by [x] = {y X : x y}. Define X/ = {[x] : x X}.Then for any surjective map f : X Y , X/ can be identified with Y .(viaX X/ Y, x 7 [x] 7 f(x)) The quotient topology on X/ is thesame as that on Y . X/ is called the quotient space.

5.3. QUOTIENT TOPOLOGY 31

Example 5.12 (HKUST Topology 13). Let X = {a, b, c, d, e} be endowedwith the topology

T = {X, , {a, b}, {c}, {a, b, c}, {d, e}, {a, b, d, e}, {c, d, e}}.

Let be the equivalence relation on X which identifies a and d (i.e. a d.).Describe the quotient topology on X/ . (i.e. write down all the open sets.)

Proof. We have X/ = {a, b, c, e}. Let f : X X/ be the naturalquotient map. Then clearly Y = X/ , TY . Let U TY .

If [a] / U , then f1(U) {b, c, e}. The only open set is {c}. HenceU = {[c]}.

If [a] U , then {a, d} f1(U) TX . Hence f1(U) = {a, b, d, e} andU = {[a], [b], [e]}.

Therefore TY = {Y, , {[c]}, {[a], [b], [e]}}.

Example 5.13. The map : R S1 = {x C : |x| = 1} defined by (t) =e2piit is surjective. Then the quotient topology on S1 means that U S1 isopen iff for any (t) U , there exists B = (t , t+ ) such that (B) U.Note that (B) is an open arc on the unit circle around S1, and such openarcs form a topological basis of S1 as a topological subspace of R2. Hence thequotient topology is the same as standard subspace topology on S1.

Theorem 5.11 further tells us that continuous periodic functions of period1 are in one-to-one correspondence with the continuous functions on the unitcircle.

Example 5.14. Let X and Y be topological spaces and let f : X Y be asurjective map that takes open sets to open sets. Prove that the topology onY is finer than the quotient topoology. Hence prove that if f is continuous,then the topology on Y is the quotient topology.

Proof. If U Y is open in the quotient topology, then f1(U) is an opensubset of X. Since f is surjective, we have f(f1(U)) = U . By the assump-tion that f takes open subsets to open subsets, we see that U is open in thegiven topology on Y . This proves that the given topology is finer than thequotient topology.

If f is continuous, then by Universal property of quotient topology, weknow that the quotient topology is the finest topology such that f is con-tinuous, the quotient topology is finer than the given topology. Combinedwith the earlier part, we conclude that the quotient topology and the giventopology are the same.


5.4 Problems

Problem 5.1. Prove Thoerem 5.7.

Chapter 6

Hausdorff Spaces

Defintion 6.1 (Hausdorff Spaces). A topological space X is Hausdorff ifany pair of distinct points have distinct neighborhoods. In this case limits areunique.

Defintion 6.2 (Metrization). A topological space X is metrizable if the topol-ogy can be induced by a metric on X.

Example 6.3. Metric Space is Hausdorff. Hence to prove that a topologicalspace X is not metrizable, it suffices to prove that X is not Hausdorff

Proof. Exercises.

Therefore Hausdorff topological spaces are in some sense not too far frombeing metric spaces.

Example 6.4. The discrete topology is Hausdorff, while the trivial topologyon any topological space containing at least two points is not Hausdorff (hencenot metrizable). In particular, if a topological space X has more than oneelement, then the discrete topology X is not homeomorphic to the trivialtopology X.

Example 6.5. R and Rl are Hausdorff, but Rf is not.

Proof. Exercise.

Theorem 6.6. If f : X Y is an injective continuous map and Y isHausdorff, then X is Hausdorff.

Proof. Let x, y X be distinct elements. By injectivity we have f(x) 6= f(y).Now by the Hausdorff property of Y , there exists disjoint U, V TY suchthat f(x) U, f(y) V . Hence f1(U) f1(V ) = f1(U V ) = and x f1(U) and y f1(V ) with f1(U), f1(V ) TX since f iscontinuous.

33

34 CHAPTER 6. HAUSDORFF SPACES

Theorem 6.7. The product of two spaces is Hausdorff iff the two spaces areHausdorff. A subspace of a Hausdorff space is Hausdorff.

Proof. Suppose X Y is Hausdorff, then for x1 6= x2, we have (x1, y) 6=(x2, y). Since X Y is Hausdorff, there exists U1 V1, U2 V2 TXY suchthat (xi, y) Ui Vi for i = 1, 2. Since y V1 V2, U1 and U2 must bedisjoint. Hence X is Hausdorff. Likewise for Y .

Conversely, if X and Y are Hausdorff, then for (x1, y1) 6= (x2, y2), theneither x1 6= x2 or y1 6= y2. Assume x1 6= x2, then there exists U1, U2 TX withU1U2 = such that x1 U1, x2 U2. Therefore (x1, y1) U1Y, (x2, y2) U2 Y and U1 Y, U2 Y are disjoint and open. Likewise for the casey1 6= y2. Hence X Y is Hausdorff. The proof for subspace is left as anexercise.

Theorem 6.8. Every finite set S in a Hausdorff Space is closed.

Proof. It suffices to show that X {x} is open. Since X is Hausdorff, thereexists U1, U2 T such that x U1, y U2 for all y X{x} and U1U2 = .Hence X {x} is open since x / U2. Finally, finite union of closed set isclosed, done.

6.1 Problems

Problem 6.1. Prove Example 6.3.

Problem 6.2. Prove that the cofinite topology is Hausdorff iff the topologicalspace is finite.

Problem 6.3. Prove that of X and Y are non-empty, then X Y is Haus-dorff iff X and Y are Hausdorff.

Problem 6.4. Prove that a subspace A of a Hausdorff space X is Hausdorff.

Problem 6.5. Prove Example 6.5.

Chapter 7

Connected Spaces

Theorem 7.1. Let X be a topological space, then TFAE:

1. If X = A unionsqB with A,B TX , then A = or B = .

2. If X = A unionsqB with A and B closed, then A = or B = .

3. The only subsets of X that are both open and closed are and X itself.

4. Every continuous function f : X {0, 1} is constant.

Proof. We only prove 1 and 4 are equivalent. It is equivalent to prove thatX = AunionsqB,A,B 6= iff there is an surjective continuous map f : X {0, 1}where {0, 1} is equipped with the discrete topology. If X = AunionsqB,A,B 6= ,then the map

f(x) =

{0 if x A1 if x B.

is surjective because A,B 6= . The map is continuous w.r.t. the discretetopology on {0, 1} since both f1({0}) = A and f1({1}) = B are open.

Conversely, suppose f : X {0, 1} is a surjective continuous map todiscrete topology. Then X = f1({0}) f1({1}) is a disjoint union of opensubsets. Moreover, the two subsets are nonempty because f is surjective andthe result follows.

Defintion 7.2 (Connected Spaces). A topological space X satisfying any oneof the conditions in Theorem 7.1 is called connected.

Here are two basic facts about connectedness, one is concrete and one isabstract. The concrete fact is:

35

36 CHAPTER 7. CONNECTED SPACES

Theorem 7.3. The interval [0, 1] is connected.

Proof. Suppose [0, 1] is the union of two disjoint open sets, U and V . Wemay assume 0 U . Let a = sup{x [0, 1] : [0, x) U}. Since U is open,a > 0. If a = 1 then we are done. Otherwise a / U, and therefore a V .But then a neighborhood of a is in V , so we cannot have [0, a) U.

The abstract fact is

Theorem 7.4 (Connected Image Theorem). If f : X Y is continuousand X is connected, then f(X) is connected.

Proof. Suppose to the contrary we can write f(X) = UV as the union of twononempty, disjoint open sets. Then X = f1(U) f1(V ), a contradiction.

Corollary 7.5 (Intermediate Value Theorem). Let X be a connected topo-logical space and f : X R be a continuous function. If f(x) f(y), andc [f(x), f(y)], then there exists a z X such that f(z) = c.Example 7.6. Prove that for any continuous map f : S1 R, there isx S1 such that f(x) = f(x).Proof. The map g(x) = f(x)f(x) is odd (i.e. g(x) = g(x)). Thereforethe image g(S1) contains both positive and negative numbers. Since S1 isconnected, the image g(S1) is a connected subset of R that contains positiveand negative numbers. Therefore the image must contain 0.

Example 7.7. R and Rf are connected, while Rl is not.

Proof. [0, 1] is connected by Theorem 7.3 and [0, 1] is homeomorphic to R,hence R is connected.

Next, the non-trivial open subsets of Rf are infinite and the non-trivialclosed subsets are finite . Hence there is no nontrivial open and closed subsetand therefore Rf is connected.

Finally, Rl = (, 0) [0,) is a separation and the result follows.Example 7.8. Prove that the only connected subsets of Rl are the singlepoints.

Proof. Single point set is clearly connected. Given A Rl with at least twopoints, say a < b, then we can find a < r < b such that

A = (A (, r)) unionsq (A [r,))is a separation of A. Therefore A is not connected.

37

Example 7.9. GLn(R) is not connected.

Proof. The determinant det : GLn(R) R is continuous with image det(GLn(R)) =R {0}, which is not connected. Hence by the contraposition of the con-nected image theorem, GLn(R) is not connected.

Example 7.10. R is not homeomorphic to R2.

Proof. Suppose f : R R2 is a homeomorphism, then g : R {0} R2{f(0)} is a homeomorphism for the induced topologies. But R2{f(0)}is connected, while R {0} is not.

Here are three more basic properties of connectdness.

Theorem 7.11. If Ai X are connected subspaces which all have a pointin common, then Y =

iAi is connected.

Proof. Let Y = U unionsq V and suppose the common point p U. Note thatU Ai and V Ai are disjoint open sets in Ai. Since Ai is connected andp U, we have Ai U . Hence Y =

iAi U and Y is connected.

Theorem 7.12. Suppose X and Y are nonempty, then X Y is connectediff X and Y are connected.

Proof. (Necessity) Suppose X Y is connected, then by the Universal prop-erty of product topology, X and Y are continuous images under the projec-tions piX and piY . Hence by the Connected Image Theorem, X and Y areconnected.

(Sufficiency) Suppose XY = UunionsqV, (a, b) U and we are given (a, b) X Y . Then X {b} is connected, so (a, b) U . Similarly, {a} Y isconnected, so (a, b) U . Hence V = .

In fact, an infinite product of connected spaces is connected:

Theorem 7.13. Given any collection of connected spaces Xi, the space

iXiis connected in the product topology.

Theorem 7.14 (Connected Sandwich Theorem). If X is connected and X Y X, then Y is connected.

Example 7.15. Prove that if any two point subspace of a topological spaceX is connected, then X is also connected. What if two points are replaced by2014 points?


Proof. Fix a X. Then for any x X, the subset Ax = {a, x} is connected.By Theorem 7.11 , X =

xX Ax is connected.

The same argument applied to 2014 points, as long as X contains at least2014 points.

Example 7.16. Suppose A1, A2, X are connected subsets such that Ai Ai+1 6= . Prove that

iAi is connected.

Proof. First we show that for n Z+, ni=1Ai is connected. The base caseis true by Theorem 7.11. Assume that Bn =

ni=1Ai is connected, for n+ 1,

Bn An+1 6= , since An An+1 6= .Hence Bn An+1 = Bn+1 is connected by Theorem 7.11. By induction, Bnis connected for all n.

Next, n

Bn = B1 = A1 6= .

Hence by Theorem 7.11,nBn =

iAi is connected.

Example 7.17 (HKUST Topology 98Final). Prove that Y = {(x, y, z) R3 : x4 + y4 + z4 4, xyz = 1} is not connected.Proof. Consider the continuous map f : Y R defined by f(x, y, z) = x.Since (1, 1, 1), (1,1, 1) Y , we have 1,1 f(Y ). Now xyz = 1 impliesx 6= 0 and hence 0 / f(Y ). Therefore f(Y ) is not connected and the resultfollows.

7.1 Path Connectedness

Defintion 7.18 (Path Connectedness). A path connecting points x and y ina topological space X is a continuous map : [0, 1] X such that (0) = xand (1) = y. A space X is path-connected if for all x, y X, there is a pathconnecting x and y.

Theorem 7.19. A path-connected topological space is connected.

Proof. Suppose X = UunionsqV and x U . Pick a path : [0, 1] X connectingx to any other point y. Then ([a, b]) is connected by Connected ImageTheorem, so y U. Hence V = .Theorem 7.20. A topological space X is path-connected iff there is x0 Xsuch that for any x X, there is a path connecting x0 to x.

7.1. PATH CONNECTEDNESS 39

Proof. Necessity is obvious. For sufficiency, if there is x0 X such that forany fixed x, y X, there is a path x connecting x0 to x and a path yconnecting x0 to y. Define the map : [0, 1] X by

(t) =

{x(1 2t) if t [0, 1/2]y(2t 1) if t [1/2, 1].

Then (0) = x, (1) = y and is continuous by the pasting lemma 4.3,hence is a path connecting x to y. Since x, y are arbitrary, X is path-connected.

Example 7.21. Prove that the space X = {(x, y, z, w) R4 : x2 + y2 =z2 + w2} R4 is path-connected.Proof. Note that 0 X. If p = (x, y, z, w) X, then f(t) = (tx, ty, tz, tw)is a path in X such that f(0) = 0 and f(1) = p. Hence X is path-connectedby Theorem 7.20.

Defintion 7.22 (Convex set). A subset A Rn is convex if whenever x, y A, the straight line from x to y also lies in A. Equivalently, A is convex ifx, y A, 0 t 1 implies (t) = (1 t)x+ ty A.Example 7.23. A convex set A is path connected.

Proof. (t) = (1 t)x+ ty is a path.Remark. Note that we do not require A to be open or closed.

Example 7.24. The L-topology on C[0, 1] is path-connected.

Proof. For any f, g C[0, 1], construct a path : [0, 1] C[0, 1] by (t) =(1 t)f(x) + tg(x). Then clearly (0) = f(x), (1) = g(x). Nowd((t1), (t2)) = max

x[0,1]|((1 t1)f(x) + t1g(x)) ((1 t2)f(x) t2g(x))|

= maxxC[0,1]

|t1 t2|(f(x) g(x)) = |t1 t2|d(f, g).

Hence by choosing = d(f, d)1 , we see that is continuous.For well-behaved spaces like manifolds, connectedness and path-connectedness

are the same. The proof is a typical use of the property of connectedness:we show the set of good points is both open and closed, so it must be thewhole space.

Theorem 7.25 (Path Connectedness in Rn). An open set U Rn is con-nected iff it is path-connected.


Proof. Suppose U is connected and x U. Let V U be the set of pointsthat can be reached from x by a polygonal path. Clearly V is open. We claimit is also closed (in U). Indeed, if y V then there is a open neighborhoodBy U of y and a z By that can be reach from x by a polygonal path.But B is convex, so y can also be reached. The converse is a general fact.

Theorem 7.26 (Path-Connected Image Theorem). The image of a path-connected space under a continuous map is path-connected.

Theorem 7.27. Suppose X and Y are nonempty, then X Y is path con-nected iff both X and Y are path connected.

Proof. Exercises.

Example 7.28. Prove that A,B,C R3 are connected but D is not con-nected, where

A = {(x, y, z) R3 : x3 + y3 = z2}B = {(x, y, z) R3 : x3 + y3 = z2, x4 + y4 1}C = {(x, y, z) R3 : x3 + y3 = z2, x4 + y4 < 1}D = {(x, y, z) R3 : x3 + y3 = z2, x+ y > 0}

Proof. For any (x, y, z) A, (t) = (tx, ty, t2/3z) is a path connecting (0, 0, 0)to (x, y, z). Hence A is path-connected by Theorem 7.20. In particular, A isconnected by Theorem 7.19. The same path works for B since x4 + y4 1and t [0, 1] implies (tx)4 + (ty)4 1. Hence B is also connected. SimilarlyC is also connected.

Now define a continuous map f : D R by f(x, y, z) = z. Since(1, 1,2) D, we have 2 f(D). On the other hand, x + y > 0 x 6= y x3 + y3 6= 0. Hence (x, y, z) D z 6= 0. i.e. 0 / f(D). The facts2 f(D), 0 / f(D) together imply f(D) is not connected. Hence D isnot connected. (by the contraposition of the connected image theorem.)

7.2 Connected Components

Defintion 7.29 (Connected Component). A subset A of a topological spaceX is a connected component if it is a maximal connected subset of X:

1. A is connected, and2. If A B X and B is connected, then A = B.

Theorem 7.30. Connected components are closed.

Proof. Exercise.

7.3. PROBLEMS 41

Theorem 7.31. The connected components form a partition of X. Hence,any topological space is the disjoint union of connected components.

Proof. First we prove that any two connected components are either identicalor disjoint. Suppose A and B are connecetd components and AB 6= , thenby Theorem 7.11, AB is connected. Since A and B are maximal connectedsubsets, we have A = A B = B. Now for any x X, let

Cx =AX

A

where A runs through all connected subsets that contain x. Then Cx isconnected by Theorem 7.11 and by definition Cx is the largest connectedsubset containing x and the result follows.

Remark. By the above theorem, we may define an equivalence relation:x y if there exists a connected subspace of X containing both x and y.The equivalence classes are precisely the connected components of X.

7.3 Problems

Problem 7.1. Prove Theorem 7.1.

Problem 7.2. Is the trivial toplogy connected?

Problem 7.3. Is the discrete topology connected?

Problem 7.4. Prove that a set A R is connected iff it is convex iff it ispath-connected.

Problem 7.5. Prove that A = {(x, y) R2 : x4+y4 = 1} R2 is connected.Problem 7.6. Prove that A = {(x, y) R2 : x4 y4 = 1} R2 is notconnected.

Problem 7.7. Suppose A,Ai X are connected subsets such that AAi 6= for all i. Prove that A (iAi) is connected.Problem 7.8. Suppose f : [0, 1] [0, 1] is a continuous map. Prove thatthere exists x [0, 1] such that f(x) = x.Problem 7.9. Prove that the closure of a connected space is connected.

Problem 7.10. Prove that X is connected does not imply X is connected.

Problem 7.11. Prove Theorem 7.27

Problem 7.12. Prove Theorem 7.30

Chapter 8

Compact Spaces

Defintion 8.1 (Open Cover). A collection U of subsets of a space X is acovering of X, if the union of the elements of U is equal to X. It is calledan open cover of X if its elements are open subsets of X.

Defintion 8.2 (Compactness). A topological space X is compact if everyopen cover U of X has a finite subcover.Example 8.3. If a toplogical space contains only finitely many open sub-sets, then the space is compact. In particular, any finite topological space iscompact. The trivial topology is also compact.

Proof. Immediate from the definition.

Example 8.4 (CUHK Topology 10). Let (X, T ) be an infinite set (say R)with the cofinite topology. Prove that X is compact. Hence prove that anysubset A of X is also compact.

Proof. Let U be an open cover of X, then there exists U0 U such thatX U0 is finite. Therefore X U0 = {p1, , pn}. Since U is a cover of X,there exists Uj U such that pj Uj for each j. Hence

X = U0 (X U0) = U0 {p1, , pn} = U0 {p1} {pn} ni=0

Ui.

Now let TA denote the induced topology on A and let TC denote thecofinite topology on A. If V TA, then V = U A for some U T .If U = , then V = TC . Otherwise, X U is finite. ThereforeA V = A (U A) = A (X U) is finite. Hence TA TC .

Take any V TC . If V = , then V TA. Otherwise A V is finite.Then U = V (X A) X and X U = AV is finite. Therefore U T .Furthermore, V = U A implies V TA. Hence TC TA. Hence TA = TCand A is compact.

43

44 CHAPTER 8. COMPACT SPACES

Example 8.5 (HKUST Topology Final 13). Prove that compact sets neednot be closed in a general topological space.

Proof. Consider the Sierpinski two point space X = {a, b} with the topology{X, , {a}}. Then {a} is compact since it is finite. It is not closed, however,since its complement {b} is not open.Example 8.6 (HKUST Topology 13 Final). A map f : X Y is said tobe a local homeomorphism if each point x X has a neighborhood Ux suchthat the restriction f |Ux : Ux f(Ux) is a homeomorphism. Let X andY be a compact topological spaces and let f : X Y be a surjective localhomeomorphism. Show that for any point y Y , the set f1({y}) is finite.Proof. By assumption, for any x X, there exists Ux TX such that therestriction f |Ux : Ux f(Ux) is a homeomorphism. Then U = {Ux : x X}is an open cover of X. By the compactness of X, we have

X = Ux1 Ux2 Uxn .Now

Y = f(U) = f

(xX

Ux

)= f

(ni=1

Uxi

)=

ni=1

f(Uxi).

Hence {f(Uxi) : i = 1, , n} is a finite cover of Y . Therefore for any y Y ,

f1({y}) =ni=1

f |1Uxi ({y}).

But f |Uxi is a homeomorphism, hence f |1Uxi ({y}) has at most one elementand the result follows.

As for connectedness, there are two basic results, one concrete, one ab-stract.

Theorem 8.7 (Heine-Borel Theorem). The interval [0, 1] is compact. Moregenerally, A subset X Rn is compact iff it is closed and bounded.Theorem 8.8 (Compact Image Theorem). The continuous image of a com-pact space is compact.

Example 8.9. Prove that the following subsets of R3 are compact.

A1 = {(x, y, z) R3 : x4 + y4 + 2z4 = 1}A2 = {(x, y, z) R3 : x4 + y4 + 2z4 1}

8.1. COMPACT HAUSDORFF SPACES 45

Proof. Define f : R3 R by f(x, y, z) = x4 +y4 + 2z4. Then A1 = f1({1}),and A2 = f

1([0, 1]). By the continuity of f , A1 and A2 are closed. Moreover,the points in A1 and A2 must satisfy |x|, |y|, |z| 1. Therefore A1 and A2are bounded. Hence by Heine-Borel Theorem, A1 and A2 are compact.

Example 8.10. Prove that the following subsets of R3 are not compact.

A3 = {(x, y, z) R3 : x4 + y4 + 2z4 < 1}A4 = {(x, y, z) R3 : x4 + y4 + 2z4 1}

Proof. A3 = f1((, 1)), hence open. We claim that A3 is not closed. If

A3 were closed, since R3 is connected, by Theorem 7.1, A3 would be eitherempty set or R3, which is a contradiction. Finally, A4 is not bounded andhence both A3 and A4 are not compact by Heine-Borel Theorem.

Example 8.11 (HKUST Topoology 98 Final). Prove that X = {(x, y, z) R3 : x4 + y4 + z4 4, xyz = 1} is compact.Proof. |x|4 x4 + y4 + z4 4, hence |x| 41/4. Likewise, |x|, |y|, |z| 41/4. Therefore X is bounded. Define the continuous map f : R3 R2by f(x, y, z) = (x4 + y4 + z4, xyz). Then we have X = f1([0, 4] {1}).Since [0, 4] {1} is closed, so is X. Hence X is compact by Heine-BorelTheorem.

Theorem 8.12. Suppose X and Y are non-empty, then X Y is compactiff X and Y are compact.

Theorem 8.13 (Extreme Value Theorem). A continuous function f : X R on a compact set achieves its maximum and minimum.

Proof. In this case f(X) R is compact by Compact Image theorem, so itis closed and bounded. Thus M = sup f(X) < ,m = inf f(X) < andM,n f(X). Hence M = f(x) and m = f(y) for some x, y X.Corollary 8.14. Let f : K R be a smooth function on a closed, boundeddomain in Rn. Then f achieves its maximum at a point x K, or at apoint x int(K) where Df = 0.

8.1 Compact Hausdorff Spaces

We now explain a few points about compact sets versus closed sets.

Theorem 8.15. If X is compact, then any closed set A X is also compact.


Proof. Suppose U = {Ui} is an open cover of A, then

X = A (X A) (

i

Ui

) (X A).

Since A is closed, X A is open, and U {X A} is an open cover of X.Hence by the compactness of X, we have

X Ui1 Ui2 Uin (X A),

This impliesA Ui1 Ui2 Uin ,

And we are done.

Theorem 8.16. If X is Hausdorff, then any compact set K X is alsoclosed.

Proof. Let x X K. For each a K, we have x 6= a. By the Hausdorffcondition, there exists disjoint open subsets U and V such that x Ua anda Va, Since

K aV

Va

and K is compact, we have

A V = Va1 Va2 Van .

We also havex U = Ua1 Ua2 Uan .

By Ua Va = , we have U V = . Hence X K is open and the resultfollows.

Remark. Note that a finite set is compact, and the above theoremgeneralizes Theorem 6.8.

Corollary 8.17. A continuous bijection f : X Y between compact spaceX and Hausdorff space Y is a homeomorphism.

Proof. It suffices to show that whenever A X is closed, so is f(A). but if Ais closed, it is compact, hence f(A) is also compact, and since Y is Hausdorff,this means f(A) is closed as well.

Corollary 8.18. Compact subsets of metric spaces are closed.

8.2. PROBLEMS 47

Proof. Metric space is Hausdorff.

Example 8.19 (HKUST Topology 13 Final). Prove that in a Hausdorffspace, any intersection of compact subsets is again compact.

Proof. Suppose Xi are compact subsets of a Hausdorff space. Then each Xiis closed by Theorem 8.16. Therefore

iXi is closed. In particular, we may

choose any index i0, theniXi Xi0 is a closed subset of a compact space,

hence compact by Theorem 8.15.

Theorem 8.20 (Tychonoff). Given any collection of compact spaces Xi, thespace

iXi is compact in the product topology.

8.2 Problems

Problem 8.1 (HKUST Topology 98Final). Is the space X = {(x, y, z, w) R4 : x2 + y2 = z2 + w2} R4 compact?Problem 8.2. Prove that A,C,D R3 are not compact but B is compact,where

A = {(x, y, z) R3 : x3 + y3 = z2}B = {(x, y, z) R3 : x3 + y3 = z2, x4 + y4 1}C = {(x, y, z) R3 : x3 + y3 = z2, x4 + y4 < 1}D = {(x, y, z) R3 : x3 + y3 = z2, x+ y > 0}

Problem 8.3 (HKUST Topology 13 Final). Let X be a topological space,and A and B compact subspaces of X. Prove that A B is compact.Problem 8.4 (HKUST Topology 13 Final). Provide a counterexample of anon Hausdorff space, in which an intersection of two compact subsets is notcompact.

Chapter 9

Metric Space Revisited

9.1 Banach Fixed Point Theorem

Proposition 9.1. For any Compact Hausdorff topological space X, (C(X), d(f, g) =||f g||) is a complete metric space. i.e. Every Cauchy sequences in C(X)converges.

Defintion 9.2 (Contraction). A map f : (X1, d1) (X2, d2) is called acontraction (or contractive mapping) if there exists 0 < < 1 such that forall x, y X1, we have d2(f(x), f(y)) d1(x, y).Theorem 9.3 (Banach Fixed Point Theorem). If X is complete and f :X X is a contraction, then f has a unique fixed point.Example 9.4 (HKUST Math Competition 2013). Prove that there exists acontinuous function f on [0, 1] satisfying the integral equation:

f(x) +

10

f(y)

2 + (xy)pidy =

x0

y0

1

2 + t2pidtdy

Proof. Define the T : C[0, 1] C[0, 1] by

(Tf)(x) =

x0

y0

1

2 + t2pidtdy

10

f(y)

2 + (xy)pidy,

Since C[0, 1] is a complete metric spaces with d(f, g) = ||f g|| and

|Tf(x) Tg(x)| = 1

0

f(y)

2 + (xy)pidy

10

g(y)

2 + (xy)pidy

12 ||f g||.Hence by Banach Fixed Point Theorem, there exists f C[0, 1] such thatTf = f , done.

49

50 CHAPTER 9. METRIC SPACE REVISITED

Example 9.5 (UC Berkeley PhD quals. 87). Define a sequence of positivenumbers as follows. Let x0 > 0 be any positive number, and let xn+1 =(1 + xn)

1. Prove that the sequence converges, and find its limit.

Proof. Let g(x) = (1 + x)1. We have g(x) = 1/(1 + x)2. Therefore,

|g(x)| 1(1 + x0/2)2

1 for x > x0.

So that

|g(y) g(x)| = yx

g(t)dt

yx

|g(t)|dt < yx

1dt = y x.

Hence, by Banach Fixed Point Theorem, the sequence given by x0 > 0, xn+1 =g(xn) converges to the unique fixed point of g in [x0,). Solving g(x) = xin that domain givess us the limit (1 +5)/2.Example 9.6. In Rn, all dp(1 p ) metrics are equivalent. Henceinduce the same topology.

Proof. Consider 1 < p, q

9.2. PROBLEMS 51

9.2 Problems

Problem 9.1 (UC Berkeley PhD Quals 88). Let g : [0, 1] R be a contin-uous function. Prove that there exists a continuous function f : [0, 1] Rsatisfying the equation

f(x) x0

f(x t)et2dt = g(x).

Problem 9.2 (UC Berkeley PhD Quals 84). Show that there is a uniquecontinuous function f : [0, 1] R such that

f(x) = sinx+

10

f(y)

ex+y+1dy.

Problem 9.3 (UC Berkeley PhD Quals 82). Let K be a continuous functionon the unit square S = {(x, y) R2 : 0 x, y 1} satisfying |K(x, y)| < 1for all x, y S. Show that there is a continuous function f on [0, 1] suchthat

f(x) +

10

K(x, y)f(y)dy = ex2

.

Can there be more than one such function f?

52 CHAPTER 9. METRIC SPACE REVISITED

Chapter 10

Separation Axioms

Defintion 10.1 (Regular Space). A topology is regular if for any closedsubset A and x / A, there are disjoint open subsets U and V , such thatx U and A V.Defintion 10.2 (Normal Space). A topology is normal, if for any disjointclosed subsets A and B, there are disjoint open subsets U and V , such thatA U and B V.Defintion 10.3 (T0, Kolmogorov). A topology is T0, or Kolmogorov if forany two distinct points x and y, there is an open subsets U , such that eitherx U and y / Y , or x / U and y U.Defintion 10.4 (T1, Frechet). A topology is T1, or Frechet if any single pointset is closed.

Defintion 10.5 (T2, Hausdorff). A topology is T2, or Hausdorff if any twodistinct points are separated by disjoint open subsets.

Defintion 10.6 (T3). A topology is T3 if it is regular and Hausdorff.

Defintion 10.7 (T4). A topology is T4 if it is normal and Hausdorff.

Theorem 10.8 (Degree of Separation). T4 T3 T2 T1 T0.

53

54 CHAPTER 10. SEPARATION AXIOMS

Chapter 11

Graph Theory

11.1 Seven Bridges of Konigsberg

Defintion 11.1 (Graph). A graph consists is an ordered pair G = (V,E),consisting of a set V of vertices together with a set E of edges connectingvertices.. Moreover,

1. A path is a sequence of vertices V1, , Vk and a sequence of edgesE1, , Ek1 such that Ei connects Vi and Vi+1. The path is said to connectV1 to Vk.

2. A cycle is a path satisfying V1 = Vk.3. A graph is connected if any two vertices can be connected by a path.4. The degree of a vertex is the number of edges that connect to it.

Example 11.2. The Konigberg graph is connected. The degree of three ver-tices are 3 and the degree of the remaining one is 5.

Theorem 11.3 (Euler). For a connected graph, TFAE:1. There exists a path in which each edge appears exactly once.2. There are at most two vertices with odd degrees.

In order to prove this theorem, we first prove a few lemmas.

Lemma 11.4. For any graph G = (V,E), we haveViV

deg(Vi) = 2#E.

Proof. Exercise.

Example 11.5. For the Konigsberg grpah, we have

deg(A) + deg(B) + deg(C) + deg(D) = 3 + 3 + 3 + 5 = 2 7.

55

56 CHAPTER 11. GRAPH THEORY

Lemma 11.6. If all vertices of a connected graph G have even degrees, thendeleting any one edge from G still produces a connected graph.

Proof. Suppose G is obtained by deleting one edge Ei from G. If G werenot connected, then G would have two connected components G1, G2, andEi connects a vertex V1 G1 to V2 G2. Also, for vertices W G1, we have

degG1(W ) =

{degG(W ) if W 6= V1degG(V1) 1 if W = V1.

In particular, G1 contains exactly one vertex of odd degree. This impliesV G1

deg(V ) 1 (mod 2),

a contradiction to lemma 11.4. Hence G is connected.

Example 11.7. Since the Konigsberg graph has 4 vertices of odd degree, byEulers Theorem 11.3, we cannot make one trip so that each bridge is crossedexactly once.

PreliminariesBasic Set TheoryUseful Inequalities

Metric SpacesDefinition and Examples of Metric SpacesTopological Properties in Metric SpacesOpen SubsetContinuityLimit PointClosed Subset

Problems

Topological SpacesTopologyBasis of a TopologyComparing TopologiesTopology as a LanguageProblems

Continuity and HomeomorphismContinuous FunctionsHomeomorphismProblems

Constructing TopologiesSubspace TopologyProduct TopologyQuotient TopologyProblems

Hausdorff SpacesProblems

Connected SpacesPath ConnectednessConnected ComponentsProblems

Compact SpacesCompact Hausdorff SpacesProblems

Metric Space RevisitedBanach Fixed Point TheoremProblems

Separation AxiomsGraph TheorySeven Bridges of Knigsberg

Documents

Introduction to Topology - Ryan Lok-Wing Pang