math201-lecturenotes metric topology

7/27/2019 Math201-Lecturenotes metric topology

1/47

Math 201Metric Topology I

Lecture notes of Dr. Hicham Gebran

[email protected]

Lebanese University, Fanar, Fall 2012-2013


2/47

2

What is topology? Topology is the study of continuity in a general context. In group theoryor vector space theory, a fundamental concept is that of isomorphism. Two groups are iso-morphic if they have the same algebraic structure. The analogous concept in topology is thatof homeomorphism. A homeomorphism is a continuous bijection whose inverse is continuous.For example a circle and a square are homeomorphic as we shall see, that is, they are topo-

logically the same. A fundamental problem of topology is the classification of spaces up to ahomeomorphism.


3/47

Contents

0 Some fundamental properties of the real line 5

1 Metric spaces and topological spaces 71.1 Definitions and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2 Open and closed sets and related concepts (neighborhood, interior, closure, bound-

ary, convergence, density) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Topological spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.4 Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.5 Product spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2 Continuous functions and homeomorphisms 21

3 Compactness 273.1 Definitions, examples and properties . . . . . . . . . . . . . . . . . . . . . . . . . 273.2 Compact subspaces of IRn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.3 Compact metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313.4 Local compactness and the Alexandroff compactification . . . . . . . . . . . . . . 33

4 Connectedness 374.1 Definitions, examples and properties . . . . . . . . . . . . . . . . . . . . . . . . . 374.2 Components and local connectedness . . . . . . . . . . . . . . . . . . . . . . . . . 41

5 Complete metric spaces 45

3


4/47

4 CONTENTS


5/47

Chapter 0

Some fundamental properties of thereal line

We start by formulating two fundamental properties of the set of real numbers that we shall

use in the sequel. This set is denoted by IR and it is naturally identified to a line. An upperbound of a subset E IR is a real number M such that x M for all x E. The set E iscalled bounded from above if it has an upper bound. A lower bound of a subset E IR is areal number m such that m x for all x E. The set E is called bounded from below if it hasa lower bound.

Definition 0.1 Let E IR be bounded from above. The number is called the least upperbound of E if the following hold.

i) is an upper bound of E.ii) If < , then is not an upper bound of E.

In this case, we write = sup E. If E is not bounded from above, we write sup E = +.Proposition 0.1 Let E IR be bounded from above. Then = sup E if and only if thefollowing hold.

i) x for all x E.ii) > 0 y E such that < y.

Definition 0.2 Let E IR be bounded from below. The number is called the greatest lowerbound of E if the following hold.

i) is a lower bound of E.ii) If > , then is not a lower bound of E.

In this case, we write = infE. If E is not bounded from below, we write infE = .

Proposition 0.2 Let E IR be bounded from below. Then = infE if and only if thefollowing hold.

i) x for all x E.ii) > 0 y E such that + > y.

Proposition 0.3 Let = A B IR. Then sup A sup B and infA infB.Exercise. Let A and B be two nonempty subsets of IR. If x y for all x A and all y B,then sup A infB.

Now we can state the two fundamental properties of IR.

5


6/47

6 CHAPTER 0. SOME FUNDAMENTAL PROPERTIES OF THE REAL LINE

Theorem 0.1 Any nonempty subset of IR which is bounded from above has a least upperbound. Any nonempty subset of IR which is bounded from below has a greatest lower bound.

Any proof of this theorem involves going back to the construction of the real numbers from therational numbers. We do not address this issue here.

Recall now that a subset E

IR is an interval if and only if it is not empty and [x, y]E whenever x and y belong to E. An interval has thus one of the following ten forms

{a}, ]a, b[, ]a, b], [a, b[, [a, b], ] , a], ] , a[, ]a, [, [a, [ and IR. An interval is calledtrivial if it is a singleton.

Theorem 0.2 Any nontrivial interval of the real line contains rational as well as irrationalnumbers.

This theorem will be proved in the exercises. We end with a useful result.

Proposition 0.4 Let a and b be given real numbers. If a b + for all > 0 then a b.Proof. Otherwise, taking = ba

2, we reach a contradiction.


7/47

Chapter 1

Metric spaces and topological spaces

1.1 Definitions and examples

Definition 1.1 Let X be a non empty set. A metric or distance over X is a function

d : X X [0, [ that satisfies the following properties1. d(x, y) = 0 if and only if x = y.

2. d(x, y) = d(y, x) for all x, y X (symmetry).3. d(x, z) d(x, y) + d(y, z) for all x,y ,z X (Triangle inequality).

A metric space is a couple (X, d) where d is a distance on X.

Examples. 1) The real line with the usual distance d(x, y) = |x y|.2) The real line with the distance (x, y) = |x3 y3|.3) The real line with the distance (x, y) = min(1, |xy|). The first two properties of a distanceare clear. We prove the triangle inequality. Observe that (x, y) 1 and (x, y) |x y|. Letx,y ,z be three real numbers. We distinguish between two cases.

Case 1. Either |x y| 1 or |y z| 1. Then (x, y) = 1 o r (y, z) = 1. Therefore,(x, y) + (y, z) 1. On the other hand, (x, z) 1. Hence the triangle inequality in this case.Case 2. |x y| < 1 and |y z| < 1. Then (x, y) = |x y| and (y, z) = |y z|. Therefore,(x, z) |x z| |x y| + |y z| = (x, y) + (y, z).4) More generally, let (X, d) be a metric space and set d(x, y) = min(1, d(x, y)). Then d is adistance on X. The proof is similar as the proof above.

5) Let a < b and X = C([a, b]) be the set of continuous functions f : [a, b]

IR. Letd(f, g) = supx[a,b] |f(x) g(x)|. Then d is a distance.

6) Again consider the set X = C([a, b]) and set d(f, g) =

ba

|f(x) g(x)| dx.

7) Let X be a set. The discrete distance d on X is defined by

d(x, y) =

0 if x = y

1 if x = yThe first two properties are clear. Let x,y ,z X. To prove the triangle inequality, we distin-guish between two cases.

Case 1. x = z. Then d(x, z) = 0 d(x, y) + d(y, z).Case 2. x = z. Then either y = x or y = z. Therefore d(x, z) = 1 d(x, y) + d(y, z).

7


8/47

8 CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES

Norms and distances on IRN

Let N be a positive integer. The set of all Ntuples of real numbers (x1, x2, . . . , xN) is denotedby IRN. Otherwise stated, IRN is the cartesian product of N copies of IR. Elements of IRN areconsidered as vectors with N components. The origin of IRN is the vector (0, . . . , 0) denotedsimply by 0.

A norm on IRN is a way to measure the length of a vector (x1, . . . , xN). It is the analog ofthe absolute value of a real number. The norm of a vector x IRN is a nonnegative numberdenoted by x that satisfies the following properties.(i) x = 0 if and only if x = 0.(ii) x = ||x for any real number and any x IRN.(iii) x + y x + y for all x, y IRN. This is known as the triangle inequality.

The most natural norm on IRN is the Euclidean norm. Let x = (x1, . . . , xN) IRN. TheEuclidean norm of x is defined by

x

= N

i=1 |

xi|2

1/2

.

It is not difficult to show that x satisfies the two properties (i) and (ii) above. The triangleinequality however is not trivial. To prove (iii), let us first observe that the Euclidean norm is

just the square root of the inner product of x by itself. The inner product of two vectors x andy of IRN is

x y = x1y1 + + xNyN =Ni=1

xiyi.

Let be real number and consider the scalar product (x + y) (x + y). We have(x + y) (x + y) = x2 + 2x y + 2y2 0.

Thus we have a second order polynomial in which is always nonnegative. Therefore itsdiscriminant (x y)2 x2y2 0 (otherwise, the polynomial would have two distinct realroots and would therefore change sign). Thus,

|x y| xy.This is known as the Cauchy-Schwartz inequality.

From the last inequality we deduce that

x + y2 = (x + y) (x + y) = x2 + 2x y + y2 x2 + 2xy + y2= (

x

+

y

)2 .

Since both members of the above inequality are nonnegative, the triangle inequality follows.

Other norms can also be defined on IRN. The sup norm of a vector x is defined by x =max(|x1|, . . . , |xN|) . It is easy to check that this norm satisfies the three properties of a norm.Another norm is defined by x1 =

Ni=1

|xi|. Here again it is not difficult to prove the threeproperties. More generally, let p 1. Set

||x||p =Ni=1

|xi|p1/p

.

We shall prove in the exercises that this is norm. Note finally that all these norms coincidewhen N = 1.

Now a norm on IRN defines a metric by d(x, y) = ||x y||.


9/47

1.2. OPEN AND CLOSED SETS AND RELATED CONCEPTS (NEIGHBORHOOD, INTERIOR, CLOSUR

Balls in metric spaces

The open ball of center x and radius r is the set

B(x, r) = {y X| d(y, x) < r}.

The closed ball of center x and radius r is the set

B(x, r) = {y X| d(y, x) r}.

Examples. 1) The ball B(a, r) in IR (with the usual distance) is the interval ]a r, a + r[. In(IR2, | | | |2) the ball B(a, r) is the usual disc of center a and radius r. In (IR3, | | | |2), the ballB(a, r) is the usual geometric ball of center a and radius r.

3) In (IR2, | | | |) the ball B(0, 1) is the square of vertices (-1,-1), (1,-1) (1,1) and (-1,1). Drawa figure and explain. The closed ball is the same square with its boundary.

4) In (IR2, | | | |1), the ball B(0, 1) is the square of vertices (1,0), (0,-1) (-1,0) and (0,1). Draw afigure and explain. The closed ball is the same square with its boundary.

1.2 Open and closed sets and related concepts (neighborhood,interior, closure, boundary, convergence, density)

Definition 1.2 Let X be a metric space. A subset O X is called open if for every x Othere is an > 0 such that B(x, ) O.Examples. ]0, 1[ is open in IR but (0, 1] is not. Any interval of the form ]a, b[, ]a, [, ], a[is open.

Proposition 1.1 An open ball is open.Proof. Let x B(a, r). Then d(x, a) < r. Set = r d(x, a). We claim that B(x, ) B(a, r). Draw a figure. Indeed, let y B(x, ). Then d(x, y) < = r d(x, a). Therefored(x, a) + d(x, y) < r. By the triangle inequality, d(y, a) < r. This means that y B(a, r). Theclaim implies that B(a, r) is open.

Proposition 1.2 The following properties hold.

(i) X and are open.(ii) An arbitrary union of open sets is open.(iii) A finite intersection of open sets is open.

The proof is straightforward. Just right it.

Remark 1.1 An arbitrary intersection of open sets need not be open. For example,

n1

1n

,1

n

= {0}

is not open.

Definition 1.3 A neighborhood of a point x in a metric space X is a set V which containsan open set containing x. The set of neighborhoods of a point x is usually denoted byU(x).

Remark 1.2 Note that the whole space is a neighborhood of x, therefore U(x) is not empty.Note also that if V U(x) and V U then U U(x).


10/47


Example. [1, 1] is a neighborhood of any point x ]1, 1[. However it is not a neighborhoodof 1 nor of1.

Proposition 1.3 A set is open if and only if it is a neighborhood of all of its points.

Proof. Suppose first that A is open and let x

A. Then A contains an open set (itself)

containing x. This means that A is a neighborhood of x. Conversely, suppose that A is aneighborhood of all its points. This means that for every x A, there exists an open set Oxsuch that x Ox A. But we can write A = xAOx. Therefore A is open being a union ofopen sets.

Remark 1.3 For some mathematicians, a neighborhood of a point is an open set containingthat point. This makes a little difference because in practise we can always assume that aneighborhood is open.

Definition 1.4 The interior of a set A is the union of all open sets that are contained in A.

It is therefore the biggest open set contained in A. It is denoted byA or intA.

Examples. a) int [0,1]=]0,1[. Indeed, ]0,1[ is open and contained in [0,1]. Therefore it iscontained in the interior of [0,1]. So we have ]0, 1[ int[0, 1] [0, 1]. Therefore int[0, 1] is either]0,1[, ]0,1] [0,1[ or [0,1]. But the only open set among these sets is ]0,1[. Hence the result.Similarly, int[0, 1[= int]0, 1] =]0, 1[.b) int Q = . This is because any open interval contains points outside Q and so Q cannotcontain any nonempty open set. Similarly int(IR\Q) = .

Proposition 1.4 A set is open if and only if it is equal to its interior.

Proof. Suppose first A =A. But

A is open. Therefore A is open. Conversely, suppose that

A is open . But A contains A. Since

A is the biggest open set containing A, We have A

A.But

A A. Hence the equality.

Definition 1.5 Let X be a metric space. A subset F X is called closed if its complementis open.

Example. 1) [a, b] is closed because its complement ], a[]b, [ is open.2) [a, [ and ] , a] are closed.3) In a metric space singletons are closed. Indeed, let {a} be a singleton (one point set) and letx X {a}. Then x = a and so d(x, a) > 0. We claim that B(x, d(x, a)) X {a}. Indeed,let y B(x, d(x, a)). Then d(x, y) < d(x, a). This implies that y = a because otherwise wewould have d(x, a) < d(x, a). This means that y X {a}. Therefore etc.Proposition 1.5 A closed ball is closed.

Proof. Let B(a, r) be a closed ball in a metric space X. We have to prove that X B(a, r)is open. Let x X B(a, r). Then d(x, a) > r. Set = d(x, a) r. We claim thatB(x, ) X B(a, r). Indeed, let y B(x, ). Then d(x, y) < = d(x, a) r. Thereforer < d(x, a) d(x, y). By the triangle inequality, d(x, a) d(x, y) d(y, a). Hence d(y, a) > rand so y X B(a, r). This proves the claim. The claim implies that X B(a, r) is open.

Proposition 1.6 In a metric space X, the following properties hold.

(i) X and are closed.(ii) An arbitrary intersection of closed sets is closed.(iii) A finite union of closed sets is closed.


11/47

1.2. OPEN AND CLOSED SETS AND RELATED CONCEPTS 11

Proof. Take complements and use the properties of open sets.

Corollary 1.1 In a metric space every finite set is closed.

Remark 1.4 An arbitrary union of closed sets need not be closed. For example for each n

2,

the set [ 1n , 1 1n ] is closed. Howevern=2

[1

n, 1 1

n] =]0, 1[

is not closed.

Remark 1.5 A set which is not open is not necessarily closed. For example ]0,1] is neitheropen nor closed. Also a set can be both open and closed. Indeed, in any metric space X, and X are closed an open. Here is another example. Let d be the discrete distance on a set Xcontaining more than one point. Then B(a, 1) =

{a}

. Hence{

a}

is open. But we know that{a} is also closed. In fact every subset of X is open (because a subset is a union of singletons).But if every subset of X is open then every subset is also closed. A subset which is both closedand open is sometimes called clopen.

Definition 1.6 The closure of a set A is the intersection of all closed sets that contain A. Itis therefore the smallest closed set containing A. It is usually denoted by A and sometimes bycl(A).

Examples. a) ]0, 1[ = [0, 1]. Indeed, [0,1] is a closed set containing ]0,1[, therefore it contains]0, 1[. Thus ]0, 1[ ]0, 1[ [0, 1]. Therefore ]0, 1[ is either ]0,1[, ]0,1], [0,1[ or [0,1]. However, theonly closed set among these four sets is [0,1]. Hence the result. Similarly, ]0, 1] = [0, 1[ = [0, 1].

b) We shall see below that Q = IR . We say that Q is dense in IR. We shall also prove that theirrationals are dense in IR.

c) We shall see in the exercises that if A is the open disk x2 + y2 < 1, then the closure of A isthe closed disk x2 + y2 1.

Proposition 1.7 A set is closed if and only if it is equal to its closure.

Proof. Suppose first that A = A. Since A is closed, A is closed. Conversely, suppose thatA is closed. Since A contains A and A is the smallest closed set containing A, we have A A.But A A. Hence the equality.

Definition 1.7 Let (xn) be a sequence in a metric space (X, d). We say that (xn) convergesto a point x if d(xn, x) converges to 0 in IR. In terms of quantifiers

> 0 n0 IN n n0 d(xn, x) < .

Proposition 1.8 Let X be a metric space and let A X. Then, the following conditions areequivalent.

(i) x A.

(ii) Every neighborhood of x intersects A.

(iii) There is a sequence in A which converges to x.


12/47


Proof. Let B denote the set of points x satisfying property (ii). We prove that (i)(ii). Weclaim that B is closed. Let x / B, then there is an open neighborhood U of x not intersectingA. Consider an element y U. Since U is a neighborhood of y not intersecting A, we deducethat y / B. This means that U B and accordingly B is open. Therefore B is closed asclaimed. Observe now that B contains A. Since A is the smallest closed set containing A, we

get A B.(ii)(i). We prove that B A. Let x / A. Then x A. But A is an open set not intersectingA since A A A A = . This means that x / B.(ii)(iii). For every positive integer n the open ball B(x, 1n) intersects A. Choose accordinglyfor each n, an element xn in the intersection. This defines a sequence of points in A whichclearly converges to x.

(iii)(ii). Let U be a neighborhood of x. Since {xn} converges to x, we have that xn U forn large enough. This means that U intersects A.

Corollary 1.2 Let A

IR and let

{xn

}be sequence in A which converges to some x. Then

x A.Definition 1.8 Let X be a metric space. A subset A X is called dense if A = X.

Corollary 1.3 Let X be a metric space and A X. Then the following are equivalent

(i) A is dense in X.

(ii) Every nonempty open set of X meets A.

(iii) For everyx X, there exists a sequence in A that converges to x.

Corollary 1.4 The set of rational numbers and the set of irrational numbers are dense in IR.

Other examples. a) IR = IR {0} is dense in IR.b) IR Z is dense in IR.

Definition 1.9 Let A be a subset of a metric space X. A point x X is called a limit pointof A if every neighborhood of x meets A at a point different from x. The point x is called anisolated point of A if x A and x is not a limit point of A.It is clear that a limit point of A belongs to A and an isolated point of A belongs to A.

Examples. a) Let A =]0, 1[{2}. Then any point in [0, 1] is a limit point of A, whereas 2 isan isolated point of A.b) Every finite set in a metric space consists of isolated points. Indeed, let A = {x1, . . . , xn}be a finite set of n distinct points. For each i = 1, . . . , n,.let ri = minj=i d(xi, xj). ThenB(xi, ri) A = {xi}.c) Let A = { 1n | n = 1, 2, 3, . . .}. Then 0 is a limit point of A, and every point in A is isolated.The following proposition will be proved in the exercises.

Proposition 1.9 Let A be a subset of a metric space X and let x be a limit point of A. Thenevery neighborhood of x contains infinitely many points of A.

Definition 1.10 Let A be a set in a metric space X. The boundary of A denoted by A is

A = A A.


13/47

1.3. TOPOLOGICAL SPACES 13

Examples. 1) [0, 1] = [0, 1]]0, 1[= {0, 1}. Similarly, ]0, 1] = [0, 1[= {0, 1}2) Q = Q

Q = IR.

3) We shall see in the exercises that if A is the open disk x2 + y2 < 1, then A is the circlex2 + y2 = 1.

Remark 1.6 The boundary of a set is always a closed set. Why?

Remark 1.7 A = A is clopen.

1.3 Topological spaces

Now that we have enough examples of metric spaces, we can move to the next level of abstrac-tion.

Definition and examples

Definition 1.11 Let X be a set. A family T of subsets of of X is called a topology on Xprovided the following conditions hold.

(i) , X T.(ii) An arbitrary union of elements of T belongs toT.

(iii) A finite intersection of elements ofT belongs toT.The elements of T are called open sets of X. A topological space is a couple (X, T).Remarks. 1) IfT is a topology on X, then T P(X).

2) To prove property (iii), it is enough to prove that the intersection of two elements of T isalso an element ofT because then the result follows by induction.3) Suppose that T satisfies property (i). To prove (ii), we may assume that the sets arenonempty, because the empty sets do not change the union and so we can remove them. Toprove (iii), we may assume that none of the sets is empty. Because otherwise the intersectionwould be empty and so belongs to T by (i).Examples. 1) A metric space is a topological space. This follows from Proposition 1.2.But the converse is not true as we shall see. The topology defined from a distance is calledthe topology generated by a distance. Let (X, T) be a topological space. We say that T ismetrizable ifT is generated by a distance.2) Let X be a set. The collection {, X} is a topology called the trivial topology or theindiscrete topology. This topology is not metrizable (see below).

3) Let X be a set. The collection P(X) of all subsets of X is a topology called the discretetopology. This topology is metrizable because it is generated by the discrete distance. Observealso that any topology on X, satisfies {, X} T P(X). The trivial topology is the smallest(coarsest) topology and the discrete topology is the biggest (finest) topology on a set.

4) Let X be a set and A X. Then {, A , X } is a topology on X. This topology is notmetrizable.

5) Let X = {a,b,c} be a three point set. Then {, {a}, {a, b}, X} is a topology on X. Thistopology is not metrizable.

6) Let X be a set. Let T be the collection of all subsets of U X such that X U is finite orequal to X. Then T is a topology on X called the finite complement topology. We prove this.


14/47


(i) T because X = X and X T because X X = is finite.(ii) Let O be a collection of nonempty elements of T. Then X O = (X O) This

set is finite because it is an intersection of finite sets.

(iii) Let O1 and O2 be nonempty elements of

T. Then X

(O1

O2) = (X

O1)

(X

O2)

is finite as a union of two finite sets.

Closed sets and related concepts

All the concepts that we learned in the previous section can be defined only in terms of opensets. A set is closed if its complement is open. A neighborhood of a p oint is a set containingan open set containing the point. The interior of a set is the biggest open set contained in theset. The closure of a set is the smallest closed set containing the set. The boundary of a set isthe difference between the closure and the interior. Most of the results that we proved in themetric case still hold in a general topological space. To be more precise,

Proposition 1.10 Propositions 1.3, 1.4, 1.6, 1.7 and the results in exercises 4,8,12,13,14,19,20hold in an arbitrary topological space.

Convergence is defined as follows.

Definition 1.12 Let (xn) be a sequence of a topological space. We say that (xn) converges tox, if every neighborhood U of x, contains xn for all n large enough. In terms of quantifiers

U U(x) n0 IN n n0 xn U.

It is not difficult to see, that in a metric space, the first definition and the present one areequivalent (check this).

Proposition 1.8 and Corollary 1.3 remains partially true. More precisely,

Proposition 1.11 Let X be a topological space and let A X. Consider the followingconditions.

(i) x A.(ii) Every neighborhood of x intersects A.

(iii) There is a sequence in A which converges to x.

Then (i)(ii) and (iii)(i).It is not true that (i)(iii) in a an arbitrary topological space.

Corollary 1.5 Let X be a topological space and A X. Consider the following conditions.(i) A is dense in X.

(ii) Every nonempty open set of X meets A.

(iii) For everyx X, there exists a sequence in A that converges to x.Then (i)(ii) and (iii)(i).

Finally Proposition 1.9 does not hold in an arbitrary topological space. In the exercisesyour are asked to give a counterexample.


15/47


Hausdorff spaces

Definition 1.13 A topological space X is called a Hausdorff space if every two distinctpoints of X can be separated by two disjoint open sets. This means that if x = y then thereexists two disjoint open sets U and V such that x U and y V.

Proposition 1.12 A metrizable space is a Hausdorff space.

Proof. Choose a distance d that generates the topology of the space. Let x = y and letr = d(x, y)/2 > 0. Then the open balls B(x, r) and B(y, r) are disjoint open sets containing xand y respectively.

Corollary 1.6 A space which is not Hausdorff is not metrizable.

Now you can see why some topologies defined above are not metrizable. The finite comple-ment topology on an infinite space X is not Hausdorff and so not metrizable. In fact there areno disjoint non empty open sets. For suppose that U V = with U, V = . Then X U andX

V are finite. Therefore X

U

X

V is finite but X

U

X

V = X

(U

V) = X.

So X is finite contrary to our assumption. The finite complement topology on a finite spacecoincides with the discrete topology.

There are however Hausdorff spaces which are not metrizable. You will encouter them inyour graduate studies.

Proposition 1.13 A finite set in a Hausdorff space is closed.

Proof. It is enough to prove that singletons are closed. Consider a singletons {x}. Lety X {x}. We can separate x and y with disjoint sets U and V respectively. Then V X U X {x}. This means that X {x} is a neighborhood of y. Since y was arbitrary,this means that X {x} is a neighborhood of all its points and so it is open.

The converse is not true as testified by the finite complement topology on a infinite set.

Proposition 1.14 In a Hausdorff space, a sequence converges to at most one point.

Proof. Suppose that there is a sequence (xn) that converges to two distinct points x and y.Let U and V be two neighborhoods of x and y respectively. Then xn U for all n n1 andxn V for all n n2. Then xn U V for all n max(n1, n2). Contradiction.

Remark 1.8 In an arbitrary topological space, a sequence may converge to many points. Con-sider for instance IR equipped with the finite complement topology and consider the sequencexn = n. Then (xn) converges to every x IR. Indeed, let U be neighborhood ofx. Then IR Uis finite. Choose n0 > max(IR

U). Then xn = n

U for all n

n0.

Comparison of topologies

Definition 1.14 Let T and T be two topologies on a set X. If T T, we say that T isfiner or stronger than T; we also say that T is coarser or weaker than T. We say that TandT are comparable if eitherT T or T T.Examples. a) Let X = {a,b,c} and consider the following topologies T1 = {, {a}, {a, b}, X},T2 = {, {a}, {b}, {a, b}, X} and T3 = {, {a}, {c}, {a, c}, X}. Then T2 is finer than T1. How-ever, T1 and T3 are not comparable. Also T2 and T3 are not comparable.b) The finite complement topology on IR is weaker than the usual topology.

Suppose that we have two distances on the same set. Since each distance generates atopology, we can ask if these topologies are comparable.


16/47


Definition 1.15 Let X be a set. Let d be distance on X that generates a topology T. Letd be distance on X that generates a topology T. We say that d and d are topologicallyequivalent if T = T, that is, if they generate the same topology.We give some examples.

Proposition 1.15 Consider on IRN the three norms

x1 =Ni=1

|xi|, ||x||2 =Ni=1

|xi|21/2

, x = max(|x1|, . . . , |xN|).

Let d1, d2 and d be respectively the distances associated with these norms. Then d1, d2 andd generate the same topology on IR

N.

Proof. We prove that d2 and d are equivalent. In the exercises, you are ask to prove thatd1 and d are equivalent. It is not difficult to see that

||x|| ||x||2 N||x||.It follows that

d(x, y) d2(x, y)

N d(x, y).

Therefore,

Bd2(x, r) Bd(x, r) and Bd(x,rN

) Bd2(x, r).

Therefore etc.

Proposition 1.16 Let (X, d) be a metric space. Set (x, y) = min(1, d(x, y)). Then and dare topologically equivalent.

Proof. We know that is a metric. Let O be open for . Let x O. We need to show thatO contains a ball (relative to d) centered at x. Since O is open for , there exists r > 0 suchthat B(x, r) O. But now observe that Bd(x, r) B(x, r) since (x, y) d(x, y). Therefore,Bd(x, r) O. This means that O is open for d. Conversely, let O be open for d and let x O.Then there exists r < 1 such that Bd(x, r) O. But then B(x, r) Bd(x, r). Indeed, lety B(x, r). If d(x, y) > 1, then (x, y) = 1 < r, contradicting the choice of r. Therefore(x, y) = d(x, y) < r. Conclusion: B(x, r) O and so O is open for .

Basis for a topology

A general way of defining a topology is by specifying a basis.

Definition 1.16 A subset B P(X) is called a basis provided the following hold.1. For allx X, there exists B B such that x B.2. IfB1, B2 B and x B1 B2, then there exists B3 B such that x B3 B1 B2.

The elements ofB are called basis elements. The topologyT generated byB is defined asfollows: O T if whenever x O, there exists B B such that x B O.We need to check that T defined in this way is indeed a topology.

1. The empty set satisfies the condition vacuously. The first condition of a basis impliesthatX T.


17/47


2. Let (O)A be a collection of elements of T. Let x O. Then there is A suchthat x O. Since O T, there exists B B such that x B O and so B O.This means that O T.

3. Let O1, O2 Tand let x O1O2. Then there exists B1, B2 B such that x B1 O1and x

B2

O2

. The second condition of a basis implies that there exists B3

B suchthat x B3 B1 B2. Therefore x B3 B1 B2 O1 O2.

Remark 1.9 Observe that B T.

Examples. 1) Let X be a metric space. The collection of balls B(x, r) (x X ,r > 0) isa basis for the metric topology. In particular, the collection of open disks and the collectionof open squares are bases for the usual topology of IR2. The first condition of a basis issatisfied. Let now x B(a1, r1) B(a2, r2). Let r = min(r1 d(x, a1), r2 d(x, a2)). ThenB(x, r) B(a1, r1) B(a2, r2).2) Let X be a set. The collection of singletons of X is basis for the discrete topology on X.

3) The collection of all open intervals ]a, b[ of IR is a basis for the usual topology of IR.

4) Let X and Y be two topological spaces. Let B be the collection of all sets of the from U Vwhere U is open in X and V is open in Y. Then B is a basis. The topology generated by thisbasis is called the product topology of X Y.

Proposition 1.17 Let B be a basis for a topology T on a set. Then T equals the collectionof all unions of elements ofB.

Proof. Since B T and T is stable under unions, a union of elements ofB belongs to T.Conversely, given O T, for each x O, there exists Bx B such that x Bx O. ThenO = xBx.

Accumulation points and subsequences

Recall that a sequence (xn) of a topological space is convergent to a limit x if every neighborhoodofx contains xn for all n large enough, that is except for a finite set of indices. Not all sequencesare convergent. A weaker notion than that limit is that of accumulation point or cluster point.

Definition 1.17 Let (xn) be a sequence of a topological space. We say that x is an accumu-lation point or a cluster point of (xn) if every neighborhood of x contains xn for infinitelymany indices n. In symbols

U U(x) n IN m n such that xm U.This is equivalent to saying that

{n

IN

|xn

U}

is infinite.

Compare with the condition of convergence

U U(x) n0 IN m n0 xm U.Examples. 1) It should be clear that a limit of a convergent sequence is an accumulationpoint of the sequence. The converse however is not true as testified by the next example.

2) Let xn = (1)n. Then (xn) is not convergent in IR but has two accumulation points 1 and1.

Proposition 1.18 Let A be the set of accumulation points of a sequence (xn). Then

A =n=1

{xk|k n} =n=1

{xn, xn+1, . . .}.


18/47


Proof. Let Fn = {xn, xn+1, . . .}. Thenx A U U(x) n IN m n xm U

n IN U U(x) U Fn =

n

IN x

Fn

x Fn.

Definition 1.18 Let {xn} be a sequence of some set and let n1 < n2 < < ni < be anincreasing sequence of positive integers. Then the sequence {yi} defined by yi = xni is called asubsequence of the sequence{xn}.Examples. a) Let xn = n. Then {x2n+1} is a subsequence of {xn}.b) Let xn = 2

n. Then the sequence 1, 4, 16, 64, . . ., i.e. {x2n} is a subsequence of {xn}.

Proposition 1.19 Let (xn) be a sequence of a topological space X.

(i) If(xn) has a subsequence converging to , then is an accumulation point of (xn).

(ii) Conversely, if X is metrizable and is an accumulation of (xn) then there exists a subse-quence of (xn) that converges to .

Proof. (i). Let (xnk) be a subsequence of (xn) that converges to . Let U be a neighborhoodof . Then, there exists k0 such that xnk U for all k k0. This means that xm U for allm {nk0 , nk0+1, nk0+2 . . .} and this set of integers is infinite.(ii). In the condition

U

U(x)

n

IN

m

n such that xm

U,

take first U = B(, 1) and n = 1. Then there exists n1 1 such that xn1 B(, 1). Nexttake U = B(, 12) and n = n1 + 1. Then there exists n2 n1 + 1 such that xn2 B(, 12). Atthe kth step, take U = B(, 1k ) and n = nk1 + 1. Then there exists nk nk1 + 1 such thatxnk B(, 1k). This procedure defines a subsequence (xnk) that converges to .

1.4 Subspaces

Proposition 1.20 Let (X, T) be a topological space and let Y X. The collectionTY = {Y O|O T }

is a topology on Y called the subspace topology.

Therefore, in the subspace topology, the open sets of Y are the intersection ofY with the opensubsets of X.

Proof.

(i) Observe that = Y and Y = Y X. It follows that , Y TY.(ii) Let (O) be a collection of elements ofTY. Then there exists a collection (U) of elements

ofT such that O = Y U. Then

O =

(U Y) = (

U) Y. Since

U T, itfollows that O TY.

(iii) Let O1, O2 TY. Then there exists U1, U2 T such that O1 = Y U1 and O2 = Y U2.Then O1 O2 = Y (U1 U2) TY.


19/47

1.4. SUBSPACES 19

Examples. 1) Consider [0, [ as a subspace of IR with the usual topology. Then although[0, 1[ is not open in IR, it is open in [0, [ because for example [0, 1[= [0, [] 1, 1[.2) Consider Z as a subspace of IR with the usual topology. Then each singleton {n} is open inZ because {n} =]n 1, n + 1[Z. It follows that that every subset ofZ is open in the subspacetopology. This implies that the subspace topology and the discrete topology on Z coincide.

Corollary 1.7 Let X a be a topological space, let Y be a subspace of X and let U Y. If Uis open in Y and Y is open in X, then U is open in X.

Proof. If U is open in Y, then U = Y V where V is open in X. Since Y is also open in X,then so is Y V.

Remark 1.10 Let X be a topological space and let Z Y X. The the topology on Zinherited from X is the same as the topology on Z inhereted from Y. Indeed ifO is open inX, then O Z = (O Y) Z.

Proposition 1.21 Let Y be a subspace of a topological space X and let A Y then A isclosed in Y if and only if A = Y F where F is closed in X.

Proof. We have to show first that if F is closed in X, then Y F is closed in Y, that is, itscomplement in Y is open in Y. Now

Y (Y F) = Y (X F).

This set is open in Y because X F is open in X.Now conversely, suppose that A is closed in Y. Then YA is open in Y and so YA = YO

where O is open in X. Then A = Y

(X

O). But X

O is closed in X. Hence the conclusion.

Example. ]0,1] is closed in ]0, [ since ]0, 1] =]0, [[0, 1].

Corollary 1.8 Let X a be a topological space, let Y be a subspace of X and let B Y. If Bis closed in Y and Y is closed in X, then B is open in X.

Let (X, d) be a metric space and let Y X. Let dY denote the restriction of d to Y Y.Then dY is a metric on Y which therefore generates a topology on Y. How does this topologyrelate to the subspace topology? It turns out that the two topologies coincide. If x Y, we

denote by BY(x, r) the ball in Y of center and radius r. It coincides with Y B(x, r).Proposition 1.22 The metric topology induced on Y coincides with the subspace topology ofY.

Proof. Let A be open relative to the subspace topology. Then, A = Y O where O is openin X. Let x A. Then x O. Since O is open there exists > 0 such that B(x, ) O. ThenB(x, ) Y O Y = A, that is, BY(x, ) A. This means that A is open relative to themetric topology on Y.

Conversely, Suppose that A is open relative to the metric topology of Y. For each x A,set x = dY(x, Y A). Then x > 0 because dY(x, Y A) = 0 x Y A = Y A sinceY

A is closed in Y. We claim that if y

Y and d(x, y) <

xthen y

A. For if y /

A, then

d(x, y) x by definition ofx. It follows that A = xABY(x, x). Let now O = xAB(x, x).Then O is open in X and O Y = A.


20/47


1.5 Product spaces

If X1, X2 . . . , X n are sets, the set of ntuples (x1, . . . , xn) where each xi Xi is denoted byX1 X2 Xn or by

ni=1 Xi.

Definition 1.19 Let X1, X2 . . . , X n be topological spaces. The product topology on X1X2 Xn is the topology having as basis the collection B of all sets of the form U1U2 Unwhere each Ui is an open set of Xi.

This means that O X1 X2 Xn is open if whenever x = (x1, . . . , xn) O, there existopen sets Ui Xi such that x

ni=1 Ui O.

Proposition 1.23 The product of open sets is open. The product of closed sets is closed.

Proof. The first part is rather trivial because a product of open sets is a basis element andtherefore belongs to the product topology. For the second part, it is enough to prove that theproduct of two closed sets is closed (then the result follows by induction). Let A X1 andB X2 be closed sets. Then X1 X2 A B = (X1 A1) X2X1 (X2 B) which isthe union of two open sets.

In the exercises, you are asked to prove the following proposition.

Proposition 1.24 Let (X1, d1), . . . , (Xn, dn) be n metric spaces. Let X = X1 Xn. Forx = (x1, . . . , xn) and y = (y1, . . . , yn), define

1(x, y) =ni=1

di(xi, yi)

2(x, y) = n

i=1

di(xi, yi)2

1/2

(x, y) = maxi=1...,n

di(xi, yi).

Then 1, 2 and are metrics on X that generate the product topology on X.


21/47

Chapter 2

Continuous functions andhomeomorphisms

Let us start by defining continuity in metric spaces. Let f : (X, dX)

(Y, dY) be a functionbetween two metric space and let x X. We say that f is continuous at x if f(y) can be madearbitrarily close to f(x) for y sufficiently close to x. In metric spaces the concept of a distancegives a meaning to closeness. This leads to the following definition. We can think of as theaccuracy in the approximation.

Definition 2.1 Let f : (X, dX) (Y, dY) be a function between two metric spaces and letx X. f is said to be continuous at x if the following condition holds.

> 0 > 0 such that dX(x, y) < dY(f(x), f(y)) < .

Remark 2.1 In the definition above, one can replace 0, > 0 such that f(B(x, )) B(f(x), ).

(iii) V U(f(x)), U U(x) such that f(U) V.(iv) If{xn} is a sequence of X that converges to x then {f(xn)} converges to f(x).

21


22/47

22 CHAPTER 2. CONTINUOUS FUNCTIONS AND HOMEOMORPHISMS

(v) If {xn} is a sequence of X that converges to x then there exists a subsequence (xnk) of(xn) such that {f(xnk)} converges to f(x).

Proof. We prove that (i)(ii)(iii)(iv)(v)(i).(i)(ii). (ii) is just a reformulation of (i). Indeed, let and be as in the definition of continuity.Let y f(B(x, )). Then y = f(z) for some z B(x, ). This means that d(z, x) < andso by the condition of continuity, d(f(z), f(x)) < , that is, d(y, f(x)) < . This means thaty B(f(x), ).(ii)(iii). Let U be a neighborhood off(x). Then there exists a ball B(f(x), ) V. Take nowU = B(x, ). Then U is a neighborhood ofx. Now (ii) can be written as f(U) B(f(x), ) V.(iii)(iv). Let (xn) be a sequence of X that converges to x. Let V be a neighborhood of f(x).Then there exists a neighborhood U of x such that f(U) V. Now convergence means thatxn U for all n large enough. Therefore f(xn) f(U) V for n large enough. This meansthat (f(xn)) converges to f(x).

(iv)(v). Is clear.(v)

(i). Suppose that f is not continuous at x. Then there exists > 0 such that for all > 0

there exists y with d(y, x) < and d(f(y), f(x)) . Taking = 1n we get a sequence (yn)converging to x and such that no subsequence of (f(yn)) converges to f(x), a contradiction.

The only condition which does no involve metrics is condition (iii) and this will be ourdefinition of continuity in a general topological space.

Definition 2.2 A function f : X Y between two topological space is said to becontinuousat x X, if for every neighborhood V of f(x), there exists a neighborhood U of x such thatf(U) V.

Remark 2.2 We can replace neighborhood by open neighborhood.

Remark 2.3 If f is continuous at x and (xn) is a sequence converging to x, then the sequence(f(xn)) converges to f(x). This follows from the definitions. The converse is however not true,i.e. if for every sequence (xn) converging to x we have (f(xn)) converges to f(x), it does notfollow that f is continuous at x. The converse is true if X is metrizable.

Definition 2.3 A function f : X Y between two topological spaces is called continuous ifit is continuous at every point of X.

Theorem 2.1 Let f : X Y be a function between two topological spaces. Then the followingare equivalent.

(i) f is continuous.

(ii) The inverse image underf of any open subset of Y is open in X.

(iii) The inverse image underf of any closed subset of Y is closed in X.

Proof. (i)(ii). Let O Y be open and let x f1(O) (so that f(x) O). Since O is aneighborhood of f(x), there exists according to the previous proposition, a neighborhood U ofx such that f(U) O. Therefore U f1(f(U)) f1(O). This means that f1(O) is open.

(ii)(i). Let x X and let V be an open neighborhood of f(x). Set U = f1(V). Then Uis a neighborhood ofx (being an open set containing x) and f(U) = f(f1(V)) V. Accordingto the previous proposition, f is a continuous at x. Since x was arbitrary this proves that f is

continuous.

The equivalence (ii)(iii) follows from the fact that f1(Y O) = X f1(O).


23/47

23

Corollary 2.1 Let X be a topological space and f : X IR be a continuous function. Leta IR. Then

(i) The sets{x X|f(x) < a} and {x X|f(x) > a} are open.(ii) The sets

{x

X

|f(x)

a}

,{

x

X|f(x)

a}

and{

x

X|f(x) = a

}are closed.

Example. Since in a metric space x d(x, a) is continuous, an open ball is open and a closedball is closed. The set {x X|d(x, a) > r} is open, the set {x X|d(x, a) r} is closed. Asphere {x X|d(x, a) = r} is closed.

Proposition 2.2 The following are true.

(i) The composition of two continuous functions is continuous.

(ii) The restriction of a continuous function is continuous.

(iii) IfA

X, the inclusion map j : A

X is continuous.

Proof. (i). Let f : X Y and g : Y Z be two continuous functions. We have to provethat g f : X Z is continuous. Observe that (g f)1(O) = f1(g1(O)) for every subsetO Z. In particular, if O is open, then g1(O) is open in Y because g is continuous. Since fis continuous, we have f1(g1(O)) is open. Hence the result.

(ii). Let f : X Y be continuous and let A X. We have to prove that f|A : A Y is alsocontinuous. Let O be open in Y. Then

f|A1

(O) = A f1(O) is open in A.(iii). Let O be open in X. Then j1(O) = A O is open in A.

Proposition 2.3 For real valued functions, the sum and product of continuous functions are

continuous. If f : X IR is continuous, then 1/f is continuous on its domain, i.e. on the setwhere f does not vanish.

Proof. (i). Let f, g : X IR be continuous at some point x0. Then, given > 0, there exists aneighborhood U1 ofx0 such that |f(x)f(x0)| < /2. Also, there exists a neighborhood U2 ofx0such that |g(x)g(x0)| < /2. Then for all x U1U2, we have |f(x)+g(x)(f(x0)+g(x0)| < .(ii). Write f(x)g(x) f(x0)g(x0) = f(x)g(x) f(x0)g(x) + f(x0)g(x) f(x0)g(x0). Then, bythe triangle inequality

|f(x)g(x) f(x0)g(x0)| |g(x)||f(x) f(x0)| + |f(x0)||g(x) g(x0)|

Let > 0 be given. The continuity of g at x0 implies that there exists a neighborhood U1 of x0such that |g(x)g(x0)| < 2(|f(x0)|+1) for all x U1. This implies that |f(x0)||g(x)g(x0)| < /2for all x U1. The continuity of g at x0 also dictates that |g(x)| < |g(x0)| + 1 for all x insome neighborhood U2 of x0. Finally, the continuity of f at x0 dictates that there exists aneighborhood U3 of x0 such that |f(x) f(x0)| < 2(|g(x0)|+1) for all x U3. Therefore forx U1 U2 U3, we have |f(x)g(x) f(x0)g(x0)| < .(iii). Let x0 be a a point where f(x0) = 0. Then |f(x0)| > 0. Then, there exists a neighborhoodU1 of x0 such that |f(x) f(x0)| < |f(x0)|/2. Then |f(x)| > |f(x0)|/2 and so |f(x)||f(x0)| >|f(x0)|2/2 for all x U1. Given > 0 there exists a neighborhood U2 of x0 such that |f(x) f(x0)| < |f(x0)|

2

2 for all x U2. It follows that if x U1 U2 then 1f(x) 1f(x0) = |f(x) f(x0)||f(x)f(x0)| < .


24/47


Given two sets X and Y, the projection 1 : X Y X is defined by 1(x, y) = x.The projection 2 : X Y X is defined by 2(x, y) = y. Observe that if U X then11 (U) = U Y and if V Y then 12 (V) = X V. Therefore we have

Proposition 2.4 Let X and Y be topological spaces then the projections on X and Y arecontinuous (X Y is equipped with the product topology).

Remark 2.4 In fact, the product topology is the smallest topology that makes the projectionscontinuous. Indeed, let X and Y be two topological spaces and let T be a topology on XY thatmakes the projections 1 and 2 continuous. We have to prove that T contains the producttopology, and for this it is enough to prove that any basis element of the product topologybelongs to T. So let U and V be open sets of X and Y respectively. We already observedthat 11 (U) = U Y. Since 1 is continuous with respect to T, it follows that U Y T.Similarly, X V T. Therefore their intersection U V also belongs to T.

Proposition 2.5 Let X ,Y,Z be topological spaces and let f : X Y Z. Let f1 and f2denote the components of f. Then f is continuous if and only if f1 and f2 are continuous.

Proof. Suppose first that f is continuous. Observe that if Y and Z denote the projectionsinto Y and Z respectively, then f1 = Yf and f2 = Zf. Since the projection are continuous,it follows that f1 and f2 are continuous.

Suppose conversely that f1 and f2 are continuous. We show that the inverse image under fof a basis element U V is open in X. This follows from the fact that

f1(U V) = f11 (U) f12 (V).

Proposition 2.6 Let f : X Y Z be continuous. Then the partial map f(, y) : X Zand f(x, ) : Y Z are continuous.Proof. Let g(x) = f(x, y) for y fixed. Let (x) = (x, y). Then : X X Y is continuous(because each component is continuous). Since g = f , it follows that that g is continuous.Similarly for the partial map f(x, ).

Remark 2.5 If f : X Y Z and the partial maps f(, y) : X Z and f(x, ) : Y Z arecontinuous, it does not follow that f is continuous. For example, let f : IR IR IR be definedby f(x, y) = xy

x2+y2if (x, y) = (0, 0) and f(0, 0) = 0. Then the partial maps are continuous.

However f is not continuous at (0,0) because it has not limit there.

Definition 2.4 A function f : X Y between two topological spaces is called a homeomor-phism if it is bijective, continuous and its inverse is also continuous.

The concept of homeomorphism is fundamental in topology as is the concept of isomorphismin algebra. Recall that two isomorphic groups have the same structure and therefore may beidentified. Similarly, two homeomorphic space have exactly the same topological properties andtherefore can be identified.

Remark 2.6 A continuous bijection need not be a homeomorphism. Indeed, let T be the usualtopology on IR and let TF be the finite complement topology on IR. Let i : (IR, T) (IR, TF)be the identity function. Then i is clearly a bijection. It is continuous because if O

TFthen

i1(O) = O T. However, i1 : (IR, TF) (IR, T) is not continuous because there exists anelement O T and O / TF.


25/47

25

Remark 2.7 Let T and T be two topologies on a set X and consider the identity map i :(X, T) (X, T) defined by i(x) = x. Then T = T if and only if i is a homeomorphism.

Proposition 2.7 All intervals of the real line of the form [a, b] with a < b are homeomorphic.

Proof.The map x (x a)/(b a) is a homeomorphism between [a, b] and [0, 1].

Proposition 2.8 All open intervals of the real line are homeomorphic.

Proof. The map x (x a)/(b a) is a homeomorphism between ]a, b[ and ]0, 1[. Thereforeall intervals of the form ]a, b[ are homeomorphic between each other. The function x tan xis a homeomorphism between ] /2, /2[ and IR. The interval ]a, +[ is homeomorphic to IRby x ln(x a). The interval ] , a[ is homeomorphic to IR by x ln(a x).

Other examples. 1) A circle and an ellipse are homeomorphic (for example by the map(x, y) (ax,by)).

2) Rotations, translations, homotheties and symmetries of the plane or space are homeomor-phisms.

3) The punctured disc {0 < x2 + y2 < 1} and the annulus {1 < x2 + y2 < 2} are homeomorphic.For X = (x, y) let f(X) = X + X

||X||= ||X||+1

||X||X. Then f1(X) = ||X||1

||X||X. Then f is a

homeomorphism.

4) A circle and the boundary of a square are homeomorphic. To be specific consider the unitcircle S1 defined by x2 + y2 = 1 and the square C, defined by |x| + |y| = 1. Let f : IR2 IR2be given by f(x, y) = (x|x|, y|y|). Then f is continuous because its components are continuous.f is bijective and its inverse is given by f1(x, y) = (g(x), g(y)) where g(x) =

x if x 0 and

g(x) =

x ifx

0. By the pasting lemma, g is continuous (draw the graph of g). It follows

that f1 is also continuous. Thus f is a homeomorphism.Next, we prove that f(S1) = C. Indeed, let first (x, y) S1, then x2 + y2 = 1 and so

|x|x|| + |y|y|| = 1. This means that f(x, y) C and so f(S1) C. Conversely, let (u, v) C.Then there exists (x, y) IR2 such that f(x, y) = (x|x|, y|y|) = (u, v). Then x2 = |u| andy2 = |v|. Since |u| + |v| = 1, then x2 + y2 = 1 i.e (x, y) S1.

In fact, f maps the open unit disc onto the open square, and the exterior of the disc ontothe exterior of the square.

4) Equipp IR2 with the Euclidean distance d((x, y), (x, y)) =

(x x)2 + (y y)21/2 (butany other equivalent distance would do). Equipp lC with the distance d(z, z) = |z z| =

(x x)2 + (y

y)21/2. Then IR2 is homeomorphic to lC. Indeed, consider the function

f : IR2 lC defined by f(x, y) = x + iy. Then f is a bijection. Next, f preserves the distances(we say that f is an isometry). Therefore f is continuous. Since f1 also preserves distances,it is continuous as well.

5) A circle with a point removed is homeomorphic to IR. Let S1 denote the unit circle in IR2

and let P denote the point with coordinates (0,1). The stereographic projection with pole Passociates to each point M S1 P the point N which is the point of intersection of the line(P M) with the horizontal line {y = 0}. If M has coordinates (x, y), then the point N hascoordinates ( x1y , 0). Indeed, let N have coordinates (X, 0). Since N belongs to (P M), there

exists IR such that P N = P M and so (X 0, 0 1) = (x 0, y 1). It follows that = 1/(1 y) and so X = x1y . This defines a continuous function F : S1 P IR given by

F(x, y) =x

1 y .


26/47


It inverse is given by

F1(X) =1

X2 + 1(2X, X2 1).

Indeed, if X = x1y , then X x = yX. Squaring and using the fact that x2 + y2 = 1, we getx = 2X

X2

+1

, and then y = X21

X2

+1

. It is clear that F1 is continuous.

6) A sphere with a point removed is homeomorphic to IR2. Let S2 denote the unit sphere inIR3 defined by x2 + y2 + z2 = 1 and let P denote the point with coordinates (0,0,1). Thestereographic projection with pole P associates to each point M S2 P the point N which isthe point of intersection of the line (P M) with the plane {z = 0}. IfM has coordinates (x,y ,z),then the point N has coordinates ( x1z ,

y1z , 0). Indeed, let N have coordinates (X,Y, 0). Since

N belongs to (P M), there exists IR such that P N = P M and so (X 0, Y 0, 0 1) =(x 0, y 0, z 1). It follows that = 1/(1 z) and so X = x1z and Y = y1z . This definesa homeomorphism F : S2 P IR2 given by

F(x,y ,z) = x1 z

,y

1 zwhose inverse is given by

F1(X, Y) =1

X2 + Y2 + 1(2X, 2Y, X2 + Y2 1).

7) More generally the unit sphere Sn (defined by x21 + x22 + x2n+1 = 1) with the north pole

removed is homeomorphic to IRn via the stereographic projection

F(x1, x2, . . . , xn+1) = x1

1

xn+1

,x2

1

xn+1

, , xn1

xn+1 .

The inverse is given by

F1(X1, X2, . . . , X n) =1n

i=1 X2i + 1

(2X1, 2X2, , 2Xn,ni=1

X2i 1).

Definition 2.5 Let X andY be two topological spaces. A topological imbedding or simplyan imbedding of X in Y, is a map f : X Y such that f : X f(X) is a homeomorphism.The existence of an imbedding of X in Y, means that we can place a copy of X inside Y.

Examples. 1) Let f : IR IR2 be defined by f(x) = (x, 0). Then f is an imbedding.2) The map g : IR IR2 defined by g(x) = (x, x3) is an imbedding.3) The map g : [0, 1] ]0, 1[ defined by g(x) = 13 + 16x is an imbedding.


27/47

Chapter 3

Compactness

Now we move to another important concept in topology. Compactness of in interval [a, b]was formulated first in terms of sequences and subsequences. However, it took some time formathematicians to formulate the modern notion of compactness in a general topological space.

This is true for many abstract concepts of Mathematics. You usually wonder how did we arriveat an abstract concept or a nontrivial proof. In fact, in the courses that we teach, we give youa synthesis of a long process of distillation of mathematical ideas.

3.1 Definitions, examples and properties

Definition 3.1 Let X be a topological space. A collection of subsets (O)L of X is called acovering of X if X =

L

O. The collection is called an open covering if its elements are

open.

Definition 3.2 A topological spaceX is called compact if every open covering of X containsa finite subfamily that also covers X. Otherwise stated, if X =

I

O, then there is a finite set

J L such that X = J

O.

Examples. a) Any finite space is compact.

b) A space having a finite topology is compact.

c) A space X with the finite complement topology is compact. Indeed, let (O) be an opencovering of X. We may assume that X = (otherwise there is nothing to prove). Then thereexists an nonempty element O0 . Then, by definition of open sets, X O0 is a finite set{x1, . . . , xn}. Now each xi belongs to an element Oi . Therefore we have X = O0 O1 On . Thus, X is covered by finitely many elements of the original covering. Since any subspaceof X has the finite complement topology, it follows that any subspace of X is compact.

d) Consider X = {0}{ 1n |n IN} as a subspace of IR. Then X is compact. Indeed let (O) bean open covering of X. Then 0 belongs to some O0. Since (

1n) converges to 0, O0 contains all

1n except finitely many terms 1,

12 , . . . ,

1n0

. Now each 1i belongs to some open set Oi . It followsthat X = O0 O1 On . The argument that we used shows that more generally, if (xn)is a sequence of a topological space that converges to x, then the subspace {xn|n IN} {x}is compact.

e) The real line is not compact. Indeed, IR =

nIN] n, n[ and no subfamily of (] n, n[)n can

cover IR.f) (0, 1] is not compact. For ]0, 1] =

n1

] 1n , 1] and no finite subfamily of (]1n , 1])n can cover ]0,1].

27


28/47

28 CHAPTER 3. COMPACTNESS

g) The subspace Y = { 1n |n IN} of IR is not compact. Indeed, { 1n}n is an open covering of Ythat has no finite subcollection that covers Y. Therefore removing a point of a compact spacecan destroy the property of compactness.

Remark 3.1 Following Bourbaki, most French mathematicians include the Hausdorff condi-

tion in the definition of compactness. This is not the case with American and Russian mathe-maticians. We follow the American and Russian defintion. In the french litterature, what wecall compact is called quasi-compact.

Compactness can also be formulated in terms of closed sets.

Definition 3.3 A collection (F)L of sets is said to have the finite intersection propertyifJF = for every finite subcollection (F)J.

Example. Set for IR, F = [, [. Then the collection (F)IR has the finite intersectionproperty. Indeed, let J = {1, . . . , n} be a finite subset of IR. Then

JF = F1 F2

Fn = [t, [= where t = max(1, . . . , n). Observe however that the whole intersectionIR F =

because there is no real number bigger than every other real number.

Taking complements in the definition of compactness, one can show the following.

Theorem 3.1 Let X be a topological space. Then X is compact if and only if for every collec-tion (F)L of closed sets of X having the finite intersection property, the whole intersectionL F is nonempty.

Corollary 3.1 Let X be topological space and let (Fn) be a decreasing sequence of nonemptyclosed subsets of X (that is, F1 F2 ). Then

Fn = .

Now we prove some facts about subspaces. If Y is a subspace of a topological space X, acollection (O) of subsets of X is a covering of Y (or covers Y) if Y

O.

Lemma 3.1 Let Y be a subspace of a topological space X. Then Y is compact if and only ifevery covering of Y by open subsets of X contains a finite subcollection that also covers Y.

Proof. This follows from the definition of the subspace topology.

Proposition 3.1 Every closed subspace of a compact space is compact.

Proof. Let Y be a closed subspace of a topological space X. Let (O)L be a coveringof Y by open subsets of X. Add to this covering the open set X Y. Otherwise stated let be an element not in L and set O = X Y and consider the new collection (O)L{}.Then this new collection is an open covering of X. Compactness of X implies that a finite

subcollection (O)J covers X and consequenlty also Y. Then the collection (O)J{} is afinite subcollection of (O)L that covers Y.

Theorem 3.2 Every compact subspace of a Hausdorff space is closed (in the space).

Proof. Let X be a Hausdorff space and let K X be compact. We prove that K is open.Let x / K. For any y K there is an open neighborhood Uy of x and an open neighborhoodVy of y such that Uy Vy = . Now K

yK

Vy . Since K is compact K Vy1 Vyn.Then U = Uy1 Uyn is a neighborhood of x not intersecting K, that is, contained in thecomplement of K.

Remark 3.2 If the Hausdorff condition is removed in the theorem above, then the conclusionmay fail. Consider an infinite set X with the finite complement topology. Then the closedsubsets of X are X and the finite sets. However any subspace of X is compact.


29/47

3.2. COMPACT SUBSPACES OF IRN 29

Theorem 3.3 The image of a compact space under a continuous map is compact. In particularcompactness is preserved under homeomorphism.

Proof. Let f : X Y be continuous with X compact. Let (O)L be a covering of f(X)by open subsets of Y. Then the collection (f1(O)) is an open covering of X (because fis continuous). Compactness of X implies that finitely many subsets f1(O1),

, f1(On

)cover X. It follows that O1, , On cover f(X).

Here is a useful criterion involving compactness for proving that a continuous bijection isactually a homeomorphism.

Theorem 3.4 Let f : X Y be a continuous bijection. If X is compact and Y is Hausdorff,then f is a homeomorphism.

Proof. We have to prove that f1 is continuous and this is equivalent to proving that fmaps closed sets into closed sets (because f = (f1)1). So let A X be closed. Then A iscompact. By the preceding theorem, f(A) is compact. Since Y is Hausdorff, f(A) is closed in

Y.

Our next theorem is that the product of compact spaces is compact. For this we need alemma.

Lemma 3.2 (The tube lemma) Let X and Y be two topological spaces with Y compact. IfN is an open set of X Y conatining the slice{x0} Y, then N contains a tubeW Y whereW is a neighborhood of x0 in X.

Proof. Ify Y, then (x0, y) N. Then, by definition of the product topology, there exists abasis element UyVy containing (x0, y) and contained in N. Then the collection (UyVy)yY isan open covering of x0 Y. But x0 Y is compact being homeomorphic to Y. Therefore, thereexist finitely many basis elements U1

V1, . . . U n

Vn that cover x0

Y. Let W = U1

Un.

Then W is an open neighborhood of x0. We claim that W Y ni=1 Ui Vi. Indeed, let

(x, y) W Y. The point (x0, y) x0 Y ni=1 Ui Vi. So y Vj for some j. But

x W Uj. It follows that (x, y) Uj Vj and the claim is proved. But by construction,Ui Vi N for all i and so

ni=1 Ui Vi N. Consequently, W Y N.

Theorem 3.5 A product of compact spaces is compact.

Proof. It is enough to prove the theorem for the product of two compact spaces since thegeneral result follows by induction on the number of spaces. So let X and Y be two compactspaces. Let (O) be open covering of X Y. Given x X, the slice x Y is compactand therefore can be covered by finitely many elements O1, . . . , On of the covering. Let

N =ni=1 Oi . By the tube lemma, N contains a tube Wx Y containing x Y. Now thecollection (Wx)xX is an open covering ofX. Compactness ofX implies that there exists a finite

subcollection Wx1, . . . , W xk that covers X. Therefore the union of the tubes Wx1Y , . . . , W xkYis X Y. Since each of these tubes is covered by finitely many O, then so is X Y.

3.2 Compact subspaces of IRn

Theorem 3.6 Any closed interval [a, b] of the real line is compact.

Proof. Let [a, b] iI

Oi. We need to prove that [a, b] is covered by finitely many Oi. We

divide the proof into four steps.

Step 1. We show that if x [a, b[, then there is y ]x, b] such that [x, y] is covered by exactlyone Oj. Thus let x [a, b[. Then x iIOi, and so there is an index j I such that x Oj .


30/47


But since Oj is open, we can find an , 0 < < b such that ]x , x + [ Oj . Choose now anelement y ]x, x + [. Then [x, y] ]x , x + [ Oj. Thus the claim is proved.Step 2. Let C be the set of all y ]a, b] such that [a, y] can be covered by finitely many Oi.Applying Step 1 to x = a, we see that C = . Since C is bounded from above by b, C has aleast upper bound c.

Step 3. We prove that c C. There is k I such that c Ok. Since Ok is open, there is > 0 such that ]c, c + [ Ok. By the property of the least upper bound, there is an elementz C such that c < z c. Then [z, c] Ok. Also, [a, z] can be covered by finitely manyOi since z C. Thus, [a, c] = [a, z] [z, c] can also be covered by finitely many Oi.Step 4. We show that c = b and the proof is finished. Suppose that c < b. Applying Step 1to x = c, we see that there is y ]c, b] such that [c, y] can be covered by one element Oj . Butthen, [a, y] = [a, c] [c, y] can be covered by finitely many elements of the cover. This meansthat y C and accordingly y should not be greater than c. We thus got a contradiction.

Recall that we defined three topologically equivalent distances on IRn:

1(x, y) =ni=1

|xi yi|

2(x, y) =

ni=1

|xi yi|21/2

the Euclidean metric

(x, y) = maxi=1...,n

|xi yi| the square metric.

We also noticed that

d(x, y)

d2(x, y)

nd(x, y).

and

d(x, y) d1(x, y) nd(x, y).Now a subset A of a metric space X is called bounded if it is contained in a ball, or equiva-lently if its dimaeter is finite. The inequalities between the usual distances on IRn imply thatboundedness in 2 is equivalent to boundednes in or boundedness in 1.

Theorem 3.7 A subspace A of IRn is compact if and only if it is closed and bounded (in anyof the usual metrics).

Proof. Suppose first that A is compact. Then A is closed by a previous theorem. Nowconsider the open covering B(0, m)mIN of IRn. Compactness ofA implies that A B(0, m1) B(0, mk) = B(0, R) where R = max(m1, . . . , mk). This means that A is bounded.

Conversely, suppose that A is closed and bounded in . Then (0, x) M for everyx A. This means that |xi| M for every x = (x1, . . . , xn) A and so A [M, M]n. Thislast set is compact as a product of compact spaces. Being closed, A is also compact.

Remark 3.3 It is important to precise the distance in which a subspace is bounded. If a setis bounded in d, and d is a distance topologically equivalent to d, then boundedness in d doesnot imply boundedness in d. Here is an example. Define on IR, d(x, y) = min(1, |x y|). Thend is a distance toplogically equivalent to the usual distance d on IR. However, IR is bounded ind and unbounded in d. Therefore, IR is closed and bounded in d, however, it is not compact(because (IR, d) is not compact). As a convention, when we do not mention the distance onIRn, it means that we consider any of the usual distances.


31/47

3.3. COMPACT METRIC SPACES 31

Examples. 1) The unit sphere Sn1 and the closed ball Bn are compact in IRn because theyare closed and bounded.

2) The set A = {(x, 1x)|0 < x 1} is closed in IR2. However it is not compact because it is notbounded.

3) The set S = {(x, sin1x)|0 < x 1} is bounded but not compact because it is not closed.

Theorem 3.8 (Extreme value theorem) Let X be a topological space and let f : X IRbe continuous. If X is compact, then there exist points c and d in X such that f(c) f(x) f(d) for every x X. Otherwise stated, f has a maximum and minumum value on a compactspace.

Proof. Suppose that X is compact. Then f(X) is a compact subspace of IR. Therefore it isclosed and bounded. Boundedness implies that f(X) has a lowest upper bound sup f(X) anda greateset lower bound inff(X). But we proved in the exercises that for a subset A IR,sup A A and infA A (whenver they exist). Therefore, if A is closed, it contains its supand inf. Consequently, sup f(X)

f(X). This measn that sup f(X) = f(d) for some d

X.

Similary, inff(X) = f(c) for some c X. But inff(X) f(x) sup f(X) for every x X.Hence the conclusion.

Remark 3.4 sup f(X) is usually denoted by supxXf(x).

3.3 Compact metric spaces

Theorem 3.9 (Total boundedness) Let X be a compact space. Then for any > 0 there

are finitely many points x1, . . . , xn in X such that X =ni=1

B(xi, ). This property is called

total boundedness.

Proof. Consider the open covering (B(x, ))xX. Compactness implies that there exists afinite subcollection {B(x1, ), . . . , B(xn, )} that covers X. This is the conclusion.

Remark 3.5 The converse is not true, that is, a totally bounded space need not be compact.For example the subspace ]0,1[ is totally bounded because it is contained in the compact space[0,1]. However, ]0,1[ is not compact because it is not closed in IR.

Lemma 3.3 (The Lebesgue number lemma) Let A be an open covering of a metric space(X, d). If X is compact, then there exists > 0 such that each subset of X having diameterless than is contained in some element ofA.

The number is called a Lebesgue number of the covering A.

Proof. If X is an element ofA, then any positive number is a Lebesgue number ofA. Sowe assume that X / A.

Choose a finite subcollection {A1, . . . , An} ofA that covers X. For each i, set Ci = X Aiand let

f(x) =1

n

ni=1

d(x, Ci).

Then f(x) > 0 for all x X. Indeed, otherwise there exists x X such that d(x, Ci) = 0 forall i. Since Ci is closed, we get x Ci. Thus x Ci = (X Ai) = , a contradiction.

Since f is continuous, it has a minumum value > 0. We show that is the required number.Let B be a subset of X of diameter less than . Choose a point x0 B. Then B B(x0, ).Now, if d(x0, Cm) is the largest of the numbers d(x0, Ci), then f(x) d(x0, Cm). It follows


32/47


that B(x0, ) B(x0, d(x0, Cm)) X Cm = Am (if d(x, x0) < d(x0, Cm), then x / Cm).Consequently, B Am.

Definition 3.4 A function f from a metric space (x, dX) to a metric space (Y, dY) is said tobe uniformly continuous if for every > 0, there exists > 0 such that for every points x, y

in X, dY(f(x), f(y)) < whenever dX(x, y) < . In symbols

( > 0) ( > 0) (x, y X) d(x, y) < dY(f(x), f(y)) < .Compare this definition with the definition of continuity

(x X) ( > 0) ( > 0) (y X) d(x, y) < dY(f(x), f(y)) < .The number in the condition of continuity depends on and x, whereas in the uniformcontiuity condition it depends only on , so it the same for all x. Hence the adjectif uniform.

Examples. 1) Consider the function f :]0, 1[ IR defined by f(x) = 1x . We know that f iscontinuous. However, it is not uniformly continuous. Let 0 < < 1 be given. Let x = and

y = /2. Then |x y| = /2 < . But | 1x 1y | = 1 > 1 This means that x y can be madearbitrarily small whereas f(x) f(y) cannot. So f is not uniformly continuous.2) The function sin is uniformly continuous. This follows from the fact that | sin xsin y| |xy|for every x, y IR.

Now we give a positive result.

Theorem 3.10 (Uniform continuity theorem) Let X andY be metric space with X com-pact. Let f : X Y be continuous. Then f is uniformly continuous.Proof. Let > 0 be given. Consider the open covering ofY by the balls B(b,/2), b Y. LetA

be the colelction of f1

(B(b,/2)). ThenA

is an open covering of X (the covering is openbecause f is continuous). Let be a Lebesgue number ofA. Let x, y be two points in X suchthat d(x, y) < , then the set {x, y} has diameter less that . Therefore it is contained in someopen set f1(B(b,/2)). This means that f(x), f(y) B(b,/2). By the triangle inequalityd(f(x), f(y)) < .

Definition 3.5 A topological space is called limit point compact if every infinite subset ofX has a limit point. A topological space is called sequentially compact if very sequence inX has a convergent subsequence.

It turns out that if X is metrizable then the three conditions of compactness are equivalent.

Theorem 3.11 Let X be a metrizable space. Then the following conditions are equivalent.

(i) X is compact.

(ii) Every infinite subset of X has a limit point (limit point compactness).

(iii) Every sequence in X has a convergent subsequence (sequential compactness).

Proof. (i)(ii). Let A X be infinite. If no point of X is a limit point of A, then eachpoint x X has an open neighborhood Vx intersecting A in at most one point. Then, (Vx)xXis a open covering of X. However, no finite subfamily of this covering can cover A since A isinfinite. So indeed no subfamily can cover X. A contradiction.

(ii)(iii). Let {xn} be a sequence in X. Let E be the range of {xn}. If E is finite then{xn} contains a constant and therefore a convergent subsequence. IfE is infinite, then E has a


33/47

3.4. LOCAL COMPACTNESS AND THE ALEXANDROFF COMPACTIFICATION 33

limit point x. Choose then n1 such that xn1 B(x, 1). Having chosen n1, n2, . . . , ni1, chooseni > ni1 such that xni B(x, 1i ). This is possible since any neighborhood of x containsinfinitely many terms of E. Then clearly, {xni} converges to x.

(iii)(i). We prove this implication in three steps.Step 1. We prove that if X is sequentially compact then the Lebesgue number lemma holds

for X. So let A be open covering of X. Suppose that there is no Lebesgue number . Then foreach n IN, there exists a set Cn of diameter less than 1n and not contained in any element ofA. Fo each n, choose a point xn in Cn. By hypothesis, the sequence (xn) has a subsequence(xnk) that converges to some point x. Now x belongs to some element A of the covering.Because A is open, there exists r > 0 such that B(x, r) A. Now for all k large enough, wehave 1nk < r/2. Then Cnk B(xnk , r/2) (because the diameter of Cnk is less than r/2). Also,for all k large enough, we have d(xnk , x) < r/2 and so B(xnk , r/2) B(x, r) A by the triangleinequality. Thus, if k is large enough, then Cnk A. But this contradicts our hypothesis.Step 2. We prove that if X is sequentially compact then it is totally bounded. So let > 0be given. Suppose that X cannot be covered by finitely many open balls. We constructa sequence (xn) of X as follows. Let x1 be an arbitrary point in X. Since B(x1, ) doesnot cover X, choose a point x2 / B(x1, ). Suppose that x1, . . . , xn are contructed. SinceB(x1, ) B(xn, ) does not cover X, choose a point xn+1 not in the union of these balls.Then d(xn, xm) for all m = n. This means that (xn) cannot contain any convergentsubsequence, contradicting our assumption.

Step 3. We prove the implication (iii)(i). Let A be an open covering of X. By the firststep, there exists a Lebesgue number > 0 for A. Let = /3. By the second step, there existfinitely many B(x1, ) B(xn, ) that cover X. Each of these balls B(xi, ) has diameter atmost 2/3. Therefore it belongs to an element Ai ofA. Then the finite collection {A1, . . . , An}covers X.

Remarks. 1) Note that we did not use the metrizability of X in the implication (i)(ii). Itfollows that any compact space is limit point compact.2) It follows from this implication that a compact and discrete space is necessarily finite. Thisis an important fact in Analysis.

3.4 Local compactness and the Alexandroff compactification

In this section, we study the notion of local compactness and prove the fundamental theoremthat a locally compact Hausdorff space can be imbedded in a compact Hausdorff space calledits one point compactification or the Alexandroff compactification.

Definition 3.6 A space X is said to be locally compact at x if x has a compact neighbor-hood. If X is locally compact at each of its points, then X is said to be locally compact.

Examples. 1) A compact space is locally compact.

2) IRn is locally comapact because the closed ball B(x, r) is compact (closed and bounded).

3) A discrete space is locally compact because {x} is a compact neighborhood of x.

Theorem 3.12 Let X be a topological space. Then the following conditions are equivalent.

(i) X is locally compact Hausdorff.

(ii) There exists a spaceY such that

(1) X is a subspace of Y


34/47


(2) Y X is a singleton.(3) Y is compact Hausdorff.

Furthermore, if Y and Y are two spaces that satisfy the above conditions, then there is ahomeomorphism from Y to Y that equals the identity map on X. Any space Y satisfying

the above conditions is called the one point compactification of X or the Alexandroffcompactification ofX.

Proof. (ii) (i). Let Y be a space satisfying conditions (1)-(3). Then X is Hausdorff as asubspace of Y. Let x X. We prove that x has a compact neighborhood. Separate x and thepoint Y X by two open sets U and V respectively. The set C = Y V is closed and thereforecompact. It is also compact as a subspace of X. But x U C. Therefore C is a compactneigborhood of x.

(i) (ii). Take an object which is not in X and let Y = X{}. Define a topology on Yas follows: a set is open in Y if either it is open in X (type 1) or it is of the form Y C whereC is a compact subspace of X (type 2). We check that this is indeed a topology.

(a) The empty set is of type 1 and the space Y = Y is of type 2.(b) Let O1 and O2 be two open sets in Y. There are three cases: 1. If O1 and O2 are of

type 1, then O1 O2 is also of type 1. 2. If O1 and O2 are of type 2, then O1 O2 =(Y C1) (Y C2) = Y (C1 C2) is of type 2. 3. If O1 is of type 1 and O2 is of type 2,then O1 (Y C2) = O1 (X C1) is of type 1.

(c) Let (O)L be a collection of open sets in Y, there are also three cases: 1. If all O are oftype 1, then

O is also of type 1. If all O are of type 2, then

O =

(YC) = YC

is of type 2. Otherwise, let L1 be the set of for which O is of type 1 and L2 be the setof for which O is of type 2. Then

L

O =L1

O L2

(Y C) = O (Y C) = Y (C O)

is of type 2 because C O is closed and therefore compact.

Next, we show that X is a subspace of Y. We have to show that the subspace topologyinherited from Y coincides with the original topology of X. Let O be open in Y. If O is oftype 1, then O X = O and so O X is open in X. If O is of type 2, then O = Y C and soO X = Y C X = X C is open in X because C is closed. Thus, we have proved thatthe subspace topology is weaker than the original topology. Conversely, if O is open in X, thenO is of type 1, and therefore belongs to the topology of Y. Now, O

X = O means that O

belongs to the subspace topology.We show now that Y is compact. Let A be an open covering of Y. The collection A

contains an element of type 2, that is, of the form Y C, because none of the sets of type1 contain the point . Take all elements ofA different from Y C and intersect them withwith X. They form a collection of open sets of X that covers C. Compactness of C impliesthat finitely many of them cover C. The corresponding finite collection ofA along with Y Ccovers Y.

We show that Y is Hausdorff. Let x and y be two distinct points of Y. If x and y belong toX, then we can separate them by two open sets of X and therefore by two open sets of Y. Ifx X and y = , we choose a comapct neighborhood C of x. Then C and Y C are dijointneigborhoods of x and y respectively.

We finally prove uniqueness up to an homeomorphism. Let Y and Y be two spaces satisfyingcondditions (1)-(3). Let {p} = Y X and {q} = Y X. Define h : Y Y by h(x) = x if


35/47

3.4. LOCAL COMPACTNESS AND THE ALEXANDROFF COMPACTIFICATION 35

x X and h(p) = q. It is clear that h is a bijection. We show that h is an open map. Thesymmetry of the situation implies that h is an homeomorphism. Suppose first that U does notcontain p. Then h(U) = U. Since U is open in Y and contained in X, it is open in X. But Xis open in Y (because Y X is a singleton and singletons in a Hausdorff space are closed).Therefore U is also open in Y. Next, suppose that U contains p. Then C := YU is closed andtherefore compact. Since C X, C is also compact as a subspace of X. But X is a subspaceof Y. Therefore C is compact as a subspace of Y. Since Y is Hausdorff, C is closed in Y.But h(U) = h(Y C) = Y h(C) = Y C. Therefore h(U) is open in Y.

Remark 3.6 Using the arguments in the last part of the proof above, one can show that twohomeomorphic locally compact Hausdorff spaces have homeomorphic one point comapctifica-tions. In the exercises, you are asked to show this.

Examples. The one point compactification ofSnP is Sn. But SnP is homeomorphic withIRn. Therefore the one point compactification of IRn is homeomorphic with Sn. In particular,the one point compactification ofC is homeomorphic to the sphere S2. In the theory of complex

variables, the one point compactification ofC is called the Riemann sphere. The point thatwe added to the space is usually denoted by .


36/47



37/47

Chapter 4

Connectedness

4.1 Definitions, examples and properties

Connectedness is an important concept in topology. Roughly speaking, a set is connected if it

cannot be split into more than one part. More precisely we have the following definition.

Definition 4.1 Let X be a topological space. A separation of X is a pair (U, V) of disjointnonempty open subsets of X whose union is X. The space X is said to beconnected if theredoes not exist a separation of X.

Examples. 1) The space X = {x IR | |x| > 1} (as a subspace of IR) is not connected sinceX =] , 1[]1, [ is a separation of X.2) The space IR is not connected since IR =] , 0[]0, [ is a separation of X.3) We shall prove that IR and all its intervals are connected.

3) A singleton is connected.

4) The only connected subsets ofQ are the singletons. Indeed, let Y Q and suppose thatY contains two rational numbers a and b. Then there exists an irrational number c between aand b. Then (Y] , c[, Y]c, [) is a separation of Y.5) Let X be a set equipped with the indiscrete topology. Then X is connected (there is noenough open sets to separate X).

Proposition 4.1 Let X be a topological space. Then the following conditions are equivalent.

(i) X is not connected.

(ii) There exist two disjoint nonempty closed subsets of X whose union is X.

(iii) There exists a proper clopen subset of X.

(iv) There exists a non constant continuous function f : X Z.Proof. (i)(ii). Let (A, B) be a separation of X. Then A is also closed because itscomplement B is open. Similarly B is closed.

(ii)(iii). If X = A B where A and B are closed disjoint nonempty subsets, then A and Bare also open. So A is a proper clopen subset of X.

(iii)(iv). Let A be a proper clopen subset of X. Let f(x) = 1 if x A and f(x) = 0 if x

X

A (f is the characteristic function of A). Then f is continuous because the inverse

image of a subset O ofZ is either or A or Ac or X. In all cases, it is open. Now clearly f isnot constant.

37


38/47

38 CHAPTER 4. CONNECTEDNESS

(iv)(i). Let a and b be two distinct values taken by a continuous non constant functionf : X Z. Now {a} is open in Z because Z has the discrete topology. Therefore f1(a) isopen in X and nonempty. On the other hand, Z {a} is also open and nonempty (becauseit contains b). Therefore f1(Z {a}) is open in X and nonempty. Now we can write X =f1(a) f1(Z {a}) where f1(a) f1(Z {a}) = .

Corollary 4.1 Let X be a topological space. Then the following conditions are equivalent.

(i) X is connected.

(ii) There do not exist two disjoint nonempty closed subsets of X whose union is X.

(iii) The only clopen subsets ofX are X and.

(iv) Iff : X Z is continuous, then f is constant.

Remark 4.1 Let X be a topological space and let Y

X. We say that Y is connected if it isconnected with respect to the subspace topology. We have also the following characterization.

Lemma 4.1 Let X be a topological space and let Y X. Then Y is not connected if andonly if there exist two nonempty sets A and B such that A B = Y and A B = A B = .

Proof. Suppose first that Y is not connected, then there exist two disjoint sets A and Bwhich are clopen in Y and whose union is Y. We proved in the exercise that the closure of Ain Y is A Y where A is the closure of A in X. Since A is closed in Y, we have A = A Y.Then A B = A B Y = A B = . A similar argument shows that B A = .

Now suppose conversely that A and B are two nonempty sets whose union is Y and suchthat A

B = A

B = . Then we claim that A

Y = A and B

Y = B. Indeed, first

A Y A. Second, let x AY. Ifx / A, then x B since AB = Y. Then x AB = ,a contradiction. The other equality is proved similarly. The claim means that A and B are twodisjoint nonempty closed subsets of Y whose union is Y. Thus Y is not connected.

The next lemma will be useful.

Lemma 4.2 If (C, D) is a separation of a topological space X and Y is a connected subspaceof X, then either Y C or Y D.

Proof. Since C and D are both open in X, the sets C Y and D Y are open in Y. Thesetwo sets are disjoint and their union is Y; if they were both nonempty, they would constitute aseparation of Y. Therefore one of them is empty. Hence Y is contained

math201-lecturenotes metric topology

Documents