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Math 201

Topology I

Lecture notes of Dr. Hicham Gebran

[email protected]

Lebanese University, Fanar, Fall 2013-2014www.fs2-fanar.com/Math

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.

References

James Munkres, Topology (Pearson, 2000).

Colin Adams and Robert Franzosa, Introduction to topology, pure and applied (Pearson,2007).

Lynn Arthur Steen and J. Arthur Seebach, Counterexamples in Topology (Dover, 1978).

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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.5 Product spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2 Continuous functions and homeomorphisms 23

3 Compactness 313.1 Definitions, examples and properties . . . . . . . . . . . . . . . . . . . . . . . . . 313.2 Compact subspaces of IRn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343.3 Compact metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.4 Local compactness and the Alexandroff compactification . . . . . . . . . . . . . . 37

4 Connectedness 414.1 Definitions, examples and properties . . . . . . . . . . . . . . . . . . . . . . . . . 414.2 Components and local connectedness . . . . . . . . . . . . . . . . . . . . . . . . . 46

5 Complete metric spaces 51

3

4 CONTENTS

Chapter 0

Some fundamental properties of thereal line

We start by formulating two fundamental properties of the set of real numbers that we shalluse 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 α is called the least upperbound of E if the following hold.

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

In this case, we write α = supE. If E is not bounded from above, we write supE = +∞.

Proposition 0.1 Let E ⊂ IR be bounded from above. Then α = supE if and only if thefollowing hold.

i) x ≤ α for all x ∈ E.ii) ∀ε > 0 ∃y ∈ E such that α − ε < y.

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

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

In this case, we write α = inf E. If E is not bounded from below, we write inf E = −∞.

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

i) x ≥ α for all x ∈ E.ii) ∀ε > 0 ∃y ∈ E such that α + ε > y.

Proposition 0.3 Let ∅ 6= A ⊂ B ⊂ IR. Then supA ≤ supB and inf A ≥ inf B.

Exercise. Let A and B be two nonempty subsets of IR. If x ≤ y for all x ∈ A and all y ∈ B,then supA ≤ inf B.

Now we can state the two fundamental properties of IR.

5

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. Note that the above theorem is not truefor the set of rational numbers Q. And this is one of the reasons we prefer to work in the IRrathre than in Q.

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 + ε for all ε > 0 then a ≤ b.

Proof. Otherwise, taking ε = b−a2 , we reach a contradiction.

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 functiond : X × X → [0,∞[ that satisfies the following properties

1. 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 ρ(x, y) = |x3 − y3|.3) The real line with the distance γ(x, y) = min(1, |x−y|). The first two properties of a distanceare clear. We prove the triangle inequality. Observe that γ(x, y) ≤ 1 and γ(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 γ(x, y) = 1 or γ(y, z) = 1. Therefore,γ(x, y)+γ(y, z) ≥ 1. On the other hand, γ(x, z) ≤ 1. Hence the triangle inequality in this case.

Case 2. |x − y| < 1 and |y − z| < 1. Then γ(x, y) = |x − y| and γ(y, z) = |y − z|. Therefore,γ(x, z) ≤ |x − z| ≤ |x − y| + |y − z| = γ(x, y) + γ(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) =

∫ b

a|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 6= y

The 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 6= z. Then either y 6= x or y 6= z. Therefore d(x, z) = 1 ≤ d(x, y) + d(y, z).

7

8 CHAPTER 1. METRIC SPACES AND TOPOLOGICAL SPACES

Norms and distances on IRN

Let N be a positive integer. The set of all N−tuples 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) ‖λx‖ = |λ|‖x‖ for any real number λ and any x ∈ IRN .(iii) ‖x + y‖ ≤ ‖x‖ + ‖y‖ for all x, y ∈ IRN . This is also 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 isjust 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 =N∑

i=1

xiyi.

Let α be real number and consider the scalar product (x + αy) · (x + αy). We have

(x + αy) · (x + αy) = ‖x‖2 + 2αx · y + α2‖y‖2 ≥ 0.

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

|x · y| ≤ ‖x‖‖y‖.This is known as the Cauchy-Schwartz inequality.

From the last inequality we deduce that

‖x + y‖2 = (x + y) · (x + y) = ‖x‖2 + 2x · y + ‖y‖2

≤ ‖x‖2 + 2‖x‖‖y‖ + ‖y‖2

= (‖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 ‖x‖1 =N∑

i=1|xi|. Here again it is not difficult to prove the three

properties. More generally, let p ≥ 1. Set

||x||p =

(

N∑

i=1

|xi|p)1/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||.

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

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 ε > 0 such that B(x, ε) ⊂ O.

Otherwise stated, a set is open if whenever it contains a point, it contains a whole ballaround this point. This is why we call it open: if one member x can enter the set, some othermembers (the friends of x) can also enter the set.

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 an open set.

Remark. This proposition is not a tautology because the adjective open has different meaningsin ”open ball” and ”open set”. An open ball is a set of the form B(a, r) = {x|d(x, a) < r},whereas an open set is a set having the property that if a point belongs to this set, then awhole ball around this point is also in the set. So we have to prove that an ”open ball” has thisproperty.

Proof. Let x ∈ B(a, r). Then d(x, a) < r. Set ρ = r − d(x, a). We claim that B(x, ρ) ⊂B(a, r). Draw a figure. Indeed, let y ∈ B(x, ρ). Then d(x, y) < ρ = 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,

⋂

n≥1

]

− 1

n,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 by U(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).

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

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 Ox

such that x ∈ Ox ⊂ A. Then, we can write A = ∪x∈AOx. Therefore A is open as 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 by◦A 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)◦Q = ∅. This is because any open interval contains points outside Q and so Q cannot

contain any open interval and therefore cannot contain any nonempty open set. Similarlyint(IR\Q) = ∅.

c) Similarly,◦Z = ∅.

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.

1.2. OPEN AND CLOSED SETS AND RELATED CONCEPTS 11

Examples. 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 6= 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 6= 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 a closed set.

This proposition is not a tautology.

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 ρ = d(x, a) − r. We claim thatB(x, ρ) ⊂ X − B′(a, r). Indeed, let y ∈ B(x, ρ). Then d(x, y) < ρ = 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.

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 [ 1

n , 1 − 1n ] is closed. However

∞⋃

n=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

∀ε > 0 ∃n0 ∈ IN ∀n ≥ n0 d(xn, x) < ε.

Proposition 1.8 Let X be a metric space. Let A ⊂ X and let x ∈ X. Then, the followingconditions are equivalent.

(i) x ∈ A.

(ii) Every neighborhood of x intersects A.

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

Proof. Let B denote the set of points t satisfying property (ii), i.e., satisfying that everyneighborhood of t meets A. Proving that (i)⇔(ii) is equivalent to proving that A = B

So let us prove first that A ⊂ B. We claim that B is closed. Let t /∈ B, then there is an openneighborhood U of t not intersecting A. Consider an element y ∈ U . Since U is a neighborhoodof y not intersecting A, we deduce that y /∈ B. This means that U ⊂ {B and accordingly {B

is open. Therefore B is closed as claimed. Observe now that B contains A. Since A is thesmallest closed set containing A, we get A ⊂ B.

Let us prove next that B ⊂ A. Let t /∈ A. Then t ∈ {A. But {A is an open set not intersectingA since {A ∩ A ⊂ {A ∩ A = ∅. This means that t /∈ B.

(ii)⇒(iii). For every positive integer n the open ball B(x, 1n) intersects A. Choose accordingly

for 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 X be a metric space. Let A ⊂ X and let {xn} be sequence in A whichconverges 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 every x ∈ 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.

Proof. Let O be an open nonempty subset of IR. Then O contains an open interval I. Butany open inerval of R contains rational and irrational numbers. Therefore O meets Q as wellas IR\Q. �

Other examples. a) IR∗ = IR − {0} is dense in IR.

b) IR −Z is dense in IR. This is because Z has an empty interior.

1.3. TOPOLOGICAL SPACES 13

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. So apoint in A is either a limit point of A or is isolated in 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 6=i d(xi, xj). ThenB(xi, ri) ∩ A = {xi}.c) Let A = { 1

n |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.

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 to T .

(iii) A finite intersection of elements of T belongs to T .

The elements of T are called open sets of X. A topological space is a couple (X,T ).

Remarks. 1) If T 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 of T 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 if T 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 provethis.

(i) ∅ ∈ T because X − ∅ = X and X ∈ T because X − X = ∅ is finite.

(ii) Let Oα be a collection of nonempty elements of T . Then X − ∪αOα = ∩α(X − Oα) Thisset 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 point 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.

Let us now define the concept of convergence in a general topological space. In a metricspace, a sequence (xn) is convergent to x if

∀ε > 0 ∃n0 ∈ IN ∀n ≥ n0 d(xn, x) < ε.

This is equivalent to

∀ε > 0 ∃n0 ∈ IN ∀n ≥ n0 xn ∈ B(x, ε).


Since a neighborhood of a point contains an open ball about this point, this condition is equiv-alent to the condition that very neighborhood U of x, contains xn for all n large enough. Interms of quantifiers

∀U ∈ U(x) ∃n0 ∈ IN ∀n ≥ n0 xn ∈ U.

Now this condition involves only neighborhoods, so it depends only on the topology of the spaceand we take as our definition of convergence in an arbitrary topological space.

Remark 1.8 In a general topological space, weird and unfamiliar situations can happen. Forexample a sequence can converge to many points. Consider for instance IR equipped with thefinite complement topology and consider the sequence xn = n. Then (xn) converges to everyx ∈ IR. Indeed, let U be neighborhood of x. Then IR − U is finite. Choose n0 > max(IR − U).Then xn = n ∈ U for all n ≥ n0. This situation does not happen in metric spaces and moregenerally in Hausdorff spaces (see below).

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 every x ∈ 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.

Hausdorff spaces

Definition 1.12 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 6= y then thereexists two disjoint open sets U and V such that x ∈ U and y ∈ V . A Hausdorff space is alsocalled a T2 space.

Proposition 1.12 A metrizable space is a Hausdorff space.

Proof. Choose a distance d that generates the topology of the space. Let x 6= 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 6= ∅. 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 so it is metrizable).

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. �

Remark 1.9 A topological space set in which singletons are closed is called a T1 space. There-fore a T2 space is a T1 space. The converse is not true. For example, let X be an infinite setequipped with the finite complement topology. Then X is a T1 space which is not T2.

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. �

Comparison of topologies

Definition 1.13 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 Tand T are comparable if either T ⊂ 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.

Definition 1.14 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 topologically

equivalent if T = T ′, that is, if they generate the same topology.

We give some examples.

Proposition 1.15 Consider on IRN the three norms

‖x‖1 =N∑

i=1

|xi|, ||x||2 =

(

N∑

i=1

|xi|2)1/2

, ‖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 IRN .


Proof. We prove that d2 and d∞ are equivalent. In the exercises, you are asked to provethat d1 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) ≤√

Nd∞(x, y).

Therefore,

Bd2(x, r) ⊂ Bd∞(x, r) and Bd∞(x,

r√N

) ⊂ Bd2(x, r).

Now let O be an open set for the topology generated by d∞ and let x ∈ O. Then there existsr > 0 such that Bd∞(x, r) ⊂ O. But Bd2

(x, r) ⊂ Bd∞(x, r) and so Bd2(x, r) ⊂ O. This means

that O is open for the topology generated by d2.

Conversely, let O be an open set for the topology generated by d2 and let x ∈ O. Then thereexists r > 0 such that Bd2

(x, r) ⊂ O. But Bd∞(x, r√N

) ⊂ Bd2(x, r) and so Bd∞(x, r√

N) ⊂ O.

This means that O is open for the topology generated by d∞. �

Remark 1.10 Let d and d′ be two two distances on the same set X. Then d generates a finertopology than d′ if the following condition holds:

∀x ∈ X ∀r > 0 ∃δ > 0 Bd(x, δ) ⊂ Bd′(x, r).

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

Proof. We know that ρ is a metric. Let x ∈ X. Observe that Bd(x, r) ⊂ Bρ(x, r) sinceρ(x, y) ≤ d(x, y). By the previous remark, d generates a finer topology than ρ.

Conversely, let O be open for d and let x ∈ O. Then there exists r < 1 such that Bd(x, r) ⊂O. But then Bρ(x, r) ⊂ Bd(x, r). Indeed, let y ∈ Bρ(x, r). If d(x, y) > 1, then ρ(x, y) = 1 < r,contradicting the choice of r. Therefore ρ(x, y) = d(x, y) < r. Conclusion: Bρ(x, r) ⊂ O and soO is open for ρ. �

Basis for a topology

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

Definition 1.15 A subset B ⊂ P(X) is called a basis provided the following hold.

1. For all x ∈ X, there exists B ∈ B such that x ∈ B.

2. If B1, B2 ∈ B and x ∈ B1 ∩ B2, then there exists B3 ∈ B such that x ∈ B3 ⊂ B1 ∩ B2.

The elements of B are called basis elements. The topology T generated by B 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 .

2. Let (Oα)α∈A be a collection of elements of T . Let x ∈ ∪Oα. Then there is β ∈ A suchthat x ∈ Oβ. Since Oβ ∈ T , there exists B ∈ B such that x ∈ B ⊂ Oβ and so B ⊂ ∪Oα.This means that ∪Oα ∈ T .


3. Let O1, O2 ∈ T and let x ∈ O1∩O2. Then there exists B1, B2 ∈ B such that x ∈ B1 ⊂ O1

and 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 1.11 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 of B.

Proof. Since B ⊂ T and T is stable under unions, a union of elements of B 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 neighborhoodof x 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 1.16 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 and−1.

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, . . .}.

1.4. SUBSPACES 19

Proof. Let Fn = {xn, xn+1, . . .}. Then

x ∈ A ⇔ ∀U ∈ U (x) ∀n ∈ IN∗ ∃m ≥ n xm ∈ U

⇔ ∀n ∈ IN∗ ∀U ∈ U (x) U ∩ Fn 6= ∅

⇔ ∀n ∈ IN∗ x ∈ Fn

⇔ x ∈ ∩Fn.

�

Definition 1.17 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 = 2n. 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 neighborhood

of `. Then, there exists k0 such that xnk∈ U for all k ≥ k0. This means that xm ∈ U for all

m ∈ {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). Next

take U = B(`, 12) and n = n1 + 1. Then there exists n2 ≥ n1 + 1 such that xn2

∈ B(`, 12 ). At

the kth step, take U = B(`, 1k ) and n = nk−1 + 1. Then there exists nk ≥ nk−1 + 1 such that

xnk∈ B(`, 1

k ). 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 collection

TY = {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 of Y with the opensubsets of X.

Proof.

(i) Observe that ∅ = Y ∩ ∅ and Y = Y ∩ X. It follows that ∅, Y ∈ TY .

(ii) Let (Oα) be a collection of elements of TY . Then there exists a collection (Uα) of elementsof T such that Oα = Y ∩Uα. Then

⋃

Oα =⋃

(Uα ∩ Y ) = (⋃

Uα)∩ Y . Since⋃

Uα ∈ T , itfollows that

⋃

Oα ∈ 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 . �


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 of Z 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 1.12 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 if O 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 Y −A is open in Y and so Y −A = Y ∩Owhere 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 closed 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 , wedenote 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 δ > 0 such that B(x, δ) ⊂ O. ThenB(x, δ) ∩ Y ⊂ O ∩ Y = A, that is, BY (x, δ) ⊂ 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 δx = dY (x, Y − A). Then δx > 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) < δx then y ∈ A. For if y /∈ A, thend(x, y) ≥ δx by definition of δx. It follows that A = ∪x∈ABY (x, δx). Let now O = ∪x∈AB(x, δx).Then O is open in X and O ∩ Y = A. �

1.5. PRODUCT SPACES 21

1.5 Product spaces

If X1,X2 . . . ,Xn are sets, the set of n−tuples (x1, . . . , xn) where each xi ∈ Xi is denoted byX1 × X2 × · · · × Xn or by

∏ni=1 Xi.

Suppose now that X1, . . . ,Xn are topological space. If Ui is an open set of Xi for eachi = 1, . . . , n, their product U1 × U2 × · · · × Un will be called an open rectangle. Now, it is easyto check that the collection B of all open rectangles is a basis (indeed, first, any x ∈ ∏n

i=1 Xi

is contained in the open rectangle∏n

i=1 Xi; second, the intersection of two open rectanglesis an open retangle). The topology generated by this basis is called the product topology ofX1 × X2 × · · · × Xn.

This means that a set O ⊂ X1 × X2 × · · · × Xn is open if whenever x = (x1, . . . , xn) ∈ O,there exist open sets Ui ⊂ Xi such that x ∈∏n

i=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) × X2

⋃

X1 × (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

ρ1(x, y) =

n∑

i=1

di(xi, yi)

ρ2(x, y) =

(

n∑

i=1

di(xi, yi)2

)1/2

ρ∞(x, y) = maxi=1...,n

di(xi, yi).

Then ρ1, ρ2 and ρ∞ are metrics on X that generate the product topology on X.

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 ε 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.

∀ε > 0 ∃δ > 0 such that dX(x, y) < δ ⇒ dY (f(x), f(y)) < ε.

Remark 2.1 In the definition above, one can replace ”<” by ”≤”.

Examples. 1) A constant function is continuous.

2) The identity function is continuous.

3) Let (X, d) be a metric space and a ∈ X. The function x 7→ d(x, a) is continuous at every pointbecause |d(x, a) − d(y, a)| ≤ d(x, y). More generally, if A ⊂ X, then the function x 7→ d(x,A)is continuous at every point because |d(x,A) − d(y,A)| ≤ d(x, y).

4) All the elementary functions of calculus are continuous on their domains. The elementaryfunctions are the rational functions (ratio of polynomials), the logarithm, the exponential, thetrigonometric and inverse trigonometric functions.

Now there are several equivalent characterizations of continuity and we shall bring out acondition which is independent of the metrics so that it will serve as a definition of continuityin general topological spaces.

Proposition 2.1 Let f : (X, dX ) → (Y, dY ) be a function between two metric spaces and letx ∈ X. Then the following are equivalent.

(i) f is continuous at x.

(ii) ∀ε > 0, ∃δ > 0 such that f(B(x, δ)) ⊂ B(f(x), ε).

(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).

23

24 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 ε and δ be as in the definition of continuity.Let y ∈ f(B(x, δ)). Then y = f(z) for some z ∈ B(x, δ). This means that d(z, x) < δ andso by the condition of continuity, d(f(z), f(x)) < ε, that is, d(y, f(x)) < ε. This means thaty ∈ B(f(x), ε).

(ii)⇒(iii). Let U be a neighborhood of f(x). Then there exists a ball B(f(x), ε) ⊂ V . Take nowU = B(x, δ). Then U is a neighborhood of x. Now (ii) can be written as f(U) ⊂ B(f(x), ε) ⊂ 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 ε > 0 such that for all δ > 0there exists y with d(y, x) < δ and d(f(y), f(x)) ≥ ε. Taking δ = 1

n 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 be continuous

at 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 under f of any open subset of Y is open in X.

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

Proof. (i)⇒(ii). Let O ⊂ Y be open and let x ∈ f−1(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 ⊂ f−1(f(U)) ⊂ f−1(O). This means that f−1(O) is open.

(ii)⇒(i). Let x ∈ X and let V be an open neighborhood of f(x). Set U = f−1(V ). Then Uis a neighborhood of x (being an open set containing x) and f(U) = f(f−1(V )) ⊂ V . Accordingto the previous proposition, f is a continuous at x. Since x was arbitrary, this proves that f iscontinuous.

The equivalence (ii)⇔(iii) follows from the fact that f−1(Y − O) = X − f−1(O). �

25

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 7→ 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. Thesphere {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) If A ⊂ X, the inclusion map j : A → X is continuous.

(iv) If f : X → Y is continuous, then f : X → f(X) is also continous.

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) = f−1(g−1(O)) for every subsetO ⊂ Z. In particular, if O is open, then g−1(O) is open in Y because g is continuous. Since fis continuous, we have f−1(g−1(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 also

continuous. Let O be open in Y . Then(

f|A)−1

(O) = A ∩ f−1(O) is open in A.

(iii). Let O be open in X. Then j−1(O) = A ∩ O is open in A.

(iv). Let O be open in f(X). Then O = O′∩f(X) where O′ is open in Y . But f−1(O) = f−1(O′)and f−1(O′) is open in X. �

Proposition 2.3 For real valued functions, the sum and product of continuous functions arecontinuous. 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 ε > 0, there exists aneighborhood U1 of x0 such that |f(x)−f(x0)| < ε/2. Also, there exists a neighborhood U2 of x0

such that |g(x)−g(x0)| < ε/2. Then for all x ∈ U1∩U2, we have |f(x)+g(x)−(f(x0)+g(x0)| < ε.

(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 ε > 0 be given. The continuity of g at x0 implies that there exists a neighborhood U1 of x0

such that |g(x)−g(x0)| < ε2(|f(x0)|+1) for all x ∈ U1. This implies that |f(x0)||g(x)−g(x0)| < ε/2

for 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)| < ε

2(|g(x0)|+1) for all x ∈ U3. Therefore for

x ∈ U1 ∩ U2 ∩ U3, we have |f(x)g(x) − f(x0)g(x0)| < ε.

(iii). Let x0 be a a point where f(x0) 6= 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 ε > 0 there exists a neighborhood U2 of x0 such that |f(x) −f(x0)| < ε|f(x0)|2

2 for all x ∈ U2. It follows that if x ∈ U1 ∩ U2 then

∣

∣

∣

∣

1

f(x)− 1

f(x0)

∣

∣

∣

∣

=|f(x) − f(x0)||f(x)f(x0)|

< ε.

�

Given two sets X and Y , the projection π1 : X × Y → X is defined by π1(x, y) = x.The projection π2 : X × Y → X is defined by π2(x, y) = y. Observe that if U ⊂ X thenπ−1

1 (U) = U × Y and if V ⊂ Y then π−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 X×Y thatmakes the projections π1 and π2 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 π−1

1 (U) = U × Y . Since π1 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 f2

denote 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 πY and πZ denote the projectionsinto Y and Z respectively, then f1 = πY ◦f and f2 = πZ ◦f . 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

f−1(U × V ) = f−11 (U) ∩ f−1

2 (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 α(x) = (x, y). Then α : X → X × Y is continuous(because each component is continuous). Since g = f ◦ α, 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+y2 if (x, y) 6= (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.

27

Examples. 1) The function cos : [0, π] → [−1, 1] is a homeomorphism whose inverse isarccos : [−1, 1] → [0, π]. Continuity of these functions is proved in the first year.

2) The function sin : [−π2 , π

2 ] → [−1, 1] is a homeomorphism whose inverse is arcsin : [−1, 1] →[−π

2 , π2 ].

3) The function tan :]− π2 , π

2 [→ IR is a homeomorphism whose inverse is arctan : IR →]− π2 , π

2 [.

4) The function f : IR → IR defined by f(x) = x3 is a homeomorphism whose inverse is givenby f−1(x) = x1/3.

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 ∈ TF theni−1(O) = O ∈ T . However, i−1 : (IR,TF ) → (IR,T ) is not continuous because there exists anelement O ∈ T and O /∈ TF .

Here is another example. Let S1 denote the unit circle. The function f : [0, 2π[→ S1 givenby f(t) = (cos t, sin t) is a continous bijection. However, it is not a homeomorphism. Indeed,one can check that its inverse is given by

f−1(x, y) =

{

arccos x if y ≥ 0

π + arccos(−x) if y < 0.

Then lim(x,y)→(1,0+) f−1(x, y) = arccos(1) = 0, whereas lim(x,y)→(1,0−) f−1(x, y) = π+arccos(−1) =2π. So f−1 is not continuous. Note that in contrast, f :]0, 2π[→ S1\{(1, 0)} is a homeomorphism(the limits above do not belong to the space ]0, 2π[).

Remark 2.7 In contrast, let I and J be intervals of IR. A continuous bijection f : I → J isautomatically a homeomorphism. This will be proved in the exercises.

Remark 2.8 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 7→ (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 7→ (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 7→ tan xis a homeomorphism between ]−π/2, π/2[ and IR. The interval ]a,+∞[ is homeomorphic to IRby x 7→ ln(x − a). The interval ] −∞, a[ is homeomorphic to IR by x 7→ ln(a − x). �

Other examples. 1) A circle and an ellipse are homeomorphic (for example by the map(x, y) 7→ (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 f−1(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 → IR2

be 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 f−1(x, y) = (g(x), g(y)) where g(x) =√

x if x ≥ 0 andg(x) = −√−x if x ≤ 0. By the pasting lemma, g is continuous (draw the graph of g). It followsthat f−1 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.

5) Equip IR2 with the Euclidean distance d((x, y), (x′, y′)) =(

(x − x′)2 + (y − y′)2)1/2

(butany other equivalent distance would do). Equipp lC with the distance d(z, z′) = |z − z′| =(

(x − x′)2 + (y − y′)2)1/2

. Then IR2 is homeomorphic to lC. Indeed, consider the functionf : 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 f−1 also preserves distances,it is continuous as well.

6) 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(PM) with the horizontal line {y = 0}. If M has coordinates (x, y), then the point N hascoordinates ( x

1−y , 0). Indeed, let N have coordinates (X, 0). Since N belongs to (PM), there

exists λ ∈ IR such that ~PN = λ ~PM and so (X − 0, 0 − 1) = λ(x − 0, y − 1). It follows thatλ = 1/(1 − y) and so X = x

1−y . This defines a continuous function F : S1 − P → IR given by

F (x, y) =x

1 − y.

It inverse is given by

F−1(X) =1

X2 + 1(2X,X2 − 1).

Indeed, if X = x1−y , then X − x = yX. Squaring and using the fact that x2 + y2 = 1, we get

x = 2XX2+1

, and then y = X2−1X2+1

. It is clear that F−1 is continuous.

7) 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 (PM) with the plane {z = 0}. If M has coordinates (x, y, z),then the point N has coordinates ( x

1−z , y1−z , 0). Indeed, let N have coordinates (X,Y, 0). Since

N belongs to (PM), there exists λ ∈ IR such that ~PN = λ ~PM and so (X − 0, Y − 0, 0 − 1) =λ(x − 0, y − 0, z − 1). It follows that λ = 1/(1 − z) and so X = x

1−z and Y = y1−z . This defines

a homeomorphism F : S2 − P → IR2 given by

F (x, y, z) =

(

x

1 − z,

y

1 − z

)

whose inverse is given by

F−1(X,Y ) =1

X2 + Y 2 + 1(2X, 2Y,X2 + Y 2 − 1).

8) More generally the unit sphere Sn (defined by x21 + x2

2 + · · · 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, · · · ,

xn

1 − xn+1

)

.

29

The inverse is given by

F−1(X1,X2, . . . ,Xn) =1

∑ni=1 X2

i + 1(2X1, 2X2, · · · , 2Xn,

n∑

i=1

X2i − 1).

Definition 2.5 Let X and Y 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. Herethe copy of IR is a straight line.

2) The map g : IR → IR2 defined by g(x) = (x, x3) is an imbedding. Here the topological copyof IR is the curve of equation y = x3.

3) The map g : IR → IR2 defined by g(x) = (e−x cos x, e−x sin x) is an imbedding. Here, thetopological copy of IR is an infinite spiral.

4) The map g : [0, 1] →]0, 1[ defined by g(x) = 13 + 1

6x is an imbedding.

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 (Oλ)λ∈L of X is called acovering of X if X =

⋃

λ∈L

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

open.

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

⋃

λ∈L

Oλ, where each Oλ is

open, then there is a finite set J ⊂ L such that X =⋃

λ∈J

Oλ.

If (Oλ)λ∈L is a covering of X, a subcollection (Oλ)λ∈J that also covers X is sometimes called asubcovering of the covering (Oλ)λ∈L. Then, compactness can be also refolmulated as follows:X is compact if from every open covering of X, we can extract a finite subcovering.

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 (Oλ) be an opencovering of X. We may assume that X 6= ∅ (otherwise there is nothing to prove). Then thereexists an nonempty element Oλ0

. Then, by definition of open sets, X − Oλ0is a finite set

{x1, . . . , xn}. Now each xi belongs to an element Oλi. Therefore we have X = Oλ0

∪Oλ1∪ · · · ∪

Oλn. Thus, X is covered by finitely many elements of the original covering. Since any subspace

of 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 (Oλ) be

an open covering of X. Then 0 belongs to some Oλ0. Since ( 1

n) converges to 0, Oλ0contains all

1n except finitely many terms 1, 1

2 , . . . , 1n0

. Now each 1i belongs to some open set Oλi

. It followsthat X = Oλ0

∪Oλ1∪ · · · ∪Oλn

. 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.

31

32 CHAPTER 3. COMPACTNESS

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

n∈IN

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

cover IR.

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

n≥1] 1n , 1] and no finite subfamily of (] 1

n , 1])n can cover ]0,1].

g) The subspace Y = { 1n |n ∈ IN∗} of IR is not compact. Indeed, { 1

n}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 (Fλ)λ∈L of sets is said to have the finite intersection property

if⋂

λ∈J Fλ 6= ∅ for every finite subcollection (Fλ)λ∈J .

Example. Set for λ ∈ IR, Fλ = [λ,∞[. Then the collection (Fλ)λ∈IR has the finite intersectionproperty. Indeed, let J = {λ1, . . . , λn} be a finite subset of IR. Then

⋂

λ∈J Fλ = Fλ1∩Fλ2

· · · ∩Fλn

= [t,∞[6= ∅ where t = max(λ1, . . . , λn). Observe however that the whole intersection⋂

λ∈IR Fλ = ∅ because there is no real number bigger than every other real number.

Taking complements in the definition of compactness, one can show the following (write theproof properly as an exercise).

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

λ∈L Fλ 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 6= ∅.

Proof. The collection (Fn) has the finite intersection property beccause Fn1∩ · · · ∩ Fnk

=Fmax ni

6= ∅. �

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

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. Write the proof properlyas an exercise. �

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 (Oλ)λ∈L be a coveringof Y by open subsets of X. Add to this covering the open set X − Y . Otherwise stated letµ be an element not in L and set Oµ = X − Y and consider the new collection (Oλ)λ∈L∪{µ}.Then this new collection is an open covering of X. Compactness of X implies that a finitesubcollection (Oλ)λ∈J covers X and consequenlty also Y . Then the collection (Oλ)λ∈J−{µ} is afinite subcollection of (Oλ)λ∈L that covers Y . �

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

3.1. DEFINITIONS, EXAMPLES AND PROPERTIES 33

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 ⊂ ⋃

y∈KVy . Since K is compact K ⊂ Vy1

∪ · · · ∪ Vyn .

Then U = Uy1∩ · · · ∩ Uyn is a neighborhood of x not intersecting K, that is, contained in the

complement 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.

Theorem 3.3 The image of a compact space under a continuous map is compact. In partic-ular compactness is preserved by homeomorphism. We say that compactness is a topologicalproperty.

Proof. Let f : X → Y be continuous with X compact. Let (Oλ)λ∈L be a covering of f(X)by open subsets of Y . Then the collection (f−1(Oλ))λ is an open covering of X (because fis continuous). Compactness of X implies that finitely many subsets f−1(Oλ1

), · · · , f−1(Oλn)

cover X. It follows that {Oλ1, . . . , Oλn

} 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 f−1 is continuous and this is equivalent to proving that fmaps closed sets into closed sets (because f = (f−1)−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 inY . �

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 tube W × Y whereW is a neighborhood of x0 in X.

Proof. Draw a figure and explain. If y ∈ Y , then (x0, y) ∈ N . Then, by definition of theproduct topology, there exists a basis element Uy × Vy containing (x0, y) and contained in N .Then the collection (Uy × Vy)y∈Y is an open covering of x0 × Y . But x0 × Y is compact beinghomeomorphic to Y . Therefore, there exist finitely many basis elements U1×V1, . . . Un×Vn thatcover x0 × Y . Let W = U1 ∩ · · · ∩ Un. Then W is an open neighborhood of x0. We claim thatW × Y ⊂ ⋃n

i=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 (Oλ) be open covering of X × Y . Given x ∈ X, the slice x × Y is compactand therefore can be covered by finitely many elements Oλ1

, . . . , Oλnof the covering. Let

N =⋃n

i=1 Oλi. By the tube lemma, N contains a tube Wx × Y containing x × Y . Now the


collection (Wx)x∈X is an open covering of X. Compactness of X implies that there exists a finitesubcollection Wx1

, . . . ,Wxkthat covers X. Therefore the union of the tubes Wx1

×Y, . . . ,Wxk×Y

is X × Y . Since each of these tubes is covered by finitely many Oλ, 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] ⊂⋃

λ∈L

Oλwhere each Oλ is open in IR . We need to prove that [a, b] is

covered by finitely many Oλ. 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 Oµ. Thus let x ∈ [a, b[. Then x ∈ ∪λ∈LOλ, and so there is an index µ ∈ L such that x ∈ Oµ.But since Oµ is open, we can find an ε, 0 < ε < b such that ]x− ε, x + ε[⊂ Oµ. Choose now anelement y ∈]x, x + ε[. Then [x, y] ⊂]x − ε, x + ε[⊂ Oµ. 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 Oλ.Applying Step 1 to x = a, we see that a ∈ C and so C 6= ∅. Since C is bounded from above byb, C has a least upper bound c ∈]a, b].

Step 3. We prove that c ∈ C. There is k ∈ L such that c ∈ Ok. Since Ok is open, there isδ > 0 such that ]c−δ, c+δ[⊂ Ok. By the property of the least upper bound, there is an elementz ∈ C such that c − δ < z ≤ c. Then [z, c] ⊂ Ok. Also, [a, z] can be covered by finitely manyOλ since z ∈ C. Thus, [a, c] = [a, z] ∪ [z, c] can also be covered by finitely many Oλ.

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:

ρ1(x, y) =

n∑

i=1

|xi − yi|

ρ2(x, y) =

(

n∑

i=1

|xi − yi|2)1/2

the Euclidean metric

ρ∞(x, y) = maxi=1...,n

|xi − yi| the square metric.

We also noticed that

ρ∞(x, y) ≤ ρ2(x, y) ≤√

nρ∞(x, y).

and

ρ∞(x, y) ≤ ρ1(x, y) ≤ nρ∞(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 diameter is finite. The inequalities between the usual distances on IRn imply thatboundedness in ρ2 is equivalent to boundednes in ρ∞ or boundedness in ρ1.

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

3.3. COMPACT METRIC SPACES 35

Proof. Suppose first that A is compact. Then A is closed by a previous theorem. Nowconsider the open covering B(0,m)m∈IN∗ of IRn. Compactness of A 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 ρ∞. Then ρ∞(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.

Examples. 1) The unit sphere Sn−1 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 not

bounded.

3) The set S = {(x, sin 1x)|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 inf f(X). But we proved in the exercises that for a subset A ⊂ IR,supA ∈ A and inf A ∈ 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, inf f(X) = f(c) for some c ∈ X. But inf f(X) ≤ f(x) ≤ sup f(X) for every x ∈ X.Hence the conclusion. �

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

3.3 Compact metric spaces

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

are finitely many points x1, . . . , xn in X such that X =n⋃

i=1B(xi, ε). This property is called

total boundedness.

Proof. Consider the open covering (B(x, ε))x∈X . Compactness implies that there exists afinite subcollection {B(x1, ε), . . . , B(xn, ε)} 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 δ > 0 such that each subset of X having diameterless than δ is contained in some element of A .


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

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

Choose a finite subcollection {A1, . . . , An} of A that covers X. For each i, set Ci = X −Ai

and let

f(x) =1

n

n∑

i=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 δ > 0. We show that δ is the required number.Let B be a subset of X of diameter less than δ. Choose a point x0 ∈ B. Then B ⊂ B(x0, δ).Now, if d(x0, Cm) is the largest of the numbers d(x0, Ci), then δ ≤ f(x) ≤ d(x0, Cm). It followsthat B(x0, δ) ⊂ 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 ε > 0, there exists δ > 0 such that for every points x, yin X, dY (f(x), f(y)) < ε whenever dX(x, y) < δ. In symbols

(∀ε > 0) (∃δ > 0) (∀x, y ∈ X) d(x, y) < δ ⇒ dY (f(x), f(y)) < ε.

Compare this definition with the definition of continuity

(∀x ∈ X) (∀ε > 0) (∃δ > 0) (∀y ∈ X) d(x, y) < δ ⇒ dY (f(x), f(y)) < ε.

The number δ in the condition of continuity depends on ε and x, whereas in the uniformcontiuity condition it depends only on ε, 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 is

continuous. However, it is not uniformly continuous. Let 0 < δ < 1 be given. Let x = δ andy = δ/2. Then |x − y| = δ/2 < δ. But | 1x − 1

y | = 1δ > 1 This means that x − y can be made

arbitrarily 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 x−sin y| ≤ |x−y|for every x, y ∈ IR.

Now we give a positive result.

Theorem 3.10 (Uniform continuity theorem) Let X and Y be metric space with X com-pact. Let f : X → Y be continuous. Then f is uniformly continuous.

Proof. Let ε > 0 be given. Consider the open covering of Y by the balls B(b, ε/2), b ∈ Y . LetA be the colelction of f−1(B(b, ε/2)). Then A is an open covering of X (the covering is openbecause f is continuous). Let δ be a Lebesgue number of A . Let x, y be two points in X suchthat d(x, y) < δ, then the set {x, y} has diameter less that δ. Therefore it is contained in someopen set f−1(B(b, ε/2)). This means that f(x), f(y) ∈ B(b, ε/2). By the triangle inequalityd(f(x), f(y)) < ε. �

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.

3.4. LOCAL COMPACTNESS AND THE ALEXANDROFF COMPACTIFICATION 37

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)x∈X

is 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. If E is infinite, then E has alimit point x. Choose then n1 such that xn1

∈ B(x, 1). Having chosen n1, n2, . . . , ni−1, chooseni > ni−1 such that xni

∈ B(x, 1i ). This is possible since any neighborhood of x contains

infinitely 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 holdsfor X. So let A be an aopen covering of X. Suppose that there is no Lebesgue number δ.Then for each n ∈ IN∗, there exists a set Cn of diameter less than 1

n and not contained in anyelement of A . Fo each n, choose a point xn in Cn. By hypothesis, the sequence (xn) has asubsequence (xnk

) that converges to some point x. Now x belongs to some element A of thecovering. Because A is open, there exists r > 0 such that B(x, r) ⊂ A. Now for all k largeenough, we have 1

nk< r/2. Then Cnk

⊂ B(xnk, r/2) (because the diameter of Cnk

is less thanr/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 triangle inequality. 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 ε > 0be given. Suppose that X cannot be covered by finitely many open ε−balls. We constructa sequence (xn) of X as follows. Let x1 be an arbitrary point in X. Since B(x1, ε) doesnot cover X, choose a point x2 /∈ B(x1, ε). Suppose that x1, . . . , xn are contructed. SinceB(x1, ε) ∪ · · · ∪ B(xn, ε) does not cover X, choose a point xn+1 not in the union of these balls.Then d(xn, xm) ≥ ε for all m 6= 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 δ > 0 for A . Let ε = δ/3. By the second step, there existfinitely many B(x1, ε) · · ·B(xn, ε) that cover X. Each of these balls B(xi, ε) has diameter atmost 2δ/3. Therefore it belongs to an element Ai of A . 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 its 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 compact 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 space Y such that

(1) X is a subspace of Y

(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 satisfyingthe above conditions is called the one point compactification of X or the Alexandroff

compactification of X.

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 ω which is not in X and let Y = X ∪{ω}. 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 oftype 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 (Oλ)λ∈L be a collection of open sets in Y , there are also three cases: 1. If all Oλ are oftype 1, then

⋃

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

Oλ =⋃

(Y −Cλ) = Y −∩Cλ

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

⋃

λ∈L

Oλ =⋃

λ∈L1

Oλ ∪⋃

λ∈L2

(Y − Cλ) = 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 in X. Thus, we have provedthat the subspace topology is weaker than the original topology. Conversely, if O is open in X,

3.4. LOCAL COMPACTNESS AND THE ALEXANDROFF COMPACTIFICATION 39

then O is of type 1, and therefore belongs to the topology of Y . Now, O ∩ X = O means thatO 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 ω. Take all elements of A 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 of A 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 = ω, we choose a compact neighborhood C of x. Then C and Y − C are disjointneigborhoods 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 ifx ∈ 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 := Y −U 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 of Sn−P is Sn. But Sn−P is homeomorphic withIRn. Therefore the one point compactification of IRn is homeomorphic with Sn. In particular,the one point compactification of C is homeomorphic to the sphere S2. In the theory of complexvariables, the one point compactification of C is called the Riemann sphere. The point ω thatwe added to the space is usually denoted by ∞.

Chapter 4

Connectedness

4.1 Definitions, examples and properties

Connectedness is an important concept in topology. Roughly speaking, a set is connected if itcannot be split into more than one part (it is topologically one piece). More precisely we havethe 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 be connected if theredoes not exist a separation of X.

Remarks. 1) If {U, V } is a separation of X, then V = X − U and so U and V are clopen.

2) A separation of a space is a partition of that space. However, not every partition is aseparation; we can always write X = A∪ (X −A), but the point in the definition above is thatA is clopen.

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 of Q 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.

41

42 CHAPTER 4. CONNECTEDNESS

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 ifx ∈ X − A (f is the characteristic function of A). Then f is continuous because the inverseimage of a subset O of Z is either ∅ or A or Ac or X. In all cases, it is open. Now clearly f isnot constant.

(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 f−1(a) isopen in X and nonempty. On the other hand, Z − {a} is also open and nonempty (becauseit contains b). Therefore f−1(Z − {a}) is open in X and nonempty. Now we can write X =f−1(a) ∪ f−1(Z− {a}) where f−1(a) ∩ f−1(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 of X are X and ∅.

(iv) If f : 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 nonempty sets Aand B which are clopen in Y and whose union is Y . We proved in the exercises that the closureof A in 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, firstA ⊂ Y ∩ A. Second, let x ∈ A∩Y . If x /∈ A, then x ∈ B since A∪B = Y . Then x ∈ A∩B = ∅,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. �

Example. Consider Q as a subspace of IR. We can write Q = A∪B where A = Q∩]−∞,√

2[and B = Q∩]

√2,+∞[. So {A,B} is a separation of Q. Note that A =] − ∞,

√2] and B =

[√

2,+∞[. So A ∩ B = A ∩ B = ∅. Note however that A ∩ B = {√

2} 6= ∅.

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 in C or D. �


Theorem 4.1 Let X be topological space. The union of connected subspaces of X that havea common point is connected.

Proof. Let (Aλ) be a collection of connected subspaces of X and let p ∈ ⋂Aλ. Suppose thatY :=

⋃

Aλ is not connected, then there exists a separation (C,D) of Y . The point p is in oneof the sets C or D. We can assume that it is in C (otherwise relable C and D). Since Aλ isconnected then by the previous lemma, it must be contained in either C or D; and it cannotbe contained in D because it contains the point p of C. Hence Aλ ⊂ C. Since λ was arbitrary,we have Y =

⋃

Aλ ⊂ C, contradicting the fact that D is not empty. �

Theorem 4.2 Let A be a connected subspace of a X. If A ⊂ B ⊂ A, then B is also connected.

Said differently: if B is obtained from a connected subspace A by adjoining some or all its limitspoints, then B is connected.

Proof. Suppose that {C,D} is a separation of B. By a previous lemma, A must lie either inC or D. Suppose that A ⊂ C. Then B ⊂ A ⊂ C. But C ∩D = ∅, therefore B cannot intersectD. This contradicts the fact that D is a nonempty subset of B. �

Theorem 4.3 The image of a connected space under a continuous map is connected. Inparticular, connectedness is preserved by a homeomophism.

Proof. Let f : X → f(X) be continuous with X connected. Suppose that {A,B} is aseparation of f(X). Then f−1(A) and f−1(B) are open (because f is continuous and A, B areopen) and disjoint because A∩B = ∅. They are also nonempty because A and B are containedin the image of f . Therefore {f−1(A), f−1(B)} is a separation of X, contradicting the fact thatX is connected. �

Theorem 4.4 A product of connected spaces is connected.

Proof. We prove the theorem for the product of two spaces X and Y . The result foran arbitrary number of spaces will follow by induction. Choose a base point (a, b) ∈ X × Y .The ”horizontal slice” X × {b} is connected because it is homeomorphic with X. Also, eachvertical slice {x} × Y is connected, being homeomorphic with Y . As a result, the space Tx =(X × {b}) ∪ ({x} × Y ) is connected as a union of two connected subspaces having the point(x, b) in common. Now form the union

⋃

x∈X Tx of all these subspaces. This union is connectedbecause these subspaces have the point (a, b) in common. But X × Y =

⋃

x∈X Tx. ThereforeX × Y is connected. �

Theorem 4.5 A subset E ⊂ IR is connected if and only if it is an interval.

Proof. Let E be connected and assume it is not an interval. Then there are two points xand y in E and z ∈]x, y[ such that z /∈ E. Then, E has the following decomposition

E = E ∩ ] −∞, z[⋃

E ∩ ]z,∞[.

The two subsets are both open in E and nonempty since the first one contains x and the secondcontains y. This means that E is not connected contrary to the assumption.

Conversely, let E be an interval and suppose that is not connected. Then it is the unionof two disjoint nonempty sets A and B each of which is open in E. Choose a ∈ A and b ∈ B.We may assume that a < b (otherwise we relabel A and B). The interval [a, b] is contained inE since E is an interval. Hence [a, b] is the union of the disjoint nonempty sets A0 = A ∩ [a, b]and B0 = B ∩ [a, b]. Let c = supA0. We distinguish between two cases.


Case 1. c ∈ A0. In this case, c 6= b, because otherwise c ∈ A0 ∩ B0. So either c = a ora < c < b. In both cases, because A0 is open in [a, b], there is some interval [c, c + ε[ containedin A0. In particular, this means that there is z ∈ A0 such that c < z < c + ε. The fact thatz ∈ A0 means that z ≤ supA0 = c. Thus, we reached a contradiction.

Case 2. c ∈ B0. In this case, c 6= a, so either c = b or a < c < b. In either case, it followsfrom the fact that B0 is open in [a, b] that there is some interval of the form ]c− ε, c] containedin B0. But since c = supA0, the interval ]c − ε, c] contains an element of A0. Again this is acontradiction since A0 ∩ B0 = ∅. �

Theorem 4.6 (Intermediate value theorem) Let X be a connected topological space andlet f : X → IR be a continuous function. If a and b are two points of X and r is a numberbetween f(a) and f(b), then there exists a point c ∈ X such that f(c) = r.

Proof. Since X is connected and f is continuous, then f(X) is connected and therefore aninterval. We can suppose that f(a) < f(b). Since f(a) and f(b) belong to the interval f(X),[f(a), f(b)] ⊂ f(X). Hence any number between f(a) and f(b) is the image of an element ofX. �

Now we come to a condition stronger than connectedness: path connectedness.

Definition 4.2 Given two points x and y of a topological space X, a path in X from x to yis a continuous map f : [a, b] → X such that f(a) = x and f(b) = y. A space X is said to bepath connected if any two points of X can be joined by a path in X.

Note that since [a, b] is homeomorphic to [0, 1], the domain of any path can be taken as [0,1].This is usually done in algebraic topology.

Examples. 1) Intervals of IR are path connected.

2) A ball B(a, r) in IRN is path connected. In fact it is convex, which means that any two pointsin B can be joined by a straight line segment. Indeed let x and y be two points in B(a, r). Setf(t) = (1− t)x + ty for t ∈ [0, 1]. Then f is a path joining x to y which is contained in B(a, r).Indeed ||f(t) − a|| = ||(1 − t)(x − a) + t(y − a)|| < (1 − t)r + tr = r. Similarly, a closed ball ispath connected.

3) The punctured Euclidean space is IRN − {0}. If N > 1, then this space is path connected.Indeed, let x and y be two points in this space. If the straight line segment joining them doesnot pass by 0, then this segment is a path in IRN −{0} joining x to y. Otherwise, we can choosea point z not on the line joining x to y. Then x and y can be joined by the broken line pathfrom x to z and then from z to y.

4) The continuous image of a path connected space is also path connected (why?). In particular,the unit sphere Sn−1 in IRn is path connected for n > 1. Indeed, let g : IRn − {0} → Sn−1 bedefined by g(x) = x

||x|| . Then g is continuous and surjective. Therefore the result follows fromthe previous example.

Proposition 4.2 A path connected space is connected.

Proof. Let X be a path connected space. Suppose that X = A ∪ B is a separation of X.Let x ∈ A and y ∈ B. Then there exists a path f : [a, b] → X joining x to y. Now f([a, b])is connected as the image of a connected space under a continuous function. It follows thatf([a, b]) is contained in either A or B, which is impossible since f(a) = x ∈ A and f(b) = y ∈ B.�

The converse of this proposition is however not true as the following example shows.


0 0.2 0.4 0.6 0.8 1

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Figure 4.1: The topologist’s sine curve

Example. Let S = {(x, sin 1x), 0 < x ≤ 1}. Then S is connected as the image of ]0, 1] under

a continuous function. Therefore its closure S is also connected. S is called the topologist’ssine curve. In fact S = ({0} × [−1, 1]) ∪ S. Let us prove this. First, we claim that {0} ×[−1, 1] ⊂ S. Indeed, if (0, u) ∈ {0} × [−1, 1], then u = sin v for some v . Then the sequence( 1

v+2nπ , sin(v + 2nπ)) belongs to S and converges to (0, u). It follows that ({0} × [−1, 1])∪ S ⊂S. Next, it should be clear that S ⊂ [0, 1] × [−1, 1]. Finally we prove the reverse inclusionS ⊂ ({0} × [−1, 1]) ∪ S. Let (x, y) ∈ S − {0} × [−1, 1]. Then x > 0. If y 6= sin 1

x , separatey and sin 1

x by two neighborhoods V1 and V2. Let g(x) = sin 1x and let U be a neighborhood

of x such that g(U) ⊂ V2. Then U × V1 is a neighborhood of (x, y) that does not intersect S,contradicting the fact that (x, y) ∈ S.

We now show that S is not path connected. Suppose there is a path f : [a, c] → S joining theorigin to a point in S. The set {t ∈ [a, c]|f(t) ∈ {0}× [−1, 1]} is closed and bounded from above.Therefore it has a biggest element b. Then f : [b, c] → S is a path that maps b into the verticalslice {0}× [−1, 1] and maps the other points of [b, c] to points of S. Let f(t) = (x(t), y(t)). Thenx(b) = 0 while x(t) > 0 and y(t) = sin(1/x(t)) for t > b. By continuity, limt→b+ x(t) = x(b) = 0and so limt→b+

1x(t) = +∞. But the sine function has no limit at infinity. Therefore y(t) has no

limit as t → b. However, it should have because f is continuous. This contradiction shows thatthere is no path joining the origin to a point in S, and so S is not path connected.

To prove that y(t) has no limit as t → b, we proceed as follows. Let b < b′ < c. Then x([b, b′])is an interval that contains 0 and a positive number x(b′). Therefore [0, x(b′)] ⊂ x([b, b′]). Nowfor every n large enough 1

π2+2nπ ∈ [0, x(b′)]. Therefore there exists a sequence tn ∈ [b, b′] such

that x(tn) = 1π2+2nπ . Then y(tn) = 1. Similarly, there exists a sequence sn ∈ [b, b′] such that

x(sn) = 1π2+(2n+1)π and so y(sn) = −1. Passing to a subsequence, we may assume that tn → t

and sn → s. Then x(tn) → x(t) = 0 and x(sn) → x(s) = 0. Therefore t = s = b becauseotherwise x(t) > 0 or x(s) > 0. So we have two sequences that converge to b, yet their imagesunder y converge to distinct limits.


4.2 Components and local connectedness

If a space is not connected, it can be divided into pieces that are connected. Indeed, given atopological space X, define the binary relation on X by setting x ∼ y if there exists a connectedsubset of X containing both x and y. This is an equivalence relation. Indeed, first, the relationis reflexive: x ∼ x because {x} is connected. Next, symmetry is evident. Last (transitivity),if x ∼ y and y ∼ z, then, there is a connected subset A containing x and y and a connectedsubset B containing y and z. Then A ∪ B contains x and z and is connected because A an Bhave y in common. Therefore x ∼ z.

Definition 4.3 The equivalence classes for the relation above are called the connected com-

ponents (or sometimes just components) of X.

Now recall that given an equivalence relation on a set, the equivalence class containing anelement x is the set of points y that are in relation with x. Two equivalence classes are eitherdisjoint or identical and the whole set is the union of the pairwise disjoint equivalence classes. Itis also possible that there is only one equivalence class (in this case, all elements are equivalent).

Examples. 1) The connected components of IR∗ are ] −∞, 0[ and ]0,∞[.

2) The connected components of a discrete space are the singletons. In particular, the connectedcomponents of Z are the singletons {n}. A space whose components are singletons is usuallycalled totally disconnected.

3) The connected components of Q are the singletons.

4) A connected space has only one component: the space itself.

Here are some important properties of connected components.

Theorem 4.7 Let X be a topological space. Then

(i) Two connected components of X are either disjoint or identical and X is the union of thepairwise disjoint components.

(ii) The connected components of X containing an element x is the biggest connected subsetof X that contains x. It is usually denoted by C(x) or [x].

(iii) The components of X are closed.

(iv) Every nonempty connected subset of X is contained in exactly one connected component.

Proof. (i) is true for any equivalence relation.(ii). Let D(x) be the union of all connected sets containing x. Then D(x) is connected as

a union of connected sets having a common point. Now if y ∈ D(x), then x ∼ y because D(x)is a connected set containing x and y. Therefore y ∈ C(x). This proves that D(x) ⊂ C(x).Conversely, let y ∈ C(x), then there is a connected set A containing x and y. Then A ⊂ D(x)and so y ∈ D(x). This proves that C(x) ⊂ D(x). Now of course, the union of all connectedsets that contain x is the biggest connected set containing x.

(iii). Let C(x) be a connected component. Then C(x) is connected. By the previous point,it is contained in C(x). This means that C(x) is closed.

(iv) We already observed this. If A is a nonempty connected set, it contains x and soA ⊂ C(x). Since distinct components are disjoint, A cannot meet another component. �

Remark 4.2 If X has finitely many components, then every connected component is also openbecause its complement is a finite union of closed sets. In general, components need not beopen. For instance the components of Q are not open.

4.2. COMPONENTS AND LOCAL CONNECTEDNESS 47

Theorem 4.8 Let f : X → Y be a homeomorphism. Then for every x ∈ X, f(C(x)) =C(f(x)). In words: a homeomorphism takes connected components into connected components.In particular, the number of connected components of a space is a topological invariant.

Proof. f(C(x)) is a connected subset containing f(x). Therefore it is contained in C(f(x)).Conversely, let y = f(z) ∈ C(f(x)). Then f(z) ∼ f(x) i.e., f(z) and f(x) belong to theconnected set C = C(f(x)). But then z and x belong to the same connected set f−1(C). Thismeans that z ∼ x and so z ∈ C(x). Therefore y = f(z) ∈ f(C(x)). �

Examples. 1) The spaces [0, 1] mand S1 are not homeomorphic. Indeed, if we remove 12

from [0, 1], the resulting space is not connected; whereas if we remove any point from S1, theresulting space is connected.

2) The letters X and Y (as compact subspaces of IR2) are not homeomorphic. Indeed, if weremove the middle point from X, the resulting space has 4 connected components; whereas ifwe remove any point from Y, the resulting space has at most 3 components.

3) The letters A and ∞ are not homeomorphic. Indeed, if we remove from A the topologicalcircle that it contains, the resulting space has two connected components; whereas if we removefrom ∞ any topological cicle, the resulting space is a circle (which is connected).

4) The sphere S2 and the torus T 2(surface of an american doughnut, chambre a air, bouee)are not homeomorphic. Indeed, if we remove an apropriate circle from the torus the resultingspace is connected, whereas if we remove from S2 any topological circle, the resulting space isnot connected.

Now we define another equivalence relation on a topological space X. Let x ∼ y if there is apath in X joining x to y. Then this is an equivalence relation. Indeed, first x ∼ x because theconstant path f(t) = x joins x to itself. Next, let f : [0, 1] → X be a path joining x to y. Letg(t) = f(1 − t). Then g is a path joining y to x. This proves the symmetry of the relation.Finally, let f : [0, 1] → X be a path joining x to y and let g : [1, 2] → X be a path joining y toz. Let

h(t) =

{

f(t) if 0 ≤ t ≤ 1

g(t) if 1 ≤ t ≤ 2.

Since f(1) = g(1) = y, it follows from the pasting lemma that h is continuous. Since h(0) = xand h(2) = z, h is a path joining x to z. This proves the transitivity of the relation.

Definition 4.4 The equivalence classes for the relation above are called the path components

of X.

Example. The topologist’s sine curve S has two path components: the curve S and thevertical segment V = {0} × [−1, 1]. Note that S has one connected component since it isconnected.

If one removes from S the points of V with rational second coordinates, the resulting spaceis still connected but has infinitely many path components.

Theorem 4.9 Let X be a topological space. Then

(i) Two path components of X are either disjoint or identical and X is the union of thepairwise disjoint components.

(ii) The path component of X containing an element x is the biggest path connected subsetof X that contains x.

(iii) Every nonempty path connected subset of X is contained in exactly one path component.


Proof. The proof is similar to that of Theorem 4.7. One has to observe that the union ofpath connected sets having a common point is path connected. �

Remark 4.3 Path components need not be closed. For instance in the topologist’s sine curve,S is a path component which is not closed.

Connectedness is a useful property for a space to possess. But for some purposes, it is moreimportant that the space satisfy a connectedness condition locally. Local connectedness meansthat each point has arbitrarily small neighborhoods that are connected. More precisely, we havethe following definition.

Definition 4.5 A space X is said to be locally connected at x if for every neighborhood Uof x, there is a connected neighborhood V of x contained in U . If X is locally connected ateach of its points, it is said to be locally connected. Similarly, a space X is said to be locally

path connected at x if for every neighborhood U of x there is a path connected neighborhoodV of x contained in U . If X is locally connected at each of its points, it is said to be locally

path connected.

Remark 4.4 Since path connected ⇒ connected then locally path connected ⇒ locally con-nected.

Examples. 1) Each interval of IR is both connected and locally connected. In fact in IR, pathconnected⇔ path connected ⇔ convex ⇔ interval.

2) IRn is path connected and locally path connected. Every open subset of IRn is locally pathconnected. This follows form the fact that an open ball in IRn is path connected.

3) IR∗ is not connected but locally connected.

4) The topologist’s sine curve is connected but not locally connected.

5) The set of rationals Q is neither connected nor locally connected.

Theorem 4.10 A space X is locally connected if and only if for every open set U of X, eachcomponent of U is open in X.

Proof. Suppose first that X is locally connected. Let U be an open subset of X and letC be a component of U . Let x ∈ C. Since U is a neighborhood of x, there exists by localconnectedness a connected neighborhood V of x such that V ⊂ U . Since V is connected andcontains x, it is contained in C. This means that C is a neighborhood of each of its points.Therefore C is open in X.

Conversely, suppose that components of open sets are open. Given a point x ∈ X and anopen neighborhood U of x, let C be the component of U containing x. Since C is open andconnected, it is a connected neighborhood of x. This means that X is locally connected. �

Corollary 4.2 The connected components of a locally connected space are clopen.

Similarly, we have the following.

Theorem 4.11 A space X is locally path connected if and only if for every open set U of X,each path component of U is open in X.

The relation between connected components and path components is given in the followingtheorem,

4.2. COMPONENTS AND LOCAL CONNECTEDNESS 49

Theorem 4.12 Let X be a topological space. Each path component is contained in exactlyone connected component. If X is locally path connected, then the connected components andthe path components are the same.

Proof. The first statement is clear because a path component is connected and so it iscontained in exactly one component. Now assume that X is locally path connected (actuallya weaker assumption is needed). Let x ∈ X and let C(x) and P (x) denote respectively thecomponent and path component containing x. We need to show that C(x) ⊂ P (x), and for thisit is enough to show that C(x) is path connected. Let A be the set of points in X that can bejoined to x by a path. Let us denote this relation by ∼p. Note that A ⊂ C(x) because a pathis connected. Then x ∈ A and so A is not empty. We show that A is clopen. Let y ∈ A andlet U be a path connected neighborhood of y. If z ∈ U then z ∼p y. But y ∼p x. Thereforez ∼p x and so z ∈ A. This proves that A is a neighborhood of each of its points and so it isopen. Now let y ∈ X − A and let V be a connected neighborhood of y. Then V ⊂ X − Abecause if z ∈ V then z ∼p y; if z ∈ A, then z ∼p x and so x ∼p y, which means that y ∈ A,a contradiction. Conclusion: A is nonempty and clopen in C(x). Since C(x) is connected, itfollows that C(x) = A and so every point of C(x) can be joined to x by a path. This meansthat C(x) is path connected. �

Corollary 4.3 Let O be an open subset of IRn. Then O is connected if and only if it is pathconnected.

We end this chapter by a useful characterization of open sets of the real line.

Theorem 4.13 Every open set O of the real line is the union of a countable family of pairwisedisjoint open intervals.

Proof. Let (Iλ)λ∈L be the collection of connected components of O. We know that each Iλ

is an open interval (recall that in a locally connected space, a connected component of an openset is open). Since the rational numbers are dense in IR, each Iλ contains a rational number.By the axiom of choice, we can choose exactly one rational number in each Iλ. This defines afunction between L and Q which is one to one because two distinct components are disjoint.Therefore L is countable. Finally we have indeed O = ∪λ∈LIλ. �

Chapter 5

Complete metric spaces

Let (xn) be a sequence of a metric space (X, d) that converges to x. Recall that this means thatfor any ε > 0, we can find an integer n0 such that d(xn, x) < ε for all n ≥ n0. Therefore, by thetriangle inequality, d(xn, xm) < 2ε for all integers n and m greater or equal to n0. Otherwisestated, the difference of any two terms can be made arbitrary small when their indices are largeenough. We are thus led to formulate the following definition.

Definition 5.1 Let (X, d) be a metric space. A sequence (xn) of X is called a Cauchy sequenceif for any ε > 0 there is a number n0 such that d(xn, xm) < ε for all integers n,m ≥ n0.

According to what we said, any convergent sequence is a Cauchy sequence. Is the conversetrue? The converse need not be true. If it is true, the space is called complete.

Definition 5.2 A metric space (X, d) is called complete if every Cauchy sequence of X isconvergent.

Counterexamples. 1) The sequence 1.4, 1.41, 1.414, 1.4142, 1.41421, . . . of rational numbersis a Cauchy sequence whose limit

√2 is not in Q. Thus Q is not complete.

2) Consider the sequence (xn) of (0,2) defined by xn = 1n . This sequence is a Cauchy sequence

which is not convergent in (0,2). Thus (0,2) is not complete. This example also shows that com-pleteness is not a topological property (that is, not invariant under homeomorphisms) because(0,2) is not complete whereas IR as we shall see is complete.

Lemma 5.1 A Cauchy sequence is bounded.

Proof. Let (xn) be a Cauchy sequence of a metric spacce(X, d). Taking ε = 1 in the definitionof a Cauchy sequence, we see that there exists n0 such that d(xn, xm) < 1 for n,m ≥ n0. LetM = max{d(xn, xm)|n,m ≤ n0}. Then, by the triangle inequality d(xn, xm) ≤ d(xn, xn0

) +d(xn0

, xm) ≤ M + 1 for all n,m ∈ IN∗. This means that the sequence (xn) is bounded. �

Lemma 5.2 If a Cauchy sequence (xn) contains a convergent subsequence, then (xn) is con-vergent.

Proof. Let (xni) be a subsequence converging to x. Let ε > 0 be given. Then, there is an

index m1 such that d(xni, x) < ε/2 for i ≥ m1. Since {xn} is a Cauchy sequence and ni ≥ i,

there is an integer m2 such that d(xni, xi) < ε/2 for i ≥ m2. Therefore, d(xi, x) < ε for all

i ≥ max(m1,m2). �

Corollary 5.1 A compact metric space is complete.

Theorem 5.1 IRn is a complete under any one of the usual distances.

51

52 CHAPTER 5. COMPLETE METRIC SPACES

Proof. Since a Cauchy sequence is bounded, it belongs to a compact set (a closed ball).Therefore, it contains a convergent subsequence. According to the previous lemma it converges.�

Remark 5.1 Thus we have established a new criterion for convergence. Note that, this crite-rion permits us to decide whether a sequence is convergent without knowing its limit. This isvery useful, especially in theoretical considerations.

Remark 5.2 Here is another proof of the completeness of IR which is a direct consequence ofthe sup property. Let {xn} be a Cauchy sequence and let ε > 0 be given. Let N be such that|xn − xm| < ε for n,m ≥ N . Define

S = {y |xn < y for finitely many n} = {y |xn ≥ y for n large enough}.

Now observe that xn − ε ∈ S for all n. In particular xN − ε ∈ S. We claim that y ∈ S ⇒ y <xN + ε. Indeed, let y ∈ S. Then xn ≥ y for n large enough. If y ≥ xN + ε, then xn ≥ xN + εfor n large enough, which contradicts the definition of N . Therefore y < xN + ε. This meansthat xN + ε is an upper bound of S. Let now b = supS. It follows from what we have said thatxN−ε ≤ b ≤ xN +ε, that is |xN−b| ≤ ε. Now if m ≥ N , then |xm−b| ≤ |xm−xN |+|xN−b| ≤ 2ε.Since ε is arbitrary, it follows that that xn → b.

Proposition 5.1 Let (X, d) be a complete metric space and let F be a subspace of X. ThenF is complete if and only if it is closed in X.

Proof. Suppose that F is complete (under d). Let x ∈ F . Then there is a sequence (xn) of Fthat converges to x. Therefore (xn) is a Cauchy sequence. Since F is complete, (xn) convergesto some y ∈ F . But (xn) converges to x. Therefore x = y ∈ F . This means that F is closed asa subspace of X.

Conversely, suppose that F is closed in X. Let (xn) be a Cauchy sequence of F . Since X iscomplete, (xn) has a limit x ∈ X. But closedness of F implies that x ∈ F . �

Definition 5.3 Let X be a set and f : X → X. A fixed point of f is a point a ∈ X suchthat a = f(a).

Definition 5.4 Let (X, dX ) and (Y, dY ) be metric space. A function f : X → Y is calleda contraction if there exists a constant k < 1 such that dY (f(x), f(y)) ≤ kdX(x, y) for allx, y ∈ X.

Example. The function f : [−1.1] → [−1, 1] defined by f(x) = x2

3 is a contraction. Indeed,f ′(x) = 2

3x and so |f ′| is bounded by 23 . By the mean value theorem, |f(x) − f(y)| ≤ 2

3 |x − y|for all x, y ∈ [−1, 1].

The following theorem is important in analysis.

Theorem 5.2 (The Banach fixed point theorem) Let X be a complete metric space andlet f : X → X be a contraction. Then f has a unique fixed point.

Proof. Let k < 1 satisfy d(f(x), f(y)) ≤ kd(x, y) for all x, y ∈ X. We first prove uniqueness.Suppose that we have two ficed points a and b. Then d(f(a), f(b)) ≤ kd(a, b) and so d(a, b) ≤kd(a, b). This implies that d(a, b) = 0 because otherwise, we get 1 ≤ k.

Next, we prove existence of a fixed point. Pick a point a ∈ X and define the sequence (xn)by x0 = a and xn+1 = f(xn). Observe that for n ≥ 1,

d(xn+1, xn) = d(f(xn), xn) = d(f(xn), f(xn−1)) ≤ d(xn, xn−1).

53

It follows by induction that d(xn+1, xn) ≤ knd(x1, x0). Let now p ≥ 1. Using the traingleinequality and the last estimate, we get

d(xn, xn+p) ≤ d(xn, xn+1) + d(xn+1, xn+2) · · · + d(xn+p−1, xn+p)

≤ knd(x1, x0) + kn+1d(x1, x0) + · · · kn+p−1d(x1, x0)

= (kn + kn+1 + · · · kn+p−1)d(x1, x0)

= kn 1 − kp

1 − kd(x1, x0) ≤

kn

1 − kd(x1, x0).

Let ε > 0 be given. Since kn → 0 as n → ∞, we have kn

1−kd(x1, x0) < ε for all n large enough.Then we have d(xn, xn+p) < ε for all n large enough and all p ≥ 1. This means that (xn) isa Cauchy sequence. Since X is complete (xn) has a limit a. Letting n → ∞ in the relationxn+1 = f(xn) and using the continuity of f we get a = f(a). �

We already noticed that a compact metric space is complete. The converse is not truebecause IR is complete but not compact. However, it turns out that if a space is complete andtotally bounded, then it is compact.

Theorem 5.3 A metric space is compact iff it is totally bounded and complete.

Proof. If X is compact, then it is totally bounded and complete. Conversely, suppose thatX is totally bounded and complete. We prove that X is sequentially compact. So let (xn) bea sequence of X. First cover X by finitely many balls of radius 1. At least one of these balls,call it B1 contains xn for infintely many indices n. Let J1 = {n ∈ IN∗|xn ∈ B1}. Next, cover Xby finitely many balls of radius 1

2 . Because J1 is infinite, at least one of these balls, call it B2,contains xn for infinitely many indices n ∈ J1. Let J2 = {n ∈ J1|xn ∈ B2}. At the kth step, letJk = {n ∈ Jk−1|xn ∈ Bk} where Bk is a ball of radius 1

k and Jk is infinite.Choose n1 ∈ J1. Given nk−1, choose nk ∈ Jk such that nk > nk−1 (this is possible because

Jk is infinite). Now if i, j ≥ k, the indices ni and nj belong to Jk (because (Jk) is a decreasingsequence of subsets of IN∗). Therefore for all i, j ≥ k, the points xni

and xnjbelong to the ball

Bk of radius 1k . Therefore d(xni

, xnj) ≤ 2

k . This means that (xni) is a Cauchy sequence. Since

X is complete (xni) is convergent and this means that X is sequentially compact. �

Corollary 5.2 In a complete metric space, a subset has compact closure if and only if it istotally bounded.

Proof. Let X be a complete metric space and let A ⊂ X. Suppose first that A is compact.Then A is totally bounded. It follows that A is also totally bounded (prove this). Conversely,suppose that A is totally bounded. Then A is also totally bounded (prove this). But since Ais closed in a complete space, it is itself complete. Being totally bounded, it is compact by theprevious theorem. �

A subset with a compact closure is also called relatively compact.

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