MATH 3650 Topology

SUN, Pak Kiu

August 29, 2011

Contents

Chapter 1 Metric Spaces 11.1 Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Continuity and open sets for metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

Chapter 2 Topological spaces 72.1 Definition of a Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Basis for a topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.3 Closed Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Chapter 3 Interior, Closure and Boundary 123.1 Interior and Closure of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.2 Limit Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.3 Boundary of a Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

Chapter 4 New Spaces from Old Ones 174.1 The Subspace Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.2 The Product Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184.3 The Quotient Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Chapter 5 Continuity and Homeomorphisms 215.1 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.2 Homeomorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

Chapter 6 Connectedness 246.1 Connected Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246.2 Intermediate Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276.3 Path Connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Chapter 7 Compactness 297.1 Compactness in metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327.2 The Extreme Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357.3 Bolzano-Weierstrass property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Chapter 1

Metric Spaces

1.1 Metric

Definition 1.1. Let X be a set and d : X2 → R a function with the following properties:-

(i) d(x, y) ≥ 0 for all x, y ∈ X.

(ii) d(x, y) = 0 if and only if x = y.

(iii) d(x, y) = d(y, x) for all x, y ∈ X.

(iv) d(x, y) + d(y, z) ≥ d(x, z) for all x, y, z ∈ X. (Triangle inequality)

Then d is a metric on X and (X, d) is a metric space.

The properties (i)-(iv) can be illustrated as follows:

(i) Distances are always positive.

(ii) Two points are zero distance apart if and only if they are the same point.

(iii) The distance from A to B is the same as the distance from B to A.

(iv) The distance from A to B via C is at least as great as the distance from A to B directly.

Example 1.2. Let d(x, y) = |x− y| for all x, y ∈ R. Show that (R, d) is a metric space.

Proof. It is easy to check that conditions (i)-(iii) satisfied. Let x, y, z ∈ R and WLOG, assume y ≤ x ≤ z.We have

d(x, y) + d(y, z) = (x− y) + (z − y)

≥ z − y

≥ z − x

= d(x, z).

This metric is called the Euclidean metric or standard metric on R.

Example 1.3. Let p, q ∈ R2 with p = (p1, p2) and q = (q1, q2). We defined

d(p, q) =√

(p1 − q1)2 + (p2 − q2)2

and d is called the standard metric on R2. Also, if we let

dT = |p1 − q1|+ |p2 − q2|,

then dT is a metric (verify yourself) and it is called the taxicab metric.

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The following exercises help you understand better the properties of metric spaces.

Exercise 1.4. If d : X2 → R is a function with the following properties:-

(ii) d(x, y) = 0 if and only if x = y,

(iii) d(x, y) = d(y, x) for all x, y ∈ X,

(iv) d(x, y) + d(y, z) ≥ d(x, z) for all x, y, z ∈ X,

show that d is a metric on X. [Thus condition (i) of the definition is redundant.]

Exercise 1.5. Let X = {a, b, c} with a, b and c distinct. Write down functions dj : X2 → R satisfyingcondition (i) of Definition 1.1 such that:-

(1) d1 satisfies conditions (ii) and (iii) but not (iv).

(2) d2 satisfies conditions (iii) and (iv) and d2(x, y) = 0 implies x = y, but it is not true that x = y impliesd2(x, y) = 0.

(3) d3 satisfies conditions (iii) and (iv) and x = y implies d3(x, y) = 0. but it is not true that d3(x, y) = 0implies x = y.

(4) d4 satisfies conditions (ii) and (iv) but not (iii). You should verify your statements.

Exercise 1.6. Let X be a set and ρ : X2 → R a function with the following properties.

(i) ρ(x, y) ≥ 0 for all x, y ∈ X.

(ii) ρ(x, y) = 0 if and only if x = y.

(iv) ρ(x, y) + ρ(y, z) ≥ ρ(x, z) for all x, y, z ∈ X.

Show that, if we set d(x, y) = ρ(x, y) + ρ(y, x), then (X, d) is a metric space.

The next example is a metric that is very important.

Definition 1.7. If X is a set and we define d : X2 → R by

d(x, y) =

{0 if x = y,

1 if x = y,

then d is called the discrete metric on X.

Lemma 1.8. The discrete metric on X is indeed a metric.

Proof. The only non-evident condition is the triangle law. But

d(x, y) + d(y, z) = d(x, y) = d(x, z) ≥ d(x, z) if y = z,

d(x, y) + d(y, z) = d(y, z) = d(x, z) ≥ d(x, z) if x = y,

d(x, y) + d(y, z) ≥ 1 + 1 = 2 ≥ 1 ≥ d(x, z) otherwise,

so we are done.

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1.2 Continuity and open sets for metric spaces

Some definitions and results transfer essentially unchanged from classical analysis on R to metric spaces.Recall the classical definition of continuity.

Definition 1.9. [Analysis definition.] A function f : R → R is called continuous if, given t ∈ R andϵ > 0, we can find a δ(t, ϵ) > 0 such that

|f(t)− f(s)| < ϵ whenever |t− s| < δ(t, ϵ).

It is not hard to extend this definition to our new, wider context.

Definition 1.10. [Metric definition.] Let (X, d) and (Y, ρ) be metric spaces. A function f : X → Y iscalled continuous if, given t ∈ X and ϵ > 0, we can find a δ(t, ϵ) > 0 such that

ρ(f(t), f(s)) < ϵ whenever d(t, s) < δ(t, ϵ).

The term ρ(f(t), f(s)) is ‘the distance from f(t) to f(s) in Y ’ and d(t, s) is ‘the distance from t to s inX’.

Lemma 1.11. [The composition law.] If (X, d) and (Y, ρ) and (Z, σ) are metric spaces and g : X → Y ,f : Y → Z are continuous, then so is the composition fg.

Proof. Let ϵ > 0 be given and let x ∈ X. Since f is continuous, we can find a δ1 > 0 (depending on ϵ andfg(x) = f(g(x)) with

σ(f(g(x)), f(y)) < ϵ whenever ρ(g(x), y) < δ1.

Since g is continuous, we can find a δ2 > 0 such that

ρ(g(x), g(t)) < δ1 whenever d(x, t) < δ2.

We now haveσ(f(g(x)), f(g(t))) < ϵ whenever d(x, t) < δ2

as required.

Exercise 1.12. Let R and R2 have their usual (Euclidean) metric.

(i) Suppose that f : R → R and g : R → R are continuous. Show that the map (f, g) : R2 → R2 iscontinuous.

(ii) Show that the map M : R2 → R given by M(x, y) = xy is continuous.

(iii) Use the composition law to show that the map m : R2 → R given by m(x, y) = f(x)g(y) is continuous.

Just as there are ‘well behaved’ and ‘badly behaved’ functions between spaces so there are ‘well behaved’and ‘badly behaved’ subsets of spaces. In classical analysis and analysis on metric spaces the notion ofcontinuous function is sufficiently wide to give us a large collection of interesting functions and sufficientlynarrow to ensure reasonable behaviourIn introductory analysis we work on R with the Euclidean metric andonly consider subsets in the form of intervals. Once we move to R2 with the Euclidean metric, the idea ofopen ball evolved.

Definition 1.13. Let (X, d) be a metric space. For x ∈ X and ϵ > 0, we define the open ball of radius ϵcentered at x to be the set

Bd(x, ϵ) = {y ∈ X | d(x, y) < ϵ}

Cantor identified two particular classes of ‘well behaved’ sets. We start with open sets.

Definition 1.14. Let (X, d) be a metric space. A subset U of it is open if for each x ∈ U , we can find anopen ball Bd(x, ϵ) such that Bd(x, ϵ) ⊆ U .

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The following example verified that all open balls are open set.

Example 1.15. Consider the two sets.

1. Let (X, d) be a metric space. If r > 0, then

Bd(x, r) = {y | d(x, y) < r}

is open.

2. If (X, d) is a discrete metric space, then

{x} = Bd(x, 1/2)

and all subsets of X are open.

Proof. The two sets are

1. If y ∈ Bd(x, r), then let δ = r − d(x, y) > 0 and, whenever d(z, y) < δ, the triangle inequality gives us

d(x, z) ≤ d(x, y) + d(y, z) < r

so z ∈ B(x, r). Thus B(x, r) is open.

2. Observe that d(x, x) = 0 < 1/2 and d(x, y) = 1 > 1/2 for x = y. If x ∈ E then d(x, y) < 1/2 impliesy = x ∈ E so E is open.

Theorem 1.16. If (X, d) is a metric space then the following statements are true.

(i) The empty set ∅ and the space X are open.

(ii) If Uα is open for all α ∈ A, then∪

α∈A Uα is open. (In other words the union of open sets is open.)

(iii) If Uj is open for all 1 ≤ j ≤ n, then∩n

j=1 Uj is open.

Proof. (i) Since there are no points e in ∅, the statement

x ∈ ∅ whenever d(x, e) < 1

holds for all e ∈ ∅. Since every point x belongs to X, the statement

x ∈ X whenever d(x, e) < 1

holds for all e ∈ X.

(ii) If e ∈∪α∈A

Uα, then we can find a particular α1 ∈ A with e ∈ Uα1 . Since Uα1 is open, we can find a

δ > 0 such thatx ∈ Uα1 whenever d(x, e) < δ.

Since Uα1 ⊆∪

α∈A Uα,

x ∈∪α∈A

Uα whenever d(x, e) < δ.

Thus∪α∈A

Uα is open.

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(iii) If e ∈n∩

j=1

Uj then e ∈ Uj for each 1 ≤ j ≤ n. Since Uj is open, we can find a δj > 0 such that

x ∈ Uj whenever d(x, e) < δj .

Setting δ = min1≤j≤n δj , we have δ > 0 and

x ∈ Uj whenever d(x, e) < δ

for all 1 ≤ j ≤ n. Thus

x ∈n∩

j=1

Uj whenever d(x, e) < δ

and we have shown that

n∩j=1

Uj is open.

It is important to realize that we place no restriction on the size of A in (ii). In particular, A could beuncountable. However, the number of sets in condition (iii) must be finite.

Example 1.17. Let us work in Rn with the usual metric. Then B(x, 1/j) is open but

∞∩j=1

B(x, 1/j) = {x}

is not.

There is a remarkable connection between the notion of open sets and continuity.

Theorem 1.18. Let (X, d) and (Y, ρ) be metric spaces. A function f : X → Y is continuous if and only iff−1(U) is open in X whenever U is open in Y .

Proof. Suppose first that f is continuous and that U is open in Y . If x ∈ f−1(U), then we can find a y ∈ Uwith f(x) = y. Since U is open in Y , we can find an ϵ > 0 such that

z ∈ U whenever ρ(y, z) < ϵ.

Since f is continuous, we can find a δ > 0 such that

ρ(y, f(w)) = ρ(f(x), f(w)) < ϵ whenever d(x,w) < δ.

Thusf(w) ∈ U whenever d(x,w) < δ.

In other words,w ∈ f−1(U) whenever d(x,w) < δ.

We have shown that f−1(U) is open.We now seek the converse result. Suppose that f−1(U) is open in X whenever U is open in Y . Suppose

x ∈ X and ϵ > 0. We know that the open ball

Bρ(f(x), ϵ) = {y ∈ Y | ρ(f(x), y) < ϵ}

is open. Thus x ∈ f−1{Bρ

(f(x), ϵ

)}and f−1

{Bρ

(f(x), ϵ

)}is open. It follows that there is a δ > 0 such that

w ∈ f−1{Bρ

(f(x), ϵ

)}whenever d(x,w) < δ,

so, in other words,ρ(f(x), f(w)) < ϵ whenever d(x,w) < δ.

Thus f is continuous.

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Note that the theorem does not work ‘in the opposite direction’.

Example 1.19. Let X = R and d be the discrete metric. Let Y = R and ρ be the usual (Euclidean) metric.

(i) If we define f : X → Y by f(x) = x, then f is continuous but there exist open sets U in X such thatf(U) is not open.

(ii) If we define g : Y → X by g(y) = y, then g is not continuous but g(V ) is open in X whenever V isopen in Y .

Proof. Since every set is open in X, we have f−1(V ) = g(V ) open for every V in Y and so, in particular, forevery open set. Thus f is continuous. We observe that U = {0} is open in X and g−1(U) = f(U) = U = {0}is not open in Y . Thus g is not continuous.

Definition 1.20. Consider a sequence xn in a metric space (X, d). If x ∈ X and, given ϵ > 0, we can findan integer N ∈ N (depending on ϵ) such that

d(xn, x) < ϵ for all n ≥ N,

then we say that xn → x as n → ∞.

Definition 1.21. Let (X, d) be a metric space. A set F in X is said to be closed if, whenever xn ∈ F andxn → x as n → ∞, it follows that x ∈ F .

The following exercise is easy but instructive.

Exercise 1.22. (i) If (X, d) is any metric space, then X and ∅ are both open and closed.

(ii) If we consider R with the usual metric and take b > a, then [a, b] is closed but not open, (a, b) is openbut not closed and [a, b) is neither open nor closed.

(iii) If (X, d) is a metric space with discrete metric d, then all subsets of X are both open and closed.

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Chapter 2

Topological spaces

2.1 Definition of a Topology

We now investigate general objects which have the structure described by Theorem 1.16.

Definition 2.1. Let X be a set and T a collection of subsets of X, each called an open set, such that

(i) The empty set ∅ ∈ T and the space X ∈ T .

(ii) If Uα ∈ T for all α ∈ A, then∪

α∈A Uα ∈ T .

(iii) If Uj ∈ T for all 1 ≤ j ≤ n, then∩n

j=1 Uj ∈ T .

Then we say that T is a topology on X and that (X, T ) is a topological space.

Theorem 2.2. If (X, d) is a metric space, then the collection of open sets forms a topology.

Proof. This is Theorem 1.16.

If (X, d) is a metric space we call the collection of open sets the topology induced by the metric.

Exercise 2.3. If (X, d) is a metric space with the discrete metric, show that the induced topology consistsof all the subsets of X.

We call the topology consisting of all subsets of X the discrete topology on X.

Exercise 2.4. If X is a set and T = {∅, X}, then T is a topology.

Exercise 2.5. We call {∅, X} the indiscrete topology on X.

(i) If F is a finite set and (F, d) is a metric space show that the induced topology is the discrete topology.

(ii) If F is a finite set with more than one point, show that the indiscrete topology is not induced by anymetric

Definition 2.6. Let X be a set and let T1 and T2 be two topologies on X. If T1 ⊆ T2 then T2 is said to befiner than T1, and T1 is said to be coarser than T2. Furthermore, if T2 is finer than T1 but not equal to T1then T2 is said to be strictly finer than T1. Strictly coarser is defined similarly.

In general, two topologies on a given set X need not be comparable, that is, neither topology would befiner than the other.

We use the following definition to simplify our future discussion on open sets.

Definition 2.7. Let X be a topological space and x ∈ X. An open set U containing x is said to be aneighborhood of x.

Theorem 2.8. Let X be a topological space and let A be a subset of X. Then A is open in X if and only iffor each x ∈ A, there is a neighborhood U of x such that x ∈ U ⊆ A.

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Before proceeding to the proof of Theorem 2.8, we introduce a very useful lemma.

Lemma 2.9. The Union Lemma LetX be a set and C be a collection of subsets of X. Assume that for

each x ∈ X, there is a set Ax in C such that x ∈ Ax. Then∪x∈X

Ax = X.

Proof. We prove that∪

Ax ⊆ X and X ⊆∪

Ax. First, since each Ax is a subset of X, it follows that∪Ax ⊆ X.Conversely, suppose that y ∈ X. There exists Ay in C such that y ∈ Ay. Thus y ∈ Ay ⊆

∪Ax. It follows

that X ⊆∪

Ax and hence∪

Ax = X.

Proof of Theorem 2.8. First suppose that A is open in X and x ∈ A. Let U = A, then U is a neighborhoodof x for which x ∈ U ⊆ A.

Conversely, for every x ∈ A there exists a neighborhood Ux of x such that x ∈ Ux ⊆ A. By the UnionLemma (Lemma 2.9), it follows that A =

∪x∈A Ux. Thus, A is a union of open sets and therefore is open.

2.2 Basis for a topology

In general, it is difficult to specify the entire collection of open sets. We usually specify a smaller collectionof open sets, a basis, which can generate the rest of the open sets.

Definition 2.10. Let X be a set and B be a collection of subsets of X. We say B is a basis if the followingstatements hold:

(i) For each x ∈ X, there is a B in B such that x ∈ B.

(ii) If B1 and B2 are in B and x ∈ B1 ∩B2, then there exists B3 in B such that x ∈ B3 ⊆ B1 ∩B2.

Example 2.11. On the real line R, let B = {(a, b) ⊆ R | a < b}, the set of open intervals in R. Certainlyevery point of R is contained in an open interval and therefore is contained in a set in B. Moreover, theintersection of two open intervals is either an open interval or empty. Thus, a point in the intersection oftwo sets in B is contained in a set in B and B is a basis.

Next, we describe how to generate a collection T from a basis B and show that T is a topology.

Definition 2.12. Let B be a basis on a set X. The topology T generated by B is obtained by definingthe open sets to be the empty set and every set that is equal to a union of basis elements.

Before showing that T is a topology, we need the following lemma.

Lemma 2.13. Let B be a basis. Assume that B1, . . . , Bn ∈ B and that x ∈∩n

i=1Bi. Then there existsB′ ∈ B such that x ∈ B′ ⊆

∩ni=1Bi.

Proof. We prove this by induction on n, starting with n = 2. The n = 2 case holds by the second conditionin the definition of a basis.

Assume that the result is true for n−1. Suppose that the sets B1, . . . , Bn are in B and that x ∈∩n

i=1Bi.Then x ∈

∩n−1i=1 Bi, and the induction hypothesis implies that there exists B∗ ∈ B such that x ∈ B∗ ⊆∩n−1

i=1 Bi. Now x ∈ B∗∩Bn; therefore by the second condition in the definition of a basis, there exists B′ ∈ Bsuch that x ∈ B′ ⊆ B∗∩Bn. Since B

∗ ⊆∩n−1

i=1 Bi, it follows that x ∈ B′ ⊆∩n

i=1Bi. Thus, if the result holdsfor n− 1, then it holds for n, and by induction the lemma follows.

Theorem 2.14. The topology T generated by a basis B is a topology.

Proof. The empty set ∅ is in T by definition. Since every point in X is contained in some basis element, itfollows that X is the union of all of these basis elements and therefore is in T .

Next we show that a finite intersection of sets in T is in T . Thus, let V = U1 ∩ · · · ∩ Un where each Ui

is in T . If one of the Ui is empty, then so is V , and thus V ∈ T . Assume that each Ui is a union of basiselements. Let x ∈ V be arbitrary. Then x ∈ Ui for all i. Since each Ui is a union of basis elements, there

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exists a basis element Bi such that x ∈ Bi ⊆ Ui for each i. Then x ∈n∩

i=1

Bi; therefore, by Lemma 2.13,

there exists a basis element Bx such that x ∈ Bx ∈n∩

i=1

Bi ⊆ V . It follows from the Union Lemma that

V =∪x∈X

Bx, and therefore V is a union of basis elements. Thus, a finite intersection of sets in T is in T .

Finally, we show that an arbitrary union of sets in T is in T . Let V =∪

Uα where each Uα is either

the empty set or a union of basis elements. If each Uα is empty, then so is V ; on the other hand, if at leastone Uα is nonempty, then V is a union of basis elements, since it is the union of all of the basis elementsmaking up the Uα’s. Therefore an arbitrary union of sets in T is in T . As a result, the collection of sets Tis a topology, and we are justified in calling it the topology generated by the basis B.

Example 2.15. In example 2.11, we showed the collection B = {(a, b) ⊆ R | a < b} is a basis on R and thetopology generated is the standard topology. On the other hand, the collection Bl = {[a, b) ⊆ R | a < b} isalso a basis on R (left as exercise) and the topology generated is the lower limit topology.

By definition, the open sets in the topology generated by a basis are the sets obtained by taking unionsof basis elements. The next theorem provides another simple way to describe the sets in such a topology.

Theorem 2.16. Let X be a set and B be a basis for a topology on X. Then U is open in the topologygenerated by B if and only if for each x ∈ U , there exists a basis element Bx ∈ B such that x ∈ Bx ⊆ U .

Proof. Suppose U is an open set in the topology generated by B and that x ∈ U . Since U is a union of basiselements, there is at least one basis element B′ among this union that contains x. Thus, we have x ∈ B′ ⊆ U .

Conversely, suppose U ⊆ X and U has the property that for each x ∈ U , there exists Bx ∈ B such that

x ∈ Bx ⊆ U . By the Union Lemma, U =∪x∈U

Bx, and therefore U is a union of basis elements. Thus, U is

an open set in the topology generated by B.

Exercise 2.17. Show that the collections B = {B(x, ϵ) | x ∈ R2, ϵ > 0} and Br = {(a, b)× (c, d) ⊆ R2 | a <b, c < d} generate the same topology (standard topology) on R2.

We now know that different basis can generate the same topology and the following theorem provide aquick method to check whether a collection of open sets is a basis.

Theorem 2.18. Let X be a set with topology T and let C be a collection of open sets in X. If, for eachopen set U in X and for each x ∈ U , there is an open set V in C such that x ∈ V ⊆ U , then C is a basisthat generates the topology T .

Proof. (i) The collection C is a basis.

Let x ∈ X. From the assumption of C, there is an open set V in C such that x ∈ V ⊆ X. Thereforeevery point in X is contained in some open set V in the collection C. Thus the first condition for abasis is satisfied.

Suppose x ∈ V1 ∩ V2, where V1 and V2 in C and so V1 ∩ V2 is an open set. According to our hypothesis,there must be an open set V3 in C such that x ∈ V3 ⊆ V1 ∩ V2. Therefore C is a basis.

(ii) The topology T ′ generated by C coincides with TSuppose U is open in T . Then by the hypothesis, for every x ∈ U there is an open set V in C such thatx ∈ V ⊆ U . So by Theorem 2.16, U is open in T ′, the topology generated by C. Therefore T ⊆ T ′.Now suppose that W is an open set in T ′. Then W is a union of open sets in C, all of which are openin T . However, every union of open sets in T is open in T , so W is also open in T . Thus T ′ ⊆ T andit follows that T = T ′.

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2.3 Closed Sets

Open sets in topological spaces is a generalization of open intervals in R. Closed interval is anotherimportant subset on R and the generalization of it can be facilitate by considering [a, b] = (−∞, a)∪ (b,∞).We focus on the concept: closed interval is the complement of union of open intervals and we have thefollowing definition.

Definition 2.19. Let (X, T ) be a topological space. A set A in X is said to be closed if its complementX \A is open.

Exercise 2.20. Show that the following subsets are closed set in R2.

(i) The closed ball B(x, ϵ) = {y ∈ R2 | d(x, y) ≤ ϵ} and

(ii) the closed rectangle [a, b]× [c, d] ⊆ R2.

Example 2.21. The following topologies illustrate some important properties about open and closed sets.

(i) Let X be a set with the discrete topology. Every subset of X is an open set. Given an arbitrary setA in X, its complement is then open since every set is open. Therefore A is closed and, since A wasarbitrary, it follows that every subset of X is also a closed set in the discrete topology.

(ii) Let the set {a, b, c, d} have the topology T = {∅, X, {a, b}, {b}, {c, d}, {b, c, d}}. Note that {b} is openand not closed, {a} is closed and not open, {a, b} is both open and closed, and {b, c} is neither opennor closed.

Closed sets are complementary to open sets and so they have similar properties.

Theorem 2.22. If (X, T ) is a topological space, then the following statements are true.

1. The empty set ∅ and the space X are closed.

2. If Fα is closed for all α ∈ A, then∩α∈A

Fα is closed.

3. If Fj is closed for all 1 ≤ j ≤ n, then

n∪j=1

Fj is closed.

Proof. Consider the complement of open sets

(i) Observe that ∅ = X \X and X = X \∅.

(ii) Since Fα is closed, X \ Fα is open for all α ∈ A. It follows that

X \∩α∈A

Fα =∪α∈A

(X \ Fα)

is open and so∩α∈A

Fα is closed.

(iii) Since Fj is closed, X \ Fj is open for all 1 ≤ j ≤ n. It follows that

X \n∪

j=1

Fj =

n∩j=1

(X \ Fj)

is open and so

n∪j=1

Fj is closed.

10

All single-point sets are closed in Rn with the standard topology. However, there are topologies wherethis is not the case. For example, consider the topology T ∗ on Rn generated by the basis of sets {[n, n+1) ⊆Rn | n ∈ Z}. It is not difficult to see that no single-point set is closed in T ∗.

Next we present a property of topological spaces that has some particularly nice consequences for spacessatisfying it. As we show below, these consequences include having single-point sets closed. We will see afew of the other convenient consequences of this property throughout the text.

Definition 2.23. A topological space (X, T ) is called Hausdorff if for every x, y ∈ X and x = y, we canfind U, V ∈ T such that x ∈ U , y ∈ V and U ∩ V = ∅.

Theorem 2.24. Every metric space is Hausdorff.

Proof. Let (X, d) be a metric space. Suppose x and y are distinct points in X with d(x, y) = ϵ. Considerthe sets U = Bd(x,

ϵ2) and V = Bd(y,

ϵ2). It follows that x ∈ U, y ∈ V , and U and V are open sets. We

claim that U and V are disjoint. Suppose U ∩ V = ∅, and z is in the intersection. Then d(x, z) < ϵ2 and

d(y, z) < ϵ2 . Therefore, by the triangle inequality,

d(x, y) < d(x, z) + d(z, y) <ϵ

2+

ϵ

2= ϵ;

that is, d(x, y) < ϵ. This contradicts d(x, y) = ϵ. Thus U ∩ V = ∅. Hence, there exist disjoint open sets Uand V containing x and y, respectively, implying that X is Hausdorff.

Theorem 2.25. If (X, T ) is a Hausdorff space, then the one point sets {x} are closed.

Proof. Let x ∈ X, it is sufficient to show that A = X \ {x} is open. Let y ∈ A and we have y = x. SinceX is Hausdorff, we can find U, V ∈ T such that x ∈ U , y ∈ V and U ∩ V = ∅. It follows that x /∈ V andtherefore y ∈ V ⊆ A. Since every y ∈ A is in an open set contained in A, Theorem 2.8 implies that A isopen.

The following exercise shows that the converse to Lemma 2.25 is false.

Exercise 2.26. Let X be infinite (we could take X = Z or X = R). We say that a subset E of X lies in Tif either E = ∅ or X \ E is finite. Show that T is a topology and that every one point set {x} is closed butthat (X, T ) is not Hausdorff. What happens if X is finite?

11

Chapter 3

Interior, Closure and Boundary

3.1 Interior and Closure of Sets

An arbitrary subset A of a topological space might be neither open nor closed. However, it is often usefulto associate a related open set or a related closed set to A. In particular, we can sandwich each set A betweenthe largest open set contained in A and the smallest closed set containing A. These sets are known as theinterior of A and the closure of A, respectively.

Definition 3.1. Let A be a subset of a topological space X. The interior of A, denoted Int(A), is theunion of all open sets contained in A. The closure of A, denoted Cl(A), is the intersection of all closed setscontaining A.

Clearly, the interior of A is open and a subset of A, and the closure of A is closed and contains A. Thuswe have Int(A) ⊆ A ⊆ Cl(A).

The following properties follow readily from the definition of interior and closure.

Theorem 3.2. Let X be a topological space and A and B be subsets of X.

(i) If U is an open set in X and U ⊆ A, then U ⊆ Int(A).

(ii) If C is a closed set in X and A ⊆ C, then Cl(A) ⊆ C.

(iii) If A ⊆ B then Int(A) ⊆ Int(B).

(iv) If A ⊆ B then Cl(A) ⊆ Cl(B).

(v) A is open if and only if A = Int(A).

(vi) A is closed if and only if A = Cl(A).

Proof. We prove (i), (iii), and (v) and (ii), (iv), and (vi) left as exercise.

(i) Suppose that U is an open set in X and U ⊆ A. Since Int(A) is the union of all of the open sets thatare contained in A, it follows that U is one of the sets making up this union and therefore is a subsetof the union. That is, U ⊆ Int(A).

(iii) Since A ⊆ B, Int(A) is an open set contained in B. Part (i) implies that every open set contained inB is contained in Int(B). Therefore Int(A) ⊆ Int(B).

(v) If A = Int(A), then A is an open set since Int(A) is an open set. Now assume that A is open. First,Int(A) ⊆ A by definition of Int(A). Conversely, since A is an open set contained in A, it follows by (i)that A ⊆ Int(A). Thus A = Int(A).

The following example demonstrates that the interior and closure of a set A depend on the topology onthe set X containing A.

12

Example 3.3. Consider A = [0, 1) as a subset of R with

(i) the standard topology. Then Int(A) = (0, 1) and Cl(A) = [0, 1].

(ii) the discrete topology. Then Int(A) = Cl(A) = [0, 1).

(iii) the lower limit topology. Here, Int(A) = A since A is an open set. Note that R\[0, 1) = (−∞, 0)∪[1,∞)is an open set in R with the lower limit topology. Therefore [0, 1) is closed in the lower limit topology,implying that Cl(A) = A as well.

Definition 3.4. A subset A of a topological space X is called dense if Cl(A) = X.

Exercise 3.5. Show that Q is dense in R with the standard topology.

The following two theorems provide a simple characterization for the interior and the closure of a givenset A.

Theorem 3.6. Let X be a topological space with A ⊆ X and y ∈ X. Then y ∈ Int(A) if and only if thereexists an open set U such that y ∈ U ⊆ A.

Proof. First, suppose that there exists an open set U such that y ∈ U ⊆ A. Then, since U is open andcontained in A, it follows that U ⊆ Int(A). Thus y ∈ U implies that y ∈ Int(A).

Conversely, if y ∈ Int(A) and we set U = Int(A), it follows that U is an open set such that y ∈ U ⊆ A.

Theorem 3.7. Let X be a topological space with A ⊆ X and y ∈ X. Then y ∈ Cl(A) if and only if everyopen set containing y intersects A.

Proof. Suppose x /∈ Cl(A). Then there is a closed set C such that A ⊆ C and x /∈ C. So X \ C is aneighborhood of x such that (X \ C) ∩A = ∅.

Conversely, suppose there is a neighborhood U of x such that U ∩ A = ∅. Then X \ U is a closed setsuch that A ⊆ X \ U and x /∈ X \ U . Hence x /∈ Cl(A).

The following theorem provides some useful relationships involving interior, closure, intersection, union,and complement:

Theorem 3.8. For sets A and B in a topological space X, the following statements hold:

(i) Int(X \A) = X \ Cl(A).

(ii) Cl(X \A) = X \ Int(A).

(iii) Int(A) ∪ Int(B) ⊆ Int(A ∪B), and in general equality does not hold.

(iv) Int(A) ∩ Int(B) = Int(A ∩B).

Proof. We prove (i) and (iii) here and left the others as exercise.

(i) We prove that X \Cl(A) ⊆ Int(X \A) and Int(X \A) ⊆ X \CI(A). First, note that Cl(A) is closedand contains A, and therefore X \Cl(A) is an open set contained in X \A. It follows by Theorem 3.2(i)that X \ Cl(A) ⊆ Int(X \A).To prove that Int(X \ A) ⊆ X \ CI(A), let x ∈ Int(X \ A) be arbitrary. Note that Int(X \ A)is disjoint from A, and therefore x is in an open set that is disjoint from A. By Theorem 3.7, itfollows that x /∈ Cl(A); hence, x ∈ X \ Cl(A). Thus, Int(X \ A) ⊆ X \ Cl(A). As a result, we haveInt(X \A) = X \ CI(A).

(iii) Since Int(A) ⊆ A, it follows that Int(A) ⊆ A ∪ B. Also, Int(A) is an open set. Similarly, Int(B)is an open set contained in A ∪ B. Every open set contained in A ∪ B is contained in Int(A ∪ B).Therefore both Int(A) and Int(B) are contained in Int(A ∪B). Hence their union, Int(A) ∪ Int(B),is also contained in Int(A ∪B).

Now we must show that there are cases where Int(A) ∪ Int(B) does not equal Int(A ∪ B). TakeA = [−1, 0] and B = [0, 1] as subsets of R with the standard topology. Then Int(A) ∪ Int(B) =(−1, 0) ∪ (0, 1), but Int(A ∪B) = (−1, 1). Thus, in this case Int(A) ∪ Int(B) = Int(A ∪B).

13

3.2 Limit Points

In the standard topology on Rn, a limit point of a subset is that there is a sequence of points in thesubset approaching the limit point. In topological spaces, limit points are defined via neighborhoods ratherthan sequences and we will encounter some unusual, and perhaps counterintuitive limit point in topologicalspaces not similar to Rn.

Definition 3.9. Let A be a subset of a topological space X. A point x in X is a limit point of A if everyneighborhood of x intersects A in a point other than x.

Notice that a limit point x of a set A may or may not lie in the set A. Also, in every topology, the pointx is not a limit point of the set {x}.

Example 3.10. Consider the set A = { 1n ∈ R | n ∈ Z+} as a subset of R with the standard topology. We

show that 0 is the only limit point of A.

Proof. If U is an open set containing 0, then there is an interval (a, b) such that 0 ∈ (a, b) ⊆ U . But(a, b) ∩ A = ∅ for every such open interval. So U ∩ A = ∅. Therefore every neighborhood of 0 intersects Aand since 0 is not in A, the intersection contains points other than 0. It follows that 0 is a limit point of A.

In fact, 0 is the only limit point of A. Given x ∈ R \ {0}, we can find an open interval containing x thatis either disjoint from A (if x /∈ A) or intersects A only in x (if x ∈ A). In either case, if x = 0, then x is nota limit point of A.

Example 3.11. Consider the rational numbers Q as a subset of R with the standard topology. Every x ∈ Ris a limit point of Q.

Proof. Given a real number x, an open set U containing x contains an open interval (a, b) that also containsx. But every open interval intersects Q in infinitely many points, and therefore (a, b) intersects Q in a pointother than x. Hence, x is a limit point of Q.

Limit points provide us with an easy means to find the closure of a set.

Theorem 3.12. Let A be a subset of a topological space X, and let A′ be the set of limit points of A. ThenCl(A) = A ∪A′.

Proof. First we show that A ∪ A′ ⊆ Cl(A) and certainly A ⊆ Cl(A). Suppose x ∈ A′. Then everyneighborhood of x intersects A. By Theorem 3.7, x ∈ Cl(A), and it follows that A′ ⊆ Cl(A).

Conversely, Suppose x ∈ Cl(A). Either x ∈ A or x ∈ Cl(A) \ A. In the former case, it follows thatx ∈ AUA′. Consider the latter case, x ∈ Cl(A) \ A. Since x ∈ Cl(A), Theorem 3.7 implies that every openset containing x intersects A and such an intersection must contain a point other than x because x /∈ A.Thus x is a limit point of A, and it follows that x ∈ A ∪A′. As a result, we have Cl(A) ⊆ A ∪A′.

Corollary 3.13. A subset A of a topological space is closed if and only if it contains all of its limit points.

Proof. By Theorem 3.2(vi), A is closed if and only if A = Cl(A). By Theorem 3.12, A = Cl(A) if and onlyif A = A∪A′ where A′ is the set of limit points of A. Finally, A = A∪A′ holds if and only if A′ ⊆ A. ThusA is closed if and only A′ ⊆ A.

A concept closely related to limit points is that of a convergent sequence.

Definition 3.14. In a topological space X, a sequence (x1, x2, . . . ) converges to x ∈ X if for everyneighborhood U of x, there is a positive integer N such that xn ∈ U for all n > N . We say that x is thelimit of the sequence (x1, x2, . . . ), and we write

limn→∞

xn = x.

The idea behind a sequence converging to a point x is that, given any neighborhood U of x, the sequenceeventually enters and stays in U .

In the standard topology on R, limit points are limits of sequences, as the following theorem indicates:

14

Theorem 3.15. Let A be a subset of R in the standard topology. If x is a limit point of A, then there is asequence of points in A that converges to x.

Proof. Let x ∈ A′ and let Bn = {(x − 1n , x + 1

n) | n ∈ N} be a collection of neighborhoods of x. It followsthat Bn ⊆ Bm for all n > m and we can choose a point xn ∈ Bn ∩ A for each n because x is a limit point.Let U be a neighborhood of x and thus U contain an open interval that contain x. Then there exists p ∈ Nsuch that Bp ⊆ U . As a result, xm ∈ U for all m ≥ p.

In a general topological space X, a limit point of a set B ⊆ X is not necessarily the limit of a sequencein B.

Our intuition might suggest that if a sequence converges to a point, then that point should be unique.This may not be true in other topologies but we can ensure the point of convergence is unique if the topologyis a Hausdorff space.

Theorem 3.16. If X is a Hausdorff space, then every convergent sequence of points in X converges to aunique point in X.

Proof. Let X be a Hausdorff space. Suppose that a sequence (x1, x2, . . . ) converges to two different points, xand y, in X. Since X is Hausdorff, x and y have disjoint neighborhoods U and V , respectively. The sequence(x1, x2, . . . ) converges to x; therefore there exists N ∈ Z+ such that xn ∈ U for all n > N . Similarly,there exists M ∈ Z+ such that xn ∈ V for all n > M . Thus, if m is greater than both N and M , thenxm ∈ U ∩V , contradicting the fact that U and V are disjoint. Therefore convergent sequences in a Hausdorffspace converge to a unique point.

3.3 Boundary of a Set

We have an intuitive idea of what we mean by the boundary of a set: the points that lie close to boththe inside and the outside of the set. But as we have seen, there are examples and topologies out there thatchallenge our intuition. For example, what is the boundary of Q as a subset of R in the standard topology?The definition of the boundary of a set has to be chosen carefully and understood clearly for its proper useas a central concept in topology.

Definition 3.17. Let A be a subset of a topological space X. The boundary of A, denoted ∂A, is the setA = Cl(A) \ Int(A).

The following examples show that, as with interior and closure, the boundary of a set A depends on thetopology on the set X containing A, not just on A itself.

Example 3.18. Let A = [−1, 1] on R with

(i) the standard topology. We have Cl(A) = [−1, 1] and Int(A) = (−1, 1), therefore ∂A = {−1, 1}.

(ii) the discrete topology. Here Int(A) = Cl(A) = [−1, 1] and so ∂A = ∅.

(iii) the lower limit topology. In this case, Cl(A) = [−1, 1] and Int(A) = [−1, 1), so ∂A = {−1}.

The next theorem gives conditions that help us to determine whether or not a point x lies in the boundaryof a set A.

Theorem 3.19. Let A be a subset of a topological space X and let x ∈ X. Then x ∈ ∂A if and only if everyneighborhood of x intersects both A and X \A.

Proof. Suppose that x ∈ ∂A. Then x ∈ Cl(A) and x /∈ Int(A). Since x ∈ Cl(A), it follows that everyneighborhood of x intersects A. Furthermore, since x /∈ Int(A), it follows that every neighborhood of x isnot a subset of A and therefore intersects X \A. Thus, every neighborhood of x intersects A and X \A. Nowsuppose that every neighborhood of x intersects A and X \A. It follows that x ∈ Cl(A) and x ∈ Cl(X \A).By Theorem 3.8(ii) implies that Cl(X \ A) = X − Int(A), and therefore x /∈ Int(A). Thus, x ∈ Cl(A) andx /∈ Int(A); that is, x ∈ Cl(A) \ Int(A) = ∂A.

15

Here are some quick facts about the boundary of a set A. They all follow readily from the definition.

Theorem 3.20. Let A be a subset of a topological space X. Then the following statements about the boundaryof A hold:

(i) ∂A is closed.

(ii) ∂A = Cl(A) ∩ Cl(X \A).

(iii) ∂A ∩ Int(A) = ∅.

(iv) ∂A ∪ Int(A) = Cl(A).

(v) ∂A ⊆ A if and only if A is closed.

(vi) ∂A ∩A = ∅ if and only if A is open.

(vii) ∂A = ∅ if and only if A is both open and closed.

Example 3.21. Consider Q in the standard topology on R. Since Cl(Q) = R, and Int(Q) = ∅, it followsthat ∂Q = R. Therefore the whole real line R is the boundary of the rational numbers.

That makes sense since every real number is arbitrarily close to the set of rational numbers and to itscomplement, the set of irrational numbers.

16

Chapter 4

New Spaces from Old Ones

4.1 The Subspace Topology

Given a subset Y of a topological space X, there is a natural way to define a topology on Y , based onthe topology on X.

Definition 4.1. Let X be a topological space and let Y be a subset of X. Define TY = {U ∩ Y | U is openin X}. This is called the subspace topology on Y and, with this topology, Y is called a subspace of X.We say that V ⊆ Y is open in Y if V is an open set in the subspace topology on Y .

Lemma 4.2. The subspace topology TY defined in Definition 4.1 is a topology.

Proof. Consider the following three types of sets.

(i) The sets ∅ and Y are open in Y since ∅ = Y ∩∅ and Y = Y ∩X.

(ii) Suppose that {Vα} is a collection of sets that are open in Y . Then for each α there exists an open setUα in X such that Vα = Uα ∩ Y . Therefore,∪

Vα =∪(

Y ∩ Uα

)= Y ∩

(∪α

Uα

).

Now,∪

Uα is open in X, implying that∪

Vα is an open set in X intersected with Y . Thus,∪

Vα isopen in Y .

(iii) Suppose that V1, . . . , Vn are open in Y . Then for each i there exists Ui, open in X, such that Vi = Ui∩Y .Now,

V1 ∩ · · · ∩ Vn =n∩

j=1

(Y ∩ Uj) = Y ∩( n∩j=1

Uj

).

Since U1 ∩ · · · ∩ Un is open in X, it follows that V1 ∩ · · · ∩ Vn is an open set in X intersected with Y ,and therefore is open in Y .

Exercise 4.3. (i) If (X, T ) is a topological space and Y ⊆ X is open, show that the subspace topology TYon Y is the collection of sets U ∈ T with U ⊆ Y .

(ii) Consider R with the standard topology T . If Y = [0, 1], show that [0, 1/2) ∈ TY but [0, 1/2) /∈ T .

Exercise 4.4. Let (X, d) be a metric space, Y a subset of X and dY the metric d restricted to Y (formally,dY : Y 2 → R is given by dY (x, y) = d(x, y) for x, y ∈ Y ). Then if we give X the topology induced by d, thesubspace topology on Y is identical with the topology induced by dY .

Definition 4.5. Let X be a topological space, and let Y ⊆ X have the subspace topology. We say that aset C ⊆ Y is closed in Y if Y \ C is open in the subspace topology on Y .

17

Theorem 4.6. Let X be a topological space, and let Y ⊆ X have the subspace topology. Then C ⊆ Y isclosed in Y if and only if C = D ∩ Y for some closed set D in X.

Proof. Suppose C is a closed subset of Y . Then there is an open subset U of X such that Y \ C = Y ∩ U .Therefore,

C = Y \ (Y \ C) = Y \ (Y ∩ U) = Y ∩ (Y \ U) = Y ∩ (X \ U).

Since X \ U is closed in X and we may call it D.Conversely, suppose C ⊆ Y and C = D ∩ Y for some closed set D in X. Then X \D is open in X and

Y \ C = Y \ (Y \D) = Y ∩ (Y \D) = Y ∩ (X \D).

Therefore Y \ C is open in Y and hence C is closed in Y .

We will find it useful to be able to produce a basis for the subspace topology on a subset of a topologicalspace, using a basis for the original space.

Theorem 4.7. Let X be a topological space and B be a basis for the topology on X. If Y ⊆ X, then thecollection

BY = {B ∩ Y | B ∈ B}

is a basis for the subspace topology on Y .

Proof. First, note that BY is a collection of open sets in the subspace topology on Y . We use Theorem 2.18to prove that BY is a basis for the subspace topology. Suppose W is an open set in the subspace topologyon Y . Let y ∈ W be arbitrary. Then W = U ∩ Y , where U is open in X. There exists a basis element B inB such that y ∈ B ⊆ U . Thus y ∈ B ∩ Y ⊆ U ∩ Y = W . Since B ∩ Y ∈ BY , it now follows by Theorem 2.18that BY is a basis for the subspace topology on Y .

Exercise 4.8. If (X, T ) is a Hausdorff topological space and Y ⊆ X, then Y with the subspace topology isalso Hausdorff.

4.2 The Product Topology

Given two topological spaces X and Y , we would like to generate a natural topology on the product,X × Y . Our first inclination might be to take as the topology on X × Y the collection C of sets of the formU ×V where U is open in X and V is open in Y . But C is not a topology since the union of two sets U1×V1

and U2 × V2 need not be in the form for U × V some U ⊆ X and V ⊆ Y . However, if we use C as a basis,rather than as the whole topology, we can proceed.

Definition 4.9. Let X and Y be topological spaces and X × Y be their product. The product topology onX × Y is the topology generated by the basis B = {U × V | U is open in X and V is open in Y }.

Of course, we must verify that B is a basis for a topology on the product X × Y .

Theorem 4.10. The collection B is a basis for a topology on X × Y .

Proof. Every point (x, y) is in X × Y , and X × Y ∈ B. Therefore the first condition for a basis is satisfied.Next assume that (x, y) is in the intersection of two elements of B. That is, (x, y) ∈ (U1×V1)∩ (U2×V2)

where U1 and U2 are open sets in X, and V1 and V2 are open sets in Y . Let U3 = U1 ∩U2 and V3 = V1 ∩ V2.Then U3 is open in X, and V3 is open in Y , and therefore U3 × V3 ∈ B. Also,

U3 × V3 = (U1 ∩ U2)× (V1 ∩ V2) = (U1 × V1) ∩ (U2 × V2),

and thus (x, y) ∈ U3×V3 ⊆ (U1×V1)∩ (U2×V2). It follows that the second condition for a basis is satisfied.Therefore B is a basis for a topology on X × Y .

18

In Definition 4.9, the basis B that we use to define the product topology is relatively large since weobtain it by pairing up every open set U in X with every open set V in Y . Fortunately, as the next theoremindicates, we can find a smaller basis for the product topology by using bases for the topologies on X andY .

Theorem 4.11. If C is a basis for X and D is a basis for Y , then

E = {C ×D | C ∈ C and D ∈ D}

is a basis that generates the product topology on X × Y .

Proof. Each set C ×D ∈ E is an open set in the product topology; therefore, by Theorem 2.18, it suffices toshow that for every open set W in X × Y and every point (x, y) ∈ W , there is a set C ×D in E such that(x, y) ∈ C×D ⊆ W . But since W is open in X×Y , we know that there are open sets U in X and V in Y suchthat (x, y) ∈ U × V ⊆ W . So x ∈ U and y ∈ V . Since U is open in X, there is a basis element C ∈ C suchthat x ∈ C ⊆ U . Similarly, since V is open in Y , there is a basis element D ∈ D such that y ∈ D ⊆ V . Thus(x, y) ∈ C ×D ⊆ U × V ⊆ W . Hence, by Theorem 2.18, it follows that E = {C ×D | C ∈ C and D ∈ D}is a basis for the product topology on X × Y .

Exercise 4.12. Let (X1, d1) and (X2, d2) be metric spaces. Let T be the product topology on X1×X2 whereXj is given the topology induced by dj [j = 1, 2].

Define ρk : (X1 ×X2)2 → R by

ρ1((x, y), (u, v)) = d1(x, u),

ρ2((x, y), (u, v)) = d1(x, u) + d2(y, v),

ρ3((x, y), (u, v)) = max(d1(x, u), d2(y, v)),

ρ4((x, y), (u, v)) = (d1(x, u)2 + d2(y, v)

2)1/2.

Establish that ρ1 is not a metric and that ρ2, ρ3 and ρ4 are. Show that each of the ρj with 2 ≤ j ≤ 4induce the product topology T on X1 ×X2.

The results above can be extended to any finite (or even infinite) product of topological spaces.

Exercise 4.13. If (X, T ) and (Y, σ) are Hausdorff topological spaces, then X × Y with the product topologyis also Hausdorff.

4.3 The Quotient Topology

The concept of a quotient topology allows us to construct a variety of additional topological spaces fromthe ones that we have already introduced. Put simply, we create a topological model that mimics the processof gluing together or collapsing parts of one or more objects.

Definition 4.14. Let X be a topological space and A be a set (that is not necessarily a subset of X). Letp : X → A be a surjective map. Define a subset U of A to be open in A if and only if p−1(U) is open in X.The resultant collection of open sets in A is called the quotient topology induced by p, and the functionp is called a quotient map. The topological space A is called a quotient space.

Now, we verify that the quotient topology is a topology:

Theorem 4.15. Let p : X → A be a quotient map. The quotient topology on A induced by p is a topology.

Proof. We verify each of the three conditions for a topology.

(i) The set p−1(∅) = ∅, which is open in X. The set p−1(A) = X, which is open in X. So ∅ and A areopen in the quotient topology.

19

(ii) Suppose each of the sets in the collection {Uβ}β∈B is open in the quotient topology on A. Thenp−1

(∪(Uβ)

)=

∪p−1(Uβ), which is a union of open sets in X, and therefore is open in X. Thus,∪

(Uβ) is open in the quotient topology, implying that the arbitrary union of open sets in the quotienttopology is an open set in the quotient topology.

(iii) Suppose each of the sets Ui, i = 1, . . . , n, is open in the quotient topology on A. Then p−1(∩

(Ui))=∩

p−1(Ui), which is a finite intersection of open sets in X, and therefore is open in X. Hence,∩

Ui isopen in the quotient topology, and it follows that the finite intersection of open sets in the quotienttopology is an open set in the quotient topology.

Hence, the quotient topology is a topology on A.

20

Chapter 5

Continuity and Homeomorphisms

5.1 Continuity

Definition 5.1. Let (X, T ) and (Y, σ) be topological spaces. A function f : X → Y is said to be continuousif and only if f−1(U) is open in X whenever U is open in Y .

This topological definition of continuity is very simple to state and Theorem 1.18 tells us that if (X, d)and (Y, ρ) are metric spaces the notion of a continuous function f : X → Y is the same whether we considerthe metrics or the topologies derived from them.

By definition, a function f : X → Y is continuous if the preimage of every open set in Y is open in X.However, checking that every open set in Y has an open preimage in X is more than we really need to do.As the following theorem indicates, to prove that f is continuous, it suffices to consider only the sets in afixed basis for Y , showing that the preimage of each basis element is open in X.

Theorem 5.2. Let X and Y be topological spaces and B be a basis for the topology on Y . Then f : X → Yis continuous if and only if f−1(B) is open in X for every B ∈ B.

Proof. Suppose f : X → Y is continuous. Then f−1(V ) is open in X for every V open in Y . Since everybasis element B is open in Y , it follows that f−1(B) is open in X for all B ∈ B.

Now, suppose f−1(B) is open in X for every B ∈ B. We show that f is continuous. Let V be an openset in Y . Then V is a union of basis elements, say V =

∪Bα. Thus,

f−1(V ) = f−l(∪

Bα) =∪

f−1(Bα).

By assumption, each set f−1(Bα) is open in X; therefore so is their union. Thus, f−1(V ) is open in X, andit follows that the preimage of every open set in Y is open in X. Hence, f is continuous.

Example 5.3. Let T denote the standard topology on R and define f : (R, T ) → (R, T ) by f(x) = x3. Thecollection B of all open intervals is a basis for T . Thus, by Theorem 5.2, it is sufficient to show that theinverse image of each open interval is open. Let (a, b) be an open interval. Then f−1((a, b)) = ( 3

√a, 3

√b), so

f is continuous.

Theorem 5.4. Assume that f : X → Y is continuous. If a sequence (x1, x2, . . . ) in X converges to a pointx, then the sequence (f(x1)), f(x2), . . . ) in Y converges to f(x).

Proof. Let U be an arbitrary neighborhood of f(x) in Y . Since f is continuous, f−1(U) is open in X.Furthermore, f(x) ∈ U implies that x ∈ f−1(U). The sequence (x1, x2, . . . ) converges to x; thus, there existsN ∈ Z+ such that xn ∈ f−l(U) for all n > N . It follows that f(xn) ∈ U for all n > N , and therefore thesequence (f(x1)), f(x2), . . . ) converges to f(x).

Definition 5.5. Let (X, T ) and (Y, σ) be topological spaces. A function f : X → Y is said to be openprovided that if U ∈ T then f(U) ∈ σ.

Note that a continuous function need not to be open. For example, the function f : R → R given byf(x) = x2 is continuous, but the image of the open set (−1, 1) is [0, 1), which is not open.

21

Example 5.6. Let D denote the discrete topology on R and let T denote the standard topology on R. Thenf : (R,D) → (R, T ) defined by f(x) = x for each x ∈ R is continuous because if U ∈ T then f−1(U) ∈ D.However, f is not open because {1} ∈ D whereas {1} /∈ T .

The proof of Theorem 1.11 given on page ?? carries over unchanged to give the following generalization.

Theorem 5.7. If (X, T ) and (Y, σ) and (Z, µ) are topological spaces and f : X → Y , g : Y → Z arecontinuous, then so is the composition g ◦ f .

Proof. Suppose that f : X → Y and g : Y → Z are continuous, and let U be an open set in Z. Then(g ◦ f)−1(U) = f−1(g−1(U)). Since g is continuous, g−1(U) is open in Y , and since f is continuous,f−1(g−1(U)) is open in X. Thus, (g ◦ f)−1(U) is open in X for an arbitrary U open in Z, implying thatg ◦ f is continuous.

Theorem 5.8. Let (X, T ) and (Y, σ) be topological spaces. A function f : X → Y is continuous if and onlyif f−1(F ) is closed in X whenever F is closed in Y .

Finally, we introduce a very useful lemma called the Pasting Lemma.

Theorem 5.9. The Pasting Lemma Let X be a topological space and let A and B be closed subsets of Xsuch that A ∪ B = X. Assume that f : A → Y and g : B → Y are continuous and f(x) = g(x) for all x inA ∩B. Then h : X → Y , defined by

h(x) =

{f(x) if x ∈ Ag(x) if x ∈ B

is a continuous function.

Proof. By Theorem 5.8, it suffices to show that if C is closed in Y , then h−1(C) is closed in X. Thus supposethat C is closed in Y . Note that h−1(C) = f−1(C)∪g−1(C). Since f is continuous, it follows by Theorem 5.8that f−1(C) is closed in A. Theorem 4.6 then implies that f−1(C) = D ∩ A where D is closed in X. Now,D and A are both closed in X, and f−1(C) = D ∩ A; therefore, f−1(C) is closed in X. Similarly, g−1(C)is closed in X. Thus, h−1(C) is the union of two closed sets in X and therefore is closed in X as well. Itfollows that h is continuous.

5.2 Homeomorphism

Two groups are the same for the purposes of group theory if they are (group) isomorphic. Two vectorspaces are the same for the purposes of linear algebra if they are (vector space) isomorphic. When are twotopological spaces (X, T ) and (Y, σ) the same for the purposes of topology? In other words, when does thereexist a bijection between X and Y in which open sets correspond to open sets, and the grammar of topology(things like union and inclusion) is preserved? A little reflection shows that the next definition provides theanswer we want.

Definition 5.10. We say that two topological spaces (X, T ) and (Y, σ) are homeomorphic, denotedX ≡ Yif there exists a bijection θ : X → Y such that θ and θ−1 are continuous. We call θ a homeomorphism.

Exercise 5.11. Show that homeomorphism is an equivalence relation on topological spaces. That is, showthe following three statements.

(i) The function id : X → X, defined by id(x) = x, is a homeomorphism.

(ii) If f : X → Y is a homeomorphism, then so is f−1 : Y → X.

(iii) If f : X → Y and g : Y → Z are homeomorphisms, then so is g ◦ f : X → Z .

Certain properties defined for topological spaces are preserved by homeomorphisms. An example isprovided by the next theorem.

22

Theorem 5.12. If f : X → Y is a homeomorphism and X is Hausdorff, then Y is Hausdorff.

Proof. Suppose that X is Hausdorff and f : X → Y is a homeomorphism. Let x and y be distinct points inY . Then f−1(x) and f−1(y) are distinct points in X. Thus, there exist disjoint open sets U and V containingf−1(x) and f−1(y), respectively. It follows that f(U) and f(V ) are disjoint open sets containing x and y,respectively. Therefore Y is Hausdorff.

A property of topological spaces that is preserved by homeomorphism is said to be a topological prop-erty. Theorem 5.12 implies that being Hausdorff is a topological property. In general, properties that aredefined in terms of open sets, like Hausdorff, are topological properties.

As we have seen in a few examples in this section, we can prove that two topological spaces are home-omorphic by defining a homeomorphism between them. Conversely, in order to prove that two topologicalspaces are not homeomorphic, it is necessary to show that none of the functions defined between them is ahomeomorphism. However, it is often too difficult to consider every function between the spaces. Instead,we find a topological property held by one space but not the other. Then we know that the two spaces arenot homeomorphic.

23

Chapter 6

Connectedness

There are several natural approaches that we could take to rigorously capture the concept of connectednessfor a topological space. One approach might be to say that a topological space is connected if it cannot bebroken down into two distinct pieces that are separated from each other. Another approach might be tosay that a topological space is connected if we can take a continuous walk in the space from any point toany other point. In this chapter we define both of these types of connectedness. Spaces of the first type arecalled connected, while spaces of the second type are called path connected. We prove that these two typesof connectedness are not the same, but that path connectedness does imply connectedness.

6.1 Connected Spaces

Definition 6.1. Let X be a topological space.

(i) We call X connected if there does not exist a pair of disjoint nonempty open sets whose union is X.

(ii) We call X disconnected if X is not connected.

(iii) If X is disconnected, then a pair of disjoint nonempty open sets whose union is X is called a separationof x.

Example 6.2. The following are some disconnected topological space

(i) The subspace X = (−1, 0) ∪ (0, 1) of R is disconnected. The pair of sets, (−1, 0) and (0, 1), is aseparation of X.

(ii) If p ∈ R, then R \ {p} is a disconnected topological space since the pair of sets, U = (−∞, p) andV = (p,∞), is a separation of R \ {p}.

(iii) If a set X consists of more than one point and it has the discrete topology, then it is disconnected. IfA is any nonempty proper subset of X, then the pair of sets, A and X −A, is a separation of X.

The following result provides an alternative formulation of connectedness:

Theorem 6.3. A topological space X is connected if and only if there are no nonempty proper subsets of Xthat are both open and closed in X.

Proof. Suppose X is not connected. Then there are nonempty, disjoint open sets U and V such thatX = U ∪ V . Thus U is a nonempty proper subset of X that is open and closed since U = X \ V .

Conversely, suppose X has a nonempty proper subset U that is both open and closed. Then U and X \Vare nonempty disjoint open sets whose union is X. Therefore X is not connected.

Example 6.4. In R with the lower limit topology, intervals [a, b) are both open and closed. ThereforeTheorem 6.3 implies that R is disconnected in this topology.

Definition 6.5. A set A contained in a topological space X is said to be connected in X if A is connectedin the subspace topology. If A is not connected in X, we say it is disconnected in X.

24

The next theorem yields an alternate characterization of disconnected sets in a topological space X.

Theorem 6.6. A set A is disconnected in X if and only if there exist open sets U and V in X such thatA ⊆ U ∪ V , U ∩A = ∅, V ∩A = ∅, and U ∩ V ∩A = ∅.

Proof. Suppose that A is disconnected in X. Then there exist nonempty sets P and Q that are open inA, disjoint, and such that P ∪ Q = A. Since P and Q are open in A there exist sets U and V that areopen in X and such that U ∩ A = P and V ∩ A = Q. Clearly, A ⊆ U ∪ V , U ∩ A = ∅, V ∩ A = ∅, andU ∩V ∩A = ∅. Now suppose that U and V are open sets in X such that A ⊆ U ∪V , U ∩A = ∅, V ∩A = ∅,and U ∩ V ∩A = ∅. If we let P = U ∩A and Q = V ∩A, then it follows that the pair of sets, P and Q, is aseparation of A in the subspace topology, and therefore A is disconnected in X.

Definition 6.7. Let A be a subspace of a topological space X. If U and V are open sets in X such thatA ⊆ U ∪ V , U ∩ A = ∅, V ∩ A = ∅, and U ∩ V ∩ A = ∅, then we say that the pair of sets, U and V , is aseparation of A in X.

Note that U and V need not be disjoint but U ∩ V should disjoint from A.It is straightforward to show that connectedness is a topological properties. However, as the following

theorem indicates, we do not need the full strength of a homeomorphism to preserve connectedness:

Theorem 6.8. If X is connected and f : X → Y is continuous, then f(X) is connected in Y .

Proof. Suppose that f(X) is not connected in Y . Then there exist open sets U and V that form a separationof f(X) in Y . The function f is continuous, and therefore f−1(U) and f−1(V ) are open in X. Both U and Vhave nonempty intersection with f(X); thus f−1(U) and f−1(V ) are nonempty. Furthermore f(X) ⊆ U ∪V ,implying that X ⊆ f−1(U)∪f−1(V ). Finally, since U ∩V ∩f(X) = ∅, it follows that f−1(U) and f−1(V ) aredisjoint. Therefore the pair of sets, f−1(U) and f−1(V ), is a separation of X, contradicting the assumptionthat X is connected. Hence, f(X) is connected in Y .

Lemma 6.9. Let C and D be subsets of a topological space X. Assume that C is connected and C ⊆ D.Further assume that U and V form a separation of D in X. Then either C ⊆ U or C ⊆ V .

Proof. Suppose that neither C ⊆ U nor C ⊆ V . Then U ∩ C = ∅ and V ∩ C = ∅. It follows that U and Vform a separation of C in X, contradicting the assumption that C is connected.

Our next theorem indicates that if C is connected in X, and we add limit points to C, then the resultingset is also connected in X.

Theorem 6.10. Let C be connected in X, and assume that C ⊆ A ⊆ Cl(C). Then A is connected in X.

Proof. Suppose that A is not connected in X, and let U and V form a separation of A in X. Then byLemma 6.9, either C ⊆ U or C ⊆ V . We may assume, without loss of generality, that C ⊆ U . HenceC ∩ V = ∅. But, since U and V form a separation of A in X, it follows that A ∩ V = ∅. Pick x ∈ A ∩ V .Now, x ∈ A and A ⊆ Cl(C) imply x ∈ Cl(C). But x ∈ V , an open set in X which is disjoint from C. So xcannot be in the closure of C, yielding a contradiction. Thus, it follows that A is connected in X.

The following theorem shows that union of connected subsets of a topological space have at least onepoint in common is connected.

Theorem 6.11. Let X be a topological space, and let {Cα}α∈A be a collection of connected subsets of X

such that∩α∈A

Cα = ∅. Then∪α∈A

Cα is connected in X.

Proof. Suppose that∪α∈A

Cα a is not connected in X. Thus there exist sets U and V that form a separation

of∪α∈A

Cα in X. Let x be in∩α∈A

Cα. Then either x ∈ U or x ∈ V , but both cannot hold. We may assume,

without loss of generality, that x lies in U and does not lie in V . Lemma 6.9 implies that for all α ∈ A,

25

either Cα ⊆ U or Cα ⊆ V . Since x ∈ U and x /∈ V , it follows that Cα ⊆ U for all α ∈ A. Thus∪α∈A

Cα ⊆ U ,

contradicting the assumption that U and V form a separation of∪α∈A

Cα in X. Therefore∪α∈A

Cα is connected

in X.

We can use the previous theorem to prove that a product of connected spaces is connected.

Theorem 6.12. Let X1, . . . , Xn be connected spaces. Then the product space X1 × · · · ×Xn is connected.

Proof. We prove the result for a product of two spaces. The general result can then be shown by induction.Assume that X and Y are connected topological spaces. We prove that X × Y is connected. First, notethat for every x ∈ X, the subspace {x} × Y of X × Y is homeomorphic to Y and is therefore connected.Similarly, for every y ∈ Y , the subspace X × {y} of X × Y is connected. Thus, by Theorem 6.11, for everyx ∈ X and y ∈ Y the set ({x} × Y ) ∪ (X × {y}) is connected in X × Y .

Now fix x0 ∈ X and let y vary. Each set ({x0} × Y ) ∪ (X × {y}) contains the set {x0} × Y . It thenfollows by Theorem 6.11 that ∪

y∈Y(({x0} × Y ) ∪ (X × {y}))

is connected in X × Y . Furthermore, ∪y∈Y

(({x0} × Y ) ∪ (X × {y}))

implying that X × Y is connected.

Definition 6.13. Let X be a topological space and let ∼C be the equivalence relation on X defined byx ∼C y if x and y lie in a connected subset of X. The components of X are the equivalence classes of theequivalence relation ∼C .

Theorem 6.14. If X be a topological space, then

(i) each component of X is connected in X.

(ii) If A is connected in X, then A is a subset of a component of X.

Proof of (i). Let X be a topological space and C be a component of X. Pick a point p ∈ C. For everyx ∈ C, x ∼C p since C is an equivalence class under the equivalence relation ∼C that determines thecomponents. Therefore, by the definition of ∼C , there exists a connected set Cx containing x and p.

We claim that Cx ⊆ C. To prove the claim, suppose y ∈ Cx. Then y and p both lie in the connected setCx, implying that y ∼C p. Therefore y is in the same equivalence class as p, namely C, and it follows that

Cx ⊆ C. Now, by the Union Lemma we have∪x∈C

Cx = C. Therefore C is the union of the connected sets Cx,

and each of these connected sets contains the point p. By Theorem 6.11, it follows that C is connected.

Theorem 6.15. Let f : X → Y be a homeomorphism. If C is a component of X, then f(C) is a componentof Y .

Proof. Suppose that f : X → Y is a homeomorphism and C is a component of X. Then C is connected inX by Theorem 6.14. Thus Theorem 6.8 implies that f(C) is connected in Y . Again by Theorem 6.14, itfollows that f(C) is a subset of a component D of Y . We claim that f(C) = D to complete the proof of thetheorem.

The component D is connected in Y , and f−1 is a continuous function; therefore f−1(D) is connected inX. Furthermore, f(C) ⊆ D implies that C ⊆ f−1(D). But C is a component of X, and C is a subset of theconnected set f−1(D). Therefore C = f−1(D). Hence, f(C) = D, as we wished to show.

Theorem 6.16. The real line R in the standard topology is a connected topological subspace.

26

Proof. Suppose it is not, and let U and V form a separation of R. Pick u ∈ U and v ∈ V . We may assumeu < v without loss of generality. Let U ′ = U ∩ [u, v] and V ′ = V ∩ [u, v]. It follows that U ′ ∪ V ′ = [u, v].Since U ′ is bounded from above (by v), U ′ has a least upper bound; call it c. We have u ≤ c ≤ v. We derivea contradiction by showing that c /∈ U ′ and c /∈ V ′.

To show that c /∈ V ′, we assume that c ∈ V ′ and derive a contradiction. Since u /∈ V ′ and V ′ is openin [u, v], it follows that there exists d such that u < d < c and (d, c] ⊆ V ′. This implies that d is an upperbound of U ′ and that d is less than the least upper bound c. This is a contradiction and thus c /∈ V ′.

Next we show that c /∈ U ′. We do this by contradiction as well. Therefore, assume that c ∈ U ′ . SinceU ′ is open in [u, v] and v /∈ U ′, there exists d such that [c, d) ⊆ U ′. For any e ∈ (c, d) it follows that e ∈ U ′

and e > c, contradicting the fact that c is an upper bound of U ′. Thus c /∈ U ′.Therefore c /∈ U ′ and c /∈ V ′. But c ∈ [u, v] and U ′ ∪ V ′ = [u, v]. With this final contradiction it now

follows that R in the standard topology is a connected topological space.

Theorems 6.12 and 6.16 imply the following corollary:

Corollary 6.17. Euclidean n-space Rn is a connected topological space.

6.2 Intermediate Value Theorem

Theorem 6.18. The Intermediate Value Theorem on [a, b]. Let f : [a, b] → R be continuous, andassume that s lies between f(a) and f(b). Then there exists at least one c ∈ [a, b] such that f(c) = s.

Theorem 6.19. The Intermediate Value Theorem (General Version). Let X be a connected topo-logical space and f : X → R be continuous. If p, q ∈ f(X) and p ≤ r ≤ q, then r ∈ f(X).

Proof. Suppose f : X → R is continuous, p, q ∈ f(X), and p ≤ r ≤ q. If r = p or r = q, then we immediatelyhave that r ∈ f(X). Therefore, we only need to consider the case where p < r < q. Note that f(X) isconnected in R since X is connected and f is continuous. We prove by contradiction that r ∈ f(X). Thus,suppose that r /∈ f(X). Then U = (−∞, r) and V = (r,∞) are disjoint open subsets of R. whose unioncontains f(X). Since p ∈ U and q ∈ V , it follows that f(X) intersects both U and V . Hence, U and V form aseparation of f(X) in R. But this contradicts the fact that f(X) is connected in R. Therefore r ∈ f(X).

Theorem 6.20. The One-Dimensional Brouwer Fixed Point Theorem. Let f : [−1, 1] → [−1, 1] becontinuous. There exists at least one c ∈ [−1, 1] such that f(c) = c.

Proof. Let f : [−1, 1] → [−1, 1] be continuous. Define a function g : [−1, 1] → R by g(x) = f(x) − x.The function g is continuous. Note that f(−1) ≥ −1, and therefore g(−1) ≥ 0. Similarly g(1) ≤ 0. TheIntermediate Value Theorem implies that there exists a value c ∈ [−1, 1] such that g(c) = O. For such c itfollows that f(c) = c. Therefore there exists at least one c in [−1, 1] such that f(c) = c.

6.3 Path Connectedness

Definition 6.21. A topological space X is path connected if for every x, y ∈ X, there exists a continuousfunction γ : [0, 1] → X with γ(0) = x and γ(1) = y. The continuous function γ is called a path from x to yin X. A subset A of a topological space X is path connected in X if A is path connected in the subspacetopology that A inherits from X.

Theorem 6.22. If X is a path connected space, then it is connected.

Proof. Proof. Let X be a path connected space. We prove that X is connected by showing that it has onlyone component, or equivalently that every pair of points x, y ∈ X is contained in some connected subset ofX. Thus, let x and y be arbitrary points in X. Since X is path connected, there is a path in X from x toy. The image of such a path is a connected subset of X containing both x and y. Therefore every pair ofpoints in X is contained in a connected subset of X, and it follows that X is connected.

The next example displays a topological space that is connected but not path connected.

27

Example 6.23. We work in R2 with the usual topology. Let

E1 = {(0, y) : |y| ≤ 1} and E2 = {(x, sin 1

x) : 0 < x ≤ 1}

and set E = E1 ∪ E2.

(i) Explain why E1 and E2 are path-connected and show that E is closed and connected.

(ii) Suppose, if possible, that x : [0, 1] → E is continuous and x(0) = (1, 0), x(1) = (0, 0). Explain why wecan find 0 < t1 < t2 < t3 < . . . such that x(tj) =

((j + 1

2)π)−1. By considering the behavior of tj and

y(tj), obtain a contradiction.

(iii) Deduce that E is not path-connected.

Proof. (i) If y1, y2 ∈ [−1, 1], the function f : [0, 1] → E1 given by

f(t) =(0, (1− t)y1 + ty2

)is continuous and f(0) = (0, y1) and f(1) = (0, y2), so E1 is path-connected.

If (x1, y1), (x2, y2) ∈ E2, then yj = sin 1xj

and setting

g(t) =

((1− t)x1 + tx2, sin

1

(1− t)x1 + tx2

)we see that g is continuous and g(0) = (x1, y1) and g(1) = (x2, y2), so E2 is path-connected.

We next show that E is closed. Suppose that (xr, yr) ∈ E and (xr, yr) → (x, y). If x = 0, then we notethat, since |yr| ≤ 1 for all r and yr → y, we have |y| ≤ 1 and (x, y) ∈ E1 ⊆ E. If x = 0, then 1 ≥ x > 0(since xr ≥ 0 for all r). We can find an N such that |x − xr| < x/2 and so xr > x/2 for all r ≥ N .Thus, by continuity,

(xr, yr) = (xr, sin 1/xr) → (x, sin 1/x) ∈ E2 ⊆ E.

Thus E is closed.

Now suppose, if possible, that E is disconnected. Then we can find U and V open such that

U ∩ E = ∅, V ∩ E = ∅, U ∪ V ⊇ E and U ∩ V ∩ E = ∅.

ThenU ∪ V ⊇ Ej and U ∩ V ∩ Ej = ∅.

and so, since Ej is path-connected, so connected, we have U∩Ej = ∅ or V ∩Ej = ∅ [j = 1, 2]. Withoutloss of generality, assume V ∩ E1 = ∅ so U ⊇ E1. Since (0, 0) ∈ E1, we have (0, 0) ∈ U . Since U isopen, we can find a δ > 0 such that (x, y) ∈ U whenever ∥(x, y)∥2 < δ. If n is large,

((nπ)−1, 0) ∈ U ∩ E2 = U ∩ V ∩ E,

contradicting our initial assumptions. By reductio ad absurdum, E is connected.

(ii) Write x(t) = (x(t), y(t)). Since x is continuous so is x. Since x(0) = 1 and x(1) = 0, the intermediatevalue theorem tells us that we can find t1 with 0 < t1 < 1 and x(t1) = (32π)

−1. Applying theintermediate value theorem again, we can find t2 with 0 < t2 < t1 and x(t2) = (52π)

−1. We continueinductively.

Since the tj form a decreasing sequence bounded below by 0, we have tj → T for some T ∈ [0, 1]. Sincey is continuous

(−1)j = sin(1/x(tj)

)= y(tj) → y(T )

which is absurd.

(iii) Part (ii) tells us that there is no path joining (0, 0) and (1, 0) in E, so E is not path-connected.

28

Chapter 7

Compactness

Several important theorems in analysis hold for closed bounded intervals. Heine used a particular idea toprove one of these. Borel isolated the idea as a theorem (the Heine-Borel theorem), essentially Theorem ??below. Many treatments of analysis use the Heine-Borel theorem as a basic tool.

Definition 7.1. Let A be a subset of a topological space X, and let O be a collection of subsets of X.

(i) The collection O is said to cover A or to be a cover of A if A is contained in the union of the sets inO.

(ii) If O covers A and each set in O is open, then we call O an open cover of A.

(iii) If O covers A, and O′ is a sub-collection of O that also covers A, then O′ is called a subcover of O.

Definition 7.2. A topological space X is compact if every open cover of X has a finite subcover.

Example 7.3. The real line R in the standard topology is not compact since

O = {. . . , (−1, 1), (0, 2), (1, 3), . . . }

is an open cover, but no finite sub-collection of O covers R.

The reader is warned that compactness is a subtle property which requires time and energy to master.Up to this point most of the proofs in this course have been simple deductions from definitions. Several ofour theorems on compactness go much deeper and have quite intricate proofs.

Exercise 7.4. Here are some simple examples of compactness and non-compactness.

(i) Show that, if X is finite, every topology on X is compact.

(ii) Show that the discrete topology on a set X is compact if and only if X is finite.

(iii) Show that the indiscrete topology is always compact.

(iv) Show that the topology described in Exercise 2.26 is compact.

(v) Let X be uncountable (we could take X = R). We say that a subset A of X lies in T if either A = ∅or X \A is countable. Show that T is a topology but that (X, T ) is not compact.

Definition 7.5. Let X be a topological space, and assume A ⊆ X. Then A is said to be compact in X if Ais compact in the subspace topology inherited from X.

The following lemma allows us to check whether or not a subspace A of a topological space X is compactby considering covers of A that are made up of open sets in X rather than covers of A that are made up ofopen sets in the subspace topology on A:

Lemma 7.6. Let X be a topological space, and assume A ⊆ X. Then A is compact in X if and only if everycover of A by sets that are open in X has a finite subcover.

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Proof. Proof. Let A be compact in X, and suppose that O is a cover of A by open sets in X. Then O′ = {U∩A | U ∈ O} is a cover of A by open sets in A. Hence, there exists a finite subcover {U1∩A,U2∩A, . . . , Un∩A}of O′. But then {U1, U2, . . . , Un} is a finite subcover of O. Therefore every cover of A by open sets in X hasa finite subcover.

Conversely, suppose every cover of A by sets that are open in X has a finite subcover. Let O = {Vβ}β∈Bbe a cover of A by open sets in A. Then, by definition of the subspace topology, for each Vβ there is an openset Uβ in X such that Vβ = Uβ ∩A. It follows that the collection O′ = {Uβ}β∈B is a cover of A by open setsin X. Since O′ has a finite subcover {Uβ1 , . . . , Uβn} it follows that {Vβ1 , . . . , Vβn} is a finite subcover of O.Thus every cover of A by open sets in A has a finite subcover, and therefore A is compact.

Example 7.7. Consider (0, 1] as a subspace of R. This space is not compact. The collection O = {( 1n , 2) |n ∈ Z+} is a cover of (0, 1] by sets that are open in R. There is no finite sub-collection of O that covers(0, 1], and therefore (0, 1] is not compact as a subspace of R.

We actually do not need the full strength of a homeomorphism to ensure that a compact space maps to acompact space. The next theorem asserts that, as with connectedness and path connectedness, compactnessis preserved by continuous functions.

Theorem 7.8. Let f : X → Y be continuous, and let A be compact in X. Then f(A) is compact in Y .

Proof. Let f : X → Y be continuous, and assume that A is compact in X. To show that f(A) is compact inY , let O be a cover of f(A) by open sets in Y . Then f−1(U) is open in X for every open set U in O. HenceO′ = {f−1(U) | U ∈ O} is a cover of A by open sets in X. Since A is compact, Lemma 7.6 implies that thereis a finite sub-collection of O′, say {f−1(U1), . . . , f

−1(Un)}, that covers A. Then the collection of open sets{U1, . . . , Un} in O covers f(A). Thus, O has a finite subcover, implying that f(A) is compact in Y .

One immediate consequence of Theorem 7.8 is that a quotient space of a compact space X is compactsince it is the image of X under a quotient map, and a quotient map is a continuous function.

The next two theorems show that being closed and being compact are closely related properties.

Theorem 7.9. Let (X, T ) be a topological space and let E be compact in X. If F is closed in X, and F ⊆ E,then F is compact in X.

Proof. Since F is closed and we have X \ F ∈ T . If Uα ∈ T and F ⊆∪α∈A

Uα, then

(X \ F ) ∪∪α∈A

Uα = X ⊇ E.

By compactness, we can find α(j) ∈ A [1 ≤ j ≤ n] such that

E ⊆ (X \ F ) ∪n∪

j=1

Uα(j).

Since (X \ F ) ∩ F = ∅ and E ⊇ F , it follows that

F ⊆n∪

j=1

Uα(j)

and we are done.

The converse of the above theorem does not generally hold. A compact set in a topological space is notnecessarily a closed set, as demonstrated in the next example.

Example 7.10. Give an example of a topological space (X, T ) and a compact set in X which is not closed.

Proof. If (X, T ) has the indiscrete topology, then, if Y ⊆ X, Y = X, ∅, we have Y compact but not closed.We can take X = {a, b} with a = b and Y = {a}.

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Theorem 7.11. If (X, T ) is Hausdorff, then every compact set is closed.

Proof. Let K be a compact set. Arbitrary choosing x ∈ X \ K and k ∈ K (we have k = x). As X isHausdorff, we can find open sets Uk and Vk such that

x ∈ Vk, k ∈ Uk and Vk ∩ Uk = ∅.

Since∪k∈K

Uk ⊇∪k∈K

{k} = K, we have an open cover ofK. By compactness, we can find k(1), k(2), . . . , k(n) ∈

K such thatn∪

j=1

Uk(j) ⊇ K.

We observe that the finite intersection V =

n∩j=1

Vk(j) is an open neighborhood of x and that

V ∩K ⊆ V ∩n∪

j=1

Uk(j) = ∅,

so V ∩ K = ∅ and we have shown that every x ∈ X \ K has an open neighborhood lying entirely withinX \K. Thus X \K is open and K is closed.

Lemma 7.12. [The Tube Lemma.] Let X and Y be topological spaces, and assume that Y is compact. Ifx ∈ X, and U is an open set in X × Y containing {x} × Y , then there exists a neighborhood W of x in Xsuch that W × Y ⊆ U .

Proof. For each y ∈ Y pick open sets Wy in X and Vy in Y such that (x, y) ∈ Wy×Vy ⊆ U . The collection ofsets {Vy}y∈Y is an open cover of Y . Since Y is compact, finitely many of these sets cover Y , say Vy1 , . . . , Vyn .

Let W =

n∩i=1

Wyi . Then W is open in X, and W contains x because each Wy contains x. Note that

W × Y ⊆n∪

i=1

(Wyi × Vyi) ⊆ U.

Therefore W × Y contains {x} × Y and is contained in U .

Theorem 7.13. If X and Y are compact topological spaces, then the product X × Y is compact.

Proof. Let O be an open cover of X × Y . For each x ∈ X, the set {x} × Y is compact in X × Y . Thereforea finite sub-collection Ox of O covers {x} × Y . Let Ux be the union of the sets in Ox. The set Ux is open inX × Y and contains {x} × Y . By the Tube Lemma, for each x ∈ X there exists an open set Wx ⊆ X suchthat x ∈ Wx and Wx × Y ⊆ Ux . Note that Ox covers Wx × Y . The collection W = {Wx | x ∈ X} is anopen cover of X. Since X is compact, it is covered by finitely many of the sets in W, say Wx1 , . . . ,Wxm . Itfollows that C = Ox1 ∪ · · · ∪ Oxm covers X × Y . The collection C is a sub-collection of O and is finite, beinga finite union of finite sets. Therefore O has a finite sub-collection that covers X × Y , implying that X × Yis compact.

Using this result and induction, we then obtain the following corollary:

Corollary 7.14. Let X1, . . . , Xn be topological spaces, and let Ai be a compact subset of Xi for each i =1, . . . , n. Then A1 × · · · ×An is a compact subset of the product space X1 × · · · ×Xn.

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7.1 Compactness in metric spaces

In real analysis, where the focus is on Rn with the standard metric and topology, sometimes a set isdefined to be compact if it is closed and bounded. In this section, we show that in Rn such a definitionis consistent with the topological definition already presented. We then address the extent to which thisequivalence carries over to general metric spaces. Following that, we present some important convergenceproperties of compact sets in a metric space.

First, we define rigorously the meaning of bounded subset in metric spaces.

Definition 7.15. Let (X, d) be a metric space. A subset A of X is said to be bounded under d if thereexists a µ > 0 such that d(x, y) ≤ µ for all x, y ∈ A.

We begin with the following lemma, which we subsequently use to show that closed and bounded intervalsin R are compact:

Lemma 7.16. [The Nested Intervals Lemma.] Let {[an, bn]}n∈Z+ be a collection of nonempty closed

bounded intervals in R such that [an+1, bn+1] ⊆ [an, bn] for each n ∈ Z+. Then

∞∩n=1

[an, bn] is nonempty.

Proof. Assume that [an+1, bn+1] ⊆ [an, bn]for each n ∈ Z+. It follows that the interval endpoints satisfy theinequalities,

a1 ≤ a2 ≤ · · · ≤ an ≤ · · · ≤ bn ≤ · · · ≤ b2 ≤ b1.

The set {an}n∈Z+ is bounded from above (for example, by each bn) and therefore has a least upper boundA. Similarly, the set {bn}n∈Z+ has a greatest lower bound B. Note that A ≤ B and therefore the interval[A,B] is nonempty.

We claim that∞∩n=1

[an, bn] = [A,B], to complete the proof of the lemma. First, we show that∞∩n=1

[an, bn] ⊆

[A,B]. Thus, let x ∈∞∩n=1

[an, bn] be arbitrary. Then x ∈ [an, bn] for all n, implying that x ≥ an and x ≤ bn

for all n. Therefore x ≥ A and x ≤ B; that is, x ∈ [A,B]. Hence

∞∩n=1

[an, bn] ⊆ [A,B].

To prove that [A,B] ⊆∞∩n=1

[an, bn], let x ∈ [A,B] be arbitrary. Then x ≥ an and x ≤ bn for all n.

Therefore x ∈∞∩n=1

[an, bn]. Thus [A,B] ⊆∞∩n=1

[an, bn], and it follows that

∞∩n=1

[an, bn] = [A,B].

Lemma 7.16 does not hold if we replace the closed intervals with open intervals. For instance, thecollection of nonempty bounded open intervals {(0, 1

n) | n ∈ Z+} satisfies the condition that (0, 1n+1) ⊆ (0, 1

n)

for each n ∈ Z+, but∩∞

n=1(0,1n) = ∅.

Theorem 7.17. Every closed and bounded interval [a, b] is a compact subset of R with the standard topology.

Proof. Let O be a cover of [a, b] by open sets in R. To derive a contradiction, assume that there is no finitesub-collection of O covering [a, b].

Consider the intervals [a, a+b2 ] and [a+b

2 , b] obtained by dividing [a, b] in half. The collection O covers bothof these intervals. For at least one of the two, there is no finite sub-collection of O that covers it. Choosesuch a half of [a, b], and denote it by [a1, b1].

In a similar manner, we can choose a half of [a1, b1] that is not covered by a finite sub-collection of Oand denote it by [a2, b2]. We repeat this process. In other words, given [an, bn], a subset of [a, b] that is notcovered by a finite sub-collection of O, choose a half of [an, bn] that is not covered by a finite sub-collectionof O, and denote it by [an+1, bn+1].

Consider the collection of intervals {[an, bn]}n∈Z+ . Based on the construction of these intervals, thefollowing statements hold for each n ∈ Z+:

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(i) [an+1, bn+1] ⊆ [an, bn]

(ii) bn − an = b−a2n

(iii) [an, bn] is not covered by a finite sub-collection of O.

By Lemma 7.16, it follows that∩∞

n=1[an, bn] is nonempty. Let x be in this intersection. Then x ∈ [a, b],and therefore there exists U ∈ O such that x ∈ U . Since U is open in R, there exists ϵ > 0 such that(x−ϵ, x+ϵ) ⊆ U . Let N be a positive integer large enough so that b−a

2N< ϵ. Since x ∈

∩∞n=1[an, bn], it follows

that x ∈ [aN , bN ]. Furthermore, since bN − aN = b−a2N

< ϵ, it follows that [aN , bN ] ⊆ (x− ϵ, x+ ϵ) ⊆ U . Butthen[aN , bN ] is covered by a single set in O, contradicting the fact that [aN , bN ] is not covered by a finitesub-collection of O.

Thus there must be a finite sub-collection of O that covers [a, b], and it follows that [a, b] is compact.

In the next theorem, we extend Theorem 7.17 to a corresponding result for products of closed boundedintervals in Rn. The proof follows directly from Theorem 7.17 and Corollary 7.14.

Theorem 7.18. Let [a1, b1], . . . , [an, bn] be closed bounded intervals in R. Then [a1, b1] × · · · × [an, bn] is acompact subset of Rn.

We now come to a point to proof the Heine-Borel Theorem, which states that in Rn, our new definitionof compact in topological spaces is equivalent to the old definition we learned in analysis courses.

Theorem 7.19. [The Heine-Borel Theorem.] Let Rn have the standard topology and the standard metricd. A set A ⊆ Rn is compact in Rn if and only if it is closed and bounded.

Proof. Let A be compact in Rn. Since Rn is Hausdorff, it follows by Theorem 7.11 that A is closed. To seethat A is bounded, consider the collection O = {B(O,n) | n ∈ Z+}, made up of open balls centered at theorigin in Rn. The collection O is an open cover of A, and since A is compact, it follows that finitely many ofthe sets in O cover A. Thus, there exists N ∈ Z+ such that A ⊆ B(O,N). Therefore, for x, y ∈ A, we haved(x, y) < 2N , implying that A is bounded.

Now assume that A is closed and bounded. Let a = (a1, . . . , an) be a point in A, and assume thatd(x, y) < M for all x and y in A. Then A is contained in the product of intervals

P = [a1 −M,a1 +M ]× · · · × [an −M,an +M ].

The set P is a compact subset of Rn by Theorem 7.18. Since A is closed and a subset of P , it follows byTheorem 7.9 that A is compact in Rn.

We know from our analysis course that a bounded sequence in Rn has a convergence subsequence.However, this is not true in general metric space.

Example 7.20. Give an example of metric space (X, d) that is bounded but there exists a sequence withno convergent subsequence.

Solution. Consider the discrete metric on Z. If xn = n and x ∈ Z, then d(x, xn) = 1 for all n with at mostone exception. Thus the sequence xn can have no convergent subsequence.

Fortunately we do have a very neat and useful true theorem.

Definition 7.21. A metric space (X, d) is said to be sequentially compact if every sequence in X has aconvergent subsequence.

Theorem 7.22. If the metric space (X, d) is compact, then it is sequentially compact.

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Proof. Let (xn) be a sequence in X. We claim that there exists x ∈ X such that every nonempty open ballcentered at x contains xn for infinitely many n. Suppose this does not hold. Then for every x ∈ X thereexists ϵx such that Bd(x, ϵx) contains members of the sequence (xn) for at most finitely many n. Considerthe collection O = {Bd(x, ϵx)}x∈X . No finite sub-collection of O can cover the sequence (xn) since each setBd(x, ϵx) contains members of the sequence (xn) for only at most finitely many n. On the other hand, O isan open cover of the compact space X and therefore there is a finite sub-collection of O that covers X, thisis a contradiction. Thus, there exists x ∈ X such that every nonempty open ball centered at x contains xnfor infinitely many n.

Now, for each m ∈ Z+ let xnm be a point in the sequence (xn) such that nm > nm−1 and xnm ∈ Bd(x,1m).

Then (xnm) is a subsequence of (xn) converging to x ∈ X.

An important convergence question for metric spaces is the following: If the points in a sequence (xn)get closer and closer to each other, then does the sequence converge? We investigate this question in theremainder of the section.

Definition 7.23. Let (X, d) be a metric space.

(i) A sequence xn in X is called a Cauchy sequence if for every ϵ > 0, there exists an N(ϵ) ∈ Z+ with

d(xn, xm) < ϵ for every n, m ≥ N0(ϵ).

(ii) If every Cauchy sequence in X converges to a limit in X, then X is called complete.

The following theorem illustrated that Rn with the standard metric is complete.

Theorem 7.24. Let (xn) be a Cauchy sequence in Rn with the standard metric d. Then (xn) converges toa limit in Rn.

Proof. Suppose that (xn) is a Cauchy sequence in Rn. Pick a µ > 0. Since (xn) is a Cauchy sequence, thereexists N ∈ Z+ such that d(xn, xm) < µ for all n,m ≥ N . Let C be the closed ball of radius µ centered atxN . Then C is a compact set since it is closed and bounded. Also, the sequence (xN , xN+1, . . . ) is in C. ByTheorem 7.22, the sequence (xN , xN+1, . . . ) has a subsequence that converges to a limit x∗ in C. We claimthat the original sequence (xn) converges to x∗.

Thus, suppose that ϵ > 0 is arbitrary. Because (xn) is a Cauchy sequence, we can choose M ∈ Z+ suchthat d(xn, xm) < ϵ

2 for every n,m ≥ M . Now, let k ≥ M be arbitrary. We show that d(xk, x∗) < ϵ, proving

that (xn) converges to x∗. Since (xN , xN+1, . . . ) has a subsequence converging to x∗, there exists j ≥ M suchthat d(xj , x

∗) < ϵ2 . Also, having j, k ≥ M implies that d(xk, xj) <

ϵ2 . Therefore, by the triangle inequality,

it follows that d(xk, x∗) < ϵ, as desired.

The previous theorem shows that Rn with the standard metric is a complete metric space. However, ifwe let X = R − 0 with the metric d(x, y) = |x − y|, then X is not a complete metric space. The sequence(1, 12 ,

13 , . . . ) is a Cauchy sequence in X that does not converge in X.

Example 7.25. Let X = R and let d be the standard metric on R. Let Y = (0, 1) (the open interval withend points 0 and 1) and let ρ be the standard metric on (0, 1). Then (X, d) and (Y, ρ) are homeomorphic astopological spaces but (X, d) is complete and (Y, ρ) is not.

Solution. We know from first year analysis that f(y) = tan(π(y− 12)) is a bijective function f : Y → X which

is continuous with continuous inverse. (Recall that a strictly increasing continuous function has continuousinverse.) Thus (X, d) and (Y, ρ) are homeomorphic. We know that (X, d) is complete by the general principleof analysis.

However ( 1n) is a Cauchy sequence in Y with no limit in Y . (If y ∈ (0, 1), then there exists an N withy > N−1. If m ≥ 2N , then | 1m − y| ≥ 1

2N so ( 1n) not converge to y.)

We say that completeness is not a topological property.In Example 1.19 we saw that the continuous image of an open set need not be open. It also easy to see

that the continuous image of a closed set need not be closed.

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Exercise 7.26. Let R have the usual metric. Give an example of a continuous injective function f : R → Rsuch that f(R) is not closed.

Hint. Look at the solution of Example 7.25 if you need a hint.

Theorem 7.27. Let (X, T ) be a compact and (Y, σ) a Hausdorff topological space. If f : X → Y is acontinuous bijection, then it is a homeomorphism.

Proof. Since f is a bijection, g = f−1 is a well defined function. If K is closed in X, then (since a closedsubset of a compact space is compact)K is compact so f(K) is compact. But a compact subset of a Hausdorffspace is closed so g−1(K) = f(K) is closed. Thus g is continuous and we are done. (If U is open in X thenX \ U is closed so Y \ g−1(U) = g−1(X \ U) is closed and g−1(U) is open.)

Exercise 7.28. (i) Give an example of a compact space (X, T ) and a topological space (Y, σ) together witha continuous bijection f : X → Y which is not a homeomorphism.

(ii) Give an example of a topological space (X, T ) and a Hausdorff space (Y, σ) together with a continuousbijection f : X → Y which is not a homeomorphism.

Finally, we give the following very useful results for compact metric spaces.

Theorem 7.29. If X is a compact metric space, then X is complete.

To prove Theorem 7.29, we need the following lemma and Theorem 7.22.

Lemma 7.30. A metric space X is complete if every Cauchy sequence in X has a convergent subsequence.

Proof. Let (xn) be a Cauchy sequence in (X, d). We show that if (xn) has a subsequence (xni) that convergesto a point x, then the sequence (xn) itself converges to x.

Given ϵ > 0, first choose N large enough that d(xn, xm) < ϵ2 for all n,m ≥ N [using the fact that (xn) is

a Cauchy sequence]. Then choose an integer i large enough that ni ≥ N and

d(xni , x) <ϵ

2

[using the fact that n1 < n2 < . . . is an increasing sequence of integers and xni converges to x]. Puttingthese facts together, we have the desired result that for n ≥ N ,

d(xn, x) ≤ d(xn, xni) + d(xni , x) < ϵ.

7.2 The Extreme Value Theorem

The Extreme Value Theorem concerns continuous real-valued functions on a compact domain. In thissection we prove a general version of the Extreme Value Theorem. To begin, we need the following lemma.

Lemma 7.31. Let A be a compact subset of R. Then there exist m,M ∈ A such that m ≤ a ≤ M for alla ∈ A.

Proof. We only prove the existence of the maximum value M . The proof of the existence of the minimumvalue m is similar. Since A is compact, it is closed and bounded in R. Therefore A is bounded from above.It follows that the set A has a least upper bound; denote it by M . Of course, a ≤ M for all a ∈ A. We claimthat M ∈ A. We prove the claim by contradiction; thus suppose that M /∈ A. Since A is closed, it followsthat there exists an ϵ > 0 such that (M − ϵ,M + ϵ) ∩ A = ∅. Then M − ϵ is an upper bound for A that issmaller than M , a contradiction. Therefore M ∈ A, and A has a maximum value.

Theorem 7.32. [The Extreme Value Theorem (General Version).] Let X be compact and f : X → Rbe continuous. Then f takes on a maximum value and a minimum value on X; that is, there exist a, b ∈ Xsuch that f(a) ≤ f(x) ≤ f(b) for all x ∈ X.

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Proof. Since X is compact and f is continuous, f(X) is a compact subset of R by Theorem 7.8. Thereforef(X) contains a maximum value M and a minimum value m by the previous lemma. The points m and Mare in f(X); therefore there exist a, b ∈ X such that f(a) = m and f(b) = M . Now, for all x ∈ X we havethat f(x) ∈ f(X), and thus ,f(a) = m ≤ f(x) ≤ M = f(b), as we wished to show.

The following corollary is the version of the Extreme Value Theorem that we encountered in a calculuscourse.

Corollary 7.33. [The Extreme Value Theorem on [a,b]]. Assume that f : [a, b] → R is continuous.Then f takes on a maximum value and a minimum value on [a, b].

Example 7.34. The publishers at K Publishing House would like to catch the new wave of interest inapplied topology. They are planning to publish an applied topology text to compete with those currently onthe market. The business department has committed a $500,000 budget to the first run of the book. Thebudget is to be allocated among several aspects of the book’s production, including an authorship contract,editorial costs, printing, advertising, and distribution. Suppose that there are n such variables, v1, . . . , vn.The profit that the publishers can expect to make on this venture depends on how the resources are allocated.For example, if they opt for printing the book in a high-cost Mobius-band format and a minimal advertisingbudget, they will realize a smaller profit than if they had used a standard book format and a larger advertisingbudget. Thus, we regard the profit, P , as a function of the variables v1, . . . , vn. It is natural to assume thatP is continuous. The domain of P is the subset of Rn given by

D = {(v1, . . . , vn) | v1 ≥ 0, . . . , vn ≥ 0, v1 + · · ·+ vn ≤ 500, 000}.

Since D is closed and bounded, it is compact. Therefore the Extreme Value Theorem indicates that there isa choice for the allocation of resources resulting in a maximum profit for the planned textbook project.

7.3 Bolzano-Weierstrass property

As indicated when we first mentioned compact sets, there are other formulations of the notion of com-pactness that are frequently useful. In this section we introduce one of them. Weaker in general thancompactness, it coincides with compactness for metric spaces.

Definition 7.35. A topological space X has the Bolzano-Weierstrass property if every infinite subsetof X has a limit point.

Example 7.36. In R the subspace A = {0}∪ { 1n | n ∈ Z+} has the Bolzano-Weierstrass property. Let B be

an infinite subset of A. Then B must contain values of the form 1n where n is arbitrarily large. Therefore

0 is a limit point of B. Thus, every infinite subset of A has a limit point, implying that A is has theBolzano-Weierstrass property.

The following theorem indicates that compactness implies Bolzano-Weierstrass property in all topologicalspaces.

Theorem 7.37. If a topological space X is compact, then it has the Bolzano-Weierstrass property.

Proof. We prove that if X does not has the Bolzano-Weierstrass property, then X is not compact. Supposethat X does not has the Bolzano-Weierstrass property. Thus, X has an infinite subset B that does not havea limit point. Since each x ∈ B is not a limit point of B, it follows that for every x ∈ B there exists aneighborhood Ux of x such that Ux ∩ B = {x}. Also, since B has no limit points, that is, it contains all itslimit points and this implies that B is closed. Hence, X \ B is open. Let O be the collection of open sets{Ux | x ∈ B}∪ {X \B}. Then O is an open cover of X. Furthermore O has no finite subcover, since each ofthe infinitely many points x in B is contained in only the open set Ux in O. Therefore X has an open coverwith no finite subcover. It follows that X is not compact.

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Example 7.38. Let Z have the topology generated by the basis

B = {(−n, n) | n ∈ Z+}.

In this topology, Z is not compact since the basis B is an open cover that has no finite subcover. We claim,however, that Z has the Bolzano-Weierstrass property in this topology. In fact, we show that every nonemptysubset of Z has a limit point. Thus, let A be a nonempty subset of Z. Every subset of Z has an elementwith minimum absolute value; let s be such an element of A. For every t ∈ Z such that |t| > |s|, it followsthat every open set containing t must contain s.

Therefore every t satisfying |t| > |s| is a limit point of A. Hence, A has a limit point, implying that Zhas the Bolzano-Weierstrass property in the topology generated by B.

Before we prove the final important result in this course, that is, different notion of compactness areequivalent in metric spaces, we need the following lemma.

Lemma 7.39. Suppose that (X, d) is a sequentially compact metric space and that the collection Uα withα ∈ A is an open cover of X. Then there exists a δ > 0 such that, given any x ∈ X, there exists an α(x) ∈ Asuch that the open ball B(x, δ) ⊆ Uα(x).

Proof. Suppose the first sentence is true and the second sentence false. Then, for each n ≥ 1 we can find anxn such that the open ball B(xn,

1n) ⊆ Uα for all α ∈ A. By sequential compactness, we can find y ∈ X and

n(j) → ∞ such that xn(j) → y.Since y ∈ X, we must have y ∈ Uβ for some β ∈ A. Since Uβ is open, we can find an ϵ such that

B(y, ϵ) ⊆ Uβ. Now choose J sufficiently large that n(J) > 2ϵ−1 and d(xn(J), y) < ϵ/2. We now have, usingthe triangle inequality, that

B(xn(J), 1/n(J)) ⊆ B(xn(J), ϵ/2) ⊆ B(y, ϵ) ⊆ Uβ,

contradicting the definition of xn(J).

Theorem 7.40. Let X be a metric space. Then the following are equivalent:

(i) X is compact.

(ii) X has the Bolzano-Weierstrass property.

(iii) X is sequentially compact.

Proof. We have already proved that (i) ⇒ (ii). To show that (ii) ⇒ (iii), assume that X has the Bolzano-Weierstrass property. Given a sequence (xn) of points of X, consider the set A = {xn | n ∈ Z+}. If the set Ais finite, then there is a point x such that x = xn for infinitely many values of n. In this case, the sequence(xn) has a subsequence that is constant, and therefore converges trivially. On the other hand, if A is infinite,then A has a limit point x. We define a subsequence of (xn) converging to x as follows: First choose n1 sothat

xn1 ∈ B(x, 1).

Then suppose that the positive integer ni−1 is given. Because the ball B(x, 1i ) intersects A in infinitely manypoints, we can choose an index ni > ni−1 such that

xni ∈ B(x,1

i).

Then the subsequence xn1 , xn2 , . . . converges to x.Finally, we show that (iii) ⇒ (i). This is the hardest part of the proof.Let (Uα)α∈A be an open cover and let δ be defined as in Lemma 7.39. The B(x, δ) form a cover of X.

If they have no finite subcover, then given x1, x2, . . .xn we can find an xn+1 /∈∪n

j=1B(xj , δ). Consider thesequence xj thus obtained. We have d(xn+1, xk) > δ whenever n ≥ k ≥ 1 and so d(xr, xs) > δ for all r = s.It follows that, if x ∈ X, d(xn, x) > δ/2 for all n with at most one exception. Thus the sequence of xn hasno convergent subsequence.

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It thus follows, by reductio ad absurdum, that the B(x, δ) have a finite subcover. In other words, we canfind an M and yj ∈ X [1 ≤ j ≤ M ] such that

X =

M∪j=1

B(yj , δ).

We thus have

X =

M∪j=1

B(yj , δ) ⊆M∪j=1

Uα(yj) ⊆ X

so X =∪M

j=1 Uα(yj) and we have found a finite subcover.Thus X is compact.

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