MAT327H1Y: Introduction to TopologyLecture Notes

INTRODUCTORY REMARKS

These notes are intended to concisely summarize the material delivered in lecture. Theydo not necessarily contain all of what is presented in any given lecture. Rather, they aredesigned to serve as a convenient study guide for students.

1. TOPOLOGICAL SPACES

1.1. Informal Motivation. If a, b ∈ R, then a is ”close” to b if |a− b| < δ for small δ > 0.Equivalently, we have a ∈ (b − δ, b + δ). In this way, we are using ”nice” subsets of R toexpress the idea that two points in R are ”close”. One might try to generalize this vagueapproach for sets X other than R. Namely, we would like to use a collection of ”nice”subsets of X to express the idea that two points in X are ”close”. Indeed, we will later callsuch a collection of ”nice” subsets a topology on X, and we will refer to each ”nice” subsetof X as an open set.

1.2. Definition of a Topology. Let us fix a set X. Formally, a collection of subsets of X isa set T such that each element of T is a subset of X.

Definition 1. A topology on a set X is a collection T of subsets of X satisfying the followingproperties.

1. ∅ ∈ T and X ∈ T

2. T is closed under pairwise intersections: If U1, U2 ∈ T, then U1 ∩U2 ∈ T.3. T is closed under arbitrary unions: to come soon

Arbitrary Unions. Let X be the set fixed in the previous section. Also, let A be any set.Suppose that for each α ∈ A, we have a corresponding subset Uα of X. We will use thenotation {Uα}α∈A for this correspondence, and we will think of {Uα}α∈A as a family of setsindexed by A.

Informally, the union⋃α∈AUα is obtained by putting the elements of all theUα together

into a single set. More precisely, we have the following definition.

Definition 2. We define the union⋃α∈AUα to be the set of all x such that x ∈ Uα for some

α ∈ A (ie. x is in at least one of the sets Uα).

Example. Suppose that X = R and A = Z. For each n ∈ Z, set Un := [1, 1 + n2]. Weclaim that

⋃n∈Z[1, 1 + n

2] = [1,∞). To prove this, first suppose x ∈⋃n∈Z[1, 1 + n

2]. Itfollows that x ∈ [1, 1+ n2] for some n ∈ Z. In particular, x ≥ 1, so that x ∈ [1,∞). Hence,⋃n∈Z[1, 1+ n

2] ⊆ [1,∞).Now, suppose that x ∈ [1,∞). There exists an integer n ∈ Z, such that 1 ≤ x ≤ 1 + n2,

so that x ∈ [1, 1 + n2]. Hence, x is in one of the sets whose union we are taking, meaningthat x ∈

⋃n∈Z[1, 1 + n

2]. This establishes that [1,∞) ⊆⋃n∈Z[1, 1 + n

2]. We conclude that⋃n∈Z[1, 1+ n

2] = [1,∞). �

We may now complete the definition of a topology.1

Definition 3. A topology on a set X is a collection T of subsets of X satisfying the followingproperties.

1. ∅ ∈ T and X ∈ T

2. T is closed under pairwise intersections: If U1, U2 ∈ T, then U1 ∩U2 ∈ T.3. T is closed under arbitrary unions: If {Uα}α∈A is a family of sets in T, then

⋃α∈AUα ∈

T.

1.3. Examples of Topologies.

Example. Suppose that X = R and define T to be the collection of all subsets U of Rsatisfying the following property:

∀x ∈ U ∃δ > 0 such that (x− δ, x+ δ) ⊆ UWe claim that T is a topology on R. To show that ∅ ∈ T, we must consider the followingstatement:

∀x ∈ ∅ ∃δ > 0 such that (x− δ, x+ δ) ⊆ ∅.Since there do not exist any x ∈ ∅, there is nothing to check, and our statement is true.Hence, ∅ ∈ T. Also, one can check that R ∈ T.

Now, suppose that U1, U2 ∈ T and that x ∈ U1 ∩ U2. Since x ∈ U1 and U1 ∈ T, thereexists δ1 > 0 such that (x − δ1, x + δ1) ⊆ U1. Similarly, there exists δ2 > 0 such that(x− δ2, x+ δ2) ⊆ U2. If we set δ := min{δ1, δ2}, then

(x− δ, x+ δ) ⊆ (x− δ1, x+ δ1) ⊆ U1and

(x− δ, x+ δ) ⊆ (x− δ2, x+ δ2) ⊆ U2.In particular, (x−δ, x+δ) is a subset of bothU1 andU2, meaning that (x−δ, x+δ) ⊆ U1∩U2.This establishes that U1 ∩U2 ∈ T.

Finally, suppose that {Uα}α∈A is a family of sets in T, and suppose that x ∈⋃α∈AUα.

By definition, x ∈ Uβ for some β ∈ A. Since Uβ ∈ T, there exists δ > 0 such that(x− δ, x+ δ) ⊆ Uβ. We then have

(x− δ, x+ δ) ⊆ Uβ ⊆⋃α∈A

Uα,

so that (x− δ, x+ δ) ⊆⋃α∈AUα. This establishes that

⋃α∈AUα ∈ T. �

Definition 4. We define the above topology T to be the standard topology on R.

Example. Suppose that X = Rn. Given a point x = (x1, x2, . . . , xn) ∈ Rn, recall that thenorm or length of x is given by

‖x‖ =(x21 + x

22 + . . .+ x

2n

) 12 .

If x ∈ Rn and δ > 0, we define

Bδ(x) := {y ∈ Rn : ‖y− x‖ < δ},the ball of radius δ centred at x. This definition allows us to introduce the standard topol-ogy on Rn in analogy with the standard topology on R.

We define T to be the collection of all subsets U of Rn satisfying the following property:

∀x ∈ U ∃δ > 0 such that Bδ(x) ⊆ U2

�

Exercise. The collection T is a topology on Rn.

Definition 5. We define the above topology T to be the standard topology on Rn.

Example. Let X be any set and let T be the collection of all subsets of X. Then, T is atopology on X, called the discrete topology. �

Example. Let X be any set and let T = {∅, X}. Then, T is a topology on X, called the trivialtopology. �

Remark. The preceding two examples illustrate a general point. If X is a set, then X canadmit different topologies.

1.4. Definition of a Topological Space.

Definition 6. A topological space consists of the following two pieces of information:• a set X• a topology T on X

In this case, we denote the topological space by (X,T).

Example. If T is the standard topology on R, then (R,T) is a topological space. �

Remark. We sometimes just write ”X is a topological space” instead of ”(X,T) is a topo-logical space”.

Definition 7. If (X,T) is a topological space, then we call the sets in T the open sets.

Exercise. Consider the topological space (R,T), where T is the standard topology on R.If a, b ∈ R and a < b, show that the open interval (a, b) is an open set.

Example. Consider again the topological space (R,T), where T is the standard topologyon R. By the previous exercise, we know that the sets U1 := (1, 3) and U2 := (5, 10) areopen (ie. they are in T). Hence,⋃

n∈{1,2}

Un = U1 ∪U2 = (1, 3) ∪ (5, 10) ∈ T,

by Property 3. of a topology. In other words, the set (1, 3) ∪ (5, 10) is open. �

Exercise (Assignment #1). Consider the topological space (Rn,T), where T is the standardtopology on Rn. If x ∈ Rn and δ > 0, show that Bδ(x) is an open set.

2. BASIS OF A TOPOLOGY

Consider (R,T), where T is the standard topology on R. Recall that the open sets arethe subsets U of R satisfying the following property:

∀x ∈ U ∃δ > 0 such that (x− δ, x+ δ) ⊆ UNote that the open interval (x − δ, x + δ) is itself an open set. We have therefore defineda general open set U in terms of the more ”basic” open sets (x − δ, x + δ). The followingdefinition formalizes the idea of building general open sets using basic ones.

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Definition 8. Let (X,T) be a topological space. A basis of T is a collection B of open setssatisfying the following condition: For each open set U and each point x ∈ U, there existsa set V ∈ B such that x ∈ V and V ⊆ U.

Example. Consider (R,T), where T is the standard topology on R. If we define B to bethe collection of all sets of the form (x− δ, x+ δ) where x ∈ R and δ > 0, then B is a basisof T. �

Example. Consider (Rn,T), where T is the standard topology on Rn. If we define B to bethe collection of all sets of the form Bδ(x) where x ∈ Rn and δ > 0, then B is a basis of T.�

Example. If (X,T) is any topological space, then T is a basis of T. �

3. CLOSED SETS

3.1. Definition and Properties. For some context, we begin by considering (R,T) with T

the standard topology on R. If a, b ∈ R and a < b, then the closed interval [a, b] is notopen. Nevertheless, given the importance of closed intervals in analysis, we would liketo include them as objects of study in topology. We will achieve this by defining a specialclass of sets of which closed intervals are examples.

Definition 9. Let (X,T) be a topological space. A subsetA of X is called a closed set if X\Ais open.

Example. Consider (R,T) where T is the standard topology on R. If a, b ∈ R and a < b,then the closed interval [a, b] is indeed a closed set.

To establish this, we must show that R \ [a, b] is open. We note that

R \ [a, b] = (−∞, a) ∪ (b,∞).

So, if we can show (−∞, a) and (b,∞) to be open, then we will have written R \ [a, b] asa union of two open sets. It will then follow that R \ [a, b] is open, as desired.

Let us show (b,∞) to be open. Suppose that x ∈ (b,∞) and set δ = 12(x− b) > 0. Note

that (x− δ, x+ δ) ⊆ (b,∞). It follows that (b,∞) is open. A similar argument establishesthat (−∞, a) is open. In light of earlier remarks, we conclude that R\[a, b] is open. Hence,[a, b] is closed. �

Example. Consider (Rn,T) where T is the standard topology on Rn. If x ∈ Rn and δ > 0,then

Dδ(x) := {x ∈ Rn : ‖y− x‖ ≤ δ}is a closed set (exercise). �

Example. If (X,T) is any topological space, then• ∅ is closed (since X \ ∅ = X is open)• X is closed (since X \ X = ∅ is open)

�

Let (X,T) be a topological space. By definition, the complement of a closed subset of Xis an open subset of X. However, if U is an open subset of X, then X \ (X \U) = U is open.This means that X \ U is closed. In other words, we have shown that the complement

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of an open set is an closed set. The process of taking complements therefore gives acorrespondence between the open and closed subsets of X.

Theorem 1 (Properties of Closed Sets). Let (X,T) be a topological space.1. ∅ and X are closed sets.2. IfW1 andW2 are closed subsets of X, thenW1 ∪W2 is closed.3. If {Wα}α∈A is a family of closed subsets of X, then

⋂α∈AWα is closed.

Proof. The proof of Property 1 was given in an example.To prove Property 2, let W1 and W2 be closed subsets of X. To prove that W1 ∪W2 is

closed, we must show X \ (W1 ∪W2) to be open. To this end, note that

X \ (W1 ∪W2) = (X \W1) ∩ (X \W2). (1)

Since W1 and W2 are closed, X \W1 and X \W2 are open. Hence, (1) is an expression ofX \ (W1 ∪W2) as an intersection of two open sets. It follows that X \ (W1 ∪W2) is open,meaning thatW1 ∪W2 is closed.

To prove Property 3, let {Wα}α∈A be a family of closed subsets of X. To prove that⋂α∈AWα is closed, we must show X \

(⋂α∈AWα

)to be open. Note that

X \

(⋂α∈A

Wα

)=⋃α∈A

(X \Wα). (2)

Since each set Wα is closed, each set X \Wα is open. Therefore, (2) is an expression ofX \

(⋂α∈AWα

)as a union of open sets. It follows that X \

(⋂α∈AWα

)is open, implying

that⋂α∈AWα is closed. �

3.2. Interiors and Closures. Property 3 of a topology states that an arbitrary union ofopen sets is open. By contrast, Property 3 from the above theorem states that an arbitraryintersection of closed sets is closed. Using these two facts, we will associate both an openset (to be called the interior) and a closed set (to be called the closure) to each subset of atopological space.

Definition 10. Let (X,T) be a topological space and let A be a subset of X.

(i) The interior of A, denoted Int(A), is defined to be the union of all open subsets ofXwhich are contained in A. In other words,

Int(A) =⋃U⊆X

U open andU⊆A

U.

(ii) The closure of A, denoted A, is defined to be the intersection of all closed subsetsof Xwhich contain A. In other words,

A =⋂U⊆X

U closed andA⊆U

U.

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A few remarks are in order. Firstly, Int(A) is defined to be a union of open sets, so thatInt(A) is itself an open set. Analogously, A is defined to be an intersection of closed sets,meaning that A is itself a closed set. Finally, we have the inclusions

Int(A) ⊆ A ⊆ A.

Example. Consider (R,T) where T is the standard topology on R. If a, b ∈ R and a < b,then Int([a, b)) = (a, b) (exercise). �

Example. Consider (R,T) where T is the standard topology on R. If a, b ∈ R and a < b,then [a, b) = [a, b] (exercise). �

Example. Consider (Rn,T) where T is the standard topology on Rn. If x ∈ Rn and δ > 0,then Bδ(x) = Dδ(x) (exercise). �

Proposition 1. Let (X,T) be a topological space and let A be a subset of X.(i) A is open ⇐⇒ A = Int(A)

(ii) A is closed ⇐⇒ A = A

Proof. This is left as an exercise. �

Theorem 2. Let (X,T) be a topological space and let A be a subset of X. If x ∈ X, then

x ∈ A⇐⇒ for every open set U containing x,U ∩A 6= ∅.

Proof. (⇒) Assume that there exists an open set U containing x such that U∩A = ∅. Notethat X \U is closed, and so is one of the closed sets whose intersection one takes to get A.It follows that A ⊆ X \ U. Since x ∈ A, we have x ∈ X \ U. This contradicts the fact thatx ∈ U, completing the proof of this direction.(⇐) Assume that x 6∈ A. We then have x ∈ X\A, an open set. By hypothesis, (X\A)∩A 6= ∅.Hence, there exists an element y ∈ (X \A) ∩A. Note that y ∈ A, but y 6∈ A, contradictingthe fact that A ⊆ A. �

Remark. If we impose certain conditions on the topological space (X,T), then we obtainmore ways of determining whether x ∈ A. One such condition is that (X,T) be Hausdorff.

Definition 11. A topological space (X,T) is called Hausdorff if for any two distinct x1, x2 ∈X, there exist open sets U1, U2 such that x1 ∈ U1, x2 ∈ U2, and U1 ∩U2 = ∅.

Example. If T is the standard topology on R, then (R,T) is Hausdorff. To see this, supposewe have two distinct points x1, x2 ∈ R. Without the loss of generality, x1 < x2. Setδ := 1

2(x2 − x1) > 0. Note that x1 is in the open set (x1 − δ, x1 + δ) and x2 is in the open set

(x2 − δ, x2 + δ). Also, (x1 − δ, x1 + δ) ∩ (x2 − δ, x2 + δ) = ∅. �

Example. If T is the standard topology on Rn, then (Rn,T) is Hausdorff (exercise). �

Example. If T = {∅,R} (the trivial topology on R), then (R,T) is not Hausdorff (exercise).�

Theorem 3. Let (X,T) be a Hausdorff topological space and let A be a subset of X.(i) If x1, x2, . . . , xn ∈ X, then {x1, x2, . . . , xn} is a closed set.

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(ii) If x ∈ X, then

x ∈ A⇐⇒ x ∈ A OR for every open set U containing x,U ∩A is an infinite set.

Proof. We leave the proof of (i) as an exercise and we proceed to the proof of (ii)(=⇒) If x ∈ A, then we are done. We may therefore suppose x 6∈ A. For a contradiction,assume that there exists an open set U containing x such that U ∩ A is a finite set. Let uswrite U ∩ A = {x1, x2, . . . , xn}. Since x 6∈ A, x cannot be one of the elements x1, x2, . . . , xn.Hence, we have

x ∈ U ∩ (X \ {x1, x2, . . . , xn}).

By (i), X \ {x1, x2, . . . , xn} is open, implying that U∩ (X \ {x1, x2, . . . , xn}) is also open. Sincex ∈ A and U ∩ (X \ {x1, x2, . . . , xn}) is an open set containing x, we have

(U ∩ (X \ {x1, x2, . . . , xn})) ∩A 6= ∅.Writing U ∩A = {x1, x2, . . . , xn}, this becomes

{x1, x2, . . . , xn} ∩ (X \ {x1, x2, . . . , xn}) 6= ∅,a contradiction.(⇐=) There are two cases to consider. In the first case, x ∈ A. SinceA ⊆ A, we have x ∈ A,as desired.In the second case, U∩A is an infinite set for every open set U containing x. In particular,U ∩A 6= ∅ for every open set U containing x. Hence, x ∈ A. �

3.3. Sequences in Hausdorff Spaces. Hausdorff topological spaces provide a naturalsetting in which to develop a theory of sequences.

Definition 12. Let (X,T) be a topological space. A sequence in X is a collection {xn}∞n=1

of points in X. We say that {xn}∞n=1 converges to x ∈ X, written xn → x, if the followingcondition is satisfied: For every open set U containing x, there exists a positive integer Nsuch that xn ∈ U for all n ≥ N.

Proposition 2. Let (X,T) be a Hausdorff topological space and let {xn}∞n=1 be a sequence in X. Ifxn → x and xn → y, then x = y.

Proof. Assume that x 6= y. Since (X,T) is Hausdorff, there exist open sets U1, U2 such thatx ∈ U1, y ∈ U2, and U1 ∩ U2 = ∅. Now, since xn → x, there exists a positive integer Nsuch that xn ∈ U1 for all n ≥ N. Analogously, there exists a positive integer M such thatxn ∈ U2 for all n ≥M. Set K := max{N,M}, noting that xK ∈ U1 ∩U2. This contradicts thefact that U1 ∩U2 = ∅. �

The following example illustrates that this proposition may not hold when (X,T) is notHausdorff.

Example. Consider the topological space (R,T = {∅,R}), which is not Hausdorff. Weclaim that the sequence { 1

n}∞n=1 converges to each x ∈ R. To see this, suppose that U is an

open set containing x. Since ∅ and R are the only open sets for our choice of topology, wemust have that U = R. Hence, 1

n∈ U = R for all n ≥ 1. We have therefore shown that

1n→ x. �

Example. Consider (R,T) where T is the standard topology on R. The sequence { 1n}∞n=1

converges only to 0 (exercise). �7

4. CONTINUOUS FUNCTIONS

4.1. General Theory. Topological spaces provide a natural context in which to generalizeour classical notions of continuity. To this end, recall the usual definition of continuity fora function f : R → R from analysis courses:

• f is continuous at a point x ∈ R if

∀ε > 0 ∃δ > 0 such that if |y− x| < δ, then |f(y) − f(x)| < ε. (3)

• f is continuous if it is continuous at each point x ∈ R.

Theorem 4. Consider (R,T) with T the standard topology on R. A function f : R → R iscontinuous (as in (3)) if and only if

f−1(U) := {x ∈ R : f(x) ∈ U} is open for all open subsets U of R. (4)

The ”(=⇒)” direction was proved on Assignment #1, and the other direction is provedsimilarly.

Note that (3) depends, in part, on having a notion of distance between points. It willtherefore be difficult to generalize (3) to functions between arbitrary topological spaces.By contrast, (4) generalizes to give the desired definition of continuity.

Definition 13. Let (X,TX) and (Y,TY) be topological spaces. A function f : X → Y is acontinuous function from (X,TX) to (Y,TY) if

f−1(U) := {x ∈ X : f(x) ∈ U}

is an open subset of X for every open subset U of Y.

Remark. We will sometimes just write ”f : X→ Y is continuous” when (X,TX) and (Y,TY)are understood from context.

Remark. f−1(U) is called the pre-image of U under f. To aid our study of continuousfunctions, we will need the following properties of the pre-image operation.

Proposition 3 (Properties of Pre-Images). Let X, Y, and Z be sets, and let f : X → Y andg : Y → Z be functions.

(i) If {Uα}α∈A is a family of subsets of Y, then

f−1

(⋃α∈A

Uα

)=⋃α∈A

f−1(Uα).

(In particular, f−1(U1 ∪U2) = f−1(U1) ∪ f−1(U2).)(ii) If {Uα}α∈A is a family of subsets of Y, then

f−1

(⋂α∈A

Uα

)=⋂α∈A

f−1(Uα).

(In particular, f−1(U1 ∩U2) = f−1(U1) ∩ f−1(U2).)(iii) If U is a subset of Y, then

f−1(Y \U) = X \ f−1(U).

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(iv) If U is a subset of Z, then

(g ◦ f)−1(U) = f−1(g−1(U)).

Proof. We leave the proof as an exercise. �

Let us now give a few examples of continuous functions.

Example. Consider (Rn1 ,T1) and (Rn2 ,T2), where T1 and T2 are the standard topologies onRn1 and Rn2 , respectively. If T : Rn1 → Rn2 is a linear transformation, then T is continuous.We would like to formally prove this, and will therefore require a lemma.

Lemma 1. Assume that everything is as stated in the above example. There exists a real numberM > 0 such that

‖T(y) − T(x)‖ ≤M‖y− x‖for all x, y ∈ Rn1 .

Proof. Suppose that x ∈ Rn1 . Letting {e1, e2, . . . , en1} be the standard basis of Rn1 , we maywrite

x =

n1∑i=1

xiei.

Hence,

‖T(x)‖ ≤ ‖n1∑i=1

xiT(ei)‖

≤n1∑i=1

‖xiT(ei)‖

=

n1∑i=1

‖T(ei)‖ · |xi|

≤

(n1∑i=1

‖T(ei)‖2) 12(

n1∑i=1

x2i

) 12

≤

( n1∑i=1

‖T(ei)‖2) 12

+ 1

‖x‖.We may therefore takeM :=

(∑n1i=1 ‖T(ei)‖2

) 12 + 1. �

Theorem 5. Consider (Rn1 ,T1) and (Rn2 ,T2), where T1 and T2 are the standard topologies onRn1 and Rn2 , respectively. If T : Rn1 → Rn2 is a linear transformation, then T is continuous.

Proof. Suppose thatU is an open subset of Rn2 . To show that T−1(U) is open, suppose thatx ∈ T−1(U). Since T(x) ∈ U, an open set, there exists δ > 0 such that Bδ(T(x)) ⊆ U.

Letting M be as in the lemma and setting δ ′ := δM> 0, we claim that Bδ ′(x) ⊆ T−1(U).

To this end, suppose y ∈ Bδ ′(x). Note that

‖T(y) − T(x)‖ = ‖T(y− x)‖≤M‖y− x‖

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< Mδ ′ = δ.

In other words, T(y) ∈ Bδ(T(x)). Since Bδ(T(x)) ⊆ U, we have T(y) ∈ U. Equivalently,y ∈ T−1(U), giving us Bδ ′(x) ⊆ T−1(U). It follows that T−1(U) is open, meaning that T iscontinuous. �

Example. Consider (Rn,Tn) and (R,T1), where Tn and T1 are the standard topologies onRn and R, respectively. The multiplication function f : Rn → R

f(x1, x2, . . . , xn) = x1x2 . . . xn

is continuous.

Example. Let (X,TX) and (Y,TY) be topological spaces and let y0 ∈ Y be a point. Theconstant function f : X→ Y, f(x) = y0, is continuous (exercise).

Example. Let (X,TX) and (Y,TY) be topological spaces.

• If TX is the discrete topology, then every f : X→ Y is continuous (exercise).• If TY = {∅, Y} is the trivial topology, then every f : X→ Y is continuous (exercise).

Proposition 4. Let (X,TX) and (Y,TY) be topological spaces. A function f : X→ Y is continuousif and only if f−1(A) is closed for all closed subsets A of Y.

Proof. (=⇒) If A ⊆ Y is closed, then

X \ f−1(A) = f−1(Y \A).

Note that Y \ A is open, so that f−1(Y \ A) = X \ f−1(A) is also open. Hence, f−1(A) isclosed.(⇐=) To show that f is continuous, suppose that U ⊆ Y is open. Note that

X \ f−1(U) = f−1(Y \U).

Note that Y \ U is closed, so that (by hypothesis) f−1(Y \ U) = X \ f−1(U) is closed. Also,f−1(U) = X \ (X \ f−1(U)) is then the complement of a closed set, and hence is open. �

Proposition 5. Let (X,TX), (Y,TY), and (Z,TZ) be topological spaces, and let f : X → Y andg : Y → Z be continuous functions.

(i) If {xn}∞n=1 is a sequence in X and xn → x, then f(xn) → f(x).(ii) The function g ◦ f : X→ Z is continuous.

Proof. We leave the proof of (i) as an exercise and proceed directly to (ii). To this end,suppose that U ⊆ Z is an open set. We have

(g ◦ f)−1(U) = f−1(g−1(U)).

The continuity of g implies that g−1(U) is open, and the continuity of f then implies thatf−1(g−1(U)) is open. Hence, g ◦ f is continuous. �

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4.2. Homeomorphisms. For some context, let us revisit an extremely central concept inlinear algebra. A vector space isomorphism is a linear transformation T : V → W with alinear inverse T−1 :W → V (where V andW are vector spaces over a field F). In particular,T creates a correspondence between the underlying vector space structures of V and W.Indeed, vectors, addition, and scalar multiplication in V all have ”mirror” images inW.

With this in mind, we would like to develop a notion of isomorphism in topology (tobe called a homeomorphism).

Definition 14. Let (X,TX) and (Y,TY) be topological spaces. A function f : X→ Y is calleda homeomorphism from (X,TX) to (Y,TY) if f is continuous and has a continuous inversef−1 : Y → X.

Remark. We will sometimes just write ”f : X → Y is a homeomorphism” when (X,TX)and (Y,TY) are understood from context.

Example. If (X,T) is any topological space, the identity function idX : X→ X is a homeo-morphism from (X,T) to (X,T) (exercise).

Example. Consider (Rn,T) with T the standard topology on Rn. If T : Rn → Rn is a vectorspace isomorphism, then T is a homeomorphism from (Rn,T) to (Rn,T).

Let us make a few comments about the extent to which a homeomorphism is an iso-morphism in topology. Let (X,TX) and (Y,TY) be topological spaces, and let f : X→ Y be ahomeomorphism. Using f, we obtain a correspondence between points in X and those inY. However, there is also a correspondence between the open subsets of X and those of Y.Indeed, if U ⊆ Y is open, then f−1(U) ⊆ X is open. In the other direction, if V ⊆ X is open,then so is f(V) := {f(x) : x ∈ V} ⊆ Y (exercise). (Note: The set f(V) is not necessarily openif one only assumes f to be continuous.)

In light of the above, a homeomorphism creates a correspondence between the under-lying topological space structures of (X,TX) and (Y,TY). It is therefore a suitable notion ofisomorphism in topology.

Proposition 6. Let (X,TX), (Y,TY), and (Z,TZ) be topological spaces, and let f : X → Y andg : Y → Z be homeomorphisms. We have:

(i) f−1 : Y → X is a homeomorphism.(ii) g ◦ f : X→ Z is a homeomorphism.

Proof. For (i), we note that f−1 is continuous. Also, f−1 has a continuous inverse of (f−1)−1 =f. Hence, f−1 is a homeomorphism.

To prove (ii), note that g◦f is continuous by virtue of being a composition of continuousfunctions. Also, g ◦ f has an inverse of (g ◦ f)−1 = f−1 ◦ g−1. This inverse is continuoussince f−1 and g−1 are continuous. We conclude that g ◦ f is a homeomorphism. �

5. THE SUBSPACE TOPOLOGY

5.1. A Definition. Let (X,TX) be a topological space. We would like to use TX to constructa topology TA on each subset A ⊆ X.

Let us fix a subset A of X. To motivate the construction of TA, consider the ”inclusionmap” iA : A → X, iA(a) = a. We would like to construct TA such that iA is a continuousfunction from (A,TA) to (X,TX). This will occur if and only if i−1A (U) ∈ TA for every open

11

subset U of X. Noting that i−1A (U) = U ∩ A, our condition becomes U ∩ A ∈ TA for everyopen subset U of X.

We tentatively define TA to be the collection of all sets of the form U ∩ A, where U isan open subset of X. This turns out to be a topology on A, as is stated in the propositionbelow.

Proposition 7. Let (X,T) be a topological space and let A be a subset of X.(i) TA is a topology on A.

(ii) iA is a continuous function from (A,TA) to (X,TX).

Proof. We leave the proof as an exercise. �

Definition 15. We call TA the subspace topology on A.

5.2. Restriction of the Domain and Codomain. We would like to produce new examplesof continuous functions using ideas related to the subspace topology. To this end, recallthat a function consists of the following data:

• a domain X• a codomain Y• a rule which associates to each x ∈ X an element f(x) ∈ Y

In this case, one denotes the data by f : X → Y. Hence, if f ′ : X ′ → Y ′ is another function,then it equals our original function if and only if X ′ = X, Y ′ = Y, and f ′(x) = f(x) for allx ∈ X ′ = X. In particular, changing the domain and/or codomain of a function results ina new function. With this in mind, we have the following definition.

Definition 16. Let X and Y be sets, and let f : X→ Y be a function.(i) [Restriction of Domain] If A ⊆ X, define the new function

f|A : A→ Y

f|A(a) = f(a).

This function is called the restriction of f to A.

(ii) [Restriction of Codomain] If B ⊆ Y and f(X) ⊆ B, define the new function

f|B : X→ B

f|B(x) = f(x).

Proposition 8. Let (X,TX) and (Y,TY) be topological spaces, and let f : X → Y be a continuousfunction.

(i) If A ⊆ X, then consider (A,TA) with TA the subspace topology on A. The functionf|A : A→ Y is continuous.

(ii) If B ⊆ Y and f(X) ⊆ B, then consider (B,TB) with TB the subspace topology on B. Thefunction f|B : X→ B is continuous.

Proof. To prove (i), suppose that U ⊆ Y is open. Note that

(f|A)−1(U) = {a ∈ A : f|A(a) ∈ U}= {a ∈ A : f(a) ∈ U}= {a ∈ A : a ∈ f−1(U)}

12

= f−1(U) ∩A.

Since f−1(U) is open, it follows that f−1(U) ∩ A = (f|A)−1(U) is an open subset of A. We

conclude that f|A is continuous.We leave the proof of (ii) as an exercise. �

Remark. As discussed in lecture, there is an alternative proof of (i). We can compose thecontinuous functions iA : A → X and f : X → Y to obtain another continuous function,f ◦ iA : A→ Y. Now, note that f ◦ iA = f|A, so that f|A is continuous.

Example. Consider (R,T) with T the standard topology on R. Consider also the sub-sets [0, 1] ⊆ R and [0, 2] ⊆ R with their subspace topologies, namely ([0, 1],T[0,1]) and([0, 2],T[0,2]), respectively.

We claim that the function f : [0, 1] → [0, 2], f(x) = 2x, is a homeomorphism. To seethis, we begin with the function F : R → R, F(x) = 2x. Since F is a linear transformation,F is continuous. Hence, the restriction F|[0,1] : [0, 1] → R is also continuous. To restrict thecodomain of F|[0,1], we note that F|[0,1]([0, 1]) ⊆ [0, 2]. Therefore,(

F|[0,1])|[0,2] : [0, 1] → [0, 2]

is continuous. Note that(F|[0,1]

)|[0,2] : [0, 1] → [0, 2] and f are both functions from [0, 1] to

[0, 2] sending x to 2x. Hence, they are equal as functions, and the continuity of(F|[0,1]

)|[0,2] :

[0, 1] → [0, 2] thus implies f is continuous.Note that f has an inverse of f−1 : [0, 2] → [0, 1], f−1(x) = 1

2x. The continuity of f−1

follows from arguments similar to those used above. We conclude that f is indeed ahomeomorphism. �

5.3. A Description of Closed Sets.

Proposition 9. Let (X,TX) be a topological space and let A be a subset of X. Consider (A,TA)with TA the subspace topology on A. If B ⊆ A, then B is a closed subset of A if and only ifB = V ∩A for some closed subset V of X.

Proof.

B is a closed subset of A⇐⇒ A \ B is an open subset of A

⇐⇒ A \ B = U ∩A for some open U ⊆ X

⇐⇒ A \ (A \ B) = A \ (U ∩A) for some open U ⊆ X

⇐⇒ B = (X \U) ∩A for some open U ⊆ X

⇐⇒ B = V ∩A for some closed V ⊆ X

The equivalence of the last and second-last lines is proved as follows: If the second-lastline is satisfied, then define V := X \ U (a closed set). This V satisfies the last line. If thelast line is satisfied, then define U =: X \ V (an open set). This U satisfies the second-lastline. �

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6. THE PRODUCT TOPOLOGY

6.1. A Definition. If X and Y are sets, their Cartesian product is the new set defined by

X× Y := {(x, y) : x ∈ X, y ∈ Y}.

Hence, if (X,TX) and (Y,TY) are topological spaces, we would like to use TX and TY toproduce a topology TX×Y on X×Y. As with our construction of the subspace topology, wewill begin with the requirement that certain nice functions be continuous.

Let (X,TX) and (Y,TY) be topological spaces. Consider the projection maps

πX : X× Y → X, πX(x, y) = x

and

πY : X× Y → Y, πY(x, y) = y.

We want to construct TX×Y such that πX and πY become continuous functions. In otherwords, we want π−1

X (U) ∈ TX×Y and π−1Y (V) ∈ TX×Y for all open subsets U of X and V of Y.

However, π−1X (U) = U × Y and π−1

Y (V) = X × V , so our condition becomes U × Y ∈ TX×Yand X× V ∈ TX×Y for all open subsets U of X and V of Y.

It is reasonable to wonder about whether the setsU×Y (U ⊆ X open) and X×V (V ⊆ Yopen) are enough to constitute our topology TX×Y . For this, it will be helpful to consideran example.

Example. Consider (R,TR) with TR the standard topology on R. Eventually, we will beable to construct the product topology TR×R on R × R = R2. Note that B1(0) is not of theform U×R or R×V . Hence, if we wish to include sets like B1(0) in TR×R, we must defineTX×Y to also include sets which are not of the form U× Y or X× V . �

We now offer a solution to the above discussion. Define B to be the collection of all setsof the formU×V , whereU and V are open subsets of X and Y, respectively. One can verifythat B satisfies Conditions (i) and (ii) from the theorem in Problem #6 of Assignment #1.The theorem then implies that B is a basis of a unique topology TX×Y on X× Y.

Definition 17. We call TX×Y the product topology on X× Y.

6.2. Two Descriptions of TX×Y . Our definition of TX×Y as the unique topology with basisB is somewhat indirect. However, Problem #6 on Assignment #1 gives us two explicitdescriptions of a topology defined via a basis. Applying these to our particular basis B,we get the following two equivalent descriptions of the sets in TX×Y.

1. IfW ⊆ X× Y, thenW ∈ TX×Y if and only if there exists a family {Uα ×Vα}α∈A, witheach Uα, Vα open in X, Y, respectively, such that

W =⋃α∈A

(Uα × Vα).

2. If W ⊆ X × Y, then W ∈ TX×Y if and only if for all (x, y) ∈ W, there exists opensubsets U of X and V of Y such that (x, y) ∈ U× V and U× V ⊆W.

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6.3. Some Properties. As desired, the product topology is such that the projection mapsare continuous.

Proposition 10. If (X,TX) and (Y,TY) are topological spaces, then the projection functions πX :X× Y → X and πY : X× Y → Y are continuous.

Proof. We leave this as an exercise. �

Example (Two topologies on R2). We have considered (R2,T) with T the standard topol-ogy on R2. For a second topology on R2, take (R,TR) with TR the standard topology onR. We can then consider the product topology on R × R = R2, giving (R2,TR×R). As anexercise, show that T = TR×R. �

Proposition 11. Let (X,TX), (Y,TY) and (Z,TZ) be topological spaces. Consider (X × Y,TX×Y),where TX×Y is the product topology. If f1 : Z→ X and f2 : Z→ Y are continuous functions, thenf : Z→ X× Y, f(z) = (f1(z), f2(z)) is continuous.

Proof. LetW ⊆ X× Y be open. For each z ∈ f−1(W), we will construct an open set Oz ⊆ Zsuch that z ∈ Oz and Oz ⊆ f−1(W). We will then have

f−1(W) =⋃

z∈f−1(W)

Oz,

implying that f−1(W) is open (and hence that f is continuous).Suppose that z ∈ f−1(W). It follows that f(z) ∈ W. Recalling our basis B of TX×Y , there

exist open sets U and V in X and Y, respectively, such that f(z) ∈ U× V and U× V ⊆ W.Note that

z ∈ f−1(U× V) = {w ∈ Z : f(w) ∈ U× V}

= {w ∈ Z : (f1(w), f2(w)) ∈ U× V}

= {w ∈ Z : f1(w) ∈ U and f2(w) ∈ V}

{w ∈ Z : w ∈ f−11 (U) and w ∈ f−12 (V)}

= f−11 (U) ∩ f−12 (V).

The continuity of f1 and f2 implies that each of f−11 (U) and f−12 (V) is open, so that f−11 (U)∩f−12 (V) = f−1(U × V) is open. Also, since U × V ⊆ W, we have f−1(U × V) ⊆ f−1(W)(exercise). Hence, Oz := f−1(U × V) satisfies each of the properties we mentioned at thebeginning of the proof. By earlier remarks, this completes the proof. �

7. THE QUOTIENT TOPOLOGY

7.1. Introduction. We begin by giving a brief overview of our objectives. Let (X,T) be atopological space and let ∼ be an equivalence relation on X (to be defined later). We willfirst construct a new set X/∼, called the quotient of X by ∼. We will then use T to producea topology TX/∼ on X/ ∼, called the quotient topology.

15

7.2. Equivalence Relations.

Definition 18. Let X be a set. An equivalence relation on X is a formal symbol x ∼ y, appliedto certain x, y ∈ X, such that the following conditions are satisfied.

(i) For all x ∈ X, x ∼ x.(ii) If x ∼ y, then y ∼ x.

(iii) If x ∼ y and y ∼ z, then x ∼ z.

Example. Suppose that X = Z and fix a positive integer n. For x, y ∈ Z, define x ∼ y if ndivides y− x. We claim that ∼ is an equivalence relation on Z.

For (i), suppose that x ∈ Z. Since n divides x− x = 0, it follows that x ∼ x.For (ii), suppose that x, y ∈ Z and x ∼ y. It follows that y − x = kn for some integer k.

Note that x− y = (−k)n, so that n also divides x− y. We conclude that y ∼ x.For (iii), suppose that x, y, z ∈ Z and that x ∼ y and y ∼ z. We have that y − x = kn for

some k ∈ Z and z − y = mn for some m ∈ Z. We thus have z − x = (z − y) + (y − x) =(m+ k)n, so that n divides z− x. Hence, x ∼ z.

We have shown ∼ to be an equivalence relation on Z. �

7.3. Equivalence Classes. Let X be a set and ∼ an equivalence relation on X.

Definition 19. If x ∈ X, we define the equivalence class of x, denoted [x], by

[x] := {y ∈ X : x ∼ y}.

Example. Recall the equivalence relation on Z given by x ∼ y if n divides y − x. If x ∈ Z,then

[x] = {y ∈ Z : n divides y− x}

= {y ∈ Z : y− x = kn for some k ∈ Z}= {y ∈ Z : y = x+ kn for some k ∈ Z}

= {x+ kn : k ∈ Z}.Let us set n = 3, so that our equivalence relation becomes x ∼ y if 3 divides y− x. Note

that[0] = {3k : k ∈ Z} = {. . . ,−6,−3, 0, 3, 6, . . .},

[1] = {1+ 3k : k ∈ Z} = {. . . ,−5,−2, 1, 4, 7, . . .}, and[2] = {2+ 3k : k ∈ Z} = {. . . ,−4,−1, 2, 5, 8, . . .}.

In particular, we see that Z = [0] ∪ [1] ∪ [2].Note also that

[3] = {3+ 3k : k ∈ Z} = {. . . ,−6,−3, 0, 3, 6, . . .} = [0],

[1] = [4], and[2] = [5].

Indeed, if x ∈ Z, then [x] = [0], [x] = [1], or [x] = [2] (exercise). Hence, [0], [1], and [2] arethe only equivalence classes. �

Certain features of the above example hold in a more general context. To this end, wehave the following proposition.

Proposition 12. Let X be a set and ∼ an equivalence relation on X.16

(i) X is a union of equivalence classes.(ii) If x, y ∈ X, then either [x] = [y] or [x] ∩ [y] = ∅.

(iii) If x, y ∈ X, then [x] = [y] ⇐⇒ x ∼ y.

Proof. We leave the proof as an exercise. �

7.4. The Set X/∼. Let X be a set and ∼ an equivalence relation on X. The equivalenceclasses in X become the elements of a new set, X/∼.

Definition 20. We define the quotient of X by ∼, written as X/∼, to be the set of all equiva-lence classes. In other words,

X/∼ = {[x] : x ∈ X}.

Example. Recall the equivalence relation on Z given by x ∼ y if 3 divides y − x. Asdiscussed earlier, [0], [1], and [2] constitute all of the equivalence classes. Therefore,

Z/∼ = {[0], [1], [2]}.

�

7.5. The Topology TX/∼. Suppose that (X,TX) is a topological space and that ∼ is an equiv-alence relation on X. One may consider the ”quotient map”

π : X→ X/∼, π(x) = [x].

We would like to construct TX/∼ such that π becomes a continuous function from (X,TX)to (X/∼,TX/∼). To this end, note that π will be continuous if and only if π−1(U) is an opensubset of X for all U ∈ TX/∼. We might therefore try the following:

TX/∼ := the collection of all subsets U of X/∼ such that π−1(U) is an open subset of X.

It turns out that this defines a topology on X/∼.

Proposition 13. (i) TX/∼ is a topology on X/∼.(ii) π : X→ X/∼ is a continuous function from (X,TX) to (X/∼,TX/∼).

Proof. We begin by proving (i). Note that π−1(∅) = ∅ and π−1(X/∼) = X. Since ∅ and X areopen subsets of X, it follows that ∅, X/∼ ∈ TX/∼.

Now, suppose that U1, U2 ∈ TX/∼. By definition, each of π−1(U1) and π−1(U2) is an opensubset of X, so that

π−1(U1 ∩U2) = π−1(U1) ∩ π−1(U2)

is also an open subset of X. It follows that U1 ∩U2 ∈ TX/∼.Finally, suppose that {Uα}α∈A is a family of sets in TX/∼. By definition, each π−1(Uα) is

open in X, so that

π−1

(⋃α∈A

Uα

)=⋃α∈A

π−1(Uα)

is also open in X. We conclude that⋃α∈AUα ∈ TX/∼.

We leave the proof of (ii) as an exercise. �

Definition 21. We call TX/∼ the quotient topology on X/∼.17

Example (Real Projective Space). Consider (Rn+1,TRn+1) with TRn+1 the standard topology.Since Rn+1 \ {0} ⊆ Rn+1, we may consider (Rn+1 \ {0},TRn+1\{0}) with TRn+1\{0} the subspacetopology.

For x, y ∈ Rn+1 \ {0}, define x ∼ y if y = λx for some λ ∈ R, λ 6= 0. It turns out that ∼is an equivalence relation on Rn+1 \ {0} (exercise). Taking a quotient by ∼, we get a newtopological space

((Rn+1 \ {0})/∼,T(Rn+1\{0})/∼).

We denote (Rn+1 \ {0})/∼ by RPn and call it real projective space. �

7.6. Constructing Continuous Functions on X/∼. Suppose that (X,TX) and (Y,TY) aretopological spaces and that ∼ is an equivalence relation on X. Of course, one may consider(X/∼,TX/∼) with TX/∼ the quotient topology. We would like to develop a concrete methodfor constructing continuous functions X/∼ → Y. This naturally leads to our consideringthe following special class of functions X→ Y (not assumed to be continuous).

Definition 22. A function f : X → Y is said to be invariant under ∼ if whenever x ∼ x ′, wehave f(x) = f(x ′).

Suppose that f is as in the above definition. We can define the new function

f : X/∼ → Y, f([x]) = f(x). (5)

However, this definition is potentially problematic. Indeed, if [x] = [x ′], then we needf(x) = f(x ′) to hold in order for (5) to give a well-defined function.

Suppose that [x] = [x ′]. It follows that x ∼ x ′, and our assumption on f then givesf(x) = f(x ′). So, f is indeed a well-defined function.

Proposition 14. Suppose that (X,TX) and (Y,TY) are topological spaces, and that ∼ is an equiv-alence relation on X. Consider (X/∼,TX/∼) with TX/∼ the quotient topology. If f : X → Y is acontinuous function which is invariant under ∼, then f : X/∼ → Y is also continuous.

Proof. Consider the quotient map π : X→ X/∼, π(x) = [x]. Note that

f(π(x)) = f([x]) = f(x)

for all x ∈ X, meaning that f ◦ π = f.Let us now show f to be continuous. Suppose that U ⊆ Y is open. Note that

π−1((f)−1(U)) = (f ◦ π)−1(U) = f−1(U),

which is an open subset of X. By the definition of the quotient topology, this meansprecisely that (f)−1(U) is an open subset of X/∼. Hence, f is continuous. �

Remark. In light of the above proposition, we have developed a method for constructingcontinuous functions X/∼ → Y. Start with a continuous function f : X → Y which isinvariant under ∼. Then take f : X/∼ → Y, a continuous function.

Example (A continuous function RP1 → R). We begin with (R2 \ {0},TR2\{0}) where TR2\{0}is the subspace topology coming from R2 with the standard topology. Recall that

RP1 = (R2 \ {0})/∼,18

where x ∼ y if y = λx for some λ ∈ R, λ 6= 0. We therefore have

(RP1,T(R2\{0})/∼)

with T(R2\{0})/∼ the quotient topology. Finally, we will also consider (R,TR) with TR thestandard topology.

Suppose we knew the function

f : R2 \ {0} → R, f(x1, x2) =x21

x21 + x22

to be continuous.1As per our method for constructing continuous functions on quotient spaces, we claim

that f is invariant under ∼. To this end, suppose that x, y ∈ R2\{0} and x ∼ y. By definition,y = λx for some λ ∈ R, λ 6= 0. Writing x = (x1, x2), we have

f(y) = f(λx) = f(λx1, λx2) =λ2x21

λ2x21 + λ2x22

=x21

x21 + x22

= f(x1, x2) = f(x).

Hence, f is invariant under ∼, and we can form the continuous function

f : RP1 → R, f([x]) = f(x).

More explicitly, our continuous function is

f : RP1 → R, f([(x1, x2)]) =x21

x21 + x22

.

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8. METRIC SPACES

8.1. A Definition. Having a notion of distance between elements of Rn fundamentallyallowed us to define the standard topology. We would therefore like to consider the mostgeneral framework in which a notion of distance gives rise to a topology, namely theframework of ”metric spaces”.

Informally, a metric space is a set, together with a notion of distance between elements.More precisely, we have the following definition.

Definition 23. A metric space consists of:• a set X, and• a function d : X× X→ R (called a metric),

such that the following conditions are satisfied.(i) ∀x, y ∈ X, d(x, y) ≥ 0 and d(x, y) = 0⇐⇒ x = y

(ii) ∀x, y ∈ X, d(x, y) = d(y, x)(iii) ∀x, y, z ∈ X, d(x, z) ≤ d(x, y) + d(y, z)

We denote this metric space by (X, d).

Example. Suppose that X = R and that dR : R×R → R is defined by dR(x, y) = |x− y|. Itturns out that (R, dR) is a metric space (exercise). �

1The proof that f is continuous is similar to certain proofs one does in Problem #4 from Assignment #3.19

Example. Suppose that X = Rn and dRn : Rn ×Rn → R is defined by dRn(x, y) = ‖x− y‖.It turns out that (Rn, dRn) is a metric space (exercise). �

8.2. Topology of Metric Spaces. Let (X, d) be a metric space.

Definition 24. If x ∈ X and δ > 0, define

Bδ(x) := {y ∈ X : d(x, y) < δ}.

Note that if (X, d) = (Rn, dRn) with dRn as in the above example, then this new defini-tion of Bδ(x) agrees with the old one.

To introduce a topology on X, we begin with a candidate for a basis. Specifically, wedefine B to be the collection of all sets of the form Bδ(x), where x ∈ X and δ > 0. Asdesired, B turns out to be a basis of a unique topology on X.

Proposition 15. The collection B is a basis of a unique topology, Td, on X.

Definition 25. We call Td the metric topology on X.

Hence, we have developed a standard method for converting a metric space (X, d) intoa topological space (X,Td). Let us now prove Proposition 15.

Proof. We must verify that B satisfies Conditions (i) and (ii) from the theorem in Problem#6 on Assignment #1, namely:

(i) If x ∈ X, then x ∈ Bδ(y) for some y ∈ X and δ > 0.(ii) If x ∈ Bδ1(y1) ∩ Bδ2(y2), then there exist y3 ∈ X and δ3 > 0 such that x ∈ Bδ3(y3)

and Bδ3(y3) ⊆ Bδ1(y1) ∩ Bδ2(y2).Verify (i): Note that x ∈ B1(x).Verify (ii): We first claim that Bδ1−d(x,y1)(x) ⊆ Bδ1(y1). To see this, suppose that z ∈Bδ1−d(x,y1)(x). Note that

d(z, y1) ≤ d(z, x) + d(x, y1) < (δ1 − d(x, y1)) + d(x, y1) = δ1,

which establishes that z ∈ Bδ1(y1). Therefore, Bδ1−d(x,y1)(x) ⊆ Bδ1(y1).Similarly, Bδ2−d(x,y2)(x) ⊆ Bδ2(y2). Now, set

δ := min{δ1 − d(x, y1), δ2 − d(x, y2)} > 0.

We then haveBδ(x) ⊆ Bδ1−d(x,y1)(y1) ⊆ Bδ1(y1)

andBδ(x) ⊆ Bδ2−d(x,y2)(y2) ⊆ Bδ2(y2).

Hence, Bδ(x) ⊆ Bδ1(y1) ∩ Bδ2(y2). We have thus verified (ii). �

8.3. Two Descriptions of Td. Recall that we defined the product topology to be the uniquetopology with a particular basis. Problem #6 from Assignment #1 then allowed us to givetwo equivalent descriptions of the sets in the product topology. Since Td is the uniquetopology with basis B as above, we may analogously provide two equivalent descrip-tions of the sets in Td (ie. the open sets for the metric topology).

Let (X, d) be a metric space and consider the topological space (X,Td).20

1. If U ⊆ X, then U is open if and only if there exists a family {Bδα(xα)}α∈A, with eachxα ∈ X and each δα > 0, such that

U =⋃α∈A

Bδα(xα).

2. If U ⊆ X, then

U is open ⇐⇒ ∀x ∈ U ∃y ∈ X and δ > 0 such that x ∈ Bδ(y) and Bδ(y) ⊆ U⇐⇒ ∀x ∈ U ∃δ > 0 such that Bδ(x) ⊆ UWe leave the proof of equivalence of the last two statements as an exercise.

Example. Consider the metric space (Rn, dRn), where dRn(x, y) = ‖x−y‖. IfU ⊆ Rn, thenU ∈ TdRn if and only if ∀x ∈ U ∃δ > 0 such that Bδ(x) ⊆ U. Noting that Bδ(x) agrees withthe old definition of a δ-ball in Rn, this is equivalent to the condition thatU belongs to thestandard topology on Rn. We conclude that TdRn equals the standard topology on Rn. �

8.4. A Useful Property.

Proposition 16. If (X, d) is a metric space, then (X,Td) is Hausdorff.

Proof. Suppose that x, y ∈ X and x 6= y. Note that d(x, y) > 0, and that we have x ∈B 12d(x,y)(x) and y ∈ B 1

2d(x,y)(y). Since each ball belongs to a basis of Td, each ball is in Td

(ie. is open). Hence, it just remains to prove that B 12d(x,y)(x) ∩ B 1

2d(x,y)(y) = ∅.

Assume that B 12d(x,y)(x) ∩ B 1

2d(x,y)(y) 6= ∅. There exists an element z ∈ B 1

2d(x,y)(x) ∩

B 12d(x,y)(y). Since z belongs to each ball, we have d(x, z) < 1

2d(x, y) and d(z, y) < 1

2d(x, y).

Hence,

d(x, y) ≤ d(x, z) + d(z, y) < 1

2d(x, y) +

1

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

a contradiction. It follows that B 12d(x,y)(x)∩B 1

2d(x,y)(y) = ∅, and we conclude that (X,Td) is

Hausdorff. �

8.5. Metrizable Spaces.

Example. Consider the topological space (R,T = {∅,R}). Does there exist a metric d :R× R → R such that T = Td? In other words, is T an example of a metric topology?

To answer this, note that (R,T = {∅,R}) is not Hausdorff (as was discussed in an earlierexample). However, Proposition 16 implies that (R,Td) is Hausdorff for every metricd : R× R → R. Therefore, there cannot exist a metric d such that T = Td. �

In general, it is reasonable to ask about whether a given topological space (X,T) is suchthat T = Td for some metric d on X. Those topological spaces for which the answer is ’yes’are called metrizable.

Definition 26. A topological space (X,T) is called metrizable if T = Td for some metricd : X× X→ R.

Example. In light of the above discussion (R,T = {∅,R}) is not metrizable. �

Example. Consider (Rn,TRn) with TRn the standard topology. Note that (Rn,TRn) ismetrizable since TRn was shown to equal TdRn , where dRn(x, y) = ‖x− y‖. �

21

8.6. Continuous Functions and Metric Spaces. Consider (R,TR) with TR the standardtopology, and let f : R → R be a function. Earlier, we mentioned that f is continuous inour topological sense if and only if it satisfies the usual ε − δ definition of continuity ateach point in its domain. More explicitly, we have:

f is continuous ⇐⇒ At each point x ∈ R :

∀ε > 0 ∃δ > 0 such that if |x ′ − x| < δ, then |f(x ′) − f(x)| < ε

Since the ε − δ definition of continuity is fundamentally a statement about distancesbetween points, we might expect a similar equivalence to hold for continuous functionsbetween metric spaces. Indeed, we have the following analogue of the above equivalence.

Theorem 6. Let (X, dX) and (Y, dY) be metric spaces and consider (X,TdX) and (Y,TdY). If f :X→ Y is a function, then

f is continuous ⇐⇒ At each point x ∈ X :

∀ε > 0 ∃δ > 0 such that if dX(x ′, x) < δ, then dY(f(x ′), f(x)) < ε.

Proof. (=⇒) Suppose that x ∈ X and ε > 0. Note that f(x) ∈ Bε(f(x)), an open subsetof Y for the topology TdY . Hence, x ∈ f−1(Bε(f(x))), an open subset of X. Recalling ourdescriptions of TdX , there exists δ > 0 such that Bδ(x) ⊆ f−1(Bε(f(x))).

We claim that if dX(x ′, x) < δ, then dY(f(x ′), f(x)) < ε. Indeed,

dX(x′, x) < δ =⇒ x ′ ∈ Bδ(x)

=⇒ x ′ ∈ f−1(Bε(f(x)))=⇒ f(x ′) ∈ Bε(f(x))=⇒ dY(f(x

′), f(x)) < ε,

as desired.(⇐=) We leave the proof as an exercise. �

9. CONNECTEDNESS

9.1. Definition and Examples. We would like to generalize the Intermediate Value The-orem (IVT) to create a theorem in topology. To this end, we give the following two equiv-alent versions of the IVT.

Theorem 7 (IVT). Consider (R,TR) with TR the standard topology. Also, consider ([a, b],T[a,b]),where T[a,b] is the subspace topology. Let f : [a, b] → R be a continuous function.

(V1) If d is a real number between f(a) and f(b) (or d equals one of f(a), f(b)), then thereexists c ∈ [a, b] such that f(c) = d.

(V2) f([a, b]) = [a ′, b ′] for some a ′, b ′ ∈ R with a ′ ≤ b ′.

Exercise. The two versions (V1) and (V2) are equivalent.

Examining (V2), note that [a, b] and [a ′, b ′] have no ”gaps”, or are ”connected” in acertain sense. This informal notion is captured by the definition of a connected topologicalspace.

Definition 27. Let (X,T) be a topological space.22

(i) A splitting of (X,T) consists of open setsU1, U2 such that X = U1∪U2 andU1∩U2 =∅.

(ii) We say that (X,T) is connected if wheneverU1, U2 form a splitting of (X,T), we haveU1 = ∅ or U2 = ∅(ie. U1 = ∅, U2 = X and U1 = X,U2 = ∅ are the only splittings of X).

Our first example of a connected topological space will be (R,T) with T the standardtopology. Establishing this as an example is surprisingly difficult, and will be done in thefollowing proposition.

Proposition 17. If T is the standard topology on R, then (R,T) is connected.

Proof. Suppose that U1 and U2 are open subsets of R which form a splitting. We want toshow that U1 = ∅ or U2 = ∅. Assume that this is false, namely that U1 6= ∅ and U2 6= ∅.Hence, there exist elements a ∈ U1 and b ∈ U2. Without the loss of generality, a < b.

Now, set V1 := [a, b] ∩ U1. Note that a ∈ V1 and that for all x ∈ V1, x ≤ b. Hence, V1 isnon-empty and bounded above. We may therefore consider

c := sup(V1),

the least upper bound of V1. Since c is an upper bound for V1 and a ∈ V1, we have a ≤ c.Also, since b is an upper bound for V1, and since c is the least upper bound, we havec ≤ b. It follows that

c ∈ [a, b].

Noting that R = U1 ∪ U2 (as U1, U2 were assumed to form a splitting), it must be thatc ∈ U1 or c ∈ U2. Let us consider each of these cases.

Case 1: c ∈ U1Since c ∈ U1, b ∈ U2 and U1 ∩ U2 = ∅, it follows that c 6= b. Noting also that c ∈ [a, b],

there must exist δ > 0 such that

[c, c+ δ] ⊆ [a, b]. (6)

Furthermore, since U1 is open, there exists δ ′ > 0 such that (c − δ ′, c + δ ′) ⊆ U1. Inparticular,

[c, c+δ ′

2] ⊆ U1 (7)

Setting δ ′′ = min{δ, δ′

2} > 0, (6) and (7) imply that

c+ δ ′′ ∈ [a, b] ∩U1 = V1.So, c+ δ ′′ ∈ V1, contradicting the fact that c is an upper bound for V1.

Case 2: c ∈ U2Since U2 is open, there exists δ > 0 such that (c − δ, c + δ) ⊆ U2. We then note that

x ≤ c − δ for all x ∈ V1 (exercise). In other words, c − δ is an upper bound for V1. Thiscontradicts the fact that c is the least upper bound for V1.

Note that we have obtained a contradiction in each of Cases 1 and 2. Hence, there isan overall contradiction, and we conclude that U1 = ∅ or U2 = ∅. It follows that (R,T) isconnected. �

23

Example (Intervals are connected.). Consider (R,T) with T the standard topology. Sup-pose that I ⊆ R is an interval, meaning that I is one of (a, b), [a, b), (a, b], [a, b], (−∞, a),(−∞, a], (a,∞), [a,∞). If TI is the subspace topology on I, then (I,TI) is connected (exer-cise). �

Example. If T is the standard topology on Rn, then (Rn,T) is connected (exercise). �

Example. Consider (R,T) with T the standard topology. Also, consider the subset A =[−1, 1] ∪ [5, 8] and let TA denote the subspace topology on A. We claim that (A,TA) is notconnected. To see this, note that

[−1, 1] = (−2, 2) ∩A.Since (−2, 2) is an open subset of R, it follows that [−1, 1] is an open subset ofA. Similarly,

[5, 8] = (4, 9) ∩Ais also an open subset ofA. So, [−1, 1] and [5, 8] form a splitting ofA into non-empty opensubsets. We conclude that (A,TA) is not connected. �

9.2. An Equivalent Description of Connectedness.

Proposition 18. A topological space (X,T) is connected if and only if the only subsets of X whichare both open and closed are ∅ and X.

Proof. We leave the proof as an exercise. �

9.3. Properties of Connectedness.

Theorem 8. Let (X,TX) be a connected topological space and let (Y,TY) be a topological space.Suppose that f : X→ Y is a continuous function, and consider (f(X),Tf(X)) with Tf(X) the subspacetopology on f(X).

The topological space (f(X),Tf(X)) is connected.

Proof. Suppose that U1 and U2 are open subsets of f(X) which form a splitting. We maywrite U1 = V1 ∩ f(X) and U2 = V2 ∩ f(X), where V1 and V2 are open subsets of Y.

Now, note that

f−1(U1) = f−1(V1 ∩ f(X))

= f−1(V1) ∩ f−1(f(X))= f−1(V1) ∩ X= f−1(V1).

Since f is continuous and V1 is an open subset of Y, we have that f−1(U1) = f−1(V1) isan open subset of X. Analogously, f−1(U2) is also an open subset of X.

We claim that the open sets f−1(U1), f−1(U2) form a splitting of (X,T). To see this, notethat

f−1(U1) ∪ f−1(U2) = f−1(U1 ∪U2) = f−1(f(X)) = Xand

f−1(U1) ∩ f−1(U2) = f−1(U1 ∩U2) = f−1(∅) = ∅.24

Hence, f−1(U1), f−1(U2) indeed form a splitting of (X,T). Since (X,T) is connected, wemust have f−1(U1) = ∅ or f−1(U2) = ∅. In our case, these facts imply that U1 = ∅ or U2 = ∅(exercise). We conclude that (f(X),Tf(X)) is connected. �

Example. Consider X := {(0, 2x, x) : x ∈ R} and let TX denote the subspace topology onX coming from the standard topology on R3. We claim that (X,TX) is connected. To thisend, consider the function

f : R → R3, f(x) = (0, 2x, x).

Since f is a linear transformation, f is a continuous function from (R,TR) to (R3,TR3) whereTR and TR3 are the standard topologies. By the theorem above, (f(R),Tf(R)) is connected.Noting that f(R) = X, it follows that (X,TX) is connected. �

We now consider a few nice corollaries of our most recent theorem.

Corollary 1. Let (X,TX) be a connected topological space, and let (Y,TY) be a topological space.If there exists a continuous surjective function f : X→ Y, then (Y,TY) is connected.

Proof. By the theorem above, (f(X),Tf(X)) is connected. Since f is surjective, f(X) = Y, sothat (Y,TY) is connected. �

Corollary 2. Let (X,TX) be a connected topological space, and let ∼ be an equivalence relation onX. The topological space (X/∼,TX/∼) is connected, where TX/∼ is the quotient topology.

Proof. The quotient map π : X → X/∼ is continuous and surjective. Since (X,TX) is con-nected, our first corollary implies that (X/∼,TX/∼) is connected. �

Corollary 3 (Intermediate Value Theorem, V1). Consider (R,TR) with TR the standard topol-ogy. Also, consider ([a, b],T[a,b]), where T[a,b] is the subspace topology. Let f : [a, b] → R be acontinuous function.

If d is a real number between f(a) and f(b) (or d equals one of f(a), f(b)), then there existsc ∈ [a, b] such that f(c) = d.

Proof. We will provide a proof in the case f(a) < f(b). The case f(a) > f(b) is handledsimilarly, while the case f(a) = f(b) is extremely brief.

Assume that no such c exists. It follows that d 6∈ f([a, b]). Hence, each element off([a, b]) is in (−∞, d) or (d,∞), meaning that

f([a, b]) = ((−∞, d) ∩ f([a, b])) ∪ ((d,∞) ∩ f([a, b])). (8)

Since (−∞, d) and (d,∞) are open subsets of R, each of the sets (−∞, d) ∩ f([a, b]) and(d,∞) ∩ f([a, b]) is open in f([a, b]). Hence, (8) gives a splitting of f([a, b]).

Now, since d 6∈ f([a, b]), we have f(a) 6= d and f(b) 6= d. Hence, f(a) < d < f(b).It follows that f(a) ∈ (−∞, d) ∩ f([a, b]) and f(b) ∈ (d,∞) ∩ f([a, b]). In particular,(−∞, d) ∩ f([a, b]) 6= ∅ and (d,∞) ∩ f([a, b]) 6= ∅. In other words, each set in the splitting(8) is non-empty, implying that f([a, b]) is not connected. However, f([a, b]) is connectedsince [a, b] is connected and f is continuous. We have therefore obtained a contradiction,completing the proof. �

Theorem 9. If (X,TX) and (Y,TY) are connected topological spaces, then (X × Y,TX×Y) is con-nected, where TX×Y is the product topology.

Proof. We leave this as an exercise. The proof will be given during a tutorial. �25

9.4. Path Connectedness.

Definition 28. Let (X,TX) be a topological space, and consider ([0, 1],T[0,1]) with T[0,1] thesubspace topology coming from the standard topology on R.

(i) A path in X is a continuous function ϕ : [0, 1] → X.(ii) We say that (X,T) is path connected if for all x, y ∈ X, there exists a path in X,

ϕ : [0, 1] → X, such that ϕ(0) = x and ϕ(1) = y.

Example. If T is the standard topology on Rn, then (Rn,T) is path connected. To see this,suppose that x, y ∈ Rn. Consider the function

ϕ : [0, 1] → Rn, ϕ(t) = (1− t)x+ ty.

Note thatϕ(0) = x andϕ(1) = y. Furthermore,ϕ is a continuous function from ([0, 1],T[0,1])to (Rn,T) (exercise). We conclude that (Rn,T) is path connected. �

Example. Consider (Rn \ {0},TRn\{0}) where TRn\{0} is the subspace topology coming fromthe standard topology on Rn. If n ≥ 2, then (Rn \ {0},TRn\{0}) is path connected (exercise).�

Theorem 10. If (X,T) is a path connected topological space, then (X,T) is connected.

Proof. Assume that (X,T) is not connected. There exists a splitting U1, U2 of X such thatU1 6= ∅ and U2 6= ∅. Hence, there exist x ∈ U1 and y ∈ U2. Since (X,T) is path connected,there exists a path in X, ϕ : [0, 1] → X, with ϕ(0) = x and ϕ(1) = y.

Now, note thatϕ([0, 1]) = (U1 ∩ϕ([0, 1])) ∪ (U2 ∩ϕ([0, 1])) (9)

Since U1, U2 are open subsets of X, the sets U1 ∩ ϕ([0, 1]) and U2 ∩ ϕ([0, 1]) are open inϕ([0, 1]). Therefore, (9) gives a splitting of ϕ([0, 1]). Noting that ϕ([0, 1]) is connected,we have that U1 ∩ ϕ([0, 1]) = ∅ or U2 ∩ ϕ([0, 1]) = ∅. In the case U1 ∩ ϕ([0, 1]) = ∅, wenote that x ∈ U1 ∩ ϕ([0, 1]) (a contradiction). In the case U2 ∩ ϕ([0, 1]) = ∅, we note thaty ∈ U2 ∩ ϕ([0, 1]) (a contradiction). Hence, we have an overall contradiction, meaningthat (X,T) is connected. �

This result has a number of interesting implications for examples of connected topolog-ical spaces.

Example. The following topological spaces are connected:(i) (Rn \ {0},TRn\{0}) for n ≥ 2, where TRn\{0} is the subspace topology coming from the

standard topology on Rn(ii) (Sn,TSn), where TSn is the subspace topology coming from the standard topology

on Rn+1(iii) (RPn,TRPn), where TRPn is the usual quotient topology

Proof. For (i), we use the fact that (Rn \ {0},TRn\{0}) is path connected for n ≥ 2 to concludeit is also connected.

For (ii), consider the function

f : Rn+1 \ {0} → Sn, f(x) =x

‖x‖.

26

Note that f is continuous (see Assignment #3) and surjective. Also, since n + 1 ≥ 2,(Rn+1 \ {0},TRn+1\{0}) is connected. We conclude that (Sn,TSn) is connected.

For (iii), let ∼ denote the equivalence relation on Rn+1 \ {0} used to construct RPn. Sincen+1 ≥ 2, we know that (Rn+1\{0},TRn+1\{0}) is connected. Noting that RPn = (Rn+1\{0})/∼,it follows that (RPn,TRPn) is connected. �

9.5. Connectedness and Homeomorphisms.

Theorem 11. Suppose that (X,TX) and (Y,TY) are homeomorphic topological spaces (ie. that thereexists a homeomorphism X→ Y). Then, (X,TX) is connected if and only if (Y,TY) is connected.

Proof. We leave this as an exercise. �

In certain cases, we can use this theorem to determine that two topological spaces arenot homeomorphic.

Example. Consider (R,TR) with TR the standard topology. Let A = [−1, 1] ∪ [5, 8] andB = [−3, 3], and let TA and TB be their respective subspace topologies. Note that (A,TA)is not connected, while (B,TB) is connected. By the theorem above, (A,TA) and (B,TB)cannot be homeomorphic. �

10. COMPACTNESS

10.1. Definition and Examples. As with the topic of connectedness, our motivation forstudying compactness will be to generalize a particularly nice theorem from analysis.This theorem is as follows.

Theorem 12. Consider (R,TR) with TR the standard topology. Suppose that A is a closed andbounded subset of R, and consider (A,TA) with TA the subspace topology.

If f : A→ R is continuous, then f(A) is also a closed and bounded subset of R.

Our initial approach to generalizing this theorem will be to define a type of topologicalspace of which (A,TA) is an example. However, before we give this definition, we willneed to discuss the concept of an ”open cover”.

Definition 29. Let (X,T) be a topological space. An open cover of X is a family {Uα}α∈A ofopen sets such that

X =⋃α∈A

Uα.

Example. Consider (R,T) with T the standard topology.(i) {R \ {a}}a∈R is an open cover of R.

(ii) {(−a, a)}a∈(0,∞) is an open cover of R.�

Definition 30. A topological space (X,T) is called compact if every open cover {Uα}α∈A ofX has a finite subcover: ie. there exist finitely many Uα1 , Uα2 , . . . , Uαn such that

X =

n⋃i=1

Uαi .

27

Example. Let us return to our most recent example of (R,T) with T the standard topology.(i) The open cover {R \ {a}}a∈R has a finite subcover since R = (R \ {0}) ∪ (R \ {1}).

(ii) We claim that the open cover {(−a, a)}a∈(0,∞) does not have a finite subcover. Tosee this, suppose that there existed a finite subcover. It would follow that

R =

n⋃i=1

(−ai, ai)

for some finitely many (−a1, a1), (−a2, a2), . . . , (−an, an). Now, consider x = maxni=1{ai}.Since x ∈ R =

⋃ni=1(−ai, ai), we see that x ∈ (−ai, ai) for some i. Hence, x < ai,

contradicting the fact that x ≥ ai.

Since we have found an open cover of R that has no finite subcover, we conclude that(R,T) is not compact. �

Our first example of a compact topological space will be ([a, b],T[a,b]) with T[a,b] thesubspace topology coming from the standard topology on R. Establishing that this is asan example is quite difficult, and will be done in the following proposition.

Proposition 19. If a, b ∈ R and a < b, consider ([a, b],T[a,b]) with T[a,b] the subspace topologycoming from the standard topology on R.

The topological space ([a, b],T[a,b]) is compact.

Proof. Suppose that {Uα}α∈A is an open cover of [a, b]. We want to show that this opencover has a finite subcover.

Consider the set

C := {x ∈ [a, b] : a < x and there exist Uα1 , Uα2 , . . . , Uαn such that [a, x] ⊆n⋃i=1

Uαi}.

Note that if b ∈ C, then we are done.

Claim #1: C 6= ∅

Proof. Since [a, b] =⋃α∈AUα, we know that a ∈ Uβ for some β ∈ A. Noting that Uβ is an

open subset of [a, b], we may write Uβ = V ∩ [a, b] for some open subset V of R. Sincea ∈ V , there exists δ > 0 such that (a − δ, a + δ) ⊆ V . This allows us to find δ ′ > 0 suchthat

[a, a+ δ ′] ⊆ V ∩ [a, b] = Uβ.

So, [a, a + δ ′] is a subset of a union of finitely many (in fact, one) Uα. We conclude thata+ δ ′ ∈ C, implying that C 6= ∅. �

Since C ⊆ [a, b], we know that C is bounded above. Noting that C 6= ∅, we mayconsider

c := sup(C).Let us make a few observations about c:

(i) Since c is an upper bound for C and a+ δ ′ ∈ C, we see that a+ δ ′ ≤ c.(ii) Since b is an upper bound for c, we know that c ≤ b.

28

By (i) and (ii), c ∈ (a, b].

Claim #2: c ∈ CProof. Since c ∈ [a, b], we know that c ∈ Uγ for some γ ∈ A. Using an argument similarto one offered in the proof of Claim #1, there exists δ ′′ > 0 such that

[c− δ ′′, c] ⊆ Uγ.

Sub-claim: There exists d ∈ C such that d ∈ (c− δ ′′, c).

Proof. If not, then c − δ ′′ is an upper bound for C (exercise). This contradicts the fact thatc is the least upper bound. �

Let d be as in the sub-claim. Since d ∈ C, we have

[a, d] ⊆n⋃i=1

Uαi

for some Uα1 , Uα2 , . . . , Uαn . Furthermore,

[a, c] = [a, d] ∪ [c− δ ′′, c]

⊆

(n⋃i=1

Uαi

)∪Uγ.

Hence, [a, c] belongs to the union of finitely many Uα, implying that c ∈ C.

�

Claim #3: c = b

Proof. Assume that c 6= b. Since c ∈ [a, b], it follows that c < b. Also, since c ∈ [a, b], wehave c ∈ Uω for some ω ∈ A. These two observations allow us to find δ ′′′ > 0 such that[c, c+ δ ′′′] ⊆ Uω (exercise). Now, note that

[a, c+ δ ′′′] = [a, c] ∪ [c, c+ δ ′′′]. (10)

Since c ∈ C, [a, c] belongs to a union of finitely many Uα. Also, as noted before, [c, c +δ ′′′] ⊆ Uω. Hence, (10) implies that [a, c + δ ′′′] is a union of finitely many Uα. It followsthat c+ δ ′′′ ∈ C, contradicting the fact that c is an upper bound for C. �

Since c ∈ C (by Claim #2), Claim #3 implies that b ∈ C. As noted earlier, this factcompletes the proof. �

Theorem 13. Let (X,TX) be a Hausdorff topological space, and letA be a subset ofXwith subspacetopology TA. If (A,TA) is compact, then A is a closed subset of X.

Proof. We want to show that X \ A is open. To this end, suppose that x ∈ X \ A. We willfind an open subset Ux of X such that x ∈ Ux and Ux ⊆ X \ A. Taking a union of all suchopen sets Ux, it will follow that X \A is open.

Suppose that x ∈ X \ A. Then for all a ∈ A, we have x 6= a. Since (X,TX) is Hausdorff,there exist open subsets Ua and Va of X such that a ∈ Ua, x ∈ Va and Ua ∩ Va = ∅. Now,note that

29

(i) for all a ∈ A, Ua ∩A is open in A, and(ii) for all a ∈ A, a ∈ Ua ∩A.

These two facts imply that {Ua ∩ A}a∈A is an open cover of A. Since (A,TA) is compact,there exist Ua1 ∩A,Ua2 ∩A, . . . , Uan ∩A such that

A =

n⋃i=1

(Uai ∩A).

We now make the following two observations.1.⋂ni=1 Vai is an open subset of X containing x

2.⋂ni=1 Vai ⊆ X \A.

Setting Ux =⋂ni=1 Vai , we have satisfied the objectives stated at the beginning of the

proof. �

10.2. Compact Subsets of R and Rn.

Theorem 14. Consider (R,TR) with TR the standard topology. Let A be a subset of R withsubspace topology TA.

Then, (A,TA) is compact if and only if A is closed and bounded.

Proof. (=⇒) Since (R,TR) is Hausdorff, the previous theorem implies that A is closed.To show that A is bounded, consider the open cover {(−b, b) ∩ A}b∈(0,∞) of A. Since

(A,TA) is compact, there exist (−b1, b1) ∩A, (−b2, b2) ∩A, . . . , (−bn, bn) ∩A such that

A =

n⋃i=1

((−bi, bi) ∩A).

SettingM := maxni=1{bi}, we see that |x| ≤M for all x ∈ A. Hence, A is bounded.(⇐=) We leave this as an exercise. The proof will be given in a tutorial. �

We have a completely analogous result for subsets of Rn. The only notable differenceis that boundedness in Rn is defined with respect to the usual norm, rather than absolutevalue.

Theorem 15. Consider (Rn,TRn) with TRn the standard topology. Let A be a subset of Rn withsubspace topology TA.

Then, (A,TA) is compact if and only if A is closed and bounded.

Example. The topological space (Sn,TSn) is compact, where TSn is the subspace topologyon Sn coming from the standard topology on Rn+1.

Proof. For all x ∈ Sn, ‖x‖ = 1. Hence, Sn is a bounded subset of Rn+1. Also, Sn is a closedsubset of Rn+1 (exercise). By the previous theorem, (Sn,TSn) is compact. �

10.3. More Properties of Compactness.

Proposition 20. Let (X,TX) be a compact topological space. Also, let A be a subset of X withsubspace topology TA. If A is closed, then (A,TA) is compact.

Proof. We leave this as an exercise. The proof will be give during a tutorial. �30

Theorem 16. Let (X,TX) be a compact topological space, and (Y,TY) a topological space. If f :X→ Y is a continuous function, then (f(X),Tf(X)) is compact, where Tf(X) is the subspace topologyon f(X) coming from the topology on Y.

Proof. Suppose that {Uα}α∈A is an open cover of f(X). Since each Uα is open in f(X), wemay write

Uα = Vα ∩ Xwith Vα an open subset of Y. Hence,

f−1(Uα) = f−1(Vα ∩ f(X))

= f−1(Vα) ∩ f−1(f(X))= f−1(Vα) ∩ X= f−1(Vα),

an open subset of X. Also,⋃α∈A

f−1(Uα) = f−1

(⋃α∈A

Uα

)= f−1(f(X)) = X.

Hence, {f−1(Uα)}α∈A is an open cover of X. Since (X,TX) is compact, there existf−1(Uα1), f

−1(Uα2), . . . , f−1(Uαn) such that

X =

n⋃i=1

f−1(Uαi).

With this fact, it follows (exercise) that

f(X) =

n⋃i=1

Uαi .

We conclude that (f(X),Tf(X)) is compact. �

Example. Consider X := {(x, 2x) ∈ R2 : x ∈ [−1, 5]}, and let TX be the subspace topologyon X coming from the standard topology on R2. We claim that (X,TX) is compact.

Note that f : [−1, 5] → R2, f(x) = (x, 2x) is a continuous function from ([−1, 5],T[−1,5])to (R2,TR2) (exercise), where T[−1,5] is the subspace topology coming from the standardtopology on R and TR2 is the standard topology. Also, ([−1, 5],T[−1,5]) is compact. By thetheorem above, (f([−1, 5]),Tf([−1,5])) is compact. However, f([−1, 5]) = X, so that (X,TX) iscompact. �

Corollary 4. Let (X,TX) be a compact topological space, and let (Y,TY) be a topological space. Ifthere exists a surjective continuous function f : X→ Y, then (Y,TY) is compact.

Proof. We know that (f(X),Tf(X)) is compact. The surjectivity of f implies that f(X) = Y, sothat (Y,TY) is compact. �

Corollary 5. Let (X,TX) be a compact topological space, and let ∼ be an equivalence relation onX. The topological space (X/∼,TX/∼) is compact, where TX/∼ is the quotient topology.

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Proof. Note that the quotient map π : X → X/∼ is surjective and continuous. By theprevious corollary, (X/∼,TX/∼) is compact. �

Having developed some machinery, we may now prove the theorem used to motivateour study of compactness in the first place.

Corollary 6. Consider (R,TR) with TR the standard topology. Suppose that A is a closed andbounded subset of R, and consider (A,TA) with TA the subspace topology.

If f : A→ R is continuous, then f(A) is also a closed and bounded subset of R.

Proof. SinceA is a closed and bounded subset of R, (A,TA) is compact. Hence, (f(A),Tf(A))is compact. Noting that f(A) ⊆ R, this implies that f(A) is closed and bounded. �

10.4. Homeomorphisms and Compactness.

Theorem 17. Suppose that (X,TX) and (Y,TY) are homeomorphic topological spaces. Then,(X,TX) is compact if and only if (Y,TY) is compact.

Proof. We leave this as an exercise. �

Example. Consider (R,TR) and (S1,TS1), where TR is the standard topology and TS1 isthe subspace topology coming from the standard topology on R2. Since (R,TR) is notcompact while (S1,TS1) is compact, these topological spaces cannot be homeomorphic. �

Example. We claim that (RPn,TRPn) is compact, where TRPn is the usual quotient topol-ogy. To see this, consider (Sn,TSn) with TSn the subspace topology coming from the stan-dard topology on Rn+1. Recall from Assignment #3 that (RPn,TRPn) is homeomorphic to(Sn/∼,TSn/∼), where x ∼ y if y = x or y = −x.

Since (Sn,TSn) is compact, it follows that (Sn/∼,TSn/∼) is also compact. Since the secondspace is homeomorphic to (RPn,TRPn), we conclude that (RPn,TRPn) is compact. �

Let us now discuss an important point encountered on Assignment #4. Consider ([0, 2π),T[0,2π)),and (S1,TS1), where T[0,2π) is the subspace topology coming from the standard topologyon R and TS1 is the subspace topology coming from the standard topology on R2. OnAssignment #4, it was remarked that

f : [0, 2π) → S1, f(x) = (cos(x), sin(x))

is a continuous bijection, but not a homeomorphism. Hence, one generally cannot provesomething to be a homeomorphism by showing it to be continuous and bijective. Usually,one must construct a continuous inverse. However, there are ”nice” assumptions underwhich a continuous bijection must be a homeomorphism.

Theorem 18. Let (X,TX) be a compact topological space, and let (Y,TY) be a Hausdorff topologicalspace. If f : X→ Y is a continuous bijection, then f is a homeomorphism.

Proof. Since f is a bijection, f−1 : Y → X exists. We want to show that f−1 is continuous. Todo this, we will show that the pre-image of a closed set under f−1 is a closed set.

Suppose that V ⊆ X is a closed set. Note that

(f−1)−1(V) = f(V)32

(exercise), where the inner ”-1” refers to the inverse, and the outer ”-1” refers to the pre-image. Hence, we must prove that f(V) is a closed subset of Y.

Since V is a closed subset of X and (X,TX) is compact, it follows that (V,TV) is compact.Furthermore, f|V : V → Y is continuous, meaning that (f(V),Tf(V)) is compact. Noting that(Y,TY) is Hausdorff, this implies that f(V) is a closed subset of Y. �

Example. Consider (R,TR) and (S1,TS1), where TR is the standard topology and TS1 is thesubspace topology coming from the standard topology on R2. In this example, we willshow that S1 is homeomorphic to a quotient of R.

The functionf : R → S1, f(x) = (cos(x), sin(x))

is continuous (see Assignment #4). Now, define an equivalence relation ∼ on R by x ∼ y ify− x = 2nπ for some n ∈ Z. Note that f is invariant under ∼ (exercise). We may thereforeconsider the continuous function

f : R/∼ → S1, f([x]) = f(x) = (cos(x), sin(x))

from (R/∼,TR/∼) to (S1,TS1). We claim that:

(i) f is a continuous bijection (exercise),(ii) (S1,TS1) is Hausdorff (see Assignment #5), and

(iii) (R/∼,TR/∼) is compact (exercise).

Hint: For (iii), find a continuous surjective function g : [0, 2π] → R/∼.

By the previous theorem, (i), (ii), and (iii) imply that f : R/∼ → S1 is a homeomorphism.�

10.5. Products and Compactness.

Theorem 19. If (X,TX) and (Y,TY) are compact topological spaces, then (X×Y,TX×Y) is compact,where TX×Y is the product topology.

Proof. Please refer to Optional Challenge Problem #6. �

11. TOPOLOGICAL MANIFOLDS

11.1. A Motivating Example. Consider (S1,TS1), where TS1 is the subspace topology com-ing from the standard topology on R2. Suppose that y0 ∈ S1 and that y0 = (cos(x0), sin(x0)).One can show that

f : (x0, x0 + 2π) → S1 \ {y0}, f(x) = (cos(x), sin(x))

is a homeomorphism from ((x0, x0 + 2π),T(x0,x0+2π)) to (S1 \ {y0},TS1\{y0}).

The main point is that, locally, S1 ”looks like” an open subset of R. With this in mind,we want to define a topological n-manifold to be a topological space which locally ”lookslike” an open subset of Rn. Let us attempt to be more precise about this.

11.2. Definition and Examples.

Definition 31. A topological space (X,T) is called a topological n-manifold if(i) (X,T) is Hausdorff, and

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(ii) for all x ∈ X, there exist an open subset U of X containing x, an open subset V ofRn, and a homeomorphism ϕ : U → V from (U,TU) to (V,TV), where TU and TVare the subspace topologies.

It is Condition (ii) which is meant to capture the idea that X locally looks like an opensubset of Rn.

Example. If T is the standard topology on Rn, then (Rn,T) is a topological n-manifold. Tosee this, recall that (Rn,T) is Hausdorff. For Condition (ii), suppose that x ∈ Rn. Considerthe identity function idRn : Rn → Rn, a homeomorphism. Setting U = Rn, V = Rn, andϕ = idRn , we see that Condition (ii) is satisfied. Hence (Rn,T) is a topological n-manifold.�

Example. If TS1 is the subspace topology coming from the standard topology on R2, then(S1,TS1) is a topological 1-manifold. For a proof, please see Assignment #5. �

Example. If TSn is the subspace topology coming from the standard topology on Rn+1,then (Sn,TSn) is a topological n-manifold. The proof is similar to the proof that (S1,TS1)is a topological 1-manifold. �

Example. If TRPn is the usual quotient topology, then (RPn,TRPn) is a topologicaln-manifold.The proof is quite involved and will be done in stages. �

We will first show that (RPn,TRPn) is Hausdorff. To begin, we will require Problem #6from the midterm.

Theorem 20 (Problem #6 from the midterm). Let (X,TX) be a topological space and ∼ anequivalence relation on X. Consider (X/∼,TX/∼) with TX/∼ the quotient topology. Let π : X→ X/∼denote the quotient map, and assume that π(U) is an open subset of X/∼ for every open subset Uof X. Finally, consider (X× X,TX×X) with TX×X the product topology.

Then, (X/∼,TX/∼) is Hausdorff if and only if

A := {(x, y) ∈ X× X : x ∼ y}

is a closed subset of X× X.

Let us translate the notation from the above theorem into the situation of interest to us.Namely:

• X = Rn+1 \ {0}, TX = the subspace topology coming from the standard topology onRn+1(Notation: Y := (Rn+1 \ {0})× (Rn+1 \ {0}), TY = the product topology)• x ∼ y if y = λx for some λ ∈ R, λ 6= 0• π : Rn+1 \ {0} → RPn quotient map

In order to apply our theorem, we will need the following result to hold.

Exercise. The quotient map π : Rn+1 \ {0} → RPn satisfies the condition that π(U) is anopen subset of RPn for every open subset U of Rn+1 \ {0}.

By our theorem, (RPn,TRPn) is Hausdorff if and only if

A := {(x, y) ∈ Y : y = λx for some λ ∈ R, λ 6= 0}is a closed subset of Y.

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Lemma 2. The set A is a closed subset of Y.

Proof. We want to show that Y \A is an open subset of Y. To this end, note that

Y \A = {(x, y) ∈ Y : {x, y} is linearly independent}

= {(x, y) ∈ Y : rank([xy

])= 2}.

Now, suppose that x = (x1, x2, . . . , xn+1) and y = (y1, y2, . . . , yn+1). If i, j ∈ {1, 2, . . . , n+ 1}and i < j, then consider the function

sij : Y → R, sij(x, y) = det([xi xjyi yj

]).

We leave it as an exercise to verify that the functions sij have the following properties.(i) For all i < j, sij : Y → R is a continuous function from (Y,TY) to (R,TR), where TR

is the standard topology.

(ii) We have rank([xy

])= 2 if and only if sij(x, y) 6= 0 for some i < j.

Continuing our proof, we have

Y \A = {(x, y) ∈ Y : rank([xy

])= 2}

= {(x, y) ∈ Y : sij(x, y) 6= 0 for some i < j}= {(x, y) ∈ Y : sij(x, y) ∈ R \ {0} for some i < j}

= {(x, y) ∈ Y : (x, y) ∈ (sij)−1(R \ {0}) for some i < j}

=⋃

1≤i<j≤n

(sij)−1(R \ {0})

Since R \ {0} is an open subset of R, each (sij)−1(R \ {0}) is an open subset of Y. We have

therefore expressed Y \A as a union of open sets, meaning that Y \A is an open subset ofY. This completes the proof of the lemma. �

Having established that A being a closed subset of Y is equivalent to (RPn,TRPn) beingHausdorff, we conclude that (RPn,TRPn) is Hausdorff. However, it remains to verify that(RPn,TRPn) satisfies Condition (ii) from the definition of a topological n-manifold.

Lemma 3. The topological space (RPn,TRPn) satisfies Condition (ii) from the definition of a topo-logical n-manifold.

Proof. Suppose that [x] = [(x1, x2, . . . , xn+1)] ∈ RPn. We must find an open subsetU of RPncontaining [x], and a homeomorphism from U to an open subset of Rn.

Since x 6= 0, we know that xi 6= 0 for some i. Now, consider

U := {[(y1, y2, . . . , yn+1)] ∈ RPn : yi 6= 0}.

We note that(1) [x] ∈ U, and(2) U is an open subset of RPn (exercise).

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We will show that there exists a homeomorphism ϕ : U → Rn by finding a homeomor-phism Rn → U and taking its inverse. To this end, consider the function

Ψ : Rn → Rn+1, Ψ(z1, z2, . . . , zn) = (z1, z2, . . . , zi−1, 1, zi, . . . , zn).

This is a continuous function from (Rn,TRn) to (Rn+1,TRn+1), where TRn and TRn+1 are thestandard topologies.

Since Ψ(Rn) ⊆ Rn+1 \ {0}, we may consider

Ψ|Rn+1\{0} : Rn → Rn+1 \ {0}.

This is a continuous function from (Rn,TRn) to (Rn+1 \ {0},TRn+1\{0}), where TRn+1\{0} isthe subspace topology. We also have the quotient map π : Rn+1 \ {0} → RPn, anothercontinuous function. Hence, the composition

α := π ◦ (Ψ|Rn+1\{0}) : Rn → RPn

is a continuous function. Since

α(z1, z2, . . . , zn) = π(Ψ(z1, z2, . . . , zn)) = [(z1, z2, . . . , zi−1, 1, zi, . . . , zn)],

we see that α(Rn) ⊆ U. Therefore,

α|U : Rn → U

is a continuous function from (Rn,TRn) to (U,TU), where TU is the subspace topology.

It turns out that α|U is a homeomorphism (exercise). Hence, ϕ := (αU)−1 : U → Rn is ahomeomorphism. This completes the verification of Condition (ii). �

From all of our work, it follows that (RPn,TRPn) is a topological n-manifold.

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