TOPOLOGY

P. Veeramani

Department of Mathematics

Indian Institute of Technology Madras

Chennai - 600 036.

Contents

1 Topological Spaces 3

1.1 Basic Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 The Metric Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3 Interior Points, Limit Points, Boundary Points, Closure of a Set . . . 15

1.4 Hausdorff Topological Spaces . . . . . . . . . . . . . . . . . . . . . . 19

1.5 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2 Product and Quotient Spaces 35

2.1 Product Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2.2 The Box Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.3 Quotient (Identification) Spaces . . . . . . . . . . . . . . . . . . . . . 44

3 Connected Topological Spaces 53

3.1 Connected Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

3.2 Connected Subsets of the Real Line . . . . . . . . . . . . . . . . . . 54

3.3 Some Properties of Connected Spaces . . . . . . . . . . . . . . . . . . 56

3.4 Connected Components . . . . . . . . . . . . . . . . . . . . . . . . . 65

4 Compact Topological Spaces 77

4.1 Compact Spaces and Related Results . . . . . . . . . . . . . . . . . . 77

4.2 Local Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

4.3 One Point Compactification of a Topological Space (X,J ) . . . . . . 88

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4.4 Tychonoff Theorem for Product Spaces . . . . . . . . . . . . . . . . . 93

5 Countability and Separation Axioms 111

5.1 First and Second Countable Topological Spaces . . . . . . . . . . . . 111

5.2 Properties of First Countable Topological Spaces . . . . . . . . . . . 119

5.3 Regular and Normal Topological Spaces . . . . . . . . . . . . . . . . 122

5.4 Urysohn Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

5.5 Tietze Extension Theorem . . . . . . . . . . . . . . . . . . . . . . . . 132

5.6 Baire Category Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 134

5.7 Urysohn Metrization Theorem . . . . . . . . . . . . . . . . . . . . . . 138

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Chapter 1

Topological Spaces

1.1 Basic Concepts

We start with the assumption that we intuitively understand what is meant

by a set. For us, set is a collection of well defined objects. We have a set X and let

J be a collection of subsets of X satisfying: (T1) J , X J , where is the empty set (or say null set). (T2) Suppose we have an arbitrary nonempty set J and to each J

we have a subset A of X such that A J , then our J has the property thatJ

A J , where J

A = {x X : x A for at least one J}. (T3) If A1, A2 are in J then A1 A2 is also in J (that is A1, A2 J implies

A1 A2 J ).In such a case, the given collection J is called a topology on X and the pair

(X, J ) is called a topological space.Remark 1.1.1. If A is a member of J and x A then we say that A is aneighbourhood (also known as open neighbourhood) of x. That is for each x in X, Jcontains the collection Nx = {U J : x U} of all open neighbourhoods of x.

Suppose we are given a set X. Now our aim is to find collections B and J ofsubsets of X satisfying:

(i) B J , (ii) J satisfies (T1), (T2), (T3), and (iii) J = {U X : x U impliesthere exists B B such that x B U}.

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In such a case, J is said to be a topology on X generated by the collection B and Bis said to be a basis for the topology J . Each member of J is called an open subsetof X and each member of B is called an essential neighbourhood or a basic open set

in X. Since X J , by (iii) for each x X there exists B B such that x B.Also note that if B1, B2 B then B1 B2 J . Hence for any x B1 B2 thereexists B3 B such that x B3 B1 B2. Therefore B satisfies the following:

(B1) For every x X there exists B B such that x B. (B2) B1, B2 B and x B1 B2 implies there exists B3 B such thatx B3 B1 B2.

x

BB

B1

32

Figure 1.1

Suppose a collection B of subsets of a given set X satisfies the conditions (B1), (B2).

Then using (iii) we can define J and such a collection J satisfies (T1), (T2),and (T3).

Let us prove the following theorem:

Theorem 1.1.2. Suppose a collection B of subsets of a given set X satisfies:

(B1) For every x X there exists B B such that x B.(B2) B1, B2 B and x B1 B2 implies there exists B3 B such thatx B3 B1 B2.

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Then the collection J defined as J = {U X : x U implies there exists B Bsuch that x B U } is a topology on X.

Proof. To prove (T1): From (B1), x X implies there exists B B such thatx B X. Hence by the definition of J , X J . Now we will have to prove thatthe null set J . How to prove? Our statement namely x U implies there existsB B such that x B U is a conditional statement. That is, we have statementssay p and q. Now consider the truth table

p q p qT T T

T F F

F T T

F F T

The so-called null set (or empty set) is a subset of X. Whether satisfies the stated

property? What is the stated property with respect to our set ? If x then thereexists B B such that x B , where are we in the truth table? Whether thereis x ? The answer is no. So our statement x is false. In such a case whetherq is true or false it does not matter and p q is true. So the conclusion is that thenull set has the stated property, therefore by the definition of J , J .To prove (T2): Suppose J is a nonempty set and for each J , A J . Now wewill have to prove that

JA J .

If J

A = , then J (follows from (T1)). So let us assume thatJ

A 6= . Let x J

A, then there exists 0 J such that x A0 .Now x A0 and A0 J therefore by the definition of J there exists B B such

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that x B A0 . But A0 J

A. Hence x J

A implies there exists B Bsuch that x B

JA (since x B A0).

Therefore by the definition of J , J

A J . That is, if J is a nonempty setand A J for all J then

JA J .

To prove (T3): Let A1, A2 J . Again if A1A2 = then by (T1), J and henceA1 A2 J .

Suppose A1 A2 6= . Now let x A1 A2 then x A1 andx A2. Now x A1, A1 J implies there exists B1 B so that x B1 A1.Also x A2, A2 J implies there exists B2 B such that x B2 A2. NowB1, B2 B are such that x B1 B2. Hence by (B2) there exists B3 B suchthat x B3 B1 B2. But B1 B2 A1 A2. Hence B3 B is such thatx B3 A1 A2.

That is x A1 A2 implies there exists B3 B such that x B3 A1 A2implies A1 A2 J (by the definition of J ). Now J satisfies (T1), (T2), (T3) andtherefore J is a topology on X.

Remark 1.1.3. The topology J defined as in theorem 1.1.2 is called the topologygenerated by B. If we want to define a topology on a set X then we search for a

collection B of subsets of X satisfying (B1), (B2) and once we know such a collection

B then we know how to get the topology generated by B. Such a collection B is

called a basis for a topology on X and the topology generated by B is normally

denoted by JB. Definition 1.1.4. If B is a collection of subsets of a given set X satisfying (B1),

(B2) then B is called a basis for a topology on X.

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1.2 The Metric Topology

Let X be a nonempty set and (x, y) X X. With each (x, y) X X weassociate a non-negative real number which we denote by d(x, y). We want to identify

d(x, y) as the distance between the elements x, y in X. So it is natural to expect that

(M1) d(x, y) = 0 if and only if x = y; (M2) d(x, y) = d(y , x ) for all x, y X; (M3) d(x, y) d(x, z) + d(z, y) for all x, y, z X.

It is to be noted that to each element (x, y) in X X we associate a unique elementd(x, y) in R+ = [0, ). That is d(x, y) is the image of (x, y) X X. Hence d is afunction from X X into R+ i.e. d : X X R+.

If X is a nonempty set and d : X X R+ is a function satisfying the aboveconditions (M1), (M2), (M3) then we say that d is a metric on X. In such a case, the

pair (X, d) is called a metric space.

Let us fix x X. Now we want to collect all those elements of the space Xwhich are not far away from x and such a set is known as a neighbourhood of x.

Well, what do you mean by not far away from x? The term not far away is a

relative term. So we fix an r > 0 (in some sense radius of our neighbourhood) and

then take an element, say y from X. If the distance between x and y is strictly less

than r, that is d(x, y) < r, then we say that y is in our neighbourhood of x. Let us

define B(x, r) = {y X : d(x, y) < r}, and call this set as one of our neighbourhoodsof x. If we change r, we get different neighbourhoods of x and B(x, r) is also known

as the ball centered at x and radius r. When X = R3 and d, the distance function,

is the usual Euclidean distance, i.e. for any x = (x1, x2, x3), y = (y1, y2, y3) R3

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d(x, y) =

(x1 y1)2 + (x2 y2)2 + (x3 y3)2, then B(x, r) is the usual Euclideanball centered at x and radius r > 0.

Remark 1.2.1. One can have different metrics on R3 (or Rn, n 1) and forx = (x1, x2, x3) R3, r > 0, B(x, r) may be a cube or a solid sphere or an ellipsoid(excluding the points on the boundary) or a singleton {x} or the whole space R3 undersuitable metrics. Now consider a subset A of X. Suppose A has the property: if

x A then there exists at least one neighbourhood of x say B(x, r) which is containedin our set A. That is, x A implies there exists r > 0 such that B(x, r) A (such ar > 0 depends on x A. i.e. same r may not work for every x A). Note. Our statement namely x A implies there exists r > 0 such that B(x, r) Ais a conditional statement. The so-called empty set (or null set) is a subset of our

space X. Whether empty set has the stated property? What is the stated property?

Well, following the same argument as given in theorem 1.1.2 we see that has the

stated property. Now it is easy (if not obvious) to prove:

X has the stated property.

A,B X such that A,B have the stated property then A B has the statedproperty.

Consider a nonempty set J . Suppose for each J,A X and A has thestated property, then

JA has the stated property. That is the collection

Jd defined as Jd = {A X : x A implies there exists r > 0 such thatB(x, r) A} is a topology on X, known as the topology induced by the givenmetric d.

In this sense we say that every metric space (X, d) is a topological space. ?

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Theorem 1.2.2. In a metric space (X, d) for each x X, r > 0, B(x, r) is an opensubset of (X, Jd).

Proof. Let y B(x, r). Then d(x, y) < r. Let s = r d(x, y). If z B(y, s),then d(y, z) < s = r d(x, y). So d(x, y) + d(y, z) < r. By the triangle inequality,d(x, z) < r. That is z B(x, r). Thus B(y, s) B(x, r). Hence B(x, r) is open.

It is interesting to note that B = {B(x, r) : x X, r > 0} is a basis for atopology on X and it is clear from the definition of Jd that the topology JB generatedby B is same as Jd.

Now let us give some important examples of topological spaces.

Let X be a set and let Jt = {,X}, JD = { A : A is subset of X },Jf = {A : XKA = Ac is a finite subset of X or Ac = X}, Jc = {A : XKA isa countable subset of A or Ac = X}. It is easy to prove that Jt, JD, Jf , Jc aretopological spaces on X, Jt is known as the trivial or indiscrete topology on X, JDis known as the discrete topology on X, Jf is known as the cofinite topology on X,Jc is known as the co-countable topology on X.

Recall that a set A is a countable set if and only if A is a finite set or A is a

countably infinite set. Also note that A is a countably infinite set if and only if there

exists a bijective function f from N to A, where N is the set of all natural numbers.

Also it is known that a nonempty set A is a countable set if and only if there exists

a surjective function say f : N A.Now let us prove that Jc = {A : XKA is a countable subset of A or Ac = X}

is a topology on X.

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Proof. Now c = X implies Jc, Xc = and is a countable set implies X Jc.Hence

, X Jc. (1.1)

Let J be a nonempty set and for each J , A J.Claim:

JA Jc.

Now ( J

A)c =

JAc. Hence we will have to prove that either

JAc = X or

J

Ac is a countable subset of X. If J

Ac = X then we are through (from (T1)).

Suppose not. This implies for at least one 0 J , Ac0 6= X, A0 Jc implies Ac0 isa countable set. Since

JAc Ac0 , J A

c is a countable set (subset of a countable

set is countable). We have proved that either J

Ac = X or J

Ac is a countable

set. Hence

J

A Jc. (1.2)

Let A1, A2 Jc implies that Ac1 is a countable set or Ac1 = X and A2 Jc impliesAc2 is a countable set or A

c2 =X. Now

(A1 A2

)c= Ac1 Ac2 = X when Ac1 or Ac2 =

X or Ac1 Ac2 is a countable set since in this case both Ac1 and Ac2 are countable sets.Hence

(A1 A2

)c= X or it is a countable set. This implies that

A1 A2 Jc. (1.3)

From Eqs. (1.1), (1.2), and (1.3), Jc is a topology on X. Now let us give someexamples to illustrate the natural way of obtaining topologies once we know bases

satisfying (B1) and (B2).

Example 1.2.3. Let X be a nonempty set and B = {{x} : x X}. Then B is abasis for a topology on X.

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(i) For every x X there exists B = {x} B such that x B.(ii) B1, B2 B and x B1 B2 implies there exists B3 = {x} B such thatx B3 B1 B2. Hence both (B1) and (B2) are satisfied. This implies that thecollection B = {{x} : x X} is a basis for a topology on X.

Now let us find out JB, the topology, generated by B. In theorem 1.1.2 wehave proved that if we define JB as JB = {U X : x U implies there existsB B such that x B U} is a topology on X.

In this case for any nonempty subset U of X, x U implies there existsB = {x} such that x B U. Hence by the definition of JB, A B whenever A isa nonempty subset of X. Also the null set JB (recall the proof given in theorem1.1.2). Hence A X implies A JB implies P(X) JB. Also by the definition,JB P(X), the collection of all subsets of X. This implies that JB is same as thediscrete topology JD defined on X.Exercise 1.2.4. Let X be a nonempty set and let d be a metric on X. That is (X, d)

is a given metric space. Then prove that the collection B defined as B = {B(x, r) :x X, r > 0} is a basis for Jd. 2

Now let us consider the special case X = R, the set of all real numbers and

d(x, y) = |x y| for x, y R. Then d is a metric on R. What is the collectionB = {B(x, r) : x X, r > 0}. Note that B(x, r) = (x r, x + r) = (a, b), wherea = x r R, b = x+ r R with a < b. That is B {(a, b) : a, b R, a < b} = B .

a b a+b2

x =

Figure 1.2

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On the other hand take a member say B B . Since B B there exista, b R, a < b such that B = (a, b). Now let x = a+b

2and r = |ab

2| = ba

2> 0. Then

B(x, r) = (x r, x+ r) = (a+b2 ba

2, a+b

2+ ba

2

)= (a, b) implies (a, b) = B(x, r) B

implies B B. Also we have B B and hence B = {B(x, r) : x R, r > 0} =

{(a, b) : a, b R, a < b} = B . That is {(a, b) : a, b R, a < b} is basis for a topologyon R and JB = Jd. This topology is called the standard or usual topology on R, andit is denoted by Js.Exercises 1.2.5. (i) Prove that BQ = {(a, b) : a, b Q, a < b}, where Q - the set ofall rational numbers is also a basis for the usual topology on R. That is JBQ is sameas the usual topology on R.

(ii) Is B0 = {B(x, 1n) : x Q, n N} a basis for the usual topology on R? Justifyyour answer.

(iii) It is given that (X, d) is a metric space. Now prove that B = {B(x, 1n) :

x X,n N} is a basis for a topology on X. Also prove that JB = Jd. 2Definition 1.2.6. A subset A of a topological space (X, d) is said to be a closed

set if the complement XKA = Ac of A is an open set.

Use the DeMorgans law to prove the following theorem.

Theorem 1.2.7. In a topological space (X, J ) we have:(i) X and are closed.

(ii) Suppose we have a nonempty index set J and to each J , A is a closed subsetof X. Then

JA is a closed subset of X. That is arbitrary intersection of closed sets

is closed.

(iii) If A1, A2 are closed sets then A1 A2 is also a closed set.Use induction to prove that finite union of closed sets is closed.

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Now let us prove the following theorem which tells us when a subcollection B

of a given topology J on X generates the topology J .Theorem 1.2.8. Let (X,J ) be a topological space and B J . Further suppose foreach A J and x A there exists B B such that x B A. Then B is a basisfor a topology on X and JB = J .

Proof. First let us prove that B is a basis for a topology on X.

(B1) Let x X. Since X J , by hypothesis, there exists B B such that x B.(B2) Let B1, B2 B and x B1 B2. It is given that B J . Hence B1, B2 Jand this implies B1B2 J . Now x B1B2 and B1B2 J implies there existsB3 B such that x B3 B1 B2. From (B1) and (B2) we see that B is a basisfor a topology on X.

b

x

B2

1

B2b1

a1 a2

Figure 1.3

Now let us prove that JB = J . Let U JB.Claim: U J . If U = then this implies that U J . Otherwise let x U. By thedefinition of JB there exists Bx B J such that x Bx U . This implies thatxU

Bx = U and hence U J .

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Conversely, let 6= U J . Then for each x U there exists B B JBsuch that x B U. This proves that

xUBx = U JB. Hence J JB. Already

we have proved that JB J and therefore J = JB.

Now it is natural to introduce the following definition.

Definition 1.2.9. If (X,J ) is a topological space and B J such that for eachA J and x A there exists B B such that x B A, then we say that B is abasis for J .Theorem 1.2.10. Let (X,J ) be a topological space and B J . Then B is a basisfor J if and only if every member A of J is the union of member of some subcollectionof B.

Proof. Left as an exercise. To have a feeling of this concept do the following exercise:

Exercise 1.2.11. For a1, a2, b1, b2 R, a1 < a2, b1 < b2 let R = {(x1, x2) R2 :a1 < x1 < a2, b1 < x2 < b2}. That is, R is an open rectangle having sides parallelto the coordinate axes. Let B be the collection of all such open rectangles. Now it

is easy to see that B is a basis for the topology Jd on R2, where d is the Euclideanmetric on R2. 2

Remark 1.2.12. Let X be a set and S be a collection of subsets of X. SupposeAS

A = X. Then we say that S is a subbasis for a topology on X. In this case, letB = {A X : A =

BFB, for a finite subcollection F of S}. Then it is easy to

prove that B is a basis for a topology on X. The topology JB generated by B iscalled the topology on X generated by the subbasis S. Exercises 1.2.13. (i) Let S1 = {(a,) : a R}. Prove that S1 is a subbasis for atopology on R. Find out the topology J1 generated by S1.

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(ii) Let S2 = {(, a) : a R}. Prove that S2 is a subbasis for a topology on R.Find out the topology J2 generated by S2. 2

1.3 Interior Points, Limit Points, Boundary Points, Closure of a Set

Let A be a nonempty subset of a topological space (X,J ) and x A. Then xis said to be an interior point of A if there exists an open set U such that x Uand U A. Also the collection of all interior points of A denoted it by int(A) or A.For a nonempty subset A of a topological space (X,J ), a point x X is said to bea limit point or an accumulation point of A if for each open set U containing x,

U (AK{x}) 6= .For A X, the derived set of A denoted by A is defined as A = {x X :

x is a limit point of A}.A point x X is said to be a boundary point of A if for each open set U

containing x, U A 6= and U Ac 6= .For A X, the boundary of A, denoted by bd(A), is defined as

bd(A) = {x X : for each open set U containing x, U A 6= and U Ac 6= }.That is bd(A) is the collection of all boundary points of A.

For A X, the closure of A denoted by A or cl(A), is defined as A = AA.Examples 1.3.1. (i) Let X = {1, 2, 3} and J = {,X, {1}, {2}, {1, 2}, {1, 3}, {2, 3}}Is J a topology on X? Let A = {1, 3}, B = {2, 3}. Here A J , B J , butA B = {3} / J . Hence J is not a topology on X.(ii) Let X = {1,2,3} and J = {,X, {1}, {2}, {1, 2}} then J is a topology on X. NowA = {2, 3} is a subset of X. 2 A and also there is an open set U = {2} such that

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2 U and U A. Hence 2 is an interior point of A. But 3 is not an interior pointof A. How to check 3 is an interior point of A or not?

Step 1: First check whether 3 A (if x is an interior point of A then it is essentialthat x A). Yes here 3 {2, 3} = A.Step 2: Now find out all the open sets containing 3. X is the only open set containing

3. But this open set is not contained in A. Hence 3 is not an interior point of A.

What will happen if the given set A is an open subset of a topological space X. Our

aim is to check whether an element x X is an interior point of A.Step 1: It is essential that x A.Step 2: Is it necessary to find out all the open sets containing x? Of course not

necessary. It is enough if we find at least an open set U such that x U and U A.In this case the given set A is an open set and hence there exists an open set U = A

such that x U and U = A A. Therefore every element x of A is an interior pointof A. That is A A. By definition A A. Hence A = A. That is if A is an openset then A = A. What about the converse? Suppose for a subset A of X,A = A.

Is A an open set? Yes, A is an open subset of X. Take x A. Then x A. Henceby the definition of A there exists at least one open set say Ux such that x Uxand Ux A. This implies that A =

xAUx. Now by the definition, J is closed under

arbitrary union. Hence for each x A,Ux J implies xA

Ux J implies A J .That is, A is an open set. Thus, we have proved:

Theorem 1.3.2. For a subset A of topological space (X, J ), A is open if and onlyif A = A.

Now let us prove that for any subset A of X, A is an open set and if B is an

open set contained in A (B A) then B A.

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Theorem 1.3.3. For any subset A of a topological space (X, J ), A is the largestopen set contained in A.

Proof. If A = then A = . For A 6= , let us prove that B = A is an openset. Due to theorem 1.3.2, it is enough to prove that (A) = A = A. If A =

then A = and we are through. Also by definition A A. Let x A. Thenby the definition, there exists an open set Ux such that x Ux A. Note that foreach y Ux, y Ux A. That is y A and there exists an open set Ux such thaty Ux and Ux A. This implies that y A. That is y Ux implies y A impliesUx A. We have the following:

x A and, there exists an open set Ux such that x Ux, Ux A.

This implies that x A. That is x A implies x A implies A A. Also wehave A A implies (A) = A. From the theorem 1.3.2, A is an open set. Alsoby definition, A A.

To prove the second part assume that B is an open subset of X such that

B A. Now we aim to prove B A. Which is obvious since for each x B thereexists an open set B such that x B and B A. Hence by definition B A.

Consider X = {1, 2, 3}, J = {, X, {1}, {1, 2}} and A = {1, 2}. What isA, the collection of all limit points of A. Is 1 A? The answer is no. Since {1} isan open set containing 1, but {1} AK{1} = {1} {2} = . Is 2 A? Again theanswer is no. Since {2} is an open set containing 2 and {2}AK{2} = {2}{1} = .Is 3 A? First find out all the open sets containing 3.

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Here the whole space X is the only open set containing 3 and X AK{3} ={1, 2, 3} {1, 2} 6= . That is for each open set U containing 3, the condition namelyUAK{3} 6= is satisfied. Hence 3 is a limit point of A. That is 3 A. Here A = {3}.What is A, the closure of A? By definition A = A A = {1, 2} {3} = {1, 2, 3}.

Now let us prove that for any subset A of a topological space X,

A is a closed set and A A. Whenever B is a closed set such that A B then A B that is we aim to

prove:

Theorem 1.3.4. For a subset A of a topological space X, A is always a closed set

containing A and it is the smallest closed set containing A.

Proof. Let us prove that (A)c = XKA is an open set. Hence we will have to prove

that interior of (A)c is itself. Let x (A)c then x / A. Hence there exists an openset U containing x such that U A = U Ac. This imply that x is an interiorpoint of Ac, but we have to prove that x is an interior point of (A)c. So it is enough

to prove that U (A)c.Suppose not. Then there exists y U such that y / (A)c implies y (A). Also

U is an open set containing y. Hence U A 6= . This is contradiction to U A = .We arrived at this contradiction by assuming that U is not a subset of (A)c. Hence

U (A)c, where U is an open set containing x and x (A)c. Therefore every pointof (A)c is an interior point. This implies that (A)c is an open set and hence A is a

closed set.

Now let B be a closed set containing A then we will have to prove that A B.That is to prove A Bc = .

18

Suppose not. Then there exists x A Bc, Bc is an open set containing x.Now if x A = A A is such that x A then x B (given that A B) and weare through. On the other hand if x A and x / A then by the definition of A,Bc AK{x} = Bc A 6= (note: x / A AK{x} = A) is a contradiction sinceA B implies A Bc B Bc = . Hence A Bc = . That is A B.

1.4 Hausdorff Topological Spaces

Definition 1.4.1. A topological space (X,J ) is said to be a Hausdorff topologicalspace (or Hausdorff space) if for x, y X, x 6= y, there exist U, V J such that(i) x U, y V, (ii) U V = .Note. In definition 1.4.1, in place of if it is also absolutely correct to use if and only

if. That is, definition 1.4.1 can also be read as:

A topological space (X,J ) is said to be a Hausdorff topological space (or Hausdorffspace) if and only if (iff) for x, y X, x 6= y, there exist U, V J such that(i) x U, y V, (ii) U V = .

What is important to note here (that is while giving a definition) is one can

use interchangeably if and if and only if. ?

Example 1.4.2. If X = R, and Js is the standard topology on R, then (R,Js) is aHausdorff space.

Example 1.4.3. Every discrete topological space (X,J ) is a Hausdorff space.Example 1.4.4. If X is a set containing at least two elements and J = {,X} then(X,J ) is not a Hausdorff space.Example 1.4.5. If X = R,B = {(a,) : a R} then B is a basis for a topologyJB on R. It is easy to see that (R,JB) is not a Hausdorff space.

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Example 1.4.6. Bl = {[a, b) : a, b R, a < b}, Jl = JBl is known as the lower limittopology on R and Js Jl. Hence (R,Jl) is a Hausdorff space.Note. Weaker topology is Hausdorff implies stronger is also Hausdorff. ?

Let X be an infinite set and Jf be the cofinite topology on X. Also letx, y X, x 6= y. If U Jf and x U then U c is finite, because U c 6= X.Also y V Jf implies V c is finite. If U V = , then X = (U V )c = U cV c andhence X is a finite set. Which gives a contradiction. Therefore U V 6= . Hence Jfis not a Hausdorff space.

Example 1.4.7. Let X = {a, b, c} and J = {,X, {a}, {b}, {a, b}}. Then (X,J ) isnot a Hausdorff space.

Example 1.4.8. B = {U1 U2 Un R R : each Ui is open in R,i = 1, 2, . . . , n, n N} is a basis for a topology J (known as product topology) onRw, where Rw = {x = (xn)n=1 : xn R n}. Now X = Rw, x = (xn) X andy = (yn) X such that x 6= y. Therefore there exists k N such that xk 6= yk.Let = |xkyk|

2> 0 and Uk = (xk , xk + ), Vk = (yk , yk + ). Let

U = R R R Uk R R and V = R R R Vk R R .Clearly, x U , y V and U V = R R R R R = . HenceX = Rw is a Hausdorff space.

Note.n=1

(1n, 1n) is not an open set in the product topology on Rw. ?

Definition 1.4.9. A sequence {xn} in a topological space (X,J ) is said to convergeto a point x X if for each open set U containing x there exists n0 N such thatxn U, n n0. In symbol we write xn x as n.Note that xn x as n if and only if for each open set U containing x thereexists n0 N such that xn U, n > n0.

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Example 1.4.10. If X 6= , J ={,X}, and {xn} is a sequence in X. Then {xn}converges to every element of X.

Example 1.4.11. If X be an infinite set, Jf = {A X : Ac is finite or Ac = X},then Jf is not Hausdorff. Let {xn} be a sequence in X and x X. Now U Jfand x U U c is finite. If U c = then U = X. Otherwise U c is nonempty andfinite and hence J = {n : xn U c} is a finite set. If J = let n0 = 1, otherwise letn0 = max{n : n J}. Then xn U, n > n0. Therefore xn x as n . So,(xn) converges to every element of X.

Theorem 1.4.12. Let (X,J ) be a Hausdorff space and let A X. Then an elementx A if and only if for each open set U containing x, U A is an infinite set.

Proof. Assume that x A and suppose for some open set U containing x,U (AK{x}) is a nonempty finite set. Let U (AK{x}) = {x1, x2, . . . , xn}. For eachi, xi 6= x and (X,J ) is a Hausdorff space implies there exist open sets Ui and Visuch that xi Ui, x Vi and Ui Vi = . Note that xi / Vi for all i = 1, 2, . . . , nand x V = n

i=1Vi, V is an open set. Also x U. Therefore x U V. But

(AK{x}) (U V ) = which is a contradiction. Hence U A is an infinite set.Conversely, if for x X, U A is an infinite set for each open set containing

x then in particular U (AK{x}) 6= , for each open set containing x. Hence x is alimit point of A. That is x A.

Exercise 1.4.13. Let (X,J ) be a topological space such that for each x in X,{x} is closed in X. Then prove that an element x A if and only if for each openset U containing x, U A is an infinite set. 2Note. If X is a Hausdorff space, and if A is a finite subset of X, then A = . ?

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Definition 1.4.14. A topological space (X,J ) is said to be metrizable if thereexists a metric d on X such that Jd = J .Theorem 1.4.15. Let (X, d) be a Hausdorff topological space. Then {x} is closedfor each x X.

Proof. If {x}c = , then X = {x} is a closed set. Suppose {x}c 6= , this impliesthat for each y {x}c, y 6= x. Now (X,J ) is a Hausdorff space implies there existopen sets Ux, Vy in X such that x Ux, y Uy and Ux Vy = . Therefore Vy {x}c

implies y ({x}c)o. This implies that {x} is closed.

Theorem 1.4.16. Let (X,J ) be a topological space, Y X and let JY = {A Y :A J } then JY is a topology on Y .

Proof. (i) JY . Now JX implies Y = JY .(ii) Since X JX X Y = Y JY .Let Ai JY for i I. Now Ai JY implies there exists Bi J = JX such thatAi = Bi Y. Now Bi J for each i I, J is a topology on X implies

iIBi J .

Hence iIAi = (

iIBi) Y is open in JY .

(iii) Let A1, A2, . . . , An JY . Then there exists Bi JX such that Ai = Bi Y .Therefore

ni=1

Bi JX . Therefore A1A2 An = (B1Y )(B2Y ) (BnY )= (

ni=1

Bi) Y JY , and this implies that A1 A2 An JY . From (i), (ii),and (iii) JY is a topology on Y.

Definition 1.4.17. Let (X,J ) be a topological space, and let JY = {AY : A J }then JY is a topology on Y . This topology JY is called the relative topology on Yinduced by J .

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Note. Let A Y X. We use A to denote the closure of A in X, and AY to denotethe closure of A in Y . ?

Result 1.4.18. For A Y, AY = A Y.

Proof. It is always true that AY A. Now AY = closure of A in Y , AY Y andhence AY AY . Let x AY. This implies that x A and x Y. Then for eachopen set U containing x, U A 6= . Hence (U Y )A = U A 6= . Thus x AY .Therefore A Y = AY .

Theorem 1.4.19. Let (X,J ) be a topological space and A,B be subsets of X. Then(i) A B = A B, (ii) A B A B, (iii) A = A, (iv) (AB) AB,(v) (A B) = A B, (vi) (A) = A.

Proof. (i) An element x A if and only if for each open set U containing x suchthat U A 6= . Let x A B. This implies that for every neighbourhood Ucontaining x,

U (A B) 6= (U A) (U B) 6= . (1.4)

Suppose x / A and x / B. Then for some open sets U, V containing x such thatUA = and V B = . Let U0 = UV . Then U0(AB) = (U0A)(U0B) = ,a contradiction to Eq. (1.4). Therefore x A or x B. This proves that

A B A B. (1.5)Now let x A B. This implies that x A or B or both. Hence for eachneighbourhood U of x, UA 6= or UB 6= . This implies that (UA)(UB) 6= implies U (A B) 6= . This shows that x A B. Therefore

A B A B. (1.6)

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From Eqs.(1.5) and (1.6) A B = A B.(ii) Let x A B. Then for every open set U containing x, U (AB) 6= . HenceU A 6= and U B 6= . This implies x A and x B. Hence A B A B.(iii) A is the smallest closed set containing A implies A A. Also A is a closed setcontaining A. But A is the smallest closed set containing A. Hence A A. So wehave A = A.

(iv) Now x A B x A or x B or both. Without loss of generalityassume that x A. Then there exists U J such that x U A. This impliesthat x U (AB). Hence x (AB). That is x AB implies x (AB).Hence (A B) A B.(v) Let x (AB). Then there exists U J such that x U AB. This impliesx A and x B. Hence A B (A B).Now let x A B. Then x A and x B. Hence there exists U J such thatx U A and V J such that x V B, hence x U V A B. Hencex (A B). This implies (A B) = A B.(vi) Now A is the largest open set contained in A implies A A. Also A is anopen set contained in A. Hence A A. So we have proved that A = A.

Example 1.4.20. For A = Q, B = Qc, A B = R and (A B) = R, A = ,B = , R 6= . Hence (A B) 6= A B.A = R, B = R, A B = = . A B = R. Hence A B 6= A B.Definition 1.4.21. Let (X,J ) be a topological space. Then a set B Y is openin Y if and only if B = A Y, for some A J .Example 1.4.22. The set [0, 1) is open in Y = [0,). Note that A = (1, 1)[0,)= [0, 1). Now Ac = Y KA = [1,) is open in (Y,JY ) if and only if for each x [1,)

24

there exists U JY such that x U [1,). But 1 is not an interior point ofAc. Hence Ac is not open and therefore A is not closed. Whereas [1,) is open in[1,) Z.Exercises 1.4.23. (i) Let X = R and Jf be the cofinite topology on R. Let Y = Qwhat is Jf/Y=?(ii) Prove that for each x R, the sequence (xn) = ( 1n) x in (R,Jf ). 2Result 1.4.24. Let (X,J ) be a topological space and B is a basis for J then foreach Y X, BY = {B Y : B B} is a basis for JY .

Proof. Let U JY and x U. U JY implies U = V Y for some V J . Nowx V, V J and B is a basis for J this implies that there exists B B such thatx B U. Therefore BY V Y = U. Now BY BY such that x BY U.Therefore BY is a basis for JY .

Note. A X and B is a basis for J , then x A if and only if there exists B Bsuch that x B A. ?

1.5 Continuous Functions

Definition 1.5.1. Let (X,J ) and (Y, J ) be topological spaces and letf : (X,J ) (Y,J ). Then f is said to be continuous at a point x X if foreach open set V containing f(x) there exists an open set U containing x such that

y U implies f(y) V . If f is continuous at each x X then we say that f is acontinuous function.

Theorem 1.5.2. Let (X,J ), (Y,J ) be topological spaces. Then a functionf : X Y is continuous if and only if for each open set V in Y, f1(V ) isopen in X.

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Proof. Let f be continuous and V be an open set in Y .

Claim: f1(V ) = {x X : f(x) V } is open in X.If f1(V ) = then f1(V ) is open in X. If f1(V ) 6= , let x f1(V ). Thenf(x) V. Now f is continuous at x, V is open containing f(x) implies there existsan open set U such that x U and x U implies f(x) V. That isf(U) V. This implies U f1(f(U)). Therefore x is an interior point of f1(V ).Hence f1(V ) is open in X.

To prove the converse, assume that f1(V ) is open in X whenever V is open

in Y. Now take x X and an open set V in Y such that f(x) V. Now V is open inY implies f1(V ) is open in X. Also f(x) V implies x f1(V ) = U. That is U isan open set in X containing x such that y U implies f(y) f(f1(V )) V. Hencef is continuous at each x X.

Theorem 1.5.3. A function f : X Y continuous if and only if f1(A) is closedin X whenever A is closed in Y .

Proof. Assume that f : X Y is a continuous function. Take a closed set A inY. Since A is a closed set in Y, Ac is an open set in Y. Therefore f is a continuous

function implies f1(Ac) = [f1(A)]c is an open set in X. This proves that f1(A) is

a closed set in X.

To prove the converse, assume that f1(A) is closed in X whenever A is closed

in Y. Take an open set V in Y. Now V is an open set in Y implies f1(V c) = [f1(V )]c

is a closed set in X. Therefore f1(V c) is a closed set in X, and hence f1(V ) is an

open set in X. This gives that f is a continuous function.

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Example 1.5.4. Let X = Rw = {(x1, x2, . . . , xn . . .) : xn R, n N} and letB = {U1 U2 Uk R R : k N, each Ui is open in R, i = 1, 2, . . . , k}.For A,B B, A B = (U1 V1) (Uk Vk) R R B. Then Bis a basis for a topology on Rw. The topology J on Rw induced by B is called theproduct topology on Rw. Bb = {U1 U2 Uk Uk+1 each Uk is open inR, k N} = {

k=1Uk : Uk is open in R k}. Then Bb is also a basis for a topology

on Rw. Let Jb be the topology on Rw induced by Bb. This topology on Rw is calledthe box topology on Rw.

Example 1.5.5. Define f : R (Rw,Jb) by f(t) = (t, t, t, . . . ) and U = (1, 1) (12, 12) (1

3, 13) =

n=1(1n, 1n). Then U Jb, f1(U) = {t R : f(t) U}

= {t R : (t, t . . .) n=1

(1n, 1n) = U} = {t R : |t| < 1

n, n N} = {0}, and

{0} is not an open set in R. Hence f is not a continuous function. But the samef : R (Rw,J ) is a continuous function, when we consider the product topology Jon Rw.

Theorem 1.5.6. A function f : X Y is continuous if and only if for every subsetA of X, f(A) f(A) (where it is understood that X, Y are topological spaces).

Proof. Now assume that f : X Y is continuous. To prove for A X, f(A) f(A). Now f(A) is a closed set in Y and f : X Y is a continuous functionimplies f1(f(A)) is a closed set in X. Also A f1(f(A)) f1(f(A)). Thatis f1(f(A)) is a closed set containing A. Hence A f1(f(A)). This gives thatf(A) f(f1(f(A))) f(A).

Conversely assume that for A X, f(A) f(A). Let F be a closed set inY and A = f1(F ). Now f(A) f(A) implies f(f1(F )) f(f1(F )) F = F.Hence f1(f(f1(F ))) f1(F ). This gives that f1(F ) f1(F ). This proves that

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f1(F ) is a closed set in X whenever F is a closed set in Y. Therefore f : X Y isa continuous function.

Remark 1.5.7. Intuitively what do we mean by a continuous function? In the above

theorem, for any subset A of X, if a point x is closer to A then the image f(x) is

closer to f(A). Here x A means x is closer to A and hence f(x) f(A). Nowwe want that f(x) is closer to f(A). That is f(x) f(A). So a function f : X Yis continuous if and only if for every subset A of X, x is closer to A implies f(x) is

closer to f(A).

Definition 1.5.8. Let X, Y be topological spaces. Then a function f : X Y issaid to be a homeomorphism if and only if

(i) f is bijective

(ii) f : X Y , f1 : Y X are continuous.Example 1.5.9. (i) f : [0, 1] [a, b] defined by f(t) = (1 t)a + tb is ahomeomorphism.

(ii) f : (0, 1) (1,) defined by f(t) = 1t

is a homeomorphism.

(iii) f : (0, 1) (0,) defined by f(t) = t1t is a homeomorphism.

(iv) Let X = (R,Js), Y = (R,Jf ). Let F 6= be a closed in Y. Then F is a finiteset or F = R. In any case f1(F ) = F is closed in X. Hence f is continuous but the

identity map f1 : Y X is not continuous.Example 1.5.10. Let X = [(1,1), (1,1)] be the line segment joining the points(1,1) and (1,1) in R2 and Y = {(x, y) R2 : 1 x 1, y 0, x2 + y2 = 1}.Then X, Y are subspaces of the Euclidean space R2.

Define f : X Y as f((x, y)) = f(x,1) = (x,1 x2) then f is a homeomorphism.That is f is bijective and U is open in X if and only if f(U) is open in Y.

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x + y 122

( 1, 1) (1, 1) ( x, 1)

x-axis

Figure 1.4

Exercise 1.5.11. Let A and B be two distinct points in R2 and be a curve joining

A and B as shown below: That is : [0, 1] R2 is a one-one continuous function.

0 1t

( )t

A

B

Figure 1.5

Then prove that : [0, 1] {(t) : t [0, 1]} is a homeomorphism. That is [0,1] and{(t) : t [0, 1]} are equivalent topological spaces. That is, there is a homeomorphismbetween these two topological spaces. 2

Exercise 1.5.12. Prove that f : X Y is a homeomorphism if and only if(i) f is bijective

(ii) f(A) = f(A), for A X. 2

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Exercises

1. Prove that the half-open interval [0, 1) is neither open nor closed in R, but is

both a union of closed sets and an intersection of open sets.

2. Prove that the set A = {0} { 1n

: n N} is closed in R.3. Find the collection of all interior points, limit points and boundary points of

(i) { 1n

: n N} (ii) { nm

: m N, 1 n m} (iii) { nm

: n,m N}(iv) {m + 1

m: m N} (v) {1 + n

m: n,m N} (vi) {m + n

m: n,m N}

(vii) {(x, y) : 0 x 2, 1 < y < 2} (viii) {(x, y) : xy = 1, x > 0, x Q}.4. Prove that A =

USU , where S = {U A : U is open }.

5. Prove that A = US

U , where S = {U : A U and U is closed }.6. Let X = {a, b, c, d} be four points. Show that the collection J := {X, ,{a}, {c}, {a, b}, {a, c}, {c, d}, {a, c, d}, {a, b, c}} defines a topology on X.Find the collection of all limit points, closures, interiors, and boundaries of all

subsets of X.

7. Prove that for any set A in a topological space bd(A) = bd(A) and bd(intA) bd(A). Give an example when all these three sets are different.

8. Find examples of sets (6= , X) in a topological space that are both open andclosed, neither open, nor closed.

9. Let X be an uncountable set with co-countable topology. Which of the following

sets are closed, open. Justify your answer.

(i) A is a finite set (ii) A is a countable set (iii) A is an uncountable set

(iv) A is a proper subset of X such that both A and Ac are uncountable.

10. Let X be the set of all real numbers with lower limit topology. Let , X, < . Find the interior and closure of (i) [, ] (ii) (, ] (iii) [, ).

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11. Let X be the set of all real numbers with K- topology. That is the topology

generated by the basis B := {(a, b), (a, b)KK : a, b R, a < b}, whereK = { 1

n: n N}. Find the closure and interior of the set E = { 1

n: n N}.

12. Let X be the set of all natural numbers with the usual topology (i.e., N is

considered as a subset of R). Find all the open and closed sets in X.13. Let A, B are nonempty subsets of a topological space X. Assume that

bd(A) bd(B) = . Prove that int(A B) = int(A) int(B).14. Let J1 and J2 be two topologies on X. Prove that identity map

id : (X,J1) (X,J2) is continuous if and only if J1 is finer then J2 (that isJ2 J1).

15. Give an example of a continuous map f from a topological space X to another

topological space Y , such that f(A) is not open (respectively not closed) for a

open (closed) subset A in X.

16. Let X be a topological space. Prove that a map f : X R is continuous ifand only if for every a R the sets f1(, a) := {x R : f(x) < a} andf1(a,) := {x R : f(x) > a} are open.

17. Let (X,JX), (Y,JY ) and (Z,JZ) be topological spaces. If functions f : X Yand g : Y Z are continuous, then show that the composition g f : X Zis continuous. Is the converse true? Justify your answer.

18. Given a function f : X Y and a basis B for Y , then f is continuous if andonly if f1(U) is open in X for each U B.

19. Given a function f : X Y and a subbasis S which generates the topology onY , then f is continuous if and only if f1(U) is open for X for each U S.

20. Suppose f : X Y is continuous, A X. Show that f |A is continuous, whereg = f |A: A Y is defined as g(x) = f(x) for all x A.

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21. Give an example of a function f : R R, such that

(a) f is continuous at exactly 10 points.

(b) f is discontinuous only at 2n number of points, n N.22. Say true or false and justify your answer. Let J1, J2 be topologies on R.

Suppose f : (R,J1) (R,J2) is defined as f(x) = |x|.

(a) if J1 and J2 are usual and lower limit topologies respectively, then f iscontinuous,

(b) if J1 and J2 are lower limit and usual topologies respectively, then f iscontinuous,

(c) if J1 and J2 both are lower limit topologies, then f is continuous.23. Say true or false and justify your answer. Let J1,J2 be two topologies on R.

Suppose f :(R,J1) (R,J2) is defined as f(x) := x 1 if x < 0,x+ 1 if x 0

(a) if J1 and J2 are usual and lower limit topologies respectively, then f iscontinuous,

(b) if J1 and J2 are lower limit and usual topologies respectively, then f iscontinuous,

(c) if J1 and J2 both are lower limit topologies, then f is continuous.24. Let (X,JX) and (Y,JY ) be topological spaces and let A, B be nonempty subsets

of X with AB = X. Suppose f : X Y is a function. Then prove or disprove

(a) if f |A and f |B are continuous, then f is continuous,(b) if f |A and f |B are continuous and bd(A)bd(B) = , then f is continuous.

25. Let f, g : (R,Js) (R,Js), (where Js is the usual (standard) topology on R)be continuous. Prove or disprove:

32

(a) the set {x R : f(x) g(x)} is closed,

(b) the function h : R R defined as h(x) := min{f(x), g(x)} for x R iscontinuous,

(c) the function h : R R, defined as h(x) := max{f(x), g(x)} for x R iscontinuous.

33

34

Chapter 2

Product and Quotient Spaces

Let (Xi,Ji), i = 1, 2, . . . , n be given topological spaces. Also note that it is possiblethat (Xi,Ji) = (Xj,Jj) even when i 6= j. What do we mean by the cartesian productX = X1 X2 Xn? We define the cartesian product as X =

n

j=1

Xj = X1X2 Xn = {f : J

jJXj : f(i) = xi Xi i J}, where J = {1, 2, . . . , n}.

Here we identify an f X with (x1, x2, . . . , xn), where xi = f(i) for all i = 1, 2, . . . , n.What is the advantage of defining the cartesian product in this way? Let J

be a nonempty set (finite or infinite) and for each J we have a topological space(X,J). Now we define the cartesian product

JX = X as X = {f : J

JX :

f() = x X, J}.

2.1 Product Space

Definition 2.1.1. For each J , define a function p : X X, known as -thprojection or coordinate function, as p(f) = f() = x.

Our aim is to define a topology J on J

X which will have the following

properties:

Each projection function p : (X,J ) (X,J) is continuous.

Whenever J is a topology on X such that each p : (X,J ) (X,J) iscontinuous then J J .

35

That is J is the smallest (or weakest topology) on X that makes each p continuous.For J, let S = {p1 (U) : U J}. Then we require that p1 (U) is openin our proposed topological space (X,J ). Hence we require that

JS J .

Note that p1 (U) = {f X : p(f) = f() = x X}. Hence if we fix an J , then p1 (U) =

JA, where A = U and A = X when 6= . If

1, 2, . . . , n, n N and Ui Ji , i = 1, 2, . . . , n, then p1i (Ui) J . Also Jis closed under finite intersections means that

ni=1

p1i (Ui) = J

A J , whereAi = Ui , i = 1, 2, . . . , n and A = X when 6= 1, 2, . . . , n. Now it is easyto see that B = {

JU : U J for all J and U = X, except for finitely

many 6= 1, 2, . . . , n J} is a basis for a topology on X. The topology J inducedby B is called the product topology on X =

JX and the topological space (X,J )

is called the product topological space (also known as product space) induced by

the topological spaces (X,J), J.

Remark 2.1.2. What will happen when J = {1, 2, . . . , n} for some natural numbern? When n = 1, X = {f : {1} X1 : f(1) = x1 X1} = X1 and X ={f : {1, 2, . . . , n} n

i=1Xi : f(i) = xi Xi}. That is f = (f(1), f(2), . . . , f(n)) =

(x1, x2, . . . , xn) X =n

i=1

Xi. Hence X = {(x1, x2, . . . , xn) : xi Xi, i = 1, 2, . . . , n}= X1 X2 Xn.

In this case, that is when J = {1, 2, . . . , n} is a finite index set containing nelements, B = { n

i=1Ui: each Ui is open in Xi, i = 1, 2, . . . , n} is a basis for the product

topology J on X = ni=1

Xi.

Now let us prove the following theorem:

36

Theorem 2.1.3. Let J 6= be an index set and (X,J), J be a collection ofHausdorff topological spaces. Then the product space

(J

X,J)

is also a Hausdorff

topological space.

Proof. Our aim is to prove that the product topological space ( J

X,J ) is aHausdorff topological space. So take two distinct elements f, g in

JX. Now

f, g J

X implies f : J J

X and g : J J

X such that

f() = x X, g() = y X, for each J . Also f 6= g there exists 0 Jsuch that x0 = f(0) 6= g(0) = y0 . We have x0 , y0 X0 and x0 6= y0 . Hence(X0 ,J0) is a Hausdorff topological space implies that there exist U0 , V0 J0satisfying:

(i) x0 U0 , y0 V0 and (ii) U0 V0 = .Now use (i) and (ii) to construct basic open sets U , V in the product space

satisfying f U , g V, and U V = . So, let U = X, V = X, whenever 6= 0. We already have U0 , V0 which are open sets in (X0 ,J0). Define U, Vas U =

JU, V =

JV, where U , V are defined as above. We have f() X,

g() X for all J and hence f U , g V (why?). Also U V = ( J

U) ( J

V) = J

U V = , since U0 V0 = . That is for f, g J

X with f 6= gthere exist basic open sets U, V in the product space such that f U, g V, andU V = . This implies that the product space (

JX,J ) is a Hausdorff space.

Note. Let (X,J ) be a topological space and B be a basis for (X,J ) (or say B isa basis for J ). Then for a subset A of X, x A if and only if for each U B withx U, U A 6= . That is x A if and only if for each basic open set Ucontaining x, U A 6= . ?

37

Theorem 2.1.4. Let (X,J), J be a collection of topological spaces andA X for each J then

JA =

JA, with respect to the product space

( J

X,J ).

Proof. First let us prove J

A J

A. Let f J

A. Then f : J J

A

such that f() = x A for all J . We aim to prove that f is in the closureof

JA in the product topological space

JX. So take a basic open set B in the

product space(

J

X,J)

containing f, say B = J

U. It is given that f J

A.

Hence f() = x A for each J . Now f B = J

U implies f() Ufor all J . That is U is an open set containing x and x A. This impliesthat U A 6= . Let z U A for all J . Define g : J

JA as

g() = z A then g B J

A. This implies that for each basic open set B

containing f,B J

A 6= . This implies f J

A. That is f J

A implies

f J

A and this proves the assertion

J

A J

A. (2.1)

Now let us prove the converse part namely J

A J

A. So let f J

A J

X = X. Our aim is to prove: f J

A. But f J

X if and only if f() Xfor each J and f

JA if and only if f() A for each J . For a fixed

0 J take an open set U0 containing f(0) = x0 . We will have to use the fact thatf

JA. To use this fact we will have to construct a basic open set containing f .

Keeping this in mind, we define B = J

U, where U = X, when 6= 0 and U0is as given above. Now this B is a basic open set containing f and hence f

JA

implies B J

A 6= . That is(

J

U) (

JA)

= J

UA 6= . This impliesthat each U A 6= . In particular U0 A0 6= implies x0 A0 . Note that

38

though our 0 J is a fixed element, there is no restriction on 0 J and the proofwill go through for any J . This gives that f() = x A for all J and thisimplies that f

JA. Hence f

JA f

JA. This implies

J

A J

A. (2.2)

Now combining Eqs. (2.1) and (2.2) we have J

A = J

A.

2.2 The Box Topology

Definition 2.2.1. Let (X,J), J, be a collection of topological spaces andBb = {

JU : U J for J}. Then Bb is a basis for a topology on X =

JX

and Jb, the topology induced by Bb, is called the box topology on X.Remark 2.2.2. From the definitions of product and box topologies, it is clear that

B Bb, where the product topology J on X is induced by B (refer the definition ofproduct topology) and the box topology Jb is induced by Bb. Now B Bb impliesJB = J Jb. That is the product topology on

JX = X is weaker than the box

topology Jb on X. It is to be noted that if a subset A of X is open with respect to the product

topology on X then it is also open with respect to the box topology on X. Note that

the set A =i=1

( 1n, 1n

)is an open set in Rw = RR with respect to the box

topology but not open with respect to the product topology on Rw. Also we have

proved that if (X,J), J , is a collection of Hausdorff topological spaces then theproduct space

(J

X,J)

is also a Hausdorff space. Since J Jb it is clear thatif(X,J

), J is a collection of Hausdorff topological spaces then (

JX,Jb

)is

also a Hausdorff space.

Note that if J is a nonempty finite index set then J = Jb on J

X.

39

Theorem 2.2.3. Let (X,J), J, be a collection of topological spaces and foreach J , let A X. Then

JA =

JA, where A denotes the closure of

A in (X,J) and J

A denotes the closure of J

A in ( J

X,Jb).

Proof. Proof of this theorem is similar to that of theorem 2.1.4.

Theorem 2.2.4. Let (X,J ), (Y,J ), (Z,J ) be topological spacesand f : (X,J ) (Y,J ), g : (Y,J ) (Z,J ) be continuous functions then thecomposite function g f : (X,J ) (Z,J ) defined as (g f)(x) = g(f(x)) is also acontinuous function.

Proof. We aim to prove g f : (X,J ) (Z,J ) is a continuous function. So startwith an open set W in (Z,J ). Now W is an open set in Z (means W J ) andg : (Y,J ) (Z,J ) is a continuous function implies g1(W ) is an open set Y .Now f : (X,J ) (Y,J ) is also a continuous function. Hence f1(g1(W )) is anopen set in X. We define for A Y , f1(A) = {x X : f(x) A}. That isx f1(A) if and only if f(x) A. Hence f1(g1(W )) = {x X : f(x) g1(W )}= {x X : g(f(x)) W} = (g f)1(W ). That is we have proved: W is an open setin (Z,J ) implies (g f)1(W ) is an open in (X,J ) implies g f : (X,J ) (Z,J )is a continuous function.

Definition 2.2.5. A sequence {xn} is a topological space (X,J ) is said to convergean element x in X if for each open set U containing x, there exists a natural number

n0 (that is n0 N) such that xn U for all n n0.The product topology R R R =

n=1Rn = Rw, where Rn = R,

n = 1, 2, 3 . . . is metrizable. That is, we will have to define a metric say d1 of Rw

such that Jd1 = J , the product topology on Rw. For x = (xn)n=1 = (xn) Rw,

40

y = (yn) Rw, let d1(x, y) = supn1

{d(xn,yn)

n

}, where d(xn, yn) = min{1, |xn yn|}.

(Exercise. Let (X, d) be a metric space and d(x, y) = min{1, d(x, y)} for all x, y X.Prove that (i) d is a metric on X, (ii) Jd = Jd.)

It is easy to prove that d1 is a metric on Rw. First let us prove that Jd1 J . So,let U Jd1 . We aim to prove that each point of U is an interior point of U with respectto the product topology J on Rw. Take x U . Now x U , U is an open set in themetric space (Rw, d1) implies there exists r > 0 such that Bd1(x, r) U . Now choosen0 N such that 1n0 < r and B = (x1 , x1+ ) (xn0 , xn0 + )RR then B is a basic open set in (Rw,J ) containing x = (xn). Now we leave it as anexercise to prove that B Bd1(x, ). Hence for each x U , there exists a basic openset B in (Rw,J ) such that x B U. This proves that U J that is

Jd1 J . (2.3)

Now let us prove that J Jd1 . To prove this statement it is enough to provethat every basic open subset V of (Rw,J ) is in Jd1 . Now V is a basic open set in theproduct topology implies there exists k N such that V = V1 V2 Vk R R . Let x = (xn)n=1 V . Hence there exist 1, 2 . . . k, 0 < i < 1 for i =1, 2, . . . , k such that (xii, xi+i) Vi. Now let = min{ ii : i = 1, 2, . . . , k}. (note:we have Ui = R for all i > k and hence it is enough to consider 1, . . . , k) and we claim

that Bd1(x, ) V . So, let y Bd1(x, ) then d1(x, y) < implies supn1

{d(xn,yn)

n

}<

implies d(xn,yn)n

< for all n = 1, 2, . . . , k implies 1n

min{1, |xn yn|} < for alln = 1, 2, . . . , k implies min{1, |xn yn| < n < n < 1 for all n = 1, 2, . . . , k implies|xn yn| < n for all n = 1, 2, . . . , k implies y = (yn) V1 Vk R = V .This proves that Bd1(x, ) V . That is for each x V there exists > 0 such that

41

Bd1(x, ). Hence every point of V is an interior point of V with respect to (Rw,Jd1).Hence

V Jd1 . (2.4)

Now if U J then there exists k N and B1, B2, . . . , Bk B (B = {n=1

Un: each

Un is open in R and Un = R for except finitely many ns} is our standard basis forthe product topology J on Rw) such that U = B1 Bk. We have proved thateach basic open set B of J belongs to Jd1 (i.e B Jd1) (refer Eq. (2.4)). NowB1, B2, . . . , Bk Jd1 and Jd1 is a topology implies B1 B2 Bk Jd1 . Thisproves that U Jd1 That is

U J U Jd1 J Jd1 . (2.5)

From Eqs. (2.3) and (2.5) we see that J = Jd1 . Hence the product spaceRw = R R with product topology J is metrizable. It is interesting to notethat if we consider the box topology say Jb on Rw, then (Rw,Jb) is not a metrizabletopological space.

How to prove that a given topological space (X,J ) is not metrizable. If (X,J )is metrizable then we will have to find a metric (finding such a metric is not at all an

easy task and this statement will become meaningful if we have patience to wait and

see the proof of the Urysohn metrization theorem) say d on X such that Jd = J . Wehave just proved that (Rw,J ) (with product topology) is metrizable and in this casewe could define a metric d on Rw such that Jd = J .

To prove that a topological space is not metrizable space is comparatively

easier. For example, if the given topological space (X,J ) is not a Hausdorfftopological space then it is clear that there cannot exist any metric d on X such

42

that Jd = J . We know that if d is a metric on X, then (X,Jd) is a Hausdorffspace. So, what we need here is to find a property a metric space has whereas the

given topological space does not have that particular property. Now let us come back

and prove that (Rw,Jb) is not metrizable. Suppose there exists a metric say d onRw such that Jd = Jb. Then we know that for A X, x A if and only if thereexists a sequence (xn) in A such that (xn) x as n (to prove this statement,observe the following: For x A,B(x, 1

n)A 6= , for each n N and hence we have

{B(x, 1n)A}n=1 = {An}n=1 = {An}nN a collection of nonempty sets. Now by axiom

of choice there exists a choice function say f : N n=1

An such that xn = f(n) An.So using axiom of choice we have got a sequence (xn) in A and now it is easy to see

that xn x as n).Note that normally we just say that B(x, 1

n) A 6= implies there exists

xn B(x, 1n)A = An. In such case it is to be understood that in fact we use axiom ofchoice to define such a sequence (xn). Now let us prove that if A = {(x1, x2, x3 . . . , ) :xk > 0 for all k N}. Then 0 = (0,0,0,. . . ,) A, but there does not exist anysequence (x(n)) in A such that x(n) (0, 0, . . .) in (Rw,Jb).Step 1: Prove that 0 A.So take an open set say U containing 0 then there exists a basic open set B =

n=1

Bn =

B1B2 such that (0, 0, . . . , ) B1B2 U . (Here each Bk is an open setin R containing 0 R) 0 Bk, k = 1, 2, 3 . . . implies there exist ak, bk R, ak < bksuch that 0 (ak, bk) Bk, bk > 0 and hence bk2 > 0 implies b = ( bk2 )k=1 A Bimplies A B 6= implies A U 6= implies 0 = (0, 0, . . . , ) A. Now we claimthat there cannot exist any sequence x(n) in A such that x(n) (0, 0, 0 . . .).Let x(n) = (a1n, a2n, . . . , ) A. Then each ain > 0 for all i = 1, 2, . . .. In particular,

43

akk > 0 for all k = 1, 2, . . .. Let U = ((a11)

2, a11

2) ( (a22)

2, a22

2) then U is an

open set in (Rw,Jb) containing 0.What will happen if U A 6= . Note that for each n, ann / ( (ann)2 , ann2 ) and

hence x(n) = (a1n, a2n, . . . , ) / U . If x(n) (0, 0, . . .) then there exists n0 N suchthat x(n) U for all n n0. But here x(n) / U for every n. Hence x(n) does notconverge to (0,0,0, . . .). So (0, 0, . . .) A but there cannot exist any sequence in Awhich converges to (0, 0, . . .) with respect to Jb. This proves that (Rw,Jb) is not ametrizable topological space.

2.3 Quotient (Identification) Spaces

We start with a given topological space (X, J ). By identifying some of thepoints of X we can produce a new topology on a new set say X. For example if

we consider the closed unit ball in R2, then our given topological space is (X,J ),where X is the closed unit ball in R2. Here we consider (X,J ) as a subspace of theEuclidean space R2. Now we get a new set X = {(x1, x2) R2 : x21 + x22 < 1} {S },where S is the unit circle (boundary) of the closed disc X. By defining a suitable

topology J on X we can show that (X,J ) is homeomorphic to the 2-sphereS2 = {(x1, x2, x3) R3 : x21 + x22 + x23 = 1}. It is to be noted that here we areconsidering S2 as a subspace of R3 (also note that if no topology on Rn, n 1 ismentioned then it is understood that we have the usual topology on Rn).

Now let us see how to construct the quotient topology. Let (X,J ) be atopological space and X be a nonempty set. Let p : X X be a surjective map.

44

Then J = {A Y : p1(A) is open in (X,J )} is a topology on X. This topologyJ on X is called the quotient topology on X induced by p.

It is easy to prove that J is a topology on X and we leave it as an exercise.Definition 2.3.1. Let (X,J ) be a topological space and X be a partition of X intodisjoint subsets whose union is X. Let p : X X be the natural map satisfying thecondition namely x p(x), for each x X. Suppose for a given x X there existA,B X such that x A and x B. Then x AB. This implies B = A. Hencefor each x X there exists a unique A X such that x A and this A is our p(x).Also

AXA = X implies that p is onto. The quotient topology J on X is induced

by p and we say that (X,J ) is a quotient topology of (X,J ).Let (X,J ) be a topological space and X be a partition of X into disjoint

subsets whose union is X. Define a relation R on X as follows:

R = {(x, y) XX : x, y A for some A X} then (i) xRx, that is (x, x) R forall x X, (ii) for x, y X, xRy implies there exists A A such that x, y A. Hencey, x A and this gives yRx that is for x, y X xRy yRx, (iii) for x, y, z X, xRyand yRz implies there exist A,B X such that x, y A and y, z B. Thereforey AB and this implies that A = B. From this we have x, z A. Hence xRz. Thatis xRy and yRz implies xRz. From (i), (ii) and (ii) we see that R is an equivalence

relation on X and hence this relation R will partition X into disjoint equivalence

classes.

For each x X, the equivalence classes determined by x is given byx = {y X : yRx}. Hence if x A, for some A X then x = A. Now it iseasy to see that for U X, U J if and only if

AUA is an open subset of X.

Let (X,J ) be a topological space and X be a family of disjoint nonempty subsets

45

of X such that X = AX

A. Define q : X X as q(x) = A, where A X issuch that x A. Then the topology Jq on X is the largest topology on X whichmakes q : (X,J ) (X,Jq) a continuous function is called the quotient topology(or identification) topology on X induced by q.

Theorem 2.3.2. Let (X,Jq) be an identification space (i.e Jq is the identificationtopology on X with respect to q) defined as above and (Y,J ) be an arbitrarytopological space. Then a function f : (X,Jq) (Y,J ) is continuous if and only iff q : (X,Jq) (Y,J ) is continuous.

Proof. Let f : (X,Jq) (Y,J ) be a continuous function. We know that bythe definition of identification space, q : (X,J ) (X,Jq) is a continuous function.Hence the composite function f q : (X,Jq) (Y,J ) is a continuous function.We will have to prove that f : (X,Jq) (Y,J ) is continuous. So start with anopen set U in Y . That is we will have to prove that f1(U) is open in (X,Jq).But the subset f1(U) is open in the identification space if and only if q1(f1(U))

is open in (X,J ). But q1(f1(U)) = (f q)1(U) an open set in (X,J ) (sincef q : (X,J ) (Y,J ) is a continuous function). This is what we wanted to proveand hence f : (X,Jq) (Y,J ) is a continuous function.

Let (X,J ) be a topological space and Y be a nonempty set. Let f : X Ybe an onto map. Then X = {f1(y) : y Y } is a family of disjoint subsets of Xsuch that

yYf1(y) = X. That is X is a partition of X. Let q : X X be the

map, known as identification map, defined as above. Let J be the largest topologyon Y for which f : (X,J ) (Y,J ) is continuous. Then it is easy to prove thefollowing:

46

Theorem 2.3.3. Let (X,J ), (Y,J ) be topological spaces and f : (X,J ) onto (Y,J )be a homeomorphism. Further suppose (Z,J1) is any topological space. Then afunction g : (Y,J ) (Z,J1) is continuous if and only if g f : (X,J ) (Z,J1) iscontinuous.

Proof. To prove (Y,J ) and (X,Jq) are homeomorphic we will have to define amap say h : X Y and prove that this map is a homeomorphism.

Let z X = {f1(y) : y Y } be any element. Then z = f1(y), for somey Y . So let h(z) = h(f1(y)) = y. The defined map h : X Y is such that(h q)(x) = h(q(x)) = h(f1(y)) (where y Y is such that x f1(y)) = y = f(x).That is h q = f . Let us prove that h is continuous. Let V be an open set in Y .

X

q

hY

X

f

*

Figure 2.1

The given topology J on Y is the largest topology on Y for whichf : (X,J ) (Y,J ) is continuous. Hence V is an open set on Y implies f1(V ) isan open set in X and hence (hq)1(V ) = q1(h1(V )) is an open set in X. Thereforeh1(V ) is an open set in X. That is V is an open set in Y implies h1(V ) is an open

set in X. This implies that h is a continuous map.

47

The TorusLet X = [0, 1] [0, 1] with the topology J on X induced by the standard

topology on R2 (that is J is the topology on X induced by the Euclidean metric).Partition X into the subsets of the type:

the set A = {(0, 0), (0, 1), (1, 0), (1, 1)} consisting of the four corner points, all the sets of the form Ax = {(x, 0), (x, 1)} for 0 < x < 1, all the sets of the form Ay = {(0, y), (1, y)} for 0 < y < 1, all singleton sets of the form {(x, y)}, 0 < x < 1, 0 < y < 1. Then the resulting

identification space is the torus.

Exercise 2.3.4. Let (X,J ), (Y,J ) be topological spaces and f : X Y be an ontomap. If f maps open sets in X to open sets in Y (that is f is an open map) then

prove that J is the quotient topology on Y induced by f.

48

Exercises

1. A function f : Z X Y is continuous if and only if its component functionsf1 : Z X and f2 : Z Y are both continuous, where f(z) = (x, y) =(f1(z), f2(z)).

2. Let X and Y be topological spaces. If B1 is basis for the topology on X and B2

is a basis for the topology on Y, then show thatB := {BC : B B1, C B2}is a basis for a topology on X Y .

3. Let X denote the set of all real numbers with lower limit topology and Y denote

the set of all real numbers with usual topology. Then

(a) find the topologies on (i) X X, (ii) X Y, (iii) Y X, (iv) Y Y,

(b) show that each of the above topologies is Hausdorff,

(c) compare the topologies.

4. Prove or disprove the following:

(a) The diagonal 4 = {x x : x R} is closed with respect to the usualtopology on R2.

(b) The diagonal 4 = {xx : x R} is closed with respect to the lower limittopology on R2 = Rl Rl.

5. Show that a topological space X is Hausdorff (also known as T2-space) if and

only if the diagonal 4 = {x x : x X} is closed in X X.

6. Let X1, X2, . . . , Xn be topological spaces and let Ai Xi, i = 1, 2, . . . , n,X =

n

i=1

Xi with product topology. Then show that int(n

i=1

Ai) = (intAi)

49

7. Let P1 : R2 R be the projection of R2 onto the x-axis. Show that P1 is openbut not closed.

8. Find the interior points, limit points and boundary points of each of the

following subsets of R2, with respect to each of topology given in question (3a).

(a) A = {(x, y) : x R, y = 0}.(b) B = {(x, y) : x > 0, y 6= 0}.(c) C = A B.(d) D = {(x, y) : x Q, y R}.(e) E = {(x, y) : x, y Q}.(f) F = {(x, y) : x Q, y Qc}.(g) G = {(x, y) : x 6= 0, y 1

x}.

(h) H = {(x, y) : 0 < x2 + y2 1}.(i) I = {(x, y) : 0 < x2 y2 1}.

9. Let Rw : {(x1, x2, x3, ) : xi R, for all i N} and let J and Jb be theproduct and box topologies on Rw respectively. Suppose f is a map on Rw,

defined as f(x1, x2, x3, ) = (a1x1, a2x2, a3x3, ), where ai > 0 for all i N.Then prove or disprove:

(a) f : (Rw,J ) (Rw,J ) is continuous.(b) f1 : (Rw,J ) (Rw,J ) is continuous.(c) f : (Rw,J ) (Rw,Jb) is continuous.(d) f1 : (Rw,J ) (Rw,Jb) is continuous.(e) f : (Rw,J1) (Rw,J ) is continuous.(f) f1 : (Rw,J1) (Rw,J ) is continuous.(g) f : (Rw,J1) (Rw,Jb) is continuous.

50

(h) f1 : (Rw,J1) (Rw,Jb) is continuous.

10. Let (Xn, dn), n N be a countable collection of metric spaces and X =n=1

Xn.

For x = (xn) X, y = (yn) Y. Let d(x, y) =n=1

ndn(xn,yn)

1+dn(xn,yn). Prove that

(i) d is a metric on X, (ii) Jd is the product topology on X and {n}n=1 is asequence of positive real numbers such that

n=1

n is a convergent series.

11. For i = 1, 2 let fi : Xi Yi be maps between topological spaces. The mapg : X1 X2 Y1 Y2 is defined by g(x1, x2) = (f1(x1), f2(x2)). Show thatg is continuous if and only if f1 and f2 are continuous.

12. Let X = R and X = {{x}, {1, 12, 13, . . .} : x RK{1, 1

2, 13, . . . , }}. Define

f : X X as f(x) = {x} for all x RK{1, 12, 13, . . . , } and f(x) = {1, 1

2, 13, . . .}

for all x {1, 12, 13, . . .}. Let JK be the K-topology on R and J be the quotient

topology on X induced by f. Prove that (X,J ) is a T1-space. Is (X,J ) aHausdorff space ? Justify your answer.

51

52

Chapter 3

Connected Topological Spaces

3.1 Connected Spaces

Definition 3.1.1. A topological space (X, J ) is said to be a disconnectedtopological space if there exist nonempty open sets A and B of X such that

(i) A B = , (ii) X = A B.In such a case B = Ac and A = Bc and hence A and B are closed sets. Also

X contains a nonempty proper subset A (that is A 6= ,X which is both open andclosed in X.

A topological space (X, J ) is said to be connected if there cannot existnonempty closed (open) subsets A and B of X such that (i) AB = , (ii) X = AB.Equivalently, (X, J ) is connected if and only if and X are the only subsets of Xwhich are both open and closed in X.

Examples 3.1.2. (i) Let X be a set containing at least two elements and A X,A 6= , X. Then J = {,X,A,Ac} is a topological space. Here A is such that A 6= ,A 6= X and A is both open and closed. Hence (X, J ) is not a connected topologicalspace.

(ii) Let X be a nonempty set and Jf = {A X,XKA = Ac finite or Ac = X} then(X,Jf ) is a topology on X, known as cofinite topology on X.

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Note. If X is a finite set containing at least two elements then Jf is the discretetopology on X. In this case (X, Jf ) is not a connected topological space.

What will happen when X is an infinite set and Jf is the cofinite topology onX. Is (X, Jf ) not connected? In other words can we find a subset A of X such thatA 6= and A 6= X but A is both open and closed? A 6= X, A is closed implies A isa finite set. Also A 6= implies such a nonempty finite set cannot be an open set.Therefore and X are the only sets which are both open and closed. This implies

(X, Jf ) is a connected topological space whenever X is an infinite set. ?

3.2 Connected Subsets of the Real Line

Keeping our intuition alive, let us prove that intervals are connected subsets

of R and they are the only connected subsets of R. Recall that a subset J of R is

an interval if and only if whenever a, b J and a < c < b, we have c J. Notethat the null set and singleton sets are also intervals. For example for x R, = (x, x) = (1, 1) and {x} = [x, x], the singleton set containing x.

We say that a subset Y of a topological space X is connected if (Y,JY ) isconnected.

First let us prove that if A R is not an interval then A is not connected.Here JA = {A U : U is an open set in R}. So, we will have to prove that thetopological space (A, JA) is not connected. The given set A is not an interval impliesthere exist x, y A and z R such that x < z < y and z / A. We know that(, z), (z,) are open sets in R. This implies that (, z) A and (z,) Aare open sets in (A, JA). Also x (, z) A = C and y (z,) A = D andA = C D. That is C,D are nonempty open sets in (A, JA) such that C D =

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and A = C D. Hence the topological space (A, JA) cannot be connected. That isthe given subset A of R (which is not an interval) is not connected.

Now let us prove:

Theorem 3.2.1. Every interval in R is connected.

Proof. Let J be an interval and let us assume that J contains at least two elements.

For if J = or a singleton set then the null set and the whole space J are the

only sets which are both open and closed in J and hence J is connected. Now let

us suppose that there exist nonempty closed sets A, B in J (that is A,B J andA, B are closed sets in (J , JJ), where JJ = {U J : U is open in R}) such thatJ = A B. Fix a A, b B and without loss of generality let us say a < b. Notethat a A J , b B J .

a z b

Figure 3.1

Since J is an interval [a, b] J. Let y = sup(A [a, b]). Now let us prove thaty A B. First let us prove that y A. Let U be an open set in J containing y(y [a, b] J). Then there exists an open set V in R such that V J = U. Hencethere exists > 0 such that (y , y+ ) J V J = U. (If y = a, we are through,otherwise take 0 < < y a and < b y when y 6= b.) Now y is not anupper bound for [a, b] A implies there exists x0 [a, b] A such that y < x0.This implies x0 (y , y + ) A (y , y + ) J U. Hence x0 U A.That is, whenever U is an open set in J containing y, then U A 6= . This provesthat y AJ = A, AJ is the closure of A in J. Also if y = b, then y B. Supposey < b, (y is the supremum of C = A [a, b] and b is an upper bound of C then take

55

y0 (y, y + ). Now y0 U B. Hence y BJ = B. Hence we have y A B.Therefore there cannot exist nonempty closed subsets of A,B in the subspace (J , JJ)such that A B = , A B = J. This proves that J is connected.

3.3 Some Properties of Connected Spaces

Now let us prove that continuous image of a connected topological space is

connected.

Theorem 3.3.1. Let (X, J ) be a connected topological space and (Y, J ) be anytopological space. Suppose f : (X, J ) (Y, J ) is a surjective continuous map thenthe image f(X) = Y is a connected topological space.

Proof. (By contradiction). Let A, B be nonempty open subsets of Y such that

A B = , A B = Y . Now f is a continuous map, and A, B are open sets in Yimplies that f1(A), f1(B) are open sets in X. Also

f1(A) f1(B) = f1(A B) = (3.1)

and

X = f1(Y ) = f1(A B) (from 3.1)

= f1(A) f1(B). (3.2)

Since A,B 6= , let y A and y B, and f is a surjective map implies that there existx, x X such that f(x) = y and f(x) = y implies f(x) A and f(x) B impliesx f1(A) and x f1(B). Hence f1(A), f1(B) are nonempty open subsets ofthe connected topological space (X, J ) satisfying Eqs. (3.1) and (3.2). This means(X, J ) cannot be a connected topological space. That is we have proved: f(X) = Yis not connected implies X is also not connected and this gives a contradiction. Hence

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our assumption that (Y , J ) is not connected is not valid. Therefore (Y , J ) is aconnected topological space.

It is to be noted that if X is a connected topological space and f : X Y isa continuous function, where Y is any topological space, then the image f(X) is also

a connected topological space. Here we will have to consider f(X) as a subspace of

the given topological space Y . Also if f : X Y is continuous then f : X f(X)is also continuous. Hence this result will follow from the previous result.

Definition 3.3.2. A subspace Y of a topological space (X, J ) is said to have aseparation if and only if there exist nonempty subsets A, B of X such that

(i) Y = A B, (ii) A B = = A B. Here A is the closure of A in X.Example 3.3.3. Let X = R and Y = [0, 2) (2, 5). Then Y has a separation.Take A = [0, 2), B = (2, 5), then (i) Y = A B is satisfied. Now (ii) A B =[0, 2] (2, 5) = , and A B = [0, 2) [2, 5] = . Hence from (i) and (ii) we see thatY has a separation.

Theorem 3.3.4. A subspace Y of a topological space (X, J ) is connected if and onlyif there does not exist any separation for Y .

Proof. First let us assume that the subspace Y is connected. This means the

subspace (Y , JY ) is a connected topological space. So, we will have to provethat Y does not admit any separation. Suppose Y has a separation. Hence there

exist nonempty subsets A,B of X such that Y = A B, A B = = A B.Now AY = A Y = A (A B) = (A A) (A B) = A = A.This implies that A is a closed subset of (Y , JY ). Similarly BY = B Y =(B (AB)) = (B A) (B B) = B. This implies that B is a closed set in(Y, JY ). Also BcY = Y KB = Y Bc = (AB)Bc = (ABc)(BcB) = ABc = A

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(since A B = ). This means complement of B with respect to Y is A. Hence Bis such that B 6= , B 6= Y (since Y = A B and A 6= ) and B is both openand closed in Y . This implies that the subspace (Y , JY ) is a disconnected space.This is a contradiction and we arrived at this contradiction by assuming that Y has

a separation. Hence there does not exist any separation for Y .

Conversely, now assume that there does not exist any separation for Y . Now

suppose that the subspace (Y , JY ) is a disconnected space. Then there existnonempty closed subsets A,B in (Y , JY ) such that Y = A B and A B = .Now A B = A (Y B) = (A Y ) B = AY B = A B = . (A is closed inY implies AY = A) Similarly, A B = A Y B = A BY = A B = . (B isclosed in Y implies BY = B.) We have nonempty subsets A, B in X such that

(i) Y = AB, (ii) AB = = AB. This means Y has a separation and this givesa contradiction. We arrived at this contradiction by assuming that the subspace

(Y , JY ) is a disconnected space. Hence the assumption is wrong. That means(Y , JY ) is connected.

Theorem 3.3.5. Let (X, J ) be a disconnected topological space and A be a subsetof X such that (i) A 6= ,X (ii) A is both open and closed in X. Suppose Y is anonempty connected subspace of X. Then either Y A or Y Ac.

Proof. X = AB, whereB = Ac implies Y = XY = (AB)Y = (AY )(BY ).Also (A Y ) (B Y ) A B = A B = (since B = Ac is a closed set in X)implies (A Y ) (B Y ) = . Similarly, (A Y ) (B Y ) A B = A B = implies (A Y ) (B Y ) = . It is given that Y is a connected subspace of X.Hence it cannot happen that A Y 6= and B Y 6= . Since Y is a connected

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subspace of X, Y cannot admit any separation. Hence A Y = or B Y = implies Y Ac or Y Bc = A. Alternate Proof. Now A Y, B Y (where B = Ac) are open sets in the subspaceY such that (i) Y = (AY ) (BY ) and (ii) (AY ) (BY ) AB = impliesA Y = or B Y = implies Y Ac or Y A. Let us prove the following:Theorem 3.3.6. Let (X, J ) be a topological space and Y1, Y2 be connected subspacesof X. Further suppose Y1 Y2 6= . Then Y1 Y2 is a connected subspace of X.

Proof. We will have to prove that the subspace Y1Y2 cannot admit any separation.Suppose A, B are subsets of X such that (i) Y1Y2 = AB, (ii) AB = = AB(here A is the closure of A in X).

From (i)

Y1 = (Y1 A) (Y1 B), (3.3)

and from (ii)

(Y1 A) (Y1 B) = and (Y1 A) (Y B) = . (3.4)

It is given that Y1 is a connected subspace. Hence Y1 cannot admit a separation.

Therefore from Eqs. (3.3) and (3.4) we conclude that Y1 A = or Y1 B = . Letus say that Y1 A = . This implies that

Y1 Ac. (3.5)

Similarly, using the fact that Y2 is a connected subspace we conclude that Y2A = or Y2 B = . If Y2 B = then

Y2 Bc. (3.6)

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From Eqs. (3.5) and (3.6) we have

Y1 Y2 Ac Bc = (A B)c. (3.7)

Also Y1 Y2 6= . Hence there exists an element say x0 Y1 Y2. Then fromEq. (3.7), x0 / A B, a contradiction to Y1 Y2 = A B. Therefore both Y1 Ac

and Y2 Ac implies Y1 Y2 Ac implies A = . (If there exists x A Y1 Y2then x Y1Y2 Ac.) What we have proved is the following: If (i) and (ii) happensthen that leads to A = or B = . (Y1 A = or Y1 B = implies Y1 Ac orY1 Bc. So if we assume Y1 Bc then we would have arrived at B = .) That iswe have proved that Y1 Y2 cannot admit any separation and this implies Y1 Y2 isa connected subspace of X.

Remark 3.3.7. If X is a topological space and Y1, Y2, . . . , Yn for some n N is afinite collection of connected topological spaces such that

ni=1

Yi 6= then by usinginduction we can prove that Y1 Y2 Yn is a connected subset of X.

In fact using exactly the same idea of proving that the subspace Y1 Y2 isconnected whenever Y1, Y2 are connected subspaces with the added condition that

Y1 Y2 6= , we prove the following theorem.Theorem 3.3.8. If {Y}J is a collection of connected subspaces of a topologicalspace X and further there exists an x0 X such that x0 Y, for each J , thenJ

Y is a connected subspace of X.

Proof. Here again we will prove that J

Y cannot admit any separation. Suppose

A, B are subsets of X such that (i) J

Y = AB, (ii) AB = AB = . Then ouraim is to prove A = or B = . For each fixed 0 J, Y0 = (Y0 A) (Y0 B)

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and (Y0 A) (Y0 B) = = (Y0 A) (Y0 B). Hence Y0 is a connectedsubspace of X implies Y0 A = or Y0 B = . Let us say Y0 A = . Thiswill imply that Y0 Ac. Note that 0 J is an arbitrary element. Hence for each J, Y Ac or Y Bc. But if Y1 Ac for some 1 J and Y2 Bc forsome 2 J then x0 Y1 Y2 Ac Bc = (A B)c =

(J

Y

)c. This means

x0 / Y for some J and that gives a contradiction. Therefore Y Ac for all J or Y Bc for all J implies

JY Ac or

JY Bc implies A = or

B = . This means J

Y cannot admit any separation and that is what we wanted

to prove.

Theorem 3.3.9. Let X be a topological space and Y1, Y2, . . . , Yn, . . . be a collection

of connected topological spaces. Further suppose Yk Yk+1 6= for all k N. Thenk=1

Yk is also a connected space.

Proof. Now Y1, Y2 are connected subspaces of X and Y1Y2 6= implies E2 = Y1Y2is a connected subspace. Now E2 is a connected subspace of X and Y3 is a connected

subspace of X. Further Y2 Y3 (Y1 Y2) Y3 = E2 Y3. Hence Y2 Y3 6= implies E2 Y3 6= implies E3 = E2 Y3 = Y1 Y2 Y3 is a connected subspace ofX. Now use induction to prove that, for each k N, Ek = Y1 Y2 Yk is aconnected subspace of X. We have a collection {Ek}k=1 of connected subspaces of Xsuch that Y1 = E1

k=1

Ek. Alsok=1

Ek 6= . Since Y1 6= . Hencek=1

Ek =k=1

Yk is

a connected subspace.

Now let us prove that if Y is a nonempty connected subspace of a topological

space X and then it remains connected after adding some (or all) of its limits points

to E.

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Theorem 3.3.10. Let Y be a nonempty connected subspace of a topological space X

and E be a subset of X such that Y E Y . Then E is also a connected subspaceof X.

Proof. Let A,B be subsets of X such that (i) E = A B (ii) A B = A B = then Y = Y E = (AY )(BY ). Also (A Y )(BY ) = = (AY )(B Y ).Hence Y is a connected subspace of X implies A Y = or B Y = . Let us sayA Y = . This implies Y = B Y Y B Y B. Since, E Y , E = A Band A B = we get A E B and A = A B = . That is we have proved thatfor subsets A,B of X, E = A B, A B = A B = . This implies A = . Wearrived at this conclusion by assuming that AY = . If we had assumed BY = ,then we would have proved that B = . Therefore E does not admit any separation

and hence E is a connected subspace of X.

Theorem 3.3.11. If (X1, J1), (X2, J2) are connected topological spaces, then theproduct space X1 X2 is also a connected space.

Proof. Fix a1 X1, a2 X2. (Note. X1 = or X2 = X1 X2 = and in thiscase X1X2 is a connected space.) For each x1 X1, fx1 : X2 X1X2 defined asfx1(x2) = (x1, x2) is a continuous function. Also we know that continuous image of a

connected space is connected. In this case, the continuous image is fx1(X2) = x1X2.That is x1 X2 = {(x1, x2) : x2 X2} is a connected subspace of the productspace X1 X2 (refer the vertical line passing through x1). Similarly, X1 a2 is aconnected subspace of the product space. Also (x1, a2) (x1 X2) (X1 a2).Hence (x1 X2) (X1 a2) is a connected subspace of the product space. LetTx1 = (x1 X2) (X1 a2). Also note that (a1, a2) Tx1 , for each x1 X1. That is

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{Tx1} is a collection of connected subspaces of the product space X1 X2. Further(a1, a2) Tx1 for all x1 X1. Hence

x1X1Tx1 = X1 X2 is a connected space.

X XX

X

a

x a

(a , a )

X

X aX

1 22

1

2

1 1

1 2 1 2

Figure 3.2

Now we use mathematical induction to prove: if (X1, J1), (X2, J2), . . . , (Xn, Jn)are a finite collection of connected topological spaces then the product space

X1 X2 Xn is also a connected space. But it is to be noted that we cannotuse (say why?) mathematical induction to prove: If (Xn, Jn), n N is a collectionof connected topological spaces then the product space X1 X2 =

n=1

Xn is

also a connected space. However, we prove the following theorem when we have a

collection (X,J), J (where J is a nonempty index set) of connected topologicalspaces.

Theorem 3.3.12. Let (X,J), J be a collection of connected topological spaces.Then the product space X =

JX is also a connected space.

Proof. When J is a finite set we have already proved this result. So let us assume

that J is an infinite set. Also let us assume that each X 6= . From each X, fix an

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element say a X. That is we have a function f : J J

X such that

f() = a X for all J . That is f J

X. We normally write f = (a)J

and just say that (a)J J

X.