BASIC CONCEPTS OF

TOPOLOGY

If a mathematician is forced to subdivide mathematics into several subjectareas, then topology / geometry will be one of them. In the first part of thiscourse we will discuss some of the characteristics that distinguish topologyfrom algebra and analysis.

Any calculus student has been exposed to continuous functions, limit, limitpoint, closed set (which is a set containing all its limits points) open interval (which is a set when it contains a point x also contains all points sufficientlynear to x.

Topology is exactly that area of mathematics which takes the concepts oflimit, limit point and continuity , gives them their most general meangfulrigorous definitions , and studies the consequences.

In this course , we shall see that the ideas of continuity and limitspresented here are the most fruitful generalizations of the ideas of continuityand limits presented in the calculus book and the metric spaces andtopological spaces, which will be studied later are just the proper settings for astudy of these ideas. The branch of topology dealt with in chapter 1 is calledgeneral topology .

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A- Topological spaces:

Definition 1.1:Let X be a non-empty set. A collection of subset of X is called a topology

on X if it meets the following requirements:1. X and belong to .2. The intersection of any two sets in belong to ( i.e. finite intersection of sets).3. The union of any number of sets in belong to .

A pair X, , X is a non-empty set and is a topology on X is called atopological space.The members of are called -open sets.

Example 1.2:Let X a,b,c,d,e and let

1 X,,a,a,c,d,c,d,b,c,d,e

2 X,,a,c,d,a,c,d,b,c,d

3 X,,a,c,d,a,c,d,a,b,d,e

Then 1 is a topology on X , while 2 is not a topology on X since

a,c,d b,c,d a,b,c,d ∉ 2.

Also 3 is not a topology on X since

a,c,d ∩ a,b,d,e a,d ∉ 3.

Remark :We can define more than one topology on X.

Example 1.3:Let X a,b,c , and

D X,,a,b,c,a,b,a,c,b,c

i.e. D is the power set of X. Then X,D is a toplogical space and is calledDiscrete topological space.

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Example 1.4:Let X be any non-empty set, and I X,.Then X, I ) is a topological space which is called indiscrete topology.

Example 1.5:Let X be any infinite set and

u ⊆ X : uc is finite .

Show that (X , ) is a topological space which is called cofinite topology.

Solution :1. Since Xc finite X ∈ . By given ∈ .Hence X , ∈ .

2. Let u1, u2 ∈ u1c and u2

c are finite sets.We want to prove that u1 ∩ u2 ∈ i.e. (u1 ∩u2c is finite.Consider

u1 ∩ u2c u1c u2

c is finite u1 ∩ u2 ∈

3. We want to show that i∈I ui ∈ where ui is any class of members of . i.e. i uicis finite.Consider

i uic ∩ i∈I uic which is finite.

X, is a topological space.

Example 1.6:Let X a,b,c,d and

1 X,,a,b , 2 X,,c

Then 1,2 are topology on X. Consider 1 2 X,,a,b,c

1 2 is not a topology on X.Remark :Union of topologies need not be a topology.

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Theorem 1.7:Let 1,2 be two topologies on X. Then 1 ∩ 2 is also a topology on X.

Proof :1. Since X , ∈ 1 , and X , ∈ 2 X, ∈ 1 ∩ 2.

2. Let u1 and u2 ∈ 1 ∩ 2 u1 , u2 ∈ 1 and u1 , u2 ∈ 2 u1 ∩ u2 ∈ 1 and u1 ∩ u2 ∈2 u1 ∩ u2 ∈ 1 ∩ 2.

3. Let {ui} be any class of members of 1 ∩ 2 {ui} ∈ 1 and {ui} ∈ 2. i∈I ui ∈ 1 and i∈I ui ∈ 2 i∈I ui ∈ 1 ∩ 2.Hence 1 ∩ 2 is a topology on X.

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B-TOPOLOGICAL SUBSPACES:

Definition 1.8:Let X, be a topological space. Let A be a non empty subset of X. A

relative topology on A is defined to be the class of all intersections of A withopen subsets of X. i.e.

A v A ∩ u : u ∈ .

Then A is a topology on A, A,A) is called a topological subspace ofX,.

Example 1.9:Let X a,b,c,d,e, and A a,d,e.Let

X,,a,c,d,a,c,d,b,c,d,e.

Then clearly X, is a topological space. Find A,A ?

Solution :

X ∩ A A, ∩ A ,a ∩ A a, c,d ∩ A d,a,c,d ∩ A a,d, b,c,d,e ∩ A d,e.

A A,,a,d,a,d,d,e

Example 1.10:Let X 1, 2, 3, 4, 5, and let A 2, 3, 4.Define

X,,1, 2,4, 5,1, 2, 4, 5.

Then A A,,2,4,2, 4 which is a topology on A.

Example 1.11:Let X,D, X, I be two topological spaces mentioned in examples 1.3, 1.4

respectively then:1. Every topological subspace of discrete topological space is discrete.2. Every topological subspace of indiscrete topological space is indiscrete.

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Example 1.12:Let X 1, 2, 3 , and X,D be a discrete topological space. i.e.

D X,,1,2,3,1, 2,1, 3,2, 3

Let A 1, 3, find DA?

DA A,,1,3.

Then A,DA is a discrete topological space.

Example 1.13:Let X 1,−1, i,−i, and A i,−i, 1.Let

X, ,i,−i,i,−i.

Then (X, is a topological space.Let

A A,,i,−i,i,−i and A′ A,,i,−i.

Then A , A′ both are topology on A. But A is a topological subspace of ,while A′ is not a topological subspace of because its members are not theintersections of members of .

Theorem 1.14:Let (X, be a topological space. Let A be a non-empty subset of X. Define

A v : v A ∩ u : u ∈

Then A,A) is a topological space which is called a topological subspace ofX,.(A is a relative topology).

Proof :1. ∵ X, ∈ , then A ∩ X A, A ∩ A, ∈ A .

2. Let v1, v2 ∈ A v1 A ∩ u1 and v2 A ∩ u2 where u1 , u2 ∈ .Consider

v1 ∩ v2 A ∩ u1 ∩ A ∩ u2 A ∩ u1 ∩ u2 A ∩ u3

v1 ∩ v2 ∈ A.

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3. Let vii∈I be a class of members of A then, vi A ∩ ui for some ui ∈

i∈I vi i∈I A ∩ ui A ∩ i∈I ui

i∈I vi ∈ A A,A) is a topological space .

Remark :Every ≠ A ⊆ X of a topological space can be given relative topology andhence can be considered as a topology of (X, ).

Definition 1.15:1. Let 1,2 be topologies on a non-empty set X. If each 1 - open subset of X is also a 2

-open subset of X , i.e. 1 ⊆ 2 , then 1 is said to be coarser (weaker) or smaller than2, and 2 is called finer (larger) than 1.

2. Two topologies on X are called incomparable if neither is coarser than the other .

Example 1.16 :Let X a,b,c,

1 X,,a,b, 2 X,,a,b,b,c,b

Then 1 , 2 are two topologies on X.∵ each member of 1 is contained in 2 thus 1 ⊆ 2 and 2 is a finer

than 1, or 1 is a coarser than 2.

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C- Open sets and closed setsDefinition 1.17:1. Let (X, ) be a topological space a subset u ⊆ X is said to be open iff u ∈ .

2. A subset A ⊆ X is called closed set if its complement Ac ∈ .

Example 1.18 :Let X a,b,c,d,e and

X,,a,c,d,a,c,d,b,c,d,e

then ( X, ) is a topological space.

X,,a,c,d,a,c,d,b,c,d,e are open sets.

,X,b,c,d,e,a,b,e,b,e,a are closed sets.

Remark 1.19 :1. X, always open and closed .2. There are subset which are both open and closed as b,c,d,e.3. There are subset which are neither open nor closed as a,b.

Remark 1. 20 :1. In a discrete topological space , every subset of X are both open and closed .2. In an indiscrete topological space, every subsets other than X, are neither open nor

closed.

Remark 1.21 :Definition 1.1 , now can be considered as follows:

1. X, are open sets.2. Intersection of any two open sets is also open.3. Union of any number of open sets is also open.

Remark Therefore to give topology to X means to define open sets in X.

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Theorem 1. 22 :Let ( X, ) be a topological space then the collection of closed sets G has

the following properties:1. The intersection of any number of closed sets is closed .2. The union of any two closed sets is closed.3. X and are closed sets.

Proof :1. Let Ai : i ∈ I be any member of closed sets belong to G (collection of closed sets).To

show that ∩i∈I Ai is closed , we have to show that its complement is open i.e. ∩i∈I Aicis open ?? Consider

∩i∈I Aic i∈I Aic

∵ each Ai is closed each Aic is open∵ ( X, ) is a topological space i∈I Aic is open.Then ∩ i∈I Ai is closed .

2 Let Ai, Aj be any two closed sets. Then Aic and Ajc are two open sets.∵ ( X, ) is a topological space Aic ∩ Ajc is open Ai Ajc Aic ∩ Ajc is open. Ai Aj is closed .

3. ∵ X, are open Xc and c are closed and X are closed.

Remark :Let (R,) be usual topology . Then all open intervals are open sets .All closedintervals are closed sets.

Definition 1.23:Let (X,) be a topological space. A point p ∈ X is an accumulation point

or limit point (cluster or derived point ) of a subset A of X iff every open set ucontains a point of A different from p.i.e.,

∀u ∈ open, p ∈ u u − p ∩ A ≠ .

The set of accumulation points of A , denoted by A′, is called thederived set of A.

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Example 1.24:Let X a,b,c,d,e.Define

X,,a,c,d,a,c,d,b,c,d,e.

Consider the subset A a,b,c of X. Find the derived set of A.

Solution :- a is not a limit point of A since a ∈ aand a − a ∩ A .- b is a limit point of A, since the open sets containing b are X and

b,c,d,e.Hence

X − b ∩ a,b,c c,d,e ≠

b,c,d,e − b ∩ a,b,c c ≠ .

- c is not a limit point ,while d,e are limit points A′ b,d,e.

Example 1.25:Let X a,b,c, I X,, then

A′

if A

X − p if A pX if A contains two or more points

Definition 1.26:Let (X,) be a topological space, and A ⊆ X. The closure of A denoted by

Ā defined to be the intersection of all closed supersets of A.i.e. iff Fi, i ∈ Iis the class of all closed subsets containing A, then the closure of A is

Ā ∩ i Fi : A ⊂ Fi, i ∈ I.

Remark :1. Ā is a closed set.2. Ā is the smallest closed set containing A.

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Example 1.27:Let X a,b,c,d,e and

X,,a,c,d,a,c,d,b,c,d,e

Take A b. Find the closure of A ?The closed sets are :

,X,b,c,d,e,a,b,e,b,e,a.

All closed sets containing A are :

X,b,c,d,e,a,b,e,b,e.

Therefore

Ā X ∩ b,c,d,e ∩ a,b,e ∩ b,e b,e.

Ā b b,e.

Similarly the closure of B a,c is B a,c X.The closure of b,d b,c,d,e.

Theorem 1.28:Let(X,) be a topological space and A ⊂ X. Then

A is closed A Ā

Proof :() Suppose A is closed. Then by the definition of the closure , Ā is the

smallest closed set containing A .

A ⊂ Ā (1)

it remains to show that Ā ⊂ A.Let x ∈ Ā x ∈ Ā ∩Fi, i ∈ I, where Fi is the class of all closed

subsets containing A. But A is closed A is one of Fi

x ∈ A Ā ⊂ A (2)

Thus Ā A.

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() Conversely , let A Ā . Since Ā is closed A is closed.

D- Interior, Exterior and BoundaryDefinition 1.29:1. Let A be a subset of a topological space X. A point p ∈ A is called an interior point of A

if p belongs to an open set u ∈ contained in A.i.e . p ∈ u ⊂ A u is open.

The set of interior points of A, denoted by A∘ or intA .

2. The exterior of A,denoted by extA is the interior of the complement of A i.e.extA intAc.

3. The boundary of A ,denoted by bA is the set of points which do not belong to theinterior or the exterior of A.

Example 1.30Consider the four intervals a,b, a,b, a,b and a,b. The interior of

each intervals a,b and the boundary of each is a,b .

Example 1.31:Let X a,b,c,d,eand

X,,a,c,d,a,c,d,b,c,d,e

Take A b,c,d then:

A∘ c,d

since c ∈ c,d ⊂ b,c,d and d ∈ c,d ⊂ b,c,d.

extA intAc inta,e a

since a ∈ a ⊂ a,e.

bA b,e

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Theorem 1.32:1. The interior of a set A is the union of all open subset of A.That is if Gi is the class of

open subset of A then A∘ i Gi.2. The interior of A is open i.e A∘is open .3. A∘is the largest open subset of A.That is if G is an open subset of A then G ⊂ A∘ ⊂ A.4. A is open A A ∘

Proof :1. Let Gi be the class of open subsets of A.

Let x ∈ A∘ x ∈ Gi∘ ⊂ A x ∈ i Gi A∘ ⊂ i Gi.Let x ∈ i Gi x ∈ Gi∘ ⊂ A x is an interior point x ∈ A∘ i Gi ⊂ A∘

Hence A∘ i Gi.

2. A∘ is open ,since A∘ is a union of open subset of A.

3. Let G be an open subset of A. Then G ∈ Gi i.e. G belong to the class of all open subsetof A. G ⊂ i G G ⊂ A∘ ⊂ A.

4. Let A be an open set ,then by 3 , A ⊂ A∘ ⊂ A A∘ AIf A A∘ A is open.

Example 1.33:Let X R (real number) .Then every open intervals a,b of real numbers

is an open set Let be the class of all open sets of real numbers .Then(R,) isa topology which is called a usual topology on R.

Solution :1. R , are open sets R, ∈ .

2. Let u,v ∈ u,v are open setevery point in u and v is an interior point we have toshow u ∩ v is open i.e (every point in u ∩ v is an interior point ).

Let p ∈ u ∩ v p ∈ u and p ∈ v

∃ open intervals u1,v1 s.t. p ∈ u1 ⊂ u ∧ p ∈ v1 ⊂ v

p ∈ w ⊂ u ∩ v where w u1 ∩ v1 u ∩ v ∈ .

3. Let ui be a class of members of ,we want to show that i∈I ui ∈ .

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Let p ∈ i∈I ui

∃ u∘ ∈ s.t. p ∈ u∘

∃ Sp s.t. p ∈ Sp ⊂ u∘ ⊂ ui

Therefore ui is open.Hence (R,) is a topology .

Example 1.34:An open disc D in the plane R2 is the set of points inside a circle

D x,y : x − a2 y − b2 r2

x,y : dp,q r

where p x,y,q a,b ∈ R2 and dp,q denoted the usual distancebetween two points p,q ∈ R2.

Let the class of all open discs in the plane R2 . Then(R2,) is atopology which is called the usual topology on R2.

Solution :1. R2, are open subsets of R2.

2. Let D1and D2 be two open discs say

D1 q ∈ R2 : dp1,q r1 and D2 q ∈ R2 : dp2,q r2

Then D1 ∩ D2 is also an open set .Let p0 ∈ D1 ∩ D2 p0 ∈ D1 ∧ p0 ∈ D2

dp1,p0 r1 , dp2,p0 r2

Let

r minr1 − dp1,p0, r2 − dp2,p0 0

and let

D q ∈ R2 : dp0,q r/2

Then p0 ∈ D ⊂ D1 ∩ D2. Thus p0 is the interior point of D1 ∩ D2.Therefore D1 ∩ D2 is open .

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3. Let A be a class of open subsets of R2 and letU u : u ∈ A

We want to show that every point of U is an interior point.Let p ∈ U ∃ u0 ∈ A s.t. p ∈ u0 .But u0 is an open set ∃ an open disc Dp s.t. p ∈ Dp ⊂ u0

p ∈ Dp ⊂ u0 ⊂ u : u ∈ A U.

Thus U is open.Hence R2, is a topology .

Theorem 1.35:Let X, be a topological space A ⊂ X .Then

A is closed A ′ ⊂ A

Proof : Suppose A is closed and A′ ⊈ A .Then

∃ p ∈ A ′ s.t. p ∉ A p ∈ Ac

But A is closed Ac is open . Hence

p ∈ Ac s.t. Ac ∩ A .

p is not a limit point of A (i.e.) p ∉ A′ →←.Hence if A is closed A′ ⊂ A.

Conversly let A ′ ⊂ A we have to show that A is closed (i.e.) Ac isopen .

To prove that Ac is open it is enough to show that if p ∈ Ac is an arbitrarypoint, then p is an interior point of Ac.

Let p ∈ Ac p ∉ A , but A ′ ⊂ A p ∉ A ′

∃ an open set G s.t.

p ∈ G and G − p ∩ A .

But p ∉ A G ∩ A .So p ∈ G ⊂ Ac. Thus p is an interior point ofAc

Ac is open A is closed .

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Theorem 1.36:Let X, be a topological space .Prove the following :

1. A ⊂ B A ′ ⊂ B ′.2. A A ′ is closed .3. A A A ′.4. A ⊂ B A ⊂ B .

Proof :1. A ⊂ B A ′ ⊂ B ′.

Let p ∈ A′ ∃ an open set G s.t. G − p ∩ A ≠ .But

A ⊂ B G − p ∩ A ⊂ G − p ∩ B

G − p ∩ B ≠ .Thus p ∈ B′ A′ ⊂ B ′.

2. A A ′ is closed .To show that A A ′ is closed we have to prove that A A ′c is open

.i.e., very point of A A′c is an interior point.Let p ∈ A A′c

p ∉ A A ′ p ∉ A ∧ p ∉ A ′

∃ an open set G s.t.

p ∈ G, G − p ∩ A .

But p ∉ A G ∩ A .Moreover ,G ∩ A′ . for if g ∈ G , then

g ∈ G : G ∩ A . g ∉ A/ G ∩ A/ .

Hence G ∩ A G ∩ A′ .Accordingly,

G ∩ A A′ G ∩ A G ∩ A′ .

p ∈ G ⊂ A A′c . Thus p is an interior point A A ′c is open A A ′ is closed .

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3. Ā A A ′

We have to show that Ā ⊂ A A ′ and A ′ A ⊂ Ā.By (2) ,since A A′ is closed containing A and Ā is the smallest closed

set containing A, then

A ⊂ Ā ⊂ A A ′ (1)

By the defintion of the closure we have A ⊂ Ā. Since Ā is closed ,then by Theorem 1.35,

Ā ′ ⊂ Ā

Also, by (1),

A ⊂ Ā A ′ ⊂ Ā ′

Hence we get

A ′ ⊂ Ā ′ ⊂ Ā , A ⊂ Ā and A ′ ⊂ Ā A A ′ ⊂ Ā (2)

Therefore Ā A A ′.

4. A ⊂ B Ā ⊂ B̄A ⊂ B A ′ ⊂ B ′ A A ′ ⊂ B B ′ Ā ⊂ B̄.

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E- Neighborhood and NeighborhoodSystem

Definition 1.37:1. Let X, be a topological space , p ∈ X , a subset N of X is called a neighborhood of p iff

∃ u ∈ s.t. p ∈ u ⊆ N.

Notation :(N is a neighborhood of p) is a reverse of the relation (p is the interior point of N)

The class of all neighborhoods of p ∈ X denoted by Np is called theneighborhood system of p.

Example 1.38:Let X a,b,c and

X,,b, b,c.

Find the neighborhood system of all points of X?

Solution :1. Since a ∈ X ⊆ X , then Na X

2. Since b ∈ X ⊆ X, b ∈ b ⊆ a,b , b ∈ b ⊆ b,c ,b ∈ b ⊆ b, then

Nb X,b,b,c,a,b

3. Since c ∈ X ⊆ X, c ∈ b,c ⊆ b,c, then

Nc X, b,c

Remark 1.39:1. In the indiscrete topological space, each point has a single neighborhood which is the setX itself.

2. A neighborhood of a point may not be an open set.3. The definition of a neighborhood enables us any one who knows all the open sets in a

topological space to recognise neighborhoods.4. Open sets determine neighborhoods and conversely.

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Theorem 1.40:Let X, be a topological space. Then a subset A ⊆ X is open A is a

neighborhood of each of its points.

Proof :() Suppose A is an open set. Then

∀ p ∈ A : p ∈ A ⊆ A

Therefore A is a neighborhood of each its point.

() Suppose A is a neighborhood of each of its points. So for each pointp ∈ A,

∃ an open set up s.t. p ∈ up ⊆ A

Therefore p is an interior point of A A is open.

Theorem 1.41:Let X, be a topolgical space.Then

1. Each x ∈ X has a neighborhood.2. If A, B are neighborhoods of p, then A ∩ B is neighborhood of p.3. If A is a neighborhood of p and A ⊆ B, then B is a neighborhood of p.

Proof :1. Since X is open, then x ∈ X ⊆ X X is a neighborhood of each point.

2. Since A, B are neighborhoods of p,

∃ u1, u2 ∈ s.t. p ∈ u1 ⊆ A, and p ∈ u2 ⊆ B

p ∈ u1 ∩ u2 ⊆ A ∩ B where u1 ∩ u2 is open

Thus A ∩ B is a neighborhood of p.

3. Since A is a neighborhood of p

∃ u ∈ s.t. p ∈ u ⊆ A

But A ⊆ B ,

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p ∈ u ⊆ A ⊆ B

Thus B is a neighborhood of p.

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PROBLEMS (I)1. Let R be the set of real numbers. A subset u ⊆ R is called usual open if the following is

satisfied :∀ x ∈ u ∃ an open interval x − ,x where 0 s.t . x

x − ,x ⊆ uLet R consists of and all usual open subsets of R .Prove that R,R is a topological space .

2. List all topologies on X a,b .

3. Let be a topology on a set X consisting of four sets , i.e. X, , A, Bwhere A,B are non-empty distinct proper subsets of X. What conditions

must A and B satisfy.

4. List all topologies on X a,b,c which consists of exactly four members . (use problem3)

5. Let f : X Y be a function from a non-empty set X into a topological space Y,2.Let 1 f−1G : G 2 .Show that X,1 is a topological space .

6. Consider the following topology on X a,b,c,d,e X,,a,a,b,a,c,d,a,b,c,d,a,b,e

List the members of the relative topology A on A a,c,e .

7. Consider the usual topology on the real line R which is discribed in pro. 1 .Describe the relative topology N where N ⊆ R is a natural numbers .

8. Let A be a -open subset of X, and let A ⊂ Y ⊂ X .Show that A is also open relativeto the relative topology on Y .i.e. A Y.

9. Consider the usual topology on the real line R. Determine whether or not each of thefollowing subsets of A 0,1 are open relative to A .

i 1/2, 1 ii 1/2, 2/3 iii 0, 1/2.

10. Prove that the closure operater has the following properties :1. ̄ 2. A ⊂ A3. A B A B

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4. A A .

11. Prove that the interior operator has the following properties :1. Xo X2. Ao ⊂ A3. A ∩ Bo Ao ∩ Bo

4. Aoo Ao .

12. Let A be any set , for each p A , let up be a subset of A such thatpup ⊂ A

Then A up : p A .

13. Consider the topology X,,a,a,b,a,c,d,a,b,c,d,a,b,e

on X a,b,c,d,e .Find the derived sets ofi A c,d,eii B b .

14. Let A , B be subsets of a topological space X, . Then A B ′ A ′ B ′ .

15. Consider the following topology on X a,b,c,d,e X,,a,a,b,a,c,d,a,b,c,d,a,b,e

i List the closed subsets of X .ii Find the closure of a,b,c,e .iii Which sets in ii are dense in X .iv Find the interior points of A a,b,c .v Find ext A and bA .vi List all the neighborhood systems of every point of X .

16. Determine the neighborhood system of a point p in an indiscrete space X .

17. Determine whether or not each of the following intervals is a neighborhood of 0 underthe usual topology for the real line R .

a −1/2, 1/2, b −1, 0, c 0, 1/2, d 0, 1

Definition :A subset A is said to be dense in X if Ā X .

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A- BasisFor each of the examples in the preceding chapter , we were able to specify

the topology by discribing the entire collection of open sets. Usually, this istoo difficult. In most cases one specifies instead a smaller collection ofsubsets of X and defines the topology in terms of that.

Definition 2.1:

Let X, be a topological space. A class of open subsets of X , i .e., ⊂ , is a base for the topology iff :

Every open set u ∈ is the union of members of .

u B ,u ∈ , B ∈ .

Equivalently, ⊂ is a base for iff for any point p ∈ u there exists

Bp ∈ s.t. p ∈ Bp ⊂ u .

Example 2.2:let X a,b,c,d,e, and let

X,,a,b,c,d,a,b,c,d

Then X, is a topology . Let

X,,a,b,c,d

a subcollection of . Then is a base for .

Solution : is a base for , since every member in is a union of members of , for

X X , , a,b a,b a,b

c,d c,d c,d a,b,c,d a,b c,d .

Example 2.3:let R, be a usual topoplogy ,where R is a real line. Let be a

collection ofopen intervals . Then forms a base for R, .

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Solution :Let u ⊂ R be open with p ∈ u ⊂ R . Then by definition ∃ open interval

a,bs.t. p ∈ a,b ⊂ u . Thus is a base .

Example 2.4 : (Topology of the plane)An open disc D in the plane R2 is the set of point inside a circle , with

center p a1,a2 and radius r 0

D x,y : x − a12 y − a22 r2

Then the class u of all open discs in the plane R2 is a topology which iscalled the usual topology on R2 .Thus the open discs form a base for theusual topology on the plane R2.

Example 2.5 :The open rectangles in the plane R2 bounded by sides parallel to the x −

axis and y − axis , form a base for the usual topology on R2.

Solution :let u ⊂ R2 be an open set and p ∈ u.Then by definition of open set ∃

open discDp centered at p, s.t, p ∈ Dp ⊂ u . Then any rectangle B ∈ whose

vertices lie on the boundary of Dp satisfies

p ∈ B ⊂ Dp ⊂ u p ∈ B ⊂ Dp

Thus form a base for the usual topology on R2.

29

Example 2.6 :Consider a discrete topology on X a,b,c,

D X,,a,b,c,a,b,a,cb,c.

Then a,b,c is a base for this topology .

Solution :

Bi, i ∈ , X a b c, a a a,

b b b , c c c, a,b a b,

a,c a c , b,c b c .

Thus form a base for discrete topology .

Remark :For any discrete topology X,D , p : ∈ x of all singleton

subsets of X is a base for X,D .

In the following example we show that there may be more then one basefor a given topology .

Example 2.7 :Consider the following topology on X a,b,c,d,e

X,,a,a,d,c,b,c,a,b,c,d,c,d,e,b,c,d,e .

Then the following

1 ,a,b,c,b,c,d,e,d,e,d,c

2 ,a,b,c,d,e,d,c .

form basis for X, , where 2 ⊂ 1.

Given a class of subsets of X ,when will be a base for some topologyon X ? Clearly , X B : B ∈ is necessary for X is an open set in every

30

topology . The following example shows that other conditions are alsoneeded .

Example 2.8 :Consider the following topology on X a,b,c

X,,a,b,b,cb

and the following class of sets

a,b,b,c

Then this class cannot be a base of X, for

a,b ∩ b,c b

which is open in X and is not a union of any members of .

Theorem 2.9 :Let be a class of subsets of a non-empty set X . Then is a base for

some topology on X iff it possesses the following two properties :1. X B : B ∈ 2. For any B1,B2 ∈ , B1 ∩ B2 is the union of members of B. or equivalently

If p ∈ B1 ∩ B2 ∃ Bp ∈ s.t .

p ∈ Bp ⊂ B1 ∩ B2 .

Corollary 2.10:Let X be a set, be a basis for a topolgy on X. Then collection of

unions of elements of .

Example 2.11 :let be the class of open - closed intervals in the real line R.

a,b : a,b ∈ R, a b

Then is a base for the topology on R .

Solution :1. X R a,b : a,b ∈ .2. The intersection of two members of is

31

a,b ∩ c,d or a,b ∩ c,d c,bWhich in either cases a union of members of .Therefore form a base .Thus the class consisting of union of open -closed intervals is a topology

on R, which is called upper limit topology on R .

Similarly , the class ∗ a,b : a,b ∈ R a b form a base for atopology ∗ on R which is called lower limit topology on R.

32

B- SubbasisA question may occur to you at this point, since the topology generated by

a base may be described as the collection of arbitrary unions of elements of what happens if you start with a given collection of sets and take finiteintersections of them as well as arbitrary unions? This question leads to thenotion of a Subbase for a topology.

Definition 2.12 :Let X, be a topological space. A class S of open subset of , i.e.,

S ⊂ is a subbasefor the topolgy on X iff finite intersection of members of S form a

base for .

Example : 2.13 :Let X a,b,c,d,e and

X,,a,b,ca,b,c,b,c,d,ed,e,a,d,e.

Then X, is a topology on X.Let

S a,a,b,c,b,c,d,e,a,d,e.

Then S is subbase for .

Solution :The finite intersection of memebers of S are

a,,b,c,a,d,ea,b,c,b,c,d,e,d,e.

We have to prove that is a base for .Since

1. X B : B ∈ a b,c a,b,c ......d,e2. Any two members of , their intersection is a union of members of , example

a,b,c ∩ b,c,d,e b,c b,c b,cand so on .Hence is a base S is a subbase for .

33

Example 2.14Let R, be a usual topology i.e. R is a real line and be the class of all

open intervals of real members .Let S collection of all intervals of the form a, and − ,b .Then S is a subbase for the usual topology on R .

Solution :The finite intersection of a, ∩ − ,b is a,b a, ∩ − ,b .By example 2.3, the collection of open interval a,b form a base for the

usual topology . Hence S is a subbase for .

Example 2.15:let R2, be a usual topology . i.e R2 is a plane and be the class of all

open discs in the plane R2.Let S vertical and horizontal infinite open strip in the plane R2.Then S is a subbase for the usual topology on R2.

Solution :The finite intersection of a vertical and horizontal infinite open strip is on

open rectangle which form a base (see ex 2.5) . Hence S is a subbase .

34

C- Topologies Genarated by classes ofsets

Definition :let A be any class of subset of a non-empty set X . As we seen , A may

not be a base for a topolgy on X . However A always generates a topology onX.

Theorem 2.16:Any class A of subsets of a non-empty set X is a subbase for a unique

topology on X. That is finite intersections of members of A form a base forthe topology on X.

Example 2.17:Let X a,b,c,d and S a,b,b,c,d

S finite intersectionof members of S

union ofmembers of

Finite intersection of members of S gives the class

X,,a,b,b,b,c,d

Taking unions of members of gives the class

X,,a,b,b,c,b,d,b,d,a,b,db,c,d,a,b,c.

It is easy to check that is a topology on X .

35

PROBLEMS (II)

1. Show that the class of open equilateral triangles form a base for the usual topology R2.

2. Consider the upper limit topolgy on the real line R which has as a base the class of open- closed interval a,b. Show that

(i) 4, and − , 2 are − open sets .(ii) Any a, and − ,b are − open sets .(iii) a,b is both − open and − closed .

3. Let X a,b,c,d,e and let A a,b,c,c,d,d,e . Find the topology generatedby A .

4. Detrmine the topology on the real line R generated by the class A of all closed intervalsa,a 1 with length 1.

5. Consider the discrete topology D on X a,b,c,d,e. Find a subbase S for D whichdoes not contain any singleton sets .

6. Let S be a subbase for a topology on X , and let A ⊆ X . Show that the classSA A ∩ S : S ∈ S

is a subbase for the relative topology A,A .

36

Continuity in Topological SpaceThe concept of continuous function is a basic to much of mathematics.

Continuous functions on the real line appear in any calculus book, andcontinuous functions in the plane and in space follow not far behind. Moregeneral kinds of continuous functions arise as one goes further inmathematics. In this chapter, we shall formulate a definition of continuity thatwill include all these as special cases; and we shall study various properties ofcontinuous functions. Many of these properties are direct generalizations ofthings you learned about continuous functions in calculus and analysis.

Definition 3.1:Let (X,1) and (Y,2) be topological spaces . A function f : X Y is

continuous relative to 1 and 2 , or 1 − 2 continuous,or simplycontinuous, iff the inverse image of every open set of Y is a 1 −open set ofX. i.e.,

∀ H ∈ 2 f−1H ∈ 1

Example 3.2 :Consider the following topologies 1, 2 on X a,b,c,d and

Y x,y, z,w respectively:

1 X,,a,a,b,a,b,c,

2 Y,,x,y,x,y,y, z,w

Define f : X → Y, g : X → Y by the following diagram :

f−12 f−1Y, f −1, f −1

x, f −1y, f −1

x,y, f −1y, z,w

X,,a

37

then the inverse image of every open set in 2 is open in X relative to1 f is a continuous function .

g−12 X,,a,b,c,d

but c,d is not open in 1 g is not a continuous function .

Example 3.2 :Let X,D be discrete topological space, and Y, be any topological

space.Then any function f : X → Y is continuous , since if H ∈ f −1H

is an open subset of X , because all subsets of X belong to discrete topology .

Example 3.3 :Let X, be any topological space and Y, I be an indiscrete topological

space , then any function f : X → Y is continuous ,since f −1Y X and

f −1 which are open .

Example 3.4 :Let f : X → Y where X and Y are topological space , and let be a base for

the topology on Y .Suppose for each member B ∈ , f −1B is an open

subset of X .Then f is a continuous function .

Solution :Let H be an open subset of Y . Since is base for the topology on Y , then

H i Bi , Bi ∈

Consider

f −1H f −1

i Bi i f−1Bi

But each f −1Bi is open i f

−1Bi is open. Hence f −1

H isopen .Accordingly , f is continuous .

Corollary 3.5 :A function is f : X → Y continuous the inverse of each member of a

base for Y is an open subset of X.

38

Theorem 3.6 :Let f : X,1 → Y,2 ,and let S be a subbase for the topology 2 on Y

.Then f is continuous the inverse of every member of the subbase S is anopen subset of X; i.e. f −1

S ∈ 1 for every S ∈ S .

Proof : Suppose f is continuous , then the inverse of all open sets , including

the member of S are open .

39

Suppose f −1S ∈ 1 for every S ∈ S .We want to show that f is

continuous , i.e. if

G ∈ 2 f −1G ∈ 1 ??

Let G ∈ 2 ,then finite intersections of members of S gives a base ,i.e.,

S1 ∩ S2 ∩ ......... ∩ Sn ∈

Then every element of 2 is a union of members of . i.e.,

G i Si1 ∩ Si2 ∩ ...... ∩ Sin , Sik ∈ S

f −1G f −1

Si1 ∩ Si2 ∩ ...... ∩ Sin

i f−1Si1 ∩ f

−1Si2 ∩ ...... ∩ f −1

Sin

f −1 Sik is open in 1 finite intersection ∈ 1 arbitrary union of

finite intersection f −1G ∈ 1.

Theorem 3.7 :A function f : X → Y is continuous iff the inverse image of every closed

subset of Y is a closed subset of X .

Proof :: Suppose f is a continouous function ,and let F be a closed subset of

Y.

Fc is open subset of Y

f−1Fc f −1Fc is open subset of X

f−1F is a closed subset of X.

Conversely , suppose F is a closed subset of Y f −1Fis closed in

X.We want to show that is f continuous ??Let G ∈ Y be an open set .Then Gc is a closed subset of Y

f −1Gc f −1

Gc is closed in X

f −1G is open in X

40

Hence f is a continuous function .

41

Definition 3.8 :1. A function f : X → Y is called open function if the image of every open set in X is open

set in Y.2. A function f : X → Y is called closed function if the image of every closed set in X is

closed set in Y .

Homeomorphic Spaces :

Definition 3.9 :Two topological spaces X and Y are called homeomorphic or

topologically equivalent if there exists a bijective function f : X → Y suchthat f and f −1 are continuous .The function f is called a homeomorphism .

Theorem 3.17 :For a 1-1 mapping f of a topological space X onto a topological space Y ,

the following conditions are equivalent :1. The mapping f is a homeomorphism .2. The mapping f is closed and continuous .3. The mapping f is open and continuous .4. The set fA is closed inY A is closed in X .5. The set fA is open in Y A is open in X .

Proof :1 2Suppose f is a homeomorphism , then f , f −1are continuous

functions. We have to prove that f is closed and continuous. By given f iscontinuous .It remains to show that if

A ⊆ X is closed , then fA is closed in Y ??.

Let A ⊆ X be a closed subset .Since f −1 : Y → X is continuous , then theinverse of every closed sets in X is closed in Y .

f−1−1A fA is closed in Y.

f is closed .

Conversely,assume (2) holds ,we want to prove that f is a

42

homeomorphism .i.e, f −1 is a continuous function .Let A ⊆ X be a closed set . Then by (2) f A is a closed set in Y . Since

f −1 −1A fA which is closed in Y

f −1 is continuous .—————————————————————————

1 3Suppose f is a homeomophism . To show that f is open , let A ⊆ X be an

open set.∵ f −1 is continuous f−1−1A is open in Y f A is open in

Y f is open.

Conversely,assume 3 holds .To prove that f −1 is a continuous function,let A ⊆ X be an open set , then (f −1

−1A fA is open by 3 f −1 isan open function .

————————————————–2 4Suppose f is a closed function which is continuous .We want to prove

that

f A is closed in Y A is closed in X

Let fA be a closed set in Y. Since f is continuous f −1

fA A is closed in X .Suppose A is closed in X fA is closed in Y.

Conversely, suppose f A is closed in Y A is closed in X.We want to prove that f is closed and f is continuous .Let A ⊆ X be a closed set in X f A is closed in Y .Since f −1

fA A the inverse of a closed set of Y is closed in X.

____________________________________________3 5 exercise.

Example 3.10:let X and Y be discreate spaces .Then by example 3.2 ,all functions from X

one to the other are continuous .Hence X and Y arehomeomorphic there exists a one -one ,onto function , from one to the

43

other .

Example 3.11 :Let f : R R defined by f x x2. Then f is not open ,since

f −1, 1 0, 1 which is not open.

44

Example 3.12 :Let X 1, 2, 3, 4, 1 X,,1, 2,3, 4 and

Y a,b,c,d, 2 Y,,a,d,b,c,

Let f : X → Y be defined by

f1 a, f4 b, f3 c , f2 d

Prove that f is a homeomorphism .

Solution :1. Clearly f is 1-1 and onto .

2. f is continuous since

f −1Y X, f −1

, f −1a,d 1, 2, f −1

b,c 3, 4.

3. f −1 : Y → X is defined byf −1

a 1, f −1b 4, f −1

c 3 , f −1d 2

Then f −1 is continuous , since

f −1−1X Y, f −1

−1 , f −1−11, 2 a,d f −1

−13, 4 b,c

Hence f is a homeomorphism .

Example 3.13 :Let f : R → R be a function defined by fx x ∀x ∈ R where R, is

a usual topology .Show that f is a homeomorphism .

Solution :

Let a,b be an open interval of R . Then f −1a,b a,b which is

open f is continuous.Moreover , f is 1-1 , onto and f f −1, thus f is a homeomorphism .Every identity map of X,x is a homeomorphism .

Example 3.14 :Let f : R → −1, 1 be defined by

45

fx x1 |x|

Prove that f is a homeomorphism . [R, −1, 1 are topologicallyequivalent]

Solution :

fx

x1 x if x 0

0 if x 0x

1 − x if x 0

1. f is 1 − 1 :Let fx1 fx2 x1 , x2 0 or x1 , x2 0 .If fx1 , fx2 0 , then consider fx1 fx2

x11 x1

x21 x2

x1 x1x2 x2 x1x2

x1 x2

Similarly ,when fx1 , fx2 0.

2. f is onto :Let y ∈ −1, 1 , we want to find x ∈ R s.t. fx y.Since y ∈ −1, 1

y 0 or y 0 or y 0

If y 0 ,then fx y

x1 x y

x y xy x − xy y x1 − y y

x y1 − y ∈ R s.t. f y

1 − y y

Similarly , when y 0 .

46

3. f is continuous:Let a,b be an open interval in −1, 1. Since fx a

x1 x a x a

1 − a a 0 f −1a x

Therefore

f −1a,b a

1 − a , b1 − b is open in R .

Similarly if a 0.

4. f is open .Let a,b ⊂ R be an open interval , then

fa,b fa, fb a1 a , b

1 b if a,b 0

fa,b a1 − a , b

1 − b if a,b 0

f is a homeomorphism .

Topological Properties :

Definition 3.15 :A property P of sets is called topological invariant or topological if

whenever a topological space X, has P then every space homeomorphicto X, also has P.

Example 3.16:As in example 3.14, R and −1, 1 are homeomorphic .Hence length , and

boundedness are not a topological property since −1, 1 is bounded but R isnot .

47

PROBLEMS(III)

1. Let f : X → Y and g : Y → Z be continuous . Prove that the composition functiong ∘ f : X → Z is also continuous .

2. A function f : X Y is continuous fĀ ⊂ fA where A is any subset of X.

3.’ Let f : X, → Y,∗ be continuous . Then

fA : A,A → Y,∗ is continuous

where A ⊂ X ,and fA f ∣ A (restriction map).

4. Let be a base for a topological space X .Show that if f : X → Y has the propertythat f() is open for every B ∈ . Then f is an open function .

5. Let f : X, → Y,∗ be open and onto , let be a base for . Show thatfB : B ∈ is a base for ∗ .

6. Let f : X, → Y,∗ be 1-1 and open, let A ⊂ X, and fA B.Show that

fA : A,A B,B∗

is also open and 1-1.

48

METRIC SPACES

In this chapter we start by defining a metric space which is a generalizationof the concept of distance in elementry geometry.

Definition 4.1 :Let X be a non-empty set. A metric (distance function) on a set X is a

real-valued function

d : X X → R

which satisfies the following conditions :1. dx,y ≥ 0 ∀ x,y ∈ X2. dx,y 0 x y3. dx,y dy,x ∀ x,y ∈ X4. dx, z ≤ dx,y dy, z ∀ x,y, z ∈ X

A metric space is a pair X,d where X is the set and d is a metric on X.

Example 4.2 :Consider the real line and define a function d : R R → R by

dx,y |x − y|

Then R,d is a metric space which is called usual metric on R.Solution :1. |x − y| ≥ 0 ∀ x,y ∈ R dx,y ≥ 02. Let x y |x − y| 0 dx,y 0 .3. dx,y |x − y| | − y − x| |y − x| dy,x4. dx,y |x − y| |x − z z − y| |x − z z − y|

≤ |x − z| |z − y| dx, z dz,y

Example 4.3 :Let X be a non-empty set. Define d by the following :

49

dx,y 0 if x y1 if x ≠ y

Then X,d is a metric space which is called (discrete) metric on X .

Solution :1. dx,y ≥ 02. dx,y 0 x y3. dx,y dy,x4. If x z ,then dx,y 1 ≤ dx, z dz,y 1

If x ≠ z ,then dx,y 1 dx, z dz,y 1 1 2Thus X,d is a metric space.

Recall that :If v x1,x2 ∈ R2 is a vector in the plane ,then the length of v denoted

by ||v|| , is defined by

||v|| x12 x2

2

Definition 4.4 :Let x x1,x2, ....,xn ∈ Rn. Then the length of x which is called the

norm of x ,and denoted by ||x|| is defined by

||x|| x12 x2

2 ...... xn2 ∑ |xi|2½

( which is the distance between x and the origin )

Remark 4.5 :If x x1,x2, ......,xn, y y1,y2, ....,yn ∈ Rn , c ∈ RThen :

1. x y x1 y1,x2 y2, ......,xn yn2. x − y x1 − y1,x2 − y2, ......,xn − yn3. cx cx1,cx2, ......,cxn

4. x ∙ y x1y1 x2y2 ...... xnyn ∑i1

nxiyi ( dot product )

50

Theorem 4.6: (Cauchy-Schwarz Inequality)Let x x1,x2,.......,xn and y y1,y2, ......,yn be elements in Rn.Then

|x .y| ≤ ‖x‖‖y‖

i.e. ∑i1

n

xiyi ≤ ∑i1

n

|xi |212

∑i1

n

|yi |212

Proof :Case 1: when x 0 or y 0Then the inequality reduces to 0 ≤ 0,and therfore true.

Case 2 : x ≠ 0 and y ≠ 0For any real numbers a,b ∈ R, we have

a − b2 ≥ 0 a2 − 2ab b2 ≥ 0 a2 b2 ≥ 2abThis relation is true for any real number , so let

a |xi |‖x‖ , b |yi |

‖y‖

|xi |2

‖x‖2 |yi |2

‖y‖2 ≥ 2 |xi ||yi |‖x‖‖y‖ (*)

So summing ∗ with respect to i and using |aibi | |ai | |bi | we get

∑i1

n|xiyi |

‖x‖‖y‖ ≤ 12 ∑

i1

n|xi |2

‖x‖2 |yi |2

‖y‖2

51

1‖x‖‖y‖ ∑

i1

n

|xiyi | ≤ 12

1‖x‖2 ∑

i1

n

|xi |2 1‖y‖2 ∑

i1

n

|yi |2

1‖x‖‖y‖ ∑

i1

n

|xiyi | ≤ 12

‖x‖2

‖x‖2 ‖y‖2

‖y‖2

∑i1

n

|xiyi | ≤ ‖x‖‖y‖

|x.y| ≤ ‖x‖‖y‖

Theorem 4.7: (Minkowski’s Inequality)Let x x1,x2, ......,xn, y y1,y2, ........,yn ∈ Rn.Then

‖x y‖ ≤ ‖x‖ ‖y‖

Proof :Let ‖x y‖ ≠ 0, for any real number xi,yi ∈ R, we have

|xi yi | ≤ |xi | |yi |. Hence

‖x y‖2 ∑i1

n

|xi yi |2

∑i1

n

|xi yi ||xi yi |

≤ ∑i1

n

|xi yi ||xi | |yi |

∑i1

n

|xi yi ||xi | ∑i1

n

|xi yi ||yi |

But by Cauchy-Schwarz Inequality ∑i1n |xi yi ||xi | ≤ ‖x y‖‖x‖

Then

‖x y‖2 ≤ ‖x y‖‖x‖ ‖x y‖‖y‖≤ ‖x y‖‖x‖ ‖y‖

since‖x y‖ ≠ 0, then ‖x y‖ ≤ ‖x‖ ‖y‖.

52

Example 4.8:

Consider the plane R2. Define d : R2 R2 → R by

dx1,x2, y1,y2 x1 − y12 x2 − y22

Then R2,d is a metric space called a usual metric on R2.In general , if x x1, ..........,xn, y y1, .......,yn ∈

Rn, then d : Rn Rn → R is defined by

dx,y ∑i1

n

xi − yi21/2

Rn,d is a metric space which is called Euclidean metric for Rn.

Open sets in a metric space:Definition 4.9:

Let X,dbe a metric space. For any point p ∈ X and any real number 0, let Sp denote the set of points with in a distance of from p :

Sp x : dp,x ⊆ X

We call Sp the open sphere or sphere with center p and radius .

Example 4.10:Let X R be a metric space , p 1 and 1/2, then

S1/21 x : d1,x 1/2 x : |1 − x| 1/2 x : −1/2 1 − x 1/2 x : x ∈ R : 1/2 x 3/2 1/2, 3/2

53

S21 x : dx, 1 2 x ∈ R : |x − 1| 2 −1 x 3 −1, 3

Thus in the usual metric R,d,da,b |a − b|, the open sphere Sp isthe open interval p − ,p .

Example 4.11 :Consider the discrete metric on X.

dx,y 0 if x y1 if x ≠ y

Find S2p and S1/2p ?

54

Solution :1. S2p x : dx,p 2 x ∈ X : 0,1 2

∵ dx,y 0 or 1 2 ∀ x ∈ X . S2p X.

2. S1/2p x : dx,p 1/2Thus dx,y 0 1/2 when x p. S1/2p p ( center)

Example 4.12 :Consider the usual metric on R2, i.e.

dx1,x2, y1,y2 x1 − y12 x2 − y22

Let 1 and p 2, 3 ,find S1p ?

Solution :S1p x x1,x2 ∈ R2 : dx,p

x ∈ R2 : x1 − 22 x2 − 32 1which is the open disc .

Example 4.13 :Let X R2 and p x1,x2, q y 1,y2 ∈ R2.Define

dp,q x1 − y1 x2 − y2

Then R2,d is a metric space . Find S10, 0 ?

Solution :R2,d is a metric space (excrise) .

S10, 0 x,y ∈ R2 : dx,y, 0, 0 1 x,y ∈ R2 : |x| |y| 1

when y 0 − 1 x 1.when x 0 − 1 y 1.

55

Definition 4.14 :LetX,d be a metric space . A subset u of X is called an open set if :for every x ∈ u there exists an open sphere Sx s.t. x ∈ Sx ⊂ u .

Remark 4.15 :1. A subset u of X,d is an open set if each point of u occurs as the center of some open

sphere which is contained in u.2. If u is an open set in X,d , then ∀x ∈ u we can find a real no. 0 s.t. Sx ⊆ u.

Theorem 4.16Let X,d be a metric space Then X and are open sets.

Proof :.Let x ∈ X , and 0 . Then for every x ∈ X, the open sphere Sx

centered at x must contained in X .

i.e. Sx ⊆ X.

Thus X is open .

To show that is open , we must show that each point in is the centerof an open sphere contained in , but since there are no point in , then isopen .

Theorem 4.17:Let X,d be a metric space . Every open sphere is an open set .

Proof :Let Sp be an open sphere in X, and let x be a point in Sp. We must

find an open sphere centered on x and contained in Sp.

∵ x ∈ Sp dx,p

− dx,p 0.

Let

1 − dx,p

We must show that

56

S1x ⊆ Sp

Let

y ∈ S1x dy,x 1

Consider

dy,p ≤ dy,x dx,p 1

− dx,p −

y ∈ Sp S is an open set .

Theorem 4.18:Let X,d be a metric space . Then u ⊆ X is open u is a union of open

spheres.

Proof : Assume u is open. We want to show that u is a union of open sphere.If u , then u is a union of the empty class of open spheres .If u ≠ . Then since it is open , each point is the center of an open sphere

contained in it .

i.e., ∀x ∈ u, ∃ Sx s.t. Sx ⊆ u

Then u is a union of open spheres .

u S where S is a class of open spheres .We must show that u isopen .

If S u and by theorem 4.16, u is open .Suppose S ≠ u ≠ .Let x ∈ u x ∈ S x belongs to an open sphere say

x ∈ Sp in S

By theorem 4.17, SP is open ∃ open sphere S1x s.t.

S1x ⊆ Sp

since S1x ⊆ u, so we have an open sphere centered at x and containd in u.

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Thus u is open.

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The fundamental properties of the open sets in a metric space are thosestated in the following theorem :

Theorem 4.19:Let X,d be a metric space.Then :

1. Any union of open sets in X is open .2. Finite intersection of open sets in X is open .

Proof :1. U i ui , where ui arbitrary class of open sets in X .

If ui U U is open .Suppose ui ≠ . Then by theorem 4.18, U is a union of open

spheres( since ui is open ). Again using theorem 4.18 , U is open .

2. U ∩i1n ui where ui class of finite open sets in X ..If ui , Then ∩ i1

n which is open .suppose u1,u2, ......,un ≠ . If it happens that ∩ i1

n ui .Then by theorem 4.16, U is open .

Assume ∩i1n ui ≠ .Then let

x ∈ U ∩ i1n ui x ∈ ui ∀i 1, 2.....,n

But ui is open , for each i there is a ve real number i 0 such that

Six ⊆ ui

Let smallest ve real number in the set 1,2, ......,n.Then

Sx ⊆ Six for each i, Sx ⊆ ui for each i Sx ⊆ U ∩ i1

n ui

Thus U is open.

Definition 4.20 :Let X,d be a metric space ,Then by Theorems 4.16,4.19, the open

subsets of X,d form a topology on X . This topology is called the topologyon X induced by d or the topology on X generated by d . Thus

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Every metric space is a topological space .Remark 4.21:

All concepts defined for topological spaces are also defined for metricspaces .

For example , we can speak about open sets , closed sets , neighborhoods, limit point,.......

Example 4.22:Let R,d be a usual metric defined in example 4.2. i.e., dx,y |x − y|.

Then the open spheres in R are precisely the open intervals .Hence theusual metric on R induces the usual topology on R. Similarly, the usualmetric R2,d induces the usual topology on R2.

Example 4.23:Consider the discrete metric defined in example 4.11. Then for any

p ∈ X, S1/2p p.Hence every singelton is open ,and so every set is open.Therefore discrete

metric on X induces the discrete topology.

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PROBLEMS(IV)1. Prove that the Euclidean metric space which is defined in example 4.8 is a metric space.

2. Let d be a metric on a non-empty set X.Show that the function e defined by

ea,b min1,da,b

where a,b ∈ X is also a metric on X.

3. Let d be a metric on a non-empty set X .Show that the function e defined by

ea,b da,b1 da,b

where a,b ∈ X , is also a metric on X.

4. Let 1,2 be real numbers such that 0 1 ≤ 2.Show that the open sphereS1p ⊆ S2p.

5. Show that if S1p, S2p are two open spheres with the same center, then one of them isa subset of the other.

6. Let S1,S2 be open spheres and p ∈ S1 ∩ S2. Show that ∃ open sphere Sp such thatp ∈ Sp ⊂ S1 ∩ S2.

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Separation AxiomsThe definition of a topological space is very general ; not many interesting

theorems can be proved about all topological spaces .So in this section weshall discuss axiom of separation which concern the ways of separating pointsand closed sets in topological spaces, and examine the relationship betweenthese axioms and the concepts introduced in this course .

Definition 5.1:A topological space X is called a T0 − space iff it satisfies the folloing

axiom:For every pair of distinct points x,y ∈ X , ∃ an open set containing exactly

one of the points but not the other.

i.e., x ≠ y ∈ X, ∃ u ⊆ X is open s.t. x ∈ u and y ∉ u.

Example 5.2:X a,b,c , X,,a,a,b.Then X, is a T0 − space.

Solution :1. a,c ∈ X s.t. a ≠ c.

∃ an open set say a ∈ and a ∈ a, c ∉ a.

2. a,b ∈ X s.t. a ≠ b , ∃ a ∈ s.t. a ∈ a and b ∉ a.

3. b ≠ c ∈ X , b ∈ a,b s.t. c ∉ a,b.

Remark 5.3:

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1. Every discrete topological space is a T0 − space.2. Every indiscrete topological space is not a T0 − space.

Definition 5.4:A topological space X is called a T1 − space iff it satisfies the following

axiom:

∀x ≠ y ∈ X ∃ u,v ⊆ X s.t. x ∈ u, y ∉ u and y ∈ v, x ∉ u.

Note that : u,v are not neceesarily disjoint.

Remark :Every T1 − space is T0 − space.

Example 5.5:X 1, 2, X,,1 then X, is a T0 − space but not T1 − space.

Solution :X is a T0 − space since 1 ≠ 2 ∈ X and 1 ∈ 1 and 2 ∉ 1.It is not a T1 − space because there is no-open set containing 2.

Theorem 5.6:A topological space X is a T1 − space every singleton subset p of X is

closed.

Proof :( Suppose X is a T1 − space and pis a singleton of X. We must show

that pc is open.i.e. every x ∈ pc is an interior point.Let x ∈ pc x ∉ p

x ≠ p ∈ X.But X is a T1 − space

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∃ ux,v ∈ s.t. : x ∈ ux and p ∉ u, p ∈ v, x ∉ v

x ∈ ux ⊆ pc

pc ux : x ∈ pc union of open sets p is a closed set.

( Suppose that ∀p ∈ X : p is a closed set. Let a,b ∈ X s.t. a ≠ bSince p is closed ac and bc are open sets s.t.

a ∉ ac and b ∈ ac,b ∉ bc and a ∈ bc

X is a T1 − space.

Definition 5.7:A topological space X is called a T2 − space or Hausdarff space iff it

satisfies the following:Each pair of distinct points x,y ∈ X belongs respectively to disjoint open

sets.

i.e. ∃ u,v ⊆ s.t. : x ∈ u, y ∈ v and u ∩ v .

Example 5.8:X 1, 2, 3 X,,1,2,3,2, 3,1, 3,1, 2. Then X, is

a T2 − space.

Solution :1 ≠ 2 ∈ X ∃ 1,2, 3 ∈ s.t. 1 ∈ 1, 2 ∈ 2, 3 and 1 ∩

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2 ≠ 3 ∈ X ∃ 2,3 ∈ s.t. 2 ∈ 2, 3 ∈ 3 and 2 ∩ 3

1 ≠ 3 ∈ X ∃1,3 ∈ s.t. 1 ∈ 1, 3 ∈ 3 and 1 ∩ 3

Remark :1. Every discrete space is a T2 − space.2. Every indiscrete space is not a T2 − space.

The following theorem gives an indication why T2 − spaces areimportant.

Theorem 5.9:Every metric space is a T2 − space.

Proof :Let X,d be a metric space , and a,b ∈ X s.t. a ≠ b.Since X,d is a metric ,then da,b 0, say i.e. , da,b .

Consider the two open spheres

u S/3a ,v S/3b

centred at a and b respectively with reduis /3.Then u ∩ v , for let c ∈ u ∩ v

c ∈ u and c ∈ v da,c /3 and db,c /3

therfore

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da,b ≤ da,c dc,b /3 /3 2/3

da,b 2/3 .Therefore a ∈ u, b ∈ v and u ∩ v . (i.e. X,d is a

T2 − space Hausdorff.

Definition 5.10 :A topological space X is said to be regular iff it satisfies the following:if F is a closed subset of X and p ∈ X, p ∉ F ,then ∃ disjoint open sets u

and v s.t.

F ⊂ u and p ∈ v u ∩ v .

regular Space

Notation A regular space need not be a T1 − space.

Example 5.11:X a,b,c , X,,a,b,cThe closed subsets are : , X, b,c, a.Take F b,c, and p a ∉ b,c.

∃ u b,c , v a s.t. u ∩ v , b,c ⊂ b,c and a ∈ a.

Thus X,is a regular space.But X, is not a T1 − space since ,b is a singleton which is not closed.

Definition 5.12:A regular space X which also satisfies the sepration axiom of T1 is called

T3 − space.

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Corollary 5.13:Every T3 − space is a T2 − space.

Proof :Let X be a T3 − space, and a,b ∈ X be a distinct points.Since X is T1 − space a is closed , (Th.5.6).Since a ≠ b b ∉ a. Then as X is regular ∃ distjoint open sets u ∩ v s.t.

a ⊂ u and b ∈ vi.e. a ∈ u and b ∈ v , u ∩ v

Hence X is a Hausdarff space.

Definition 5.14:A topological space X is said to be normal iff X satisfies the following:

If F1and F2 are disjoint closed subsets of X, then ∃ disjoint open sets uand v s.t.

F1 ⊆ u and F2 ⊆ v .

Notation A normal space need not be regular or T1 − space.

Example 5.15:Let X a,b,c , X,,a,b,a,b.

The closed sets are ,X,b,c,a,c,c.Since a,b are not closed sets X, is not a T1 − space.X, is a normal space : For ,X are disjoint closed sets

s.t. ⊆ ,X ⊆ X.X, is not a regular − space : For a ∉ c, but the only open sets

containing c is X which also contains a .

Definition 5.16:

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A normal space which also satisfies the separation axiom T1 is calledT4 − space.

Corollary 5.17:A T4 − space is a T3 − space.

Proof :Let X, be a T4 − space X, is a normal T1 − space.So we have to show that X, is a regular space.Suppose F is a closed subset of X and p ∈ X s.t.

F ⊆ X and p ∉ F.

Since X is a T1 − space p is closed. i.e. F,p are two disjoint closedsets.

But X, is normal ∃ two disjoint open sets u, v s.t.

F ⊆ u and p ⊆ v u ∩ v .

i.e., F ⊂ u , p ∈ v. X, is a regular − space.

Theorem 5.18:Every metric space is a normal space.

Proof :Let X,d be a metric space, and A,B be disjoint closed subsets of X.

i.e. A ∩ B , A,B ⊆ X.

clearly da,B a 0 and db,A b 0

for each a ∈ A, choose a so that the sphere Sa Sa/2a does notintersect B,

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and for b ∈ B,Sb Sb/2b does not intersect A.Define

U Sa : a ∈ A , V Sb : b ∈ B.

Then clearly U, V are open set containing A, B respectively.We assert that U ∩ V . For, if

U ∩ V ≠ ∃ p ∈ X s.t. p ∈ U ∩ V p ∈ U and p ∈ V p ∈ Sa/2a and p ∈ Sb/2b for some a ∈ A, b ∈ B

i.e. da,p a/2 and db,p b/2Then by triangle inequality we get

da,b da,p dp,b a b2

If a ≤ b da,b b. Thus the sphere Sb/2b contains the pointa

If b ≤ a da,b a . Thus the sphere Sa/2a contains the pointb

Thus U ∩ V . Hence every metric space is normal.

Remark 5.19:1. A metric space is both normal space and T1 − space. i.e. A metric space is a T4 − space.2. The following diagram shows the relationship between spaces discussed in this section.

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PROBLEMS (V)1. Show that the property of being a T1 − space is herditary? i.e. Every subspace of aT1 − space is also T1 − space?

2. Show that the property of being a Hausdarff space is herditary? i.e. Every subspace of aT2 − space is a T2 − space?

3. Let be the topology generated by open - closed intervals a,b.Show that R, is ahausdarff?

4. Show that a property of being a space Hausdarff is a topological property.(i.e if f : X Y 1 − 1,onto, homeomorphism and X,x is a hausdarff space Y,yis a hausdarff space??)

5. Regularity is a topological property??

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Compactness andConnectedness

In the study of calculus, there are three basic theorems about continuonsfunctions and on these theorems the rest of calculus depends. They are thefollowing:

1. Intermediate value theorem: If f : a,b → R is continuous and if r ∈ s.t .r ∈ fa andfb then ∃ c ∈ a,b s.t. fc r.

2. Maximum value theorem: If f : a,b → R is continuous there exists an elementc ∈ a,b s.t. fx ≤ fc ∀ x ∈ a,b

3. Uniform continuity theorem : If f : a,b → R is continuous,then given 0∃ 0 s.t. |fx1 − fx2| ∀ pair of numbers x1,x2 of a,b for which |x1 − x2| .

We have spoken of theorems as theorems about continuous function. Butthey can also be considered as theorems about the closed interval a,b of realnumbers. The theorems depend not only on the continuity of f but also onproperties of the topological space a,b.

The property of the space a,b on which the intermediate value theoremdepends is the property called connectedness.

The property on which the Maximum value and theuniform continuity theorems depend is the property called compactness.

As the three theorems are fundamental for the theory of calculus,so are thenotions of compactness and connectedness fundamental in higheranalysis,geometry,and topology indeed,in almost any subject for which thenotion of topological space itself is relevant.

A- CompactnessIn this chapter we state the definition of compact spaces, prove serveral

simple theorems about compactness,and give a few examples.

Definition 6.11. Let A Ai : i ∈ I be a class of subsets of X. Then A is called a cover of X if

X i∈I Ai .2. A cover of X is called a finite cover if it has only a finite numbers of subsets of X.i.e.

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A A1,A2, .........,An s.t. X i1n Ai.

3. A cover of X is called an open (closed) cover if every members of the cover is open(closed) set.

4. A class of subset of X A Ai, i ∈ I is called a cover of A ⊆ X if A ⊆ i Ai.5. Let A1,A2, be two covers of X.Then A1 is a subcover of A2A1⊆ A2 if every members

of A1 is also a members of A2.

Example 6.2:Let X 1, 2, 3, 4, 5 X,,1,2,1, 4,1, 2,1, 2, 3,4, 5,1, 2, 4,1, 2, 3, 4,1, 4, 5Then

1. A1 1,2,3,4,5 is an open cover of X (finite -open cover)2. A2 1,2,3,4,4,5 is an open-cover.3. A3 1,2,4,5 is not a cover of X.

Example 6.3:1. Let X 1,2, ........,n, ..... and

A1 1, A2 2, A3 3, ........, An n, An1 n 1, ......

∵ X n1 n A Ai, i ∈ X is an infinite cover of X.

2. Let X a1,a2,a3, ........,an.

A1 a1, A2 a1,a2, A3 a1,a2,a3, ......., An X.

X i1n Ai, X is a finite cover of X.

Example 6.4:Let A 0, 1 ⊆ R .Then the following class

A Ai 1i2 , 1

i : i ∈ N of open intervals is an open covers of A. i.e.

0, 1 ⊆ 13 , 1 1

4 , 12 1

5 , 13 …

Example 6.5:Consider the class A Dp : p ∈ Z Z where Dp is the open disc in the

plane R2 with radius 1 and center p x,y ∈ Z Z .

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Then A is the cover of R2.i.e , every point in R2 belong to at least onemember of A.

On the other hand , the class of open discs B Dp∗ : p ∈ Z Z whereDp∗ is the open disc with radius 1/2 and center p ∈ Z Z , Then B is not acover of R2 .

For example , 12 , 1

2 ∈ R2,which does not belong to any member of B.

Definition 6.6:1. A topological space X, is said to be compact if every open cover of X has a finite

subcover which cover X.i.e., if A Ai, i ∈ I s.t. X i Ai then ∃ Ai1 ,Ai2 , ......Ain of open sets

s.t. X Ai1 Ai2 ... Ain .2. A subset A ⊆ X of a topological space X is compact if every open cover of A has a finite

subcover . i.e., if A ⊆ i Ai ∃ Ai1 ,Ai2 , ......Ain s.t. A ⊆ Ai1 Ai2 .... Ain .

Example 6.7:Every indiscrete space is compact.

Example 6.8:Every finite topological space is compact.

Solution :Let X a1,a2, .....an be a finite set .Then every on X has a finite

number of open sets in X.Let A Ai, i ∈ I be an open cover of X i.e., X i Ai.Then each point a1 ∈ X ∃ at least Ai1 s.t. a1 ∈ Ai1 , similarly,

a2 ∈ Ai2 , ..., ain ∈ Ain X ⊂ Ai1 Ai2 ... Ain , i.e., X Ai1 Ai2 ... Ain . Thus X is

compact.

To prove that a space is not compact we only have to find one open coverof the space with no-finite subcover.

Example 6.9:Every infinite discrete topological space is not compact.

Solution :Let X, be a discrete topological space such that X is infinite set. i.e.,

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X a1,a2,a3......

Consider the class of singleton subsets of X.

A a : a ∈ X

Then1. A is a cover of X.2. A is an open cover of X.3. No proper subclass of A is a cover of X.4. A is infinite ,since X is infinite.

Hence the open cover A of X contains no finite subcover , so X is notcompact.

Example 6.10:The real line R is not compact.

Solution :Let A −n,n ∀ n ∈ N

−1, 1, −2, 2, −3, 3.......Then A is an open cover of R . But A contains no finite subcover.

Theorem (Heine-Borel):Let A a,b be a closed and bounded interval of the real line R. Let

A Ai : i ∈ I

be a class of open intervals which covers A . i.e., A ⊂ i Ai . ThenA contains a finite subclass, say

Ai1, Ai2 , ... , Ain which also covers A.

Proof :Not required , it will be taken in course 414.

Example 6.11:Every closed bounded interval a,b is compact . ex. : 0, 1 is compact .

Example 6.12:The open interval A 0, 1 ⊂ R is not compact .

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Solution :From example 6.4 , A Ai 1

i2 , 1i i ∈ N is the class of open

intervals which covers A . i.e. ,

0, 1 ⊆ 13 , 1 1

4 , 12 1

5 , 13 ...

But A contains no finite subcover. For let

A∗ a1,b1, a2,b2, ......, an,bn

be any finite subclass of A .If mina1, .......,an , then 0 and

a1,b1 a2,b2 ...... an,bn ⊂ 0, 1

But 0, and , 1 are disjoint . Hence A∗ is not a cover of A and so A isnot compact.

Exercise :Is the closed infinite interval [1,) compact ? justify your answer .

Theorem 6.13:.The continuous image of a compact set is compact.

Proof :Let f : X → Y be continuous and A ⊆ X be compact . We want to show

that fA is compact ??Suppose

A∗ Ai : i ∈ I

is an open cover of fA, i.e. fA ⊂ i Ai

A ⊂ f−1fA ⊂ f−1 i Ai i f−1Ai

∵ f is continuous,then the inverse image of every open set of Y is open. f−1Ai is open.

Hence A f−1Ai : i ∈ I is an open cover of A.But A is compact A contains a finite subcover which cover A. i.e.,

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A ⊂ f−1Ai1 f−1Ai2 ..... f−1Ain fA ⊂ f f−1Ai1 ..... f−1Ain fA ⊂ Ai1 Ai2 ...... Ain

Hence fA is compact .Therefore continuous images of compact sets are compact.

Theorem 6.14:.Let A ⊆ X, X, is a topological space .Then the following are

equivelant:(i) A is compact with respect to .(ii) A is compact with respect to relative topology A on A .

Proof :i iiSuppose A is compact w.r to . Let Aibe a A -open which cover A. i.e.,

A ⊂ i Ai Ai ∈ A.

Now ,each Ai A ∩ Hi where Hi ∈

Ai A ∩ Hi ⊂ Hi i Ai ⊂ i Hi A ⊂ i Hi

But Hi is open in , which cover A , since A is compact w.r. to ,then Hi contains finite subcover of A , say

A ⊂ Hi1 Hi2 ... Him A ⊂ A ∩ Hi1 .... Him

Thus A ⊂ A ∩ Hi1 A ∩ Hi2 .... A ∩ Him, i.e,A ⊂ Ai1 Ai2 .... Ain .

Then Ai contains a finite subcover Ai1 , ....,Ain and A,A iscompact.

(ii) (i)let {Hi} be a -open cover of A. i.e . , A ⊂ i Hi

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A ⊂ A ∩ i H i A ∩ Hi i Ai

But Ai ∈ A Ai is a A - open cover of A.By given {Ai} contains afinite subcover of A.

i.e. A ⊂ Ai1 Ai2 ........ Aim A ⊂ A ∩ Hi1 ....... A ∩ Him A ⊂ A ∩ Hi1 Hi2 ...... Him

Therefore A ⊂ Hi1 Hi2 ...... Him .Hence A is compact w.r. to .

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B- ConnectednessThe definition of connectedness for a topological space is a quite natural

one. One says that a space can be a (separated) if it can be broken up into two(globs) disjoint open sets . Otherwise , one says that it is connected . Fromthis simple idea we get the following definition:

Definition 6.15:1. Let X, be a topological space. Two subsets u,v of X are said to be separated if u,v are

disjoint non-empty open subsets of X whose union is X. i.e.,(i) u ∩ v , u,v ∈ (ii) X u v.

2. X is said to be connected if it is not the union of two non-empty separated sets (open sets).

3. X is said to be disconnected if it is not connected.(X is a union of disjoint non-empty opensets)

i.e. X u v, u,v ∈ ,u,v ≠ ,u ∩ v

u,v are called a disconnection of X.

4. Let X, be a topological space . A subset A ⊆ X is disconnected if ∃ open sets u,v of Xsuch that A ∩ u,A ∩ v are disjoint non-empty open sets whose union is A. i.e.,

A A ∩ u A ∩ v where A ∩ u ≠ , A ∩ v ≠ and A ∩ u ∩ A ∩ v .

In this case we say u v is a disconnection of A.

5. A ⊆ X is a connected, if A is not disconnected.i.e. A is not a union of two non-empty disjoint open sets of A.

Remark 6.17:1. If X u v , u,v ∈ ,u ∩ v uc X − v , vc X − v are closed subset of X .

But uc v , vc u.Hence we can say that u,v are two disjoint non-empty closed and open

subsets.

2. Hence deffinition (2) can be state as X, is connected if it is not a uonion of two disjointnon-empty closed sets.

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A space X is connected the only subsets of X that are both open and closed are X and .

3. From definition (4) observe that

A A ∩ u A ∩ v, A ∩ u ∩ A ∩ v

A ⊂ u v, u ∩ v ⊂ AcTherefore u v is a disconnection iff A ∩ u ≠ , A ∩ v ≠

A ⊂ u v , u ∩ v ⊂ Ac.

Example 6.18:1. is connected.2. Any space with only one element p is connected.3. X, I indiscrete space is connected.

Example 6.19:Let

X 1, 2, 3, 4, 5, 1 X,,1,1, 2, 2 X,,1, 2,3, 4, 5Then 1 is connected and 2 is disconnected.

Example 6.20:Let X a,b,c,d,e, X,,a,b,c,c,d,e,c.Take A a,d,e. Then let u a,b,c,v c,d,e A ∩ u a ≠ and A ∩ v d,c ≠ such that

A A ∩ u A ∩ v.Thus A is disconnected.

Theorem 6.21:X, is connected the only non- empty sets which is open and closed

are X, itself.

Proof :Assume X, is connected. Let u be a non-empty set such that u is open

and closed .We have to provethat X u??∵ u is closed uc is open.Hence X u uc , u ∩uc X is a union of two disjoint open sets X is not connected

.Therefore X u.

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Conversely ,suppose X is the only set ( non- empty set) which is bothopen and closed .

Assume X is disconnected. ∃ two non- empty disjoint open sets s.t.

X u v ,u ∩ v ,u,v ∈ uc v is closed v X and u Hence X must be connected.

Example 6.22:1. Let X a,b,c,d,e, X,,a,c,d,a,c,d,b,c,d,e

The closed sets are

,X ,b,c,d,e,a,b,e,b,e,a

∵ X a b,c,d,e ( a,b,c,d,e are both open -and closed ).Thus X is disconnected .

2. Let A b,d,e A A, ,d∵ A, are the only open and closed sets A is connected .

Example 6.23:R , are connected .Since R, are the only subsets of R which are both open and closed .

Theorem 6.24:Continuous images of connected sets are connected .

Proof :Let f : X Y be a continuous map from a connected space X into a

topological space Y . Thusf : X fX is continuous (where fX ⊆ Y has the relative topology )

.We want to prove that fX is connected .Suppose fX is disconnected . ∃ two disjoint open sets of fX such that

fX u v, u ∩ v

∵ f is continuous f−1u , f−1 v are open subsets of X.Consider f−1u f−1v f−1u v f−1 fX X.

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and f−1u ∩ f−1v f−1u ∩ v f−1 Thus X is a union of two disjoint open subsets X is disconnected

Therefore fX is Connected .

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