TOPOLOGY

DR. ZAFFAR IQBAL

1. C HAPTER 1: S ETS AND R ELATIONS

Definition 1. A space is a nonempty set which possesses some type of mathematical structure.

Relations

Def. A relation R from A to B is a subset of A × B.

If ∗ is a relation R then (a, b) ∈ R iff a ∗ b.

• Dom(R) = {a : (a, b) ∈ R} ,

• Ran(R) = {b : (a, b) ∈ R}

• R−1 = {(b, a) : (a, b) ∈ R}.

Example: Let A = {1, 2, 3}, B = {2, 3, 4} and R = < , then

R−1 = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)}.

Definition 2. (Composition of Relations) Let R be a relation from A to B and S be a relation

from B to C. Then the relation S ◦ R from A to C is defined by

S ◦ R = {(x, y) : x ∈ A, y ∈ C} so that ∃ b ∈ B : (x, b) ∈ R, (b, y) ∈ S}.

Example: Let A = C = {1, 2, 3, 4}, B = {a, b, c, d} and the relations R and S be defined as

R = {(1, a), (1, c), (2, b), (3, d), (4, d)} and S = {(b, 1), (b, 2), (c, d), (d, 3)}. Then find S ◦ R

and R ◦ S

Definition 3. (Equivalence Relations) A relation R in a set A is said to be an equivalence

relation iff it satisfies the following axioms.

1
2 Dr. Zaffar Iqbal

(1) ∀ a ∈ A, (a, a) ∈ R (reflexive).

(2) If (a, b) ∈ R, then (b, a) ∈ R (symmetric)

(3) If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R (transitive)

Examples:

(1) A = R, R =<.

(2) A = Set of triangles, R = ∼ (similar).

(3) A = Set of all sets , R = ⊆.

Definition 4. (Equivalence classes:)

If R is an equivalence relation in A Then the equivalence class of any element a ∈ A, denoted

by [a] is defined by

[a] = {x : (a, x) ∈ R}

Definition 5. (Quotient of a set A by R) The collection of equivalence classes of A , denoted

by A/R is called the quotient of A by R, i.e.,

A/R = {[a] : a ∈ A}

Definition 6. (Partition of a set.) A collection C of a nonempty set A is called a partition of

A iff

(1) Each a ∈ A must belongs to some member of C.

(2) ki ∩ kj = φ ∀ ki , kj ∈ C, i 6= j.

Theorem: (Fundamental Theorem of equivalence relations)

Let R be an equivalence relation in A. Then the quotient set A/R is a partition of A.

Example. Let A = Z and R is the congruence mod 4 (a ≡ b(mod4)). Then

 Topology 3

[0] = {. . . , −4, 0, 4, . . .},

[1] = {. . . , −3, 1, 5, . . .},

[2] = {. . . , −2, 2, 6, . . .},

[3] = {. . . , −1, 3, 7, . . .}.

Therefore A/R = {[0], [1], [2], [3]}.

4 Dr. Zaffar Iqbal

2. C HAPTER 2: F UNCTIONS

Def. By a function f : A → B we mean a rule which assigns to each element a of A a unique

element b (= f (a)) of B.

• The set A is called domain of f and B is called co-domain of f .

• A function f whose codomain is R is called a real value f .

A subset of B consisting of all f-images of elements of A is called range of f denoted by

f (A).

• Let f, g are defined on A and if f (a) = g(a) ∀a ∈ A, the functions f and g are equal, i.e.,

f = g.

• If f (A) ⊂ B Then f is a function of A into B.

• If f (A) = B Then f is a function of A onto B.

f is onto if for every b ∈ B ∃ a ∈ A s.t. b = f (a).

• f : A → B is one to one iff f (a) = f (b) ⇒ a = b

or a 6= b ⇒ f (a) 6= f (b).

• f : A → A with f (x) = x ∀ x ∈ A is called an identity function and it is denoted by IA .

An identity function is bijective.

 Topology 5

Composition of Two Functions

Def: Let f : A → B and g : B → C be to functions then their composition is defined by

(g ◦ f )(a) = g(f (a)).

Theorem. If f : X → Y and g : Y → Z are bijective functions. Then g ◦ f : X → Z is also

bijective.

• If f : A → B. Then IB ◦ f = f , f ◦ IA = f .

• Let f : X → Y , A ⊂ X , B ⊂ Y . Then

f −1 (y) = {x ∈ X : f (x) = y}

or

x ∈ f −1 (y) ⇔ f (x) = y.

• f −1 (B) = {x ∈ X : f (x) ∈ B}, inverse images of B. Therefore

x ∈ f −1 (B) ⇔ f (x) ∈ B.

• f (X) = {y ∈ Y : y = f (x) for some x ∈ X}. Therefore

y ∈ f (X) ⇔ y = f (x) f orsome x.

Remark: Let f : X → Y and g : Y → X satisfy g ◦ f = IX , f ◦ g = IY . Then f −1 : Y → X

exists and g = f −1 .

Theorem: If f : A → B and g : B → C are bijective functions, then (g ◦ f )−1 = f −1 ◦ g −1 .

Remark. Let f : X → Y , A, B ⊆ X, U, V ⊆ Y . Then the following are true:

1: A ⊆ B ⇒ f (A) ⊆ f (B).

2: U ⊆ V ⇒ f −1 (U ) ⊆ f −1 (V ).

3: f (A ∪ B) = f (A) ∪ f (B).

4: f (A ∩ B) ⊂ f (A) ∩ f (B).
6 Dr. Zaffar Iqbal

5: f −1 (Y − U ) = X − f −1 (U ).

6: f −1 (U ∪ V ) = f −1 (U ) ∪ f −1 (V )

7: f −1 (U ∩ V ) = f −1 (U ) ∩ f −1 (V )

8: f −1 (U \V ) = f −1 (U )\f −1 (V )

9: f (A\B) ⊃ f (A)\f (B).

 Topology 7

Theorem: Let f : X → Y be a function, A ⊆ X and B ⊆ Y , then

(i) A ⊆ (f −1 ◦ f )(A)

(ii) (f −1 ◦ f )(B) ⊆ B.

Proof: (i) Let x ∈ A

⇒ f (x) ∈ f (A)

⇒ x ∈ f −1 (f (A))

⇒ x ∈ f −1 ◦ f (A)

i.e. A ⊆ f −1 ◦ f (A)

(ii) Let y ∈ f ◦ f −1 (B)

⇒ y ∈ f (f −1 (B))

⇒ ∃x ∈ f −1 (B) s.t. y = f (x)

⇒ f (x) ∈ B s.t. y = f (x)

i.e., y ∈ B

=⇒ f ◦ f −1 (B) ⊆ B

Theorem: Let f : X → Y be a function and A ⊂ X, B ⊂ Y , then

1. A = f −1 ◦ f (A) iff f is one to one.

2. B = f ◦ f −1 (B) iff f is onto.

Proof.

1: Let x ∈ A

⇒ f (x) ∈ f (A)

⇒ x ∈ f −1 (f (A))

⇒ x ∈ f −1 ◦ f (A).

Therefore A ⊂ (f −1 ◦ f )(A).

Conversely, let x ∈ (f −1 ◦ f )(A) , i.e.

8 Dr. Zaffar Iqbal

x ∈ f −1 (f (A))

⇒ f (x) ∈ f (A). Hence

∃ x1 ∈ A s.t., f (x1 ) = f (x).

But f is one to one, so x1 = x, i.e. x ∈ A

⇒ (f −1 ◦ f )(A) ⊂ A

∴ A = (f −1 ◦ f )(A).

2: Let y ∈ (f ◦ f −1 )(B)

⇒ y ∈ f (f −1 (B))

⇒ ∃ x ∈ f −1 (B) s.t. y = f (x)

⇒ f (x) ∈ B s.t. y = f (x)

i.e. y ∈ B

⇒ (f ◦ f −1 )(B) ⊂ B

Conversely, let y ∈ B. Since f is onto, so ∃ x ∈ X s.t. y = f (x), i.e.

f (x) = y ∈ B or f (x) ∈ B

⇒ x ∈ f −1 (B)

⇒ f (x) ∈ f (f −1 (B))

i.e., y ∈ f (f −1 (B)). Hence B ⊂ (f ◦ f −1 )(B). Therefore B = (f ◦ f −1 )(B).

 Topology 9

∗ Finite Sets

Def: A set A is said to be finite if it empty or if there is a bijection f : A → {1, . . . , n}, and we

say that A has cardinality n.

• A set A is said to be infinite if it is not finite.

Def: A set A is said to be countable infinite if there is a bijection f : A → N.

Example: Z is countably infinite ; the function f : Z → N defined below is a bijection



 2n,
 n > 0,
f (n) =

 −2n + 1, n ≤ 0.

.
10 Dr. Zaffar Iqbal

∗ Partially Ordered Sets

Def: A relation ≤ in a set A if partial order on A if, for ∀ a, b, c ∈ A:

1. a ≤ a (reflexive).

2. If a ≤ b and b ≤ a then a = b (antisymmetric).

3. If a ≤ b and b ≤ c then a ≤ c (transitive).

The pair (A, ≤) is called a partially ordered set (POSET).

Def: A partially ordered set A is said to be totally ordered set (TOS) if, for every a, b, ∈ A,

either a ≤ b or b ≤ a.

EXAMPLES:

1. (R, ≤) is a TOS.

2. Let A = set of all sets and ≤ = ⊆. Then (A, ≤) is a POSET and not a TOS.

3. Let A = {a, b, c, d, e} with the following diagram:

aH d
HH
yH q
 
H
cHHy
q
 HH
b  He

order : x ≤ y iff either x = y or go from x to y through diagram. Then (A, ≤) is a POSET.

Also a subset {a, c, d} is TOS and {a, b, c} is not a TOS.

 Topology 11

3. T OPOLOGY

Def: If X is any non-empty set then a collection τ of subsets of X is called a topology on X if

it satisfies the following conditions:

T1 . X and φ belong to τ .

T2 . The union of any number of members of τ belong to τ .

T3 . The intersection of finite number of members of τ belong to τ .

The pair (X, τ ) is called a topological space.

• Elements of X are called points of X.

• Elements of τ are called open sets of X.

Remark: Why finite number of intersection is needed in topology:

∞
An = {(− n1 , n1 ) : n ∈ N} and ∩ An = {0} is not an interval.
n=1

Def: If X is any set then the collection τ of all subsets of X is a topology on X and it is called

the discrete topology, and (X, τ ) is called discrete topological space.

Def: (X, τ ) with τ = {φ, X} is called indiscrete topological space and τ is called indiscrete

topology.

Example 1. Let X = {a, b, c, d, e} and

τ1 = {X, φ, {a}, {c, d}, {a, c, d}, {b, c, d, e}} is a topology on X.

τ2 = {X, φ, {a}, {c, d}, {a, c, d}, {b, c, d}} is not a topology on X.
12 Dr. Zaffar Iqbal

Theorem: If τ1 and τ2 are any two topologies on a set X , then τ1 ∩ τ2 is also a topology on X.

Proof: T1 : As τ1 and τ2 are topologies. Therefore φ, X ∈ τ1 and φ, X ∈ τ2 . Hence

X, φ ∈ τ1 ∩ τ2 .

T2 : Let {Aα : α ∈ I} be any collection of members of τ1 ∩ τ2 ,i.e.

Aα ∈ τ1 ∩ τ2 ∀ α, then

Aα ∈ τ1 and Aα ∈ τ2 . As τ1 and τ2 are topologies.

∪ Aα ∈ τ1 and ∪ Aα ∈ τ2 i.e, ∪ Aα ∈ τ1 ∩ τ2 .
α α α

T3 : Let A1 , . . . , An be any collection of members of τ1 ∩ τ2 .

⇒ A1 , . . . , An ∈ τ1 and A1 , . . . , An ∈ τ2
n n
⇒ ∩ Ai ∈ τ1 and ∩ Ai ∈ τ2
i=1 i=1
n
⇒ ∩ ∈ τ1 ∩ τ2 .
i=1

Therefore τ1 ∩ τ2 is also a topology on X.

Theorem (Cofinite Topology): If X is any non-empty set , then the collection τ of φ and all

those subsets of X whose complements are finite is a topology on X and it is called a cofinite

topology.

Proof. T1 : φ ∈ τ and X c = φ is finite ∴ X ∈ τ.

T2 : Let {Aα : α ∈ I} is any collections of members of τ i.e,

Aα ∈ τ for every α. Then Acα is finite. By Demorgan’s laws: (∪ Aα )c = ∩ Acα

α α

But Acα is finite ∴ ∩ Acα is finite. Hence ∪ Aα ∈ τ .

α α

T3 : Let A1 , . . . , Ak ∈ τ . Then Ac1 , . . . , Ack are finite.

n n k
Since ∪ Aci = ( ∩ Ai )c is finite. This implies ∩ Ai ∈ τ .
i=1 i=1 i=1

Example: Let X = N. Then show that τ = {φ, En }, where En = {n, n+1, n+2, · · · }, n ∈ N

is a topology on X.
 Topology 13

Theorem: Let f : X → Y be a function from a nonempty set X into a topology space (Y, U ).

Let τ = {f −1 (G) : G ∈ U }. Then τ is a topology on X.

Proof. T1 : Since ∅ = f −1 (∅) implies ∅ ∈ τ and X = f −1 (Y ) implies X ∈ τ .

T2 : Let {Aα : α ∈ I} is any collections of members of τ i.e, Aα ∈ τ for every α.

⇒ ∃ Gα ∈ U s.t. Aα = f −1 (Gα ) ∀ α.

⇒ ∪ Aα = ∪{f −1 (Gα )} = f −1 (∪ Gα ).
α α α

But ∪ Gα ∈ U . Hence ∪ Aα ∈ τ .
α α

T3 : exr

Theorem: Let A ⊆ X and τ be a topology on X. Then S = {U ∪ (V ∩ A) : U, V ∈ τ } is also

a topology on X.

Proof. T1 : Since ∅ = ∅ ∪ (∅ ∩ A) and X = X ∪ (X ∩ A). Hence X, ∅ ∈ S.

T2 : Let H1 , H2 ∈ S such that H1 = U1 ∪ (V1 ∩ A) ,H2 = U2 ∪ (V2 ∩ A), then

H1 ∪ H2 = U1 ∪ (V1 ∩ A) ∪ U2 ∪ (V2 ∩ A)

= (U1 ∪ U2 ) ∪ (V1 ∩ A) ∪ (V2 ∩ A)

= (U1 ∪ U2 ) ∪ [(V1 ∪ V2 ) ∩ A]
2 2 2
∴ ∪ Hi = ∪ Ui ∪ ( ∪ Vi ∩ A). Hence ∪ Hi = ∪ Ui ∪ (∪ Vi ∩ A) ∈ S.
i=1 i=1 i=1 i i i

T3 : H1 ∩ H2 = [U1 ∪ (V1 ∩ A)] ∩ [U2 ∪ (V2 ∩ A)]

= (U1 ∪ V1 ) ∩ (U1 ∪ A) ∩ (U2 ∪ V2 ) ∩ (U1 ∪ A)

= (U1 ∪ V1 ) ∩ (U2 ∪ V2 ) ∩ (U1 ∪ A) ∩ (U1 ∪ A)

= [(U1 ∪ V1 ) ∩ (U2 ∪ V2 )] ∩ [(U1 ∩ U2 ) ∪ A)]

= [(U1 ∪ V1 ) ∩ (U2 ∪ V2 ) ∩ (U1 ∩ U2 ] ∪ [(U1 ∪ V1 ) ∩ (U2 ∪ V2 ) ∩ A]

= {(U1 ∪ V1 ) ∩ U1 } ∩ {(U2 ∪ V2 ) ∩ U2 } ∪ [(U1 ∪ V1 ) ∩ (U2 ∪ V2 ) ∩ A]

= (U1 ∩ U2 ) ∪ [(U1 ∪ V1 ) ∩ (U2 ∪ V2 ) ∩ A]

14 Dr. Zaffar Iqbal

2 2 n n n
 
= ( ∩ Ui ) ∪ ( ∩ (Ui ∪ Vi ) ∩ A) ∈ S. Hence ∩ Hi = ( ∩ Ui ) ∪ ∩ (Ui ∪ Vi ) ∩ A ∈ S.
i=1 i=1 i=1 i=1 i=1

Example 1: Let X = R2 and n ∈ N and τ = {X, φ, Gn (∀ n ∈ N)}, where

Gn = {(x, y) ∈ R2 : y > n}. Then τ is a topology on X.

Solution. T2 : From Gn we have G1 = {(x, y) ∈ R2 : y > 1}, G2 = {(x, y) ∈ R2 : y > 2}

and so on. Therefore G1 ∪ G2 = G1 ∈ τ . In general ∪ Gi = G1 ∈ τ .

i≥1
n
T3 : G1 ∩ G2 = G2 ∈ τ . In general ∩ Gi = Gn ∈ τ . Hence τ is a topology on X.
i=1

Example 2: Let X = R2 and n ∈ N and τ = {X, φ, Gn (∀ n ∈ N)}, where

Gn = {(x, y) ∈ R2 : y > x + n}. Then τ is a topology on X.

Solution. T2 : From Gn we have G1 = {(x, y) ∈ R2 : y > x + 1}, G2 = {(x, y) ∈ R2 : y >

x + 2} and so on. Therefore G1 ∪ G2 = G1 ∈ τ . In general ∪ Gi = G1 ∈ τ .

i≥1
n
T3 : G1 ∩ G2 = G2 ∈ τ . In general ∩ Gi = Gn ∈ τ . Hence τ is a topology on X.
i=1

Example 3: Let X = R2 and n ∈ N and τ = {X, φ, Gn (∀ n ∈ N)}, where

Gn = {(x, y) ∈ R2 : y > nx}. Is τ a topology on X?

Solution. T2 : From Gn we have G1 = {(x, y) ∈ R2 : y > x}, G2 = {(x, y) ∈ R2 : y > 2x}

and so on.
 Topology 15

o








 G1 , x ≥ 1;

Therefore G1 ∪ G2 =

 G2 , x < 1.


This implies G1 ∪ G2 ∈
/ τ . Hence τ is not a topology on X.
16 Dr. Zaffar Iqbal

∗ Closed Sets

Def (closed set): If (X, τ ) is a T.S, then any subset F of X is called closed set , if complement

of F is open i.e., F is closed ⇔ F c is open.

Example: Let X = {a, b, c}, τ = {X, φ, {a}, {c}, {b, c}, {a, c}}. Then the closed subsets of

X are φ, X, {b, c}, {a, b}, {a}, {b}.

———————————–

Theorem: If (X, τ ) is a T.S then,

(i) X and φ are closed.

(ii) The intersection of any number of closed sets is closed.

(iii) The union of finite number of closed sets is closed.

Proof.

(i) Since φc = X is open so φ is closed and X c = φ is open so X is closed.

(ii) Let {Aα : α ∈ I} is any collection of closed sets i.e, Aα is closed for every α. Therefore

Acα is open ∀α. Then by DeMorgan’s laws (∩ Aα )c = ∪ Acα .

α α

But Acα is open so ∪ Acα is open. Hence (∩ Aα )c is open.

α α

⇒ ∩ Aα closed.
α

(iii) Let A1 , . . . , An be any closed sets in X.

⇒ Ac1 , . . . , Acn are open. Since

n n n
( ∪ Ai )c = ∩ Aci is open. Therefore ∪ Ai is closed.
i=1 i=1 i=1
 Topology 17

∗ Neighborhood and Neighborhood Systems

Def: If (X, τ ) is a topological space and x ∈ X, then a subset N of X is called a neighborhood

of x iff ∃ an open set U such that x ∈ U ⊆ N .

Def: The collection of all neighborhoods of a ∈ X is called a neighborhood system of a denoted

by N (a).

Example: Let X = {a, b, c, d} , τ = {X, φ, {c}, {c, d}, {a, c, d}}

(i) neighborhoods of a.

a ∈ {a, c, d} ⊆ {a, c, d} and a ∈ {a, c, d} ⊆ X.

∴ N (a) = {{a, c, d}, X}

(ii) neighborhoods of b.

b∈X⊆X

∴ N (b) = {X}

(iii) neighborhoods of c.

1. c ∈ {c} ⊆ {c}

2. c ∈ {c} ⊆ {a, c}

3. c ∈ {c} ⊆ {b, c}

4. c ∈ {c} ⊆ {c, d}

5. c ∈ {c} ⊆ {a, b, c}

6. c ∈ {c} ⊆ {a, c, d}

7. c ∈ {c} ⊆ {b, c, d}

8. c ∈ {c} ⊆ X

∴ N (c) = {{c}, {a, c}, {b, c}, {c, d}, {a, b, c}, {a, c, d}, {b, c, d}, X}

(iv) neighborhoods of d.

d ∈ {c, d} ⊆ {c, d}
18 Dr. Zaffar Iqbal

d ∈ {c, d} ⊆ {a, c, d}

d ∈ {c, d} ⊆ {b, c, d}

d ∈ {c, d} ⊆ X

∴ N (d) = {{c, d}, {a, c, d}, {b, c, d}, X}

Usual Topology on R: The topology τ = {R, ∅, union of open intervals} is called the usual

topology on R.

Usual Topology on R2 : The topology τ = {R2 , ∅, union of open disks} is called the usual

topology on R2 .

Usual Topology on R3 : The topology τ = {R3 , ∅, union of open balls} is called the usual

topology on R3 .

Theorem: If (X, τ ) is a topological space. Then

(i) Each x ∈ X has a neighborhood.

(ii) If A is a neighborhood of x and A ⊆ B, then B is also a neighborhood of x.

(iii) If A, B are neighborhoods of x, then A ∩ B is also a neighborhood of x.

Proof. (i) Since X is open set then x ∈ X ⊆ X for ∀ x. i.e, each x ∈ X has atleast one

neighborhood which is X itself.

(ii) Since A is a neighborhood of x then ∃ open set U such that x ∈ U ⊆ A. But A ⊆ B

⇒x∈U ⊆B

∴ B is a neighborhood of x.

(iii) Since A and B are neighborhoods of x then ∃ open sets U and V such that

x ∈ U ⊆ A and x ∈ V ⊆ B

⇒ x ∈ U ∩ V ⊆ A ∩ B. Since U ∩ V is open set

∴ A ∩ B is a neighborhood of x.
 Topology 19

Def: If (X, τ ) is a topological space. Then a subset A of X is open iff ∃ an open set U such

that x ∈ U ⊆ A ∀ x ∈ A.

Theorem: A set G is open iff G is a neighborhood of each of its points.

Proof. Suppose G is open. Hence ∀ g ∈ G we have g ∈ G ⊆ G.

∴ G is a neighborhood of each of its points.

Conversely, suppose G is neighborhood of each of its points.

⇒ ∀ g ∈ G, ∃ open set Gg such that g ∈ Gg ⊆ G. Hence

G = ∪{Gg : g ∈ G} is open as union of open sets is open.

20 Dr. Zaffar Iqbal

4. R ELATIVE T OPOLOGY AND S UBSPACES

Def: If (X, τ ) is topological space and A ⊆ X, then the collection

τA = {A ∩ G : G ∈ τ } is topology on A, called the relative topology for A.

In this case the subset A is called the subspace of X.

H ∈ τA ⇔ ∃ an open set G ⊆ X such that H = A ∩ G.

Examples: Let X = {a, b, c, d, e}, τ = {X, φ, {a}, {c, d}, {a, c, d}, {b, c, d, e}},

A = {a, d, e} ⊆ X. Then τA = {A, φ, {a}, {d}, {a, d}, {d, e}}.

Theorem: Let A be a subset of a topological space (X, τ ). Then τA = {A ∩ G : G ∈ τ } is a

topology on A.

Proof: T1 : Since τ is a topology so X and φ ∈ τ . Hence A ∩ X = A and A ∩ φ = φ.

∴ A, φ ∈ τA .

T2 : Let {Hα : α ∈ I} be a collection of members of τA .

⇒ ∀ α ∈ I, ∃ Gα ∈ τ such that Hα = A ∩ Gα . Since ∪ Gα ∈ τ . Thus

α

∪ Hα = ∪(A ∩ Gα ) = A ∩ (∪ Gα ) ∈ τA .
α α α

T3 : Suppose that H1 , . . . , Hn ∈ τA . then ∃ G1 , . . . , Gn ∈ τ such that Hi = A ∩ Gi ∀ i. Since

n
∩ Gi ∈ τ . Therefore
i=1
n n n
∩ Hi = ∩ (A ∩ Gi ) = A ∩ ( ∩ Gi ) ∈ τA .
i=1 i=1 i=1

Hence (A, τA ) is a topological space.

Theorem: Let (X, τ ) be a subspace of (Y, U ) and (Y, U ) be a subspace of (Z, V ). Then (X, τ )

is a subspace of (Z, V ).

Proof: Since X ⊆ Y ⊆ Z. Then (X, τ ) is a subspace of (Z, V ) iff VX = τ .

Let G ∈ τ . Since (X, τ ) is a subspace of (Y, U ). Therefore τ = UX

⇒ G ∈ UX .
 Topology 21

⇒ ∃ H ∈ U such that G = X ∩ H. But U = VY (∵ (Y, U ) is a subspace of (Z, V )).

Therefore H ∈ VY

⇒ ∃ K ∈ V such that H = Y ∩ K. Therefore

G=X ∩H

= X ∩ (Y ∩ K)

= (X ∩ Y ) ∩ K

=X ∩K

⇒ G ∈ VX

⇒ τ ⊆ VX .

Conversely, let G ∈ VX

⇒ ∃ L ∈ V such that G = X ∩ L. But

Y ∩ L ∈ VY = U i.e, Y ∩ L ∈ U

⇒ X ∩ (Y ∩ L) ∈ UX = τ and

X ∩ (Y ∩ L) = (X ∩ Y ) ∩ L = X ∩ L ∈ τ

⇒G∈τ

⇒ VX ⊆ τ. Hence VX = τ .
22 Dr. Zaffar Iqbal

Limit Points:

Def: Let X be a topological space and A ⊆ X. A point p ∈ X is a limit point of A iff every

open set G containing p contains a point of A other than p. i.e,

p ∈ G (open) implies (G \ {p}) ∩ A 6= φ.

Def: The set of all limit points of A, denoted by A0 is called the derived set of A.

Example 1: Let X = {a, b, c, d, e}, τ = {X, φ, {a}, {c, d}, {a, c, d}, {b, c, d, e}} and

A = {a, b, c}. Then b, d, e are the limit points of A where as a and c are not limit points of A.

Therefore A0 = {b, d, e}.

Example 2: Let X = N, τ = {φ, Gn }, where Gn = {n, n + 1, n + 2, · · · }

A = {3, 5}. Find A0 .

Theorem: A subset A of a top space X is closed iff A contain each of its limit points i.e.,

A is closed ⇔ A0 ⊂ A.

Proof: Suppose that A is closed.

⇒ Ac is open.

/ A i.e. p ∈ Ac . Also p ∈ Ac and (Ac \ {p}) ∩ A = ∅ (Always true)

Let p ∈

/ A0 . Thus A0 ⊆ A.
⇒p∈

Conversely, let A0 ⊆ A (we show that A is closed and Ac is open).

Let p ∈ Ac ⇒ p ∈ / A0
/ A⇒p∈

⇒ ∃ an open set G s.t. p ∈ G and (G \ {p}) ∩ A = φ.

/ A. Hence G ∩ A = φ ⇒ G ⊂ Ac , i.e.,
But p ∈

p ∈ G ⊂ Ac .

⇒ Ac is open or A is closed.
 Topology 23

Theorem: Let (X, τ ) be a top space. Then A ∪ A0 is closed, where A ⊆ X.

Proof: let p ∈ (A ∪ A0 )c

/ A ∪ A0
⇒p∈

⇒p∈ / A0
/ A and p ∈

⇒ ∃ an open set G s.t. p ∈ G and G ∩ A = ∅ or {p}. But p ∈

/ A.

⇒ G ∩ A = ∅.

• G ∩ A0 = ∅:

If any p ∈ G, then

p ∈ G and G ∩ A = ∅ or p ∈ G and (G \ {p}) ∩ A = ∅

/ A0
⇒p∈

⇒ G ∩ A0 = ∅. Hence

G ∩ (A ∪ A0 ) = (G ∩ A) ∪ (G ∩ A0 ) = ∅ ∪ ∅ = ∅.

Therefore G ⊆ (A ∪ A0 )c , i.e,

p ∈ G ⊆ (A ∪ A0 )c

⇒ (A ∪ A0 )c is open or A ∪ A0 is closed.

Remark: Let A ⊆ X. When will a point p ∈ X not be a limit point of A?

/ A0 iff ∃ G ∈ τ s. t. p ∈ G ⇒ (G \ {p}) ∩ A = ∅.
p∈

⇒ p ∈ G and G ∩ A = ∅ or G ∩ A = {p}.

⇒ p ∈ G and G ∩ A ⊆ {p}.

Theorem: Let A and B are subsets of a topological space X. Then

i. ∅0 = ∅.

ii. A ⊆ B ⇒ A0 ⊆ B 0 .

iii. (A ∩ B)0 ⊆ A0 ∩ B 0 .

iv. (A ∪ B)0 = A0 ∪ B 0 .
24 Dr. Zaffar Iqbal

Proof: (i) Since ∅ is closed. Therefore ∅0 ⊆ ∅. Also ∅ ⊆ ∅0 . Thus ∅0 = ∅.

(ii) Since p ∈ A0 ⇐⇒ (G \ {p}) ∩ A 6= ∅ ∀ G(open) containing p. But B ⊇ A. Therefore

(G \ {p}) ∩ B ⊇ (G \ {p}) ∩ A 6= ∅

⇒ p ∈ B 0 ,i.e, A0 ⊆ B 0 .

(iii) Since A ∩ B ⊆ A and A ∩ B ⊆ B

⇒ (A ∩ B)0 ⊆ A0 and (A ∩ B)0 ⊆ B 0

⇒ (A ∩ B)0 ⊆ A0 ∩ B 0 .

(iv) Since A ⊆ A ∪ B and B ⊆ A ∪ B.

⇒ A0 ⊆ (A ∪ B)0 and B 0 ⊆ (A ∪ B)0

⇒ A0 ∪ B 0 ⊆ (A ∪ B)0 .

/ A0 ∪ B 0 . Then ∃ G, H ∈ τ such that

Conversely, let p ∈

p ∈ G and G ∩ A ⊆ {p} , p ∈ H and H ∩ B ⊆ {p}.

But G ∩ H is open and p ∈ G ∩ H. Hence

(G ∩ H) ∩ (A ∪ B) = (G ∩ H ∩ A) ∪ (G ∩ H ∩ B)

⊆ (G ∩ A) ∪ (H ∩ B)

⊆ {p} ∪ {p} = {p}

/ (A ∪ B)0 . So (A ∪ B)0 ⊆ A0 ∪ B 0 .
⇒p∈

Thus (A ∪ B)0 = A0 ∪ B 0 .

Example: (A ∩ B)0 6= A0 ∩ B 0 .

Let A = [0, 1) , B = [1, 2) Then A0 = [0, 1] , B 0 = [1, 2], A0 ∩ B 0 = {1}, Now A ∩ B = ∅

∴ (A ∩ B)0 = ∅.
 Topology 25

Closure of a Set

Def: Let A be a subset of topological space X. Then the closure of A denoted by A is the

intersection of all closed supersets of A.i.e., If {Fi : i ∈ I} is the set of all closed subsets of X

containing A then

A = ∩ Fi .
i∈I

Exercise. A set A is closed ⇔ A = A.

Example. Let X = {a, b, c, d, e} , τ = {X, ∅, {a}, {c, d}, {a, c, d}, {b, c, d, e}}

Then the closed subsets of X are

∅, X, {b, c, d, e}, {a, b, e}, {b, e}, {a}. Then

{b} = {b, e} , {a, c} = X , {b, d} = {b, c, d, e}

•A is a closed set since it is intersection of closed sets.

• If F is a closed set containing A. Then A ⊆ A ⊆ F

Def: A subset A of X is called a dense subset of X iff A = X

Example. let X = N ,τ = {∅ , En } : En = {n, n + 1, n + 2, · · · }. Then find

(i) closed subsets of N

(ii) closure of sets {3, 6} and {2, 4, 6, . . .}.

Solution. (i) N, ∅, {1}, {1, 2}, {1, 2, 3}, · · · are closed subset of N

(ii) {3, 6} = {1, 2, 3, 4, 5, 6} , {2, 4, 6, . . .} = N

Remark: If F is a closed set containing A, then A0 ⊆ F .

Proof: Since A ⊆ F ⇒ A0 ⊆ F 0 . But

F0 ⊆ F (F is closed iff F 0 ⊆ F ). Therefore

A0 ⊆ F 0 ⊆ F or A0 ⊆ F.
26 Dr. Zaffar Iqbal

Theorem: A = A ∪ A0

Proof: Since A ⊆ A (i)

⇒ A0 ⊆ (A)0 .

Now A is closed

⇒ (A)0 ⊆ A i.e., A0 ⊆ (A)0 ⊆ A or

A0 ⊆ A. (ii)

by (i) and (ii) we have A ∪ A0 ⊆ A.

Also A ∪ A0 is a closed set containing A. So A ⊆ A ∪ A0 . Hence A ∪ A0 = A.

Theorem:

i: If A ⊆ B then A ⊆ B.

ii: A ∪ B = A ∪ B.

iii: A ∩ B ⊆ A ∩ B.

Proof. i. If A ⊆ B then A0 ⊆ B 0 . So A ∪ A0 ⊆ B ∪ B 0 ⇒ A ⊆ B.

ii. Since A ⊆ A ∪ B and B ⊆ A ∪ B

⇒ A ⊆ A ∪ B and B ⊆ A ∪ B. Taking union we have A ∪ B ⊆ A ∪ B.

As A ∪ B ⊆ A ∪ B (A ∪ B is closed containing A ∪ B)

⇒ A ∪ B ⊆ A ∪ B ⊆ A ∪ B,i.e.,

A ∪ B ⊆ A ∪ B. Hence A ∪ B = A ∪ B.

(iii) Since A ∩ B ⊆ A and A ∩ B ⊆ B

⇒ A ∩ B ⊆ A and A ∩ B ⊆ B or A ∩ B ⊆ A ∩ B.

Example. A ∩ B 6= A ∩ B:

Let A = (1, 2) , B = (2, 3). Then A = [1, 2] , B = [2, 3]. Also

A ∩ B = ∅ while A ∩ B = {2}.
 Topology 27

5. C HAPTER 5: I NTERIOR , E XTERIOR AND B OUNDARY

Def. Let A be a subset of a topological space (X, τ ). A point p ∈ A is called an Interior point

of A if ∃ an open set G(∈ τ ) containing p contained A , i.e, p ∈ G ⊆ A.

Def: The set of all the interior points of A is called an interior of A and is denoted by A◦ .

Def.

? A◦ is the union of all open sets contained in A.

? A◦ is open.

? A◦ is the largest open subset of A i.e., if G is an open subset of A then G ⊂ A◦ ⊂ A

? A is open iff A = A◦

Def: The exterior of A is the interior of Ac . i.e., ext(A) = (Ac )◦

Def: The boundary of A is the set of points of X which do not belong to A◦ or ext(A) and is

denoted by b(A) i.e.,

b(A) = X − (A◦ ∪ ext(A))

Example 1: Let X = R , A = [0, 1]. Then A◦ = (0, 1),

Ac = (−∞, 0) ∪ (1, ∞) = (Ac )◦ = ext(A), b(A) = {0, 1}.

Example 2: Let X = {a, b, c, d, e}, A = {c, d, e},

τ = {X, ∅, {a}, {c, d}, {a, c, d}, {b, c, d, e}}. Then find A◦ , ext(A) , b(A).

Solution:

(i) As c ∈ {c, d} ⊂ A ⇒ c ∈ A◦ .

d ∈ {c, d} ⊂ A ⇒ d ∈ A◦ .

A◦ = {c, d}
28 Dr. Zaffar Iqbal

(ii) ext(A)

Ac = {a, b}

a ∈ {a} ⊆ Ac ⇒ a ∈ ext(A) = (Ac )◦

(iii)

b(A) = X − (A◦ ∪ ext(A))

= {a, b, c, d, e} − {a, d, c}

= {b, e}

Proposition: Let X be a topological space and A ⊆ X, then

(i) X ◦ = X.

(ii) ∅◦ = ∅.

(iii) A◦ is largest open subset of A.

(iv) A◦ is the union of all open subsets contained in A.

(v) (A◦ )◦ = A◦ .

Theorem: Let A be a subset of a topological space X. Then A = A◦ ∪ b(A).

Proof: Since X = A◦ t b(A) t ext(A). Therefore

(A◦ ∪ b(A))c = ext(A). We show that (A)c = ext(A).

Let p ∈ ext(A) ⇒ p ∈ (Ac )◦ (by Def.)

⇒ ∃ an open set G containing p such that p ∈ G ⊂ Ac

⇒ p ∈ G and G ∩ A = ∅ or p ∈ G and (G − {p}) ∩ A = ∅.

/ A0 and p ∈
⇒p∈ / A (∵ p ∈ Ac )

/ A ∪ A0 = A
i.e., p ∈

/ A ⇒ p ∈ (A)c . So ext(A) ⊆ (A)c .

i.e., p ∈

Conversely, let p ∈ (A)c

 Topology 29

⇒p∈
/A

/ (A ∪ A0 )
⇒p∈

⇒p∈ / A0
/ A and p ∈

⇒ ∃ an open set G such that.

p ∈ G and (G − {p}) ∩ A = ∅

p ∈ G and G ∩ A = ∅ (∵ p ∈
/ A)

⇒ p ∈ G and G ⊂ Ac or p ∈ G ⊆ Ac

⇒ p ∈ (Ac )◦ = ext(A)

i.e., (A)c ⊆ ext(A)

∴ (A)c = ext(A) also we have

(A◦ ∪ b(A))c = ext(A)

∴ A = A◦ ∪ b(A)

Theorem: Let A, B ⊆ X then

a. If A ⊆ B then A◦ ⊆ B ◦

b. A◦ ∪ B ◦ ⊆ (A ∪ B)◦

c. (A ∩ B)◦ = A◦ ∩ B ◦

Proof. a. Let p ∈ A◦ ⇒ ∃ G ∈ τ s.t. p ∈ G ⊆ A. But A ⊆ B. Hence

p ∈ G ⊆ A ⊆ B or p ∈ G ⊆ B

⇒ p ∈ B◦

⇒ A◦ ⊆ B ◦ .

b. A◦ ∪ B ◦ ⊆ (A ∪ B)◦

A⊆A∪B ,B ⊆A∪B

A◦ ⊆ (A ∪ B)◦ , B ◦ ⊆ (A ∪ B)◦
30 Dr. Zaffar Iqbal

⇒ A◦ ∪ B ◦ ⊆ (A ∪ B)◦

c. (A ∩ B)◦ = A◦ ∩ B ◦

A ∩ B ⊆ A and A ∩ B ⊆ B

⇒ (A ∩ B)◦ ⊆ A◦ and (A ∩ B)◦ ⊆ B ◦

⇒ (A ∩ B)◦ ⊆ A◦ ∩ B ◦

Conversely, A◦ ⊆ A , B ◦ ⊆ B

⇒ A◦ ∩ B ◦ ⊆ A ∩ B

Since A◦ ∩ B ◦ is the open subset of A ∩ B

⇒ A◦ ∩ B ◦ ⊆ (A ∩ B)◦

Hence, (A ∩ B)◦ = A◦ ∩ B ◦

Example: (A ∪ B)◦ 6= A◦ ∪ B ◦ .

Let A = [0, 1], B = [1, 2], then A◦ = (0, 1), B ◦ = (1, 2),

A◦ ∪ B ◦ = (0, 1) ∪ (1, 2). But (A ∪ B)◦ = (0, 2).

Theorem: Let A be a subset of of a topological space X, then

i. b(A) = b(Ac ).

ii A◦ = A − b(A).

iii. A is closed ⇔ b(A) ⊆ A.

iv. A is open ⇔ A ∩ b(A) = ∅.

Proof. i.

b(A) = X − (A◦ ∪ (Ac )◦ )

= X − (((Ac )c )◦ ∪ (Ac )◦ )

= b(Ac )
 Topology 31

ii. As we have A − B = A ∩ B c . Then putting B = b(A) we have

A − b(A) = A ∩ (b(A))c

= A ∩ (A◦ ∪ (Ac )◦ )

= (A ∩ A◦ ) ∪ (A ∩ (Ac )◦ )

= A◦ ∪ ∅

= A◦

iii. Let A be closed ⇒ Ac is open

⇒ (Ac )◦ = Ac . (Y is open ⇔ Y ◦ = Y ). Therefore

b(A) = X − A◦ − (Ac )◦

= X − A◦ − Ac

= X − Ac − A◦

= A − A◦ ⊆ A

i.e., b(A) ⊆ A.

Conversely, let b(A) ⊆ A

b(A) ∪ A = A ⊆ A ∪ A0 = A, i.e.,

b(A) ∪ A ⊆ A

Also, A = A◦ ∪ b(A) ⊆ A ∪ b(A)

or A ⊆ A ∪ b(A), i.e.,

A ⊆ A ∪ b(A) = A

or A = A ⇒ A is closed.
32 Dr. Zaffar Iqbal

iv.

A is open ⇔ Ac is closed

⇔ b(Ac ) ⊆ Ac

⇔ b(A) ⊆ Ac

⇔ b(A) ∩ A = ∅
 Topology 33

Bases and Subbases

In the case of empty class ∅ of subsets of a universal set U .

(i) ∪(A : A ∈ ∅) = ∅

(ii) ∩(A : A ∈ ∅) = ∪

Base for a topology.

Def: Let (X, τ ) be a topological space. A collection B of open subsets of X(B ⊆ τ ) is a base

for the topology τ iff every open set G ∈ τ is the union of members of B.

Examples:

(1) Let X = {a, b, c, d}, τ = {X, ∅, {c}, {d}, {c, d}, {d, e}, {a, b, c}, {a, b, c, d}, {c, d, e}}

B = {X, ∅, {c}, {c, d}, {d, e}, {d}, {a, b, c}}

(2) X = {a, b, c, },τ = {X, ∅, {b}, {a, b}, {b, c}}.Then B = {{a, b}, {b, c}} is not a base for τ .

(3) The intervals from a base for τ in R : if G ⊆ R is open and p ∈ G, then ∃ (a, b) with

p ∈ (a, b) ⊆ G. open disks from a base for τ in R2 .

• If A = {A1 , A2 , · · · } such that ∪Ai = X. Then

(i) Empty union of members of A is ∅ = ∪{Ak : Ak ∈ ∅ ⊆ A}, and

(ii) Empty intersection of members of A is X = ∩{Ak : Ak ∈ ∅ ⊆ A}

34 Dr. Zaffar Iqbal

Theorem: Let B be a collection of subsets of a nonempty set X. Then B is a base for same

topology τ on X. If it satisfies the following properties:

(i) X = ∪{B : B ∈ B}

(ii) For any B1 , B2 ∈ B, B1 ∩ B2 is union of members of B.

Proof. Let B be a base for a topology τ on X. Since X is open. Therefore, by definition, X is

the union of members of B i.e., X = ∪{B : B ∈ B}

Also let B1 , B2 ∈ B

Since B1 , B2 are open ⇒ B1 ∩ B2 is open

⇒ B1 ∩ B2 is the union of members of B.

Conversely,

Let B be a class of X satisfying (i) and (ii). Let τ be the collection of all subsets of X which

are union of members of B.

We show that τ is a topology on X. From

(i) X = ∪{B : B ∈ B} ∴ X ∈ τ. Also ∅ = ∪{B : B ∈ ∅} hence ∅ ∈ τ .

By definition, each Gi is the union of members of B

∴ ∪ Gi is also union of members of B. Hence ∪ Gi ∈ τ.

i i

If G1 , G2 ∈ τ then ∃ {Bi : i ∈ I} and {Bj : j ∈ J} such that G1 = ∪ Bi and G2 = ∪ Bj .

i j

Therefore

G1 ∩ G2 = (∪ Bi ) ∩ (∪ Bj )
i j

= ∪{Bi ∩ Bj : i ∈ I, j ∈ J}

Since Bi ∩ Bj is union of members of B. (given)

k
Thus G1 ∩ G2 ∈ τ. Generally ∩ Gi ∈ τ . Hence τ is a topology on X with base B.
i=1
 Topology 35

Examples:

B1 = {(a, b) : a, b ∈ R, a < b} is a base for u-topology on R

B2 = {[a, b)}

B3 = {(a, b]}

B4 = {[a, b]} is not a base for u-topology on R. For example

B = {(−∞, a), (b, ∞) : a, b ∈ R} is not a base for any topology on R because

(−∞, 1) ∩ (0, ∞) = (0, 1) but (0, 1) cannot be written as union of members of B.

Remark:

Let (X, τ ) be a discrete topological space. Then B = {{p} : p ∈ X}, the set of singletons

from a base for τ : as every A ⊆ X is open and it can be written as union of singleton sets.

Subbases: Let (X, τ ) be a topological space A collection S of open subsets of X(S ⊆ τ ) is a

subbase for the topology τ on X iff finite intersections of members of S from a base B for τ .

Examples:

1. Let X = {a, b, c, d} and S = {{a, b}, {b, c}, {d}}

Taking the finite intersections of members of S we get

B = {X, ∅, {b}, {d}, {a, b}, {b, c}} (a base for a topology.)

Taking the finite unions of members of B we get

τ = {X, ∅, {b}, {a, b}, {b, c}, {d}, {b, d}, {a, b, d}, {b, c, d}, {a, b, c}}

2. X = R , S = {(−∞, a), (b, ∞) : a, b ∈ R} is a subbase for the usual topology on R.

3. X = R2 , S = {∅, (a, b) × R, R × (c, d) : a, b, c, d ∈ R} is a collection of vertical and

horizontal strips, is a subbase for τ .

Lemma: If B be a base for a topology τ on X and B ⊆ C ⊆ τ. Then C is also a base for τ.

36 Dr. Zaffar Iqbal

Proof: Let G be an open subset of X. Since B is a base for τ . Therefore G is the union of

members of B. i.e., G = ∪ Bi , Bi ∈ B. But B ⊆ C. Hence each Bi ∈ C. Therefore

i

G = ∪ Bi , Bi ∈ C i.e,
i

G is union of members of C. Hence C is a base for τ.

Examples 2: Let X = R and S = {[a, a + 1]}. Then show that the topology τ on X is discrete.

Solution: Let p ∈ R then the closed intervals [p − 1, p] and [p, p + 1] belongs to S and their

intersection

[p − 1, p] ∩ [p, p + 1] = {p}, a singleton belongs to B.i.e,

The set of singletons contained in B.

∴ (X, τ ) is discrete space.

Remark: If X = {a, b, c, d, e}. Then the collection

S = {{a, b}, {b, c}, {c, d}, {d, e}, {e, a}} is a subbase for the discrete space (X, D).

Theorem : Let S be a subbase for a topology τ on X and A ⊆ X. Then the collection

SA = {A ∩ S : s ∈ S} is a subbase for the relative topology τA on A.

Proof: Let H be a τA − open subset of A. Then H = A ∩ G where G is open in X. Since S is

a subbase for τ . Therefore G = ∪(Sα1 ∩ Sα2 ∩ · · · ∩ Sαk ), Sαi ∈ S. Hence

H = A∩G

= A ∩ [∪(Sα1 ∩ Sα2 ∩ · · · ∩ Sαk )]

α

= ∩ ∪[(A ∩ Sα1 ) ∩ · · · ∩ (A ∩ Sαk )]

α

H is the union of finite intersections of members of SA . Therefore SA is a subbase for τA .

Theorem: Any collection S of subsets of a nonempty set. X is the subbase for some topology

τ on X.
 Topology 37

Proof: We show the collection B of finite intersections of members of S satisfies the two

conditions for a base:

(i) X = ∪{B : B ∈ B}

(ii) G, H ∈ B then G ∩ H is union of members of B.

Since X is the empty intersection of members of S.

Therefore X ∈ B

X = ∪{B : B ∈ B}

Also, Let B1 = S1 ∩· · ·∩Sm , B2 = T1 ∩· · ·∩Tn be two elements of B. Then the intersection

B1 ∩ B2 = (S1 ∩ · · · ∩ Sm ) ∩ (T1 ∩ · · · ∩ Tn ) is also a finite intersection of elements of S

Projection maps: Let X and Y be nonempty sets. Then the functions

π1 : X × Y −→ X defined by π1 (x, y) = x

and

π2 : X × Y −→ Y defined by π2 (x, y) = y

are called projections of X × Y onto X and Y respectively.

• If X and Y are topological spaces, ∪ an open subset of X then

π1−1 (U ) = U × Y similarly

π2−1 (V ) = X × V , V open in Y .
38 Dr. Zaffar Iqbal

6. C HAPTER 6: C ONTINUITY IN T OPOLOGICAL S PACES

Continuity in R:

Definition: Let A ⊆ R , C ∈ A. Let f : A −→ R be a function. We say that f (x) is continuous

at c iff for each  > 0 , ∃ a δ > 0 such that |x − c| < δ ⇒ |f (x) − f (c)| < 

• f is continuous ⇔ lim f (x) = f (c)

x→c

Continuity in Topological Spaces:

Definition: Let X and Y be topological spaces and f : X → Y be a mapping. Then f is

continuous at x◦ ∈ X iff for each neighborhood V of f (x◦ ) in Y , there is a neighborhood U

of x◦ in X such that f (U ) ⊆ V . We say that f is continuous on X iff f is continuous at each

x◦ ∈ X.

Theorem (used as a definition): Let (X, τ1 ) and (Y, τ2 ) be topological spaces. A function

f : X −→ Y is said to be continuous iff for each open subset V of Y , the set f −1 (V ) is an open

subset of X. i.e,

f is continuous iff V ∈ τ2 ⇒ f −1 (V ) ∈ τ1

Example 1: Let (X, τ1 ) and (Y, τ2 ) be two topological spaces, where

X = {a, b, c, d} , τ1 = {X, ∅, {a}, {a, b}, {a, b, c}} and

Y = {w, x, y, z} , τ2 = {Y, ∅, {x}, {y}, {x, y}, {w, y, z}}. Consider the functions f : X → Y

and g : X → Y defined by
a .Z .w a .aa . w

b. Z b. a. x
%
 .
Z x ,
a %
f : .Z g:
c ZZZ. y c .a% .y
 %
aa
d. Z. z d .% a .z

Then f is continuous but g is not continuous.

 Topology 39

Example 2: Let (R, u) be a topological space and f : R −→ R be defined by



 x,
 x 6 1;
f (x) =

 x + 2, x > 1.


Then f is not continuous , as for A = (0, 2) , f −1 (A) = (0, 1].

Remark: Let X be a discrete space and Y be any topological space, then every function

f : X → Y is continuous.

Lemma: Let (X, τ1 ) and (Y, τ2 ) be topological spaces and f : X −→ Y be a function, and B be

a base for τ2 . Suppose for each B ∈ B , f −1 (B) is an open subset of X. Then f is continuous.

Proof: Let H be open subset of Y since B is a base for τ2 . Therefore

H = ∪ Bi , Bi ∈ B
i

Since f −1 (H) = f −1 (∪ Bi ) = ∪ f −1 (Bi )

i i

and each f −1 (Bi ) is open (given). Therefore f −1 (H) is open . Hence f is continuous.

Theorem : 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 continuous , and let F be any closed subset of Y .

Since Y − F is open in Y . Therefore f −1 (Y − F ) is open in X.(∵ f is continuous )

and f −1 (Y − F ) = X − f −1 (F ) = (f −1 (F ))c is open

⇒ f −1 (F ) is closed.

Conversely,
40 Dr. Zaffar Iqbal

Let f −1 (F ) be closed in X for every closed subset F of Y .

Let G be any open subset of Y . Then Y − G is closed in Y

⇒ f −1 (Y − G) is closed (by hypothesis ) in X and f −1 (Y − G) = X − f −1 (G) is closed

So f −1 (G) is open in X.

Theorem: A function f : X → Y is continuous iff f (A) ⊆ f (A) for every A ⊆ X.

Proof: Let f be continuous, Since f (A) is closed so f −1 (f (A)) is closed in X.

Now f (A) ⊆ f (A)

⇒ A ⊆ f −1 (f (A))

⇒ A ⊆ f −1 (f (A))

⇒ A ⊆ f −1 (f (A)) ∵ f −1 (f (A)) is closed.

⇒ f (A) ⊆ f (A)

Conversely, Let f (A) ⊆ f (A) for every A ⊆ X. Let F be any closed set in Y .

∴F =F

Since f −1 (F ) is a subset of X. So f (f −1 (F )) ⊆ f (f −1 (F )) i.e.,

f (f −1 (F )) ⊆ f (f −1 (F )) ⊆ F = F

⇒ f −1 (F ) ⊆ f −1 (F ) But

f −1 (F ) ⊆ f −1 (F ). Therefore

f −1 (F ) = f −1 (F ).

Hence f −1 (F ) is closed in X

⇒ f is continuous.

Theorem: A function f : X → Y is continuous iff f −1 (B) ⊆ f −1 (B) for every B ⊆ Y .

Proof: Let f be continuous. Let A = f −1 (B)

⇒ f (A) = B

⇒ f (A) = B
 Topology 41

but f (A) ⊆ f (A) (∵ f is continuous)

∴ f (A) ⊆ f (A) = Bi.e.,

f (A) ⊆ B

⇒ A ⊆ f −1 (B)

⇒ f −1 (B) ⊆ f −1 (B).

Conversely, let f −1 (B) ⊆ f −1 (B) for every B ⊆ Y. Let F be a closed subset of Y . Then

F = F . By hypothesis,

f −1 (F ) ⊆ f −1 (F ) = f −1 (F ) i.e,

f −1 (F ) ⊆ f −1 (F ) But

f −1 (F ) ⊆ f −1 (F ).

∴ f −1 (F ) = f −1 (F ).

Hence f −1 (F ) is closed in X. Therefore f is continuous.

Theorem: Let f : X → Y and g : Y → Z be continuous functions. Then g ◦ f is also

continuous.

Proof: Since g ◦ f : X → Z. Let G be an open set in Z. Then

g −1 (G) is open in Y .

⇒ f −1 (g −1 (G)) is open in X.

⇒ (f −1 ◦ g −1 )(G) is open in X.

⇒ (g ◦ f )−1 (G) is open in X.

⇒ g ◦ f is continuous.
42 Dr. Zaffar Iqbal

Open and Closed functions:

Open function:

A function f : X → Y is called an open function if the image of every open set is open. i.e.,

f is open ⇔ for every open set G in X we have f (G) open in Y .

Closed function: A function f : X → Y is called a closed function if the image of every

closed set is closed.

Examples: 1. f : R → R , f (x) = x2 is continuous and closed but not open.

as A = (−1, 1) , f (A) = [0, 1).

2. p1 : R2 → R , p1 (x, y) = x is continuous and open but not closed; as

A = {(x, y) : y > x1 , x > 0} is a closed set but p1 (A) = (0, ∞) not closed.

Exercise: f : X → Y, A ⊆ X, B ⊆ Y . Then f (A ∩ B) = f (A) ∩ f (B) if f is 1 − 1.

Example: Let X = {a, b, c, d} , τ = {X, ∅, {a}, {b}, {a, b}, {b, c, d}} and
a .aa . a
b. aa .
!! b
f : c .Z
!Z ! . c
!
!!Z
d .! Z . d

Then f is not continuous at c but continuous at d

 Topology 43

Homeomorphism:

Def: Let X and Y be topological spaces. A function f : X → Y is said to be a homeomorphism

if

(i) f is bijective.

(ii) f is continuous.

(iii) f −1 is continuous.

If a homeomorphism from X to Y exists, then we say that X and Y are homeomorphic and we

write X ' Y .

Examples:

(1) [0, 1] and [0, 2] are homeomorphic (f (x) = 2x).

(2) (−1, 1) and R are homeomorphic (f (x) = tan πx

2
).

(3) f : R → R by f (x) = 2x + 1 is a homeomorphism.

(4) Exercise: (a, b) ≈ R.

(5) [0, 1] w [a, b] , f (x) = (b − a)x + a.

• A homeomorphism f : X → Y gives a bijective correspondence not only between X and

Y but between the collection of open sets of X and Y .

• Homeomorphism preserves the topological structure.

Remark: The relation ' is an equivalence relation (exercise).

Theorem: Let f : X → Y be a bijective function from X to Y . Then following are equivalent.

(a) f is a homeomorphism.

(b) The set G ⊆ X is open iff f (G) is open.

(c) The set F ⊆ X is closed iff f (F ) is closed.

Proof: (a) ⇒ (b):

Let f be a homeomorphism and G ⊆ X be open. By continuity of f −1 ,

44 Dr. Zaffar Iqbal

(f −1 )−1 (G) is open in Y.i.e.,

f (G) is open in Y .

Conversely,

Let f (G) be open in Y . Then by continuity of f ,

f −1 (f (G)) is open in X.

But f −1 (f (G)) = G as f is 1 − 1 i.e.,

G is open in X.

(b) ⇒ (c):

Let F ⊆ X be closed.Then F c is open in X.

⇒ f (F c ) is open in Y .

but f is bijective. Therefore

f (F c ) = (f (F ))c is open.

⇒ f (F ) is closed.

Conversely, (⇐ to above)

Exercise (c) ⇒ (a):

 Topology 45

7. C HAPTER 7: S EPARATION A XIOMS

Topological property:

Let X and Y be topological spaces and f : X → Y be a homeomorphism. A property P for a

topological space is said to be a topological property if X has the property P then Y has also

the property P .

• Being open and being closed are topological properties.

Definitin A topological space X is called a T0 -space iff for any pair x, y ∈ X, x 6= y, ∃ an open

set U containing one and not the other. i.e.,

x ∈ U but y ∈
/U

or

y ∈ U but x ∈
/U


U 
U
y
x y or x
 

Example: Let X = {a, b, c, d}, τ = {∅, X, {a}, {b}, {a, b}}, then

a ∈ {a} but b ∈
/ {a}

b ∈ {b} but c ∈
/ {b}

a ∈ {a} but c ∈
/ {a}. Hence X is a T0 -space.

Remark: An indiscrete topological space X is not a T0 -Space.

T1 -Space:

Def: A topological space is a T1 -space iff for any pair of distinct points x, y of X, each belongs

to an open set which does not contains the other. i.e.,

For any pair x, y ∈ X, x 6= y, ∃ two open sets U and V in X such that

x ∈ U, y ∈
/ U and y ∈ V , x ∈
/ V.
46 Dr. Zaffar Iqbal

U'$
V
'$

x y
&%
&%

N ote : In this case U and V are not necessarily disjoint .

• Every T1 -space is T0 .

• Converse is not true.

Example: Let X = {a, b} , τ = {X, ∅, {a}} is a T0 − space but not T1 .

(Exercise)

• T2 or Hausdorff Space:

Def: A topological space X is called a T2 −space (Hausdorff) iff wherever x and y are distinct

points of X , ∃ disjoint open sets U and V in X with x ∈ U and y ∈ V . i.e.,

x ∈ U, y ∈ V and U ∩ V = ∅.

U
'$V
'$

x y
&%
&%

• Every T2 is T1 (and T◦ ).

Example: X = {a, b, c, d}, τ = {{a, b}, {a, c}, {c, d}, {b, d}, · · · } is a T2 −space.

Regular space: A topological space X is said to be a regular space if for any closed subset F

of X and p ∈ X, p ∈
/ F , ∃ two open sets G and H with G ∩ H = ∅ such that F ⊆ G and

p ∈ H.

G
'$ H
'$

F p

&%&%

Def: A regular space X which is also T1 -space is called a T3 -space.

 Topology 47

Remark: Every T1 -space may not be T2 −space.

Example: Consider R with cofinite topology. Then R is T1 but not T2 :

Let U = R − {y} , V = R − {x}. Then U and V are open sets in R ∵ their complements U c

and V c are finite. Suppose on contrary that R is T2 . Then for any pair x, y ∈ R, x 6= y ∃ open

sets U and V in R such that

x ∈ U, y ∈ V and U ∩ V = ∅. Hence

(U ∩ V )c = R: an infinite set.

But (U ∩ V )c = U c ∪ V c is finite , contradiction.

Theorem: Every subspace of a Huasdorff space is Hausdorff.

Proof: Let (X, τ ) be a Huasdorff space and (Y, τY ) be a subspace of X.

Let a, b ∈ Y with a 6= b.

Since X is Hausdorff; hence ∃ open sets G, H ∈ τ such that,

a ∈ G, b ∈ H and G ∩ H = ∅.

Now Y ∩ G and Y ∩ H ∈ τY (relative topology) and

a ∈ G, a ∈ Y ⇒ a ∈ Y ∩ G

b ∈ H, b ∈ Y ⇒ b ∈ Y ∩ H

Since G ∩ H = ∅. Therefore

(Y ∩ G) ∩ (Y ∩ H) = Y ∩ (G ∩ H)

= Y ∩∅

= ∅

∴ Y is also Hausdorff.

Remark: A regular space need not be a T1 −space.

48 Dr. Zaffar Iqbal

Example: Let X = {a, b, c} , τ = {X, ∅, {a}, {b, c}}. Then closed sets are X, ∅, {a}, {b, c}.

Then X is regular (prove). But X is not T1 as {b} is not closed.

Theorem: A subspace of a T1 −space is T1 .

Proof. (exercise)

Normal space: A topological space X is said to be normal iff whenever F1 and F2 are disjoint

closed subsets of X. Then ∃ disjoint open sets G and H such that F1 ⊆ G and F2 ⊆ H.

G
'$ H
'$
 
1
F
2
F
&% &%

Def: A normal space X which is also a T1 is called T4 -space.

Example: Let X = {a, b, c}, τ = {X, ∅, {a}, {b}, {a, b}}

F1 = ∅, F2 = X, F1 ⊆ ∅, F2 ⊆ X, i.e, X is normal. X is not a T1 -space since {a} is not

closed. Also X is not regular.(a ∈

/ {c}).

Theorem: A topological space X is a T1 -space iff each singleton is closed.

Proof. Suppose that X is a T1 -space. Let {x} be any singleton in X. Let y ∈ X −{x} ⇒ y 6= x.

As X is T1 , ∃ open sets Ux and Vy such that

x ∈ Ux , y ∈
/ Ux and y ∈ Vy , x ∈
/ Vy i.e,

y ∈ Vy ⊆ X − {x}

⇒ X − {x} = ∪ {y} ⊆ ∪ Vy ⊆ X − {x} i.e,

y∈X−{x} y∈X−{x}

X − {x} = ∪Vy , union of open sets is open.

Conversely, suppose that each singleton is closed.Let x, y ∈ X such that x 6= y, and {x}, {y}

are singleton sets which are closed in X. Then X − {x} and X − {y} are open.

Let U = X − {x}, V = X − {y}. Then clearly

x ∈ U, y ∈
/ U and y ∈ V, x ∈
/ V . Hence X is a T1 -space.