Lecture 17: Continuous Functions

1 Continuous Functions

Let (X, T

) and (Y, T

) be topological spaces.

Deﬁnition 1.1 (Continuous Function). A function f : X → Y is said to be

continuous if the inverse image of every open subset of Y is open in X. In other

words, if V ∈ T

, then its inverse image f

−1

(V ) ∈ T

Proposition 1.2. A function f : X → Y is continuous iﬀ for each x ∈ X and

each neighborhood N of f(x) in Y , the set f

−1

(N) is a neighborhood of x in X.

Proof. Let x be an arbitrary element of X and N an arbitrary neighborhood of

f(x) in Y . Then, f

−1

(N) and contains x and by deﬁnition, is open in X. Hence,

for each x ∈ X and each neighborhood N of f(x) in Y , the set f

−1

(N) is a

neighborhood of x in X. Conversely, let for each x ∈ X and each neighborhood N

of f(x) in Y , the set f

−1

(N) is a neighborhood of x in X. Let V be an arbitrary

open subset of Y .

i) If V ∩ f(X) = ∅, where f (X) is the range of f, then f

−1

(V ) = ∅ and hence is

open in X.

ii) If V ∩ f(X) 6= ∅, then V is a neighborhood of each of its points (let f(x)

be one such point for some x ∈ X). By assumption, f

−1

(V ) (⊆ X) must be

a neighborhood of each of its points (including the said x) in X and hence,

−1

(V ) is open in X.

Note 1. Continuity of a function depends not only on f but also on its domain

and co-domain topologies X and Y .

Example 1.3. Let (X, T

) and (Y, T

) be topological spaces and f : X → Y be

a function.

i) If f is a constant map, i.e., f(x) = y for all x ∈ X and some y ∈ Y , then f

is continuous for all topologies on X and Y because for any open subset V of

Y , f

−1

(V ) = ∅ (if y /∈ V ) or X (if y ∈ V ), both of which are always open in

any topology on X.

ii) If T

= P(X), i.e., (X, T

) is the discrete topology, then f is continuous for

any topology on Y because for any open subset V of Y , f

−1

(V ) is in P(X)

and hence is open in X.

iii) If T

= {∅, Y }, i.e., (Y, T

) is the trivial topology, then f is continuous for

any topology on X because f

−1

(∅) = ∅ and f

−1

(Y ) = X, both of which are

always open in any topology on X.

iv) The identity mapping from (X, T

) to (X, T

) is continuous because for any

U ∈ T

(co-domain topology), f

−1

(U) = U ∈ T

(domain topology).

Example 1.4.

Let (X, T

) and (Y, T

) be topological spaces deﬁned as follows:

X = {R, G, B} T

= {∅, {R}, {B}, {R, G}, {R, B}, X}

Y = {1, 2, 3} T

= {∅, {1}, {1, 2}, Y }

Let f and g be bijective mapping deﬁned as f(R) = 1, f(G) = 2 and f(B) = 3.

Then, f is continuous since

−1

(∅) = ∅, f

−1

({1}) = {R}, f

−1

({1, 2}) = {R, G}, f

−1

(Y ) = X

all of which are open in X. However, its inverse map g, with g(1) = R, g(2) = G

and g(3) = B, is not continuous since

−1

({B}) = {3} /∈ T

and g

−1

({R, B}) = {1, 3} /∈ T

Example 1.5. The unit step function u : R → {0, 1} is given by

u(x) =

(

0 if x < 0

1 if x > 0.

f(x)

Let R be equipped with the standard topology, i.e., all open intervals are open,

and the set {0, 1} be equipped with the discrete topology. Then, u

−1

(0) = (−∞, 0)

is open in the standard topology on R, but u

−1

(1) = [0, ∞) is not. Hence, the unit

step function is discontinuous.

Example 1.6. Let R and R

denote the set of real numbers equipped with the

standard and lower limit topology respectively, and f : R → R

and g : R

→ R

be identity functions, i.e., f(x) = g(x) = x, for every real number x. Then, f

is not continuous because the inverse image of the open set [a, b) in R

is [a, b)

which is not open in the standard topology. But g is continuous because the

inverse image of open interval (a, b) in the standard topology on R is open in R

−1

((a, b)) = (a, b) = ∪

n∈N

[a +

, b) and countable union of open sets is open).

Example 1.7. A function f : R → R is said to be continuous at x

∈ R if

∀  > 0, ∃ δ > 0 such that |x − x

| < δ → |f(x) − f(x

)| < ,

where both the domain and co-domain topologies are the standard topology on

R. The equivalence of this deﬁnition of continuity to the open-set deﬁnition of

continuity at x

is shown below.

Let f be continuous at x

by the open set deﬁnition, i.e., inverse image of

every open set containing x

is open. Given any  > 0, the interval V = (f(x

) −

, f(x

) + ) is open in the co-domain topology and hence, f

−1

(V ) is open in the

domain topology. Since f

−1

(V ) contains x

, it contains a basis (a, b) about x

(since for every open set S and every s ∈ S, there exists a basis B

such that

s ∈ B

⊆ S). Let δ be minimum of x

− a and b − x

. Then if |x − x

| < δ,

x must be in (a, b) and f

−1

(V ) (since (a, b) ⊆ f

−1

(V )). Hence f(x) ∈ V and

|f(x) − f(x

)| <  as required.

Now, let f be −δ−continuous at x ∈ R and V be an open set in the co-domain

topology containing f(x). Since V is open and f(x) ∈ V , there exists some  > 0,

such that (f(x) − , f(x) + ) ⊆ V . By continuity at x, there exists some δ > 0

such that (x − δ, x + δ) ⊆ f

−1

(V ). Since (x − δ, x + δ) is open in the domain

topology and the choices of  and V were arbitrary, inverse image of every open

set containing x is open as required by the open set deﬁnition of continuity at x.

Note that if an open set V in co-domain topology does not intersect the range of

f, then f

−1

(V ) = ∅, which is open in the domain topology.

Following are some properties of continuity.

1. For two topologies T

and T

on X, the identity map 1

from (X, T

) to

(X, T

) is continuous iﬀ T

is ﬁner than T

Proof. Let f = 1

. Since the map is identity, f

−1

(S) = S for any subset S

of X. Let the identity map be continuous. Then, for any V in T

, f

−1

(V ) is

in T

. Since f

−1

(V ) = V , this means that V is also in T

. Thus, T

⊆ T

i.e., (X, T

) is ﬁner than (X, T

). Conversely, let T

is ﬁner than T

. Then,

any set S in T

is also in T

. For any V in T

, f

−1

(V ) is in T

because

−1

(V ) = V and V is in T

. Thus, the identity map is continuous.

2. A continuous map remains continuous if the domain topology becomes ﬁner

or the co-domain topology becomes coarser.

Proof. Let (X, T

), (X, T

), (Y, S

) and (Y, S

) be topologies with T

and S

ﬁner than T

and S

respectively. Let f be a continuous map from (X, T

)

to (Y, S

i) Let V be in S

. Then, f

−1

(V ) is in T

, since f is continuous, and in T

since it is ﬁner than T

. Thus, f is also a continuous map from (X, T

)

to (Y, S

ii) Let V be in S

. Since, S

is ﬁner than S

, it contains V . Also T

contains

−1

(V ) since f is a continuous. Thus, f is also a continuous map from

(X, T

) to (Y, S

Note 2. i) From Property 1, it can be inferred that, continuity of a bijective

function f : X → Y does not guarantees continuity of its inverse (cf. Exam-

ples 1.4 and 1.6).

ii) In Example 1.6, had f been the identity map from R to itself then it would

have been continuous but replacing the co-domain topology with a ﬁner topol-

ogy (R

) renders it discontinuous.

To test the continuity of a map from a topological space on X to that on Y ,

checking whether inverse image of each open set in Y is open in X is not necessary.

Theorem 1.8. Let (X, T

) and (Y, T

) be topological spaces and f : X → Y be a

function. Then, the following statements are equivalent:

1. f is continuous.

2. Inverse image of every basis element of T

is open.

3. Inverse image of every subbasis element of T

is open.

Thus, to test the continuity of a function it suﬃces to check openness of inverse

images of elements of only a subset of T

, viz., its subbasis.

Proof. (1)→(2) Let f be continuous. Since every basis element of T

is open, its

inverse image will be open.

(2)→(1) Let B

be a basis for T

and let the inverse image of every basis element

B ∈ B

be open in X, i.e., f

−1

(B) ∈ T

. Note that any open set V

in Y can be written as a union of the basis elements, i.e., V = ∪

j∈J

−1

(V ) = ∪

j∈J

−1

), for some {B

, . . . , B

|J|

} ⊆ B

. Since union of opens

sets is open, f

−1

(V ) is open.

(2)→(3) Since every subbasis element is in the basis it generates, inverse image

of every subbasis element of Y is open in X.

(3)→(2) Let S

be subbasis of Y which generates the basis B

. Let the inverse

image of every subbasis element S ∈ S

be open in X, i.e., f

−1

(S) ∈ T

Since any basis element can be written as a ﬁnite intersection of subbasis

elements, i.e., B = ∩

i=1

, f

−1

(B) = ∩

i=1

−1

). Since ﬁnite intersection

of open sets is open, f

−1

(B) is open in X.

Theorem 1.9. Let f be a map from a topological space on X to a topological space

on Y . Then, the following statements are equivalent:

1. f is continuous.

2. Inverse image of every closed set of Y is closed in X.

3. For each x ∈ X and every neighborhood V of f(x), there is a neighborhood

U of x such that f(U) ⊆ V .

4. For every subset A of X, f(A) ⊆ f(A).

5. For every subset B of Y , f

−1

(B) ⊆ f

−1

(B).

Proof. (1)→(2) Let a subset C of Y be closed. Then, its complement Y \C is open

and the inverse image of the complement f

−1

(Y \C) = f

−1

(Y )\f

−1

X\f

−1

(2)→(1) Let V be open in Y . Then, its complement Y \V is closed and the

inverse image of the complement f

−1

(Y \V ) = f

−1

(Y )\f

−1

(V ) = X\f

−1

(V )

is closed in X. Hence, f

−1

(V ) is open in X.

(1)→(3) Since f

−1

(V ) is an open neighborhood of x, choose U = f

−1

(V ).

(3)→(4) Let A ⊆ X and x ∈ A. Let V be a neighborhood f(x) and U be a

neighborhood of x such that f(U) ⊆ V . Since x ∈ A, U ∩ A 6= ∅ and

hence ∅ 6= f(U ∩ A) ⊆ f(U) ∩ f(A) ⊆ V ∩ f (A) (cf. Lecture 5, Theorem

2.3(vii)). Since the choice of V neighborhood of f(x) was arbitrary, every

neighborhood of f(x) intersects f(A). Hence, f(x) ∈ f(A) and f(A) ⊆ f(A).

(4)→(5) Let A = f

−1

(B). Then, by (4), f(A) ⊆ f(A) = f(f

−1

(B)) = B. Hence,

−1

(B) ⊆ f

−1

(B).

(5)→(2) Let B ⊆ Y be closed; then, B = B since a set is closed iﬀ it is equal to

its closure. Then, by (5), f

−1

(B) ⊆ f

−1

(B) and since f

−1

(B) ⊆ f

−1

(B) is

always true, f

−1

(B) = f

−1

(B). Hence, f

−1

(B) is closed (being equal to its

closure).

2 Homeomorphism

Deﬁnition 2.1 (Homeomorphism). Let (X, T

) and (Y, T

) be topological

spaces and f : X → Y be a bijection. If both f and its inverse f

−1

: Y → X are

continuous, then f is called a homeomorphism.

The two spaces are said to be homeomorphic and each is a homeomorph of the

other. If a map is a homeomorphism, then so is its inverse. Composition of any

two homeomorphisms is again a homeomorphism.

The requirement the f

−1

be continuous means that for any U open in X, its

inverse image under f

−1

be open in Y . But since the inverse image of U under f

−1

is same as the image of U under f (cf. Lecture 5, Remark 2(vi)), another way to

deﬁne a homeomorphism is to say that it is a bijective map f : X → Y such that

f(U) is open iﬀ U is open. Thus, a homeomorphism is a bijection between T

and T

. Consequently, any property of X expressed in terms of T

(or the open

sets), yields, via f, the corresponding property for Y . Such a property is called a

topological property of X.

Let f : X → Y be an injective continuous map and Z = f(X) ⊂ Y be its

range, considered as a subspace of Y . Then, the map obtained by restricting Y to

Z, f

: X → Z is a bijection. If f

happens to be a homeomorphism, then we say

that f : X → Y is a topological imbedding, or simply an imbedding, of X in Y .

Example 2.2. Let R be equipped with the trivial, standard or discrete topology.

For every pair of real numbers m and c, the function f

m,c

: R → R deﬁned by

m,c

(x) = mx + c, ∀x ∈ R is a homeomorphism.

Example 2.3. i) The identity map from a topological space to itself is a home-

omorphism (Example 1.3(iv)).

ii) The map f in Examples 1.4 and the map g in1.6 are both continuous and

bijective but not homeomorphic because their inverse maps are not continuous.

Example 2.4. i) Two discrete spaces are homeomorphic iﬀ there is a bijection

between them, i.e., iﬀ they have the same cardinality. This is true because

every function on a discrete space is continuous, no matter the co-domain

topology (Example 1.3(ii)).

ii) Two trivial topologies are homeomorphic iﬀ there is a bijection between them.

This holds because every function to a trivial topology is continuous regardless

of the domain topology (Example 1.3(iii)).

Proposition 2.5. Let (X, T

) and (Y, T

) be topological spaces and f : X → Y

be a function. Then, the following statements are equivalent:

i) f is a homeomorphism.

ii) U is open in X iﬀ f(U) is open in Y .

iii) C is closed in X iﬀ f(C) is closed in Y .

iv) V is open in Y iﬀ f

−1

(V ) is open in X.

v) D is closed in Y iﬀ f

−1

(D) is closed in X.

3 Constructing Continuous Functions

Some rules for constructing continuous functions are given below.

Theorem 3.1. Let X, Y and Z be topological spaces.

1. (Constant function) If f : X → Y deﬁned as f(x) = y for all x ∈ X and

some y ∈ Y , then f is continuous.

2. (Inclusion) If A is a subspace of X, then the inclusion function j : A → X

is continuous. (j(a) = a, ∀ a ∈ A)

3. (Composites) If f : X → Y and g : Y → Z are continuous, then so is their

composition g ◦ f : X → Z.

4. (Restricting the domain) If f : X → Y is continuous and A is a subspace of

X, then the restriction of f to A, f|A : A → Y is also continuous.

5. (Restricting or expanding the range) Let f : X → Y be continuous.

a) If Z is subspace of Y containing the range f(X), then the function g :

X → Z obtained by restricting the co-domain topology is continuous.

b) If Z is space containing Y as a subspace, then the function h : X → Z

obtained by expanding the co-domain topology is continuous.

6. (Local formulation of continuity)The map f : X → Y is continuous if X cna

be written as the union of open sets U

such that f|U

is continuous for each

α.

Proof. i) See Example 1.3(i).

ii) If U is open in X, then j

−1

(U) = U ∩ A is open in A by deﬁnition of subspace

topology.

iii) If W is open in Z, then g

−1

(W ) is open in Y since g is continuous. Since

f is continuous, f

−1

(W )) is open in X. Thus, g ◦ f is continuous (since

−1

(W )) = (g ◦ f)

−1

(W )).

iv) f|A = j ◦f, both of which are continuous and composition of continuous maps

is continuous.

v) (a) Let W be open in Z. Then, B = Z ∩ U for some U open in Y . Since

f(Z) ⊆ Z, f

−1

(B) = f

−1

(U) and is open in X because f

−1

(U) is open in X.

(b) Let j : Y → Z be the inclusion map. Then, h = f ◦ j.

vi) Let V be open in Y . Then,

−1

(V ) ∩ U

= (f|U

)

−1

(V )

and is open in U

and hence open in X. But

−1

(V ) =

[



−1

(V ) ∩ U



so that V is also open in X.

Theorem 3.2 (The Pasting Lemma). Let X = A ∪ B, where A and B are

closed in X. Let f : A → Y and g : B → Y e continuous maps. If f(x) = g(x)

for every x ∈ A ∩ B, the f and g combine to give a continuous map h : X → Y ,

deﬁned as

h(x) =

(

f(x) if x ∈ A

g(x) if x ∈ B.

Proof. Let C be a closed subset of Y . Then, h−1(C) = f

−1

closed in X since each of f

−1

The pasting lemma hold even if A and B are open in X and is a special case

of Theorem 3.1(vi).

Theorem 3.3 (Maps into Products). Let f : A → X × Y be deﬁned as f(a) =

(a), f

(a)). Then f is continuous iﬀ both the co-ordinate functions f

: A → X

and f

: A → Y are continuous.

Proof. The projection maps π

: X × Y → X and π

: X × Y → Y onto the ﬁrst

and second factor space are continuous since π

−1

(U) = U ×Y and π

−1

(V ) = X ×V

are open if U and V are open in X and Y respectively. Note that f

= π

◦ f and

= π

◦ f. If f is continuous, then so are f

and f

(composites of continuous

functions). Conversely, let f

and f

are continuous. Let U × V be a basis element

of X × Y . A point a is in f

−1

(U × V ) iﬀ f(a) ∈ U × V , i.e., iﬀ f

(a) ∈ U and

(a) ∈ V . Hence, f

−1

(U × V ) = f

−1

(U) ∩ f

−1

(V ) and is open in A since both

−1

(U) and f

−1

(V ) are open. Thus, since inverse image of every basis element is

open, f is continuous (by Theorem 1.8(2))