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Of course we are already familiar¹ with the notion of a continuous function from R into R.

A function f : R → R is said to be continuous if for each a ∈ R and each positive real number ε, there exists a positive real number δ such that | x − a |< δ implies | f (x) − f (a) |< ε.

It is not at all obvious how to generalize this definition to general topological spaces where we do not have “absolute value
or “subtraction
. So we shall seek another (equivalent) definition of continuity which lends itself more to generalization.

It is easily seen that f : R→ R is continuous if and only if for each a ∈ R and each interval (f (a)− ε, f (a) + ε), for ε > 0, there exists a δ > 0 such that f (x) ∈ (f (a) − ε , f (a) + ε) for all x∈ (a − δ , a + δ).

This definition is an improvement since it does not involve the concept “absolute value
but it still involves “subtraction
. The next lemma shows how to avoid subtraction.

1The early part of this section assumes that you have some knowledge of real analysis and, in particular, the ε–δ definition of continuity. If this is not the case, then proceed directly to Definition 5.1.3.

90

5.1. CONTINUOUS MAPPINGS 91

5.1.1 Lemma. Let f be a function mapping R into itself. Then f is continuous if and only if for each a ∈ R and each open set U containing f (a), there exists an open set V containing a such that f (V )⊆ U .

Proof. Assume that f is continuous. Let a ∈ R and let U be any open set containing f (a).

Then there exist real numbers c and d such that f (a) ∈ (c, d) ⊆ U . Put ε equal to the smaller of the two numbers d− f (a) and f (a) − c, so that

(f (a)− ε , f (a) + ε) ⊆ U.

As the mapping f is continuous there exists a δ > 0 such that f (x)∈ (f (a) − ε , f (a) + ε) for all x∈ (a − δ , a + δ). Let V be the open set (a − δ , a + δ). Then a ∈ V and f (V ) ⊆ U , as required.

Conversely assume that for each a∈ R and each open set U containing f (a) there exists an open set V containing a such that f (V ) ⊆ U . We have to show that f is continuous. Let a ∈ R and ε be any positive real number. Put U = (f (a)− ε , f (a) + ε). So U is an open set containing f (a). Therefore there exists an open set V containing a such that f (V ) ⊆ U . As V is an open set containing a, there exist real numbers c and d such that a∈ (c, d) ⊆ V . Put δ equal to the smaller of the two numbers d− a and a − c, so that (a − δ , a + δ) ⊆ V . Then for all x ∈ (a − δ , a + δ),

f (x) ∈ f (V ) ⊆ U , as required. So f is continuous. 

We could use the property described in Lemma 5.1.1 to define continuity, however the following lemma allows us to make a more elegant definition.

5.1.2 Lemma. Let f be a mapping of a topological space (X,

τ

) into a topological space (Y,

τ

⁰). Then the following two conditions are equivalent:

(i) for each U ∈

τ

⁰, f⁻¹(U )∈

τ

,

(ii) for each a∈ X and each U ∈

τ

⁰ with f (a) ∈ U , there exists a V ∈

τ

such that a ∈ V and f (V )⊆ U .

Proof. Assume that condition (i) is satisfied. Let a ∈ X and U ∈

τ

^{0 with f (a)} ∈ U . Then f⁻¹(U ) ∈

τ

. Put V = f⁻¹(U ), and we have that a ∈ V, V ∈

τ

, and f (V )⊆ U . So condition (ii) is satisfied.

Conversely, assume that condition (ii) is satisfied. Let U ∈

τ

⁰. If f⁻¹(U ) = Ø then clearly f⁻¹(U ) ∈

τ

^{. If f−1}(U )6= Ø, let a ∈ f⁻¹(U ). Then f (a)∈ U . Therefore there exists a V ∈

τ

^such

that a∈ V and f (V ) ⊆ U . So for each a ∈ f⁻¹(U ) there exists a V ∈

τ

such that a ∈ V ⊆ f⁻¹(U ).

By Corollary 3.2.9 this implies that f⁻¹(U )∈

τ

. So condition (i) is satisfied. 

Putting together Lemmas 5.1.1 and 5.1.2 we see that f : R→ R is continuous if and only if for each open subset U of R, f⁻¹(U ) is an open set.

This leads us to define the notion of a continuous function between two topological spaces as follows:

5.1.3 Definition. Let (X,

τ

) and (Y,

τ

1) be topological spaces and f a function from X into Y . Then f : (X,

τ

⁾ → (Y,

τ

1) is said to be a continuous mapping if for each U ∈

τ

1, f⁻¹(U ) ∈

τ

^.

From the above remarks we see that this definition of continuity coincides with the usual definition when (X,

τ

) = (Y,

τ

1) = R.

5.1. CONTINUOUS MAPPINGS 93 Let us go through a few easy examples to see how nice this definition of continuity is to apply in practice.

5.1.4 Example. Consider f : R → R given by f (x) = x, for all x ∈ R; that is, f is the identity function. Then for any open set U in R, f⁻¹(U ) = U and so is open. Hence f is continuous. 

5.1.5 Example. Let f : R → R be given by f (x) = c, for c a constant, and all x ∈ R. Then let U be any open set in R. Clearly f⁻¹(U ) = R if c∈ U and Ø if c 6∈ U . In both cases, f⁻¹(U ) is

open. So f is continuous. 

5.1.6 Example. Consider f : R→ R defined by

f (x) =

(x− 1, if x≤ 3

1

2(x + 5), if x > 3.

..................................................................................

.............................................

1 2 3 4 5

1 2 3 4 5

Recall that a mapping is continuous if and only if the inverse image of every open set is an open set.

Therefore, to show f is not continuous we have to find only one set U such that f⁻¹(U ) is not open.

Then f⁻¹((1, 3)) = (2, 3], which is not an open set. Therefore f is not continuous. 

Note that Lemma 5.1.2 can now be restated in the following way.²

5.1.7 Proposition. Let f be a mapping of a topological space (X,

τ

) into a space (Y,

τ

^0).

Then f is continuous if and only if for each x ∈ X and each U ∈

τ

⁰ with f (x) ∈ U , there

exists a V ∈

τ

such that x∈ V and f (V ) ⊆ U . 

5.1.8 Proposition. Let (X,

τ

), (Y,

τ

1) and (Z,

τ

2) be topological spaces. If f : (X,

τ

)→ (Y,

τ

1) and g : (Y,T1) → (Z,

τ

2) are continuous mappings, then the composite function g◦ f : (X,

τ

⁾→ (Z,

τ

2) is continuous.

Proof.

To prove that the composite function g◦ f : (X,

τ

)→ (Z,

τ

2) is continuous, we have to show that if U ∈

τ

2, then (g◦ f )⁻¹(U ) ∈

τ

.

But (g◦ f )⁻¹(U ) = f⁻¹(g⁻¹(U )).

Let U be open in (Z,

τ

2). Since g is continuous, g⁻¹(U ) is open in

τ

1. Then f⁻¹(g⁻¹(U )) is open in

τ

as f is continuous. But f⁻¹(g⁻¹(U )) = (g◦ f )⁻¹(U ). Thus g◦ f is continuous. 

The next result shows that continuity can be described in terms of closed sets instead of open sets if we wish.

5.1.9 Proposition. Let (X,

τ

) and (Y,

τ

1) be topological spaces. Then f : (X,

τ

) → (Y,

τ

1) is continuous if and only if for every closed subset S of Y, f⁻¹(S) is a closed subset of X.

Proof. This results follows immediately once you recognize that

f⁻¹(complement of S) = complement of f⁻¹(S). 

2If you have not read Lemma 5.1.2 and its proof you should do so now.

5.1. CONTINUOUS MAPPINGS 95 5.1.10 Remark. There is a relationship between continuous maps and homeomorphisms: if f : (X,

τ

) → (Y,

τ

1) is a homeomorphism then it is a continuous map. Of course not every continuous map is a homeomorphism.

However the following proposition, whose proof follows from the definitions of “continuous

and “homeomorphism
tells the full story.

5.1.11 Proposition. Let (X,

τ

) and (Y,

τ

⁰) be topological spaces and f a function from X into Y . Then f is a homeomorphism if and only if

(i) f is continuous,

(ii) f is one-to-one and onto; that is, the inverse function f⁻¹ : Y → X exists, and

(iii) f⁻¹ is continuous. 

A useful result is the following proposition which tells us that the restriction of a continuous map is a continuous map. Its routine proof is left to the reader – see also Exercise Set 5.1 #8.

5.1.12 Proposition. Let (X,

τ

) and (Y,

τ

1) be topological spaces, f : (X,

τ

)→ (Y,

τ

1) a continuous mapping, A a subset of X, and

τ

2 the induced topology on A. Further let g : (A,

τ

2)→ (Y,

τ

1) be the restriction of f to A; that is, g(x) = f (x), for all x∈ A. Then g is continuous.

Exercises 5.1

1. (i) Let f : (X,

τ

⁾→ (Y,

τ

1) be a constant function. Show that f is continuous.

(ii) Let f : (X,

τ

⁾→ (X,

τ

) be the identity function. Show that f is continuous.

2. Let f : R → R be given by

f (x) =

(−1, x≤ 0 1, x > 0.

(i) Prove that f is not continuous using the method of Example 5.1.6.

(ii) Find f⁻¹{1} and, using Proposition 5.1.9, deduce that f is not continuous.

3. Let f : R → R be given by

f (x) =

(x, x≤ 1 x + 2, x > 1.

Is f continuous? (Justify your answer.)

4. Let (X,

τ

) be the subspace of R given by X = [0, 1]∪ [2, 4]. Define f : (X,

τ

)→ R by

f (x) =

(1, if x∈ [0, 1]

2, if x∈ [2, 4].

Prove that f is continuous. (Does this surprise you?)

5. Let (X,

τ

) and (Y,

τ

1) be topological spaces andB1 a basis for the topology

τ

1. Show that a map f : (X,

τ

)→ (Y,

τ

1) is continuous if and only if f⁻¹(U ) ∈

τ

, for every U ∈ B1.

6. Let (X,

τ

^{) and (Y,}

τ

1) be topological spaces and f a mapping of X into Y . If (X,

τ

^{) is a}

discrete space, prove that f is continuous.

7. Let (X,

τ

^{) and (Y,}

τ

1) be topological spaces and f a mapping of X into Y . If (Y,

τ

1) is an indiscrete space, prove that f is continuous.

8. Let (X,

τ

^{) and (Y,}

τ

1) be topological spaces and f : (X,

τ

⁾→ (Y,

τ

1) a continuous mapping.

Let A be a subset of X,

τ

2 the induced topology on A, B = f (A),

τ

3 the induced topology on B and g : (A,

τ

2)→ (B, T3) the restriction of f to A. Prove that g is continuous.

5.2. INTERMEDIATE VALUE THEOREM 97

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