The following is a fairly general definition of what it means for a function to be continuous. It includes everything seen in typical calculus classes as a special case.

Definition 11.1.26 Let f : X → Y be a function where

and

are metric spaces. Then
f is continuous at x ∈ X if and only if the following condition holds. For every ε > 0, there exists
δ > 0 such that if d

< δ, then ρ

< ε. If f is continuous at every x ∈ X we say
that f is continuous on X.

For example, you could have a real valued function f

defined on an interval

. In this case you
would have X =

and
Y = ℝ with the distance given by d

=

. Then the following theorem
is the main result.

Theorem 11.1.27 Let f : X → Y where

and

are metric spaces. Then the following
are equivalent.

- f is continuous at x.
- Whenever x
_{n}→ x, it follows that f→ f.

Also, the following are equivalent.

- f is continuous on X.
- Whenever V is open in Y, it follows that f
^{−1}≡is open in X. - Whenever H is closed in Y, it follows that f
^{−1}is closed in X.

Proof: a

b: Let
f be continuous at x and suppose x

< δ, then ρ

< ε. Since x

< δ and so, if n ≥ N, it follows that ρ

< ε. Since
ε > 0 is arbitrary, it follows that f

→ f

.

b

a: Suppose b holds but
f fails to be continuous at x. Then there exists ε > 0 such that for all
δ > 0, there exists

such that
d

< δ but ρ

≥ ε. Letting δ = 1∕n, there exists x

< 1∕n but ρ

≥ ε. Now this is a contradiction because by assumption, the fact
that x

→ f

. In particular, for large enough
n, ρ

< ε contrary to
the construction.

c

d: Let
V be open in Y . Let x ∈ f

so that
f

∈ V. Since V is open, there exists ε > 0
such that B

⊆ V . Since f is continuous at x, it follows that there exists δ > 0 such
that if

∈ B

, then f

∈ B

⊆ V.

In other words,
B

⊆ f

⊆ f

which shows that, since
x was an arbitrary point of f

, every
point of f

is an interior point which implies
f

is open.

d

e: Let
H be closed in Y . Then f

which is open by assumption. Hence
f

is closed because its complement is open.

e

d: Let
V be open in Y. Then f

which is assumed to be closed. This is
because the complement of an open set is a closed set.

d

c: Let
x ∈ X be arbitrary. Is it the case that f is continuous at x? Let ε > 0 be given. Then
B

is an open set in
V and so x ∈ f

which is given to be open. Hence there exists
δ > 0 such that x ∈ B

⊆ f

. Thus, f

⊆ B

so
ρ

< ε.
Thus f is continuous at x for every x. ■

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