Continuous Functions Between Topological Spaces

Table of Contents

Continuous Functions

Definition: Let

$E$

and

$F$

be topological spaces. A function

f : E \to V

is Continuous at $$x$$ if for every neighbourhood

V_{f(x)}

$f(x)$

$F$

there exists a neighbourhood

$U_x$

$x$

$E$

such that

f(u) \in V

for all

u \in U

, or equivalently,

f(U_x) \subseteq V_{f(x)}

. The function

$f$

is said to be Continuous on $$E$$ if it is continuous at every point

x \in E

Proposition 1: Let

$E$

and

$F$

be topological spaces and let

f : E \to F

. Then

$f$

is continuous on

$E$

if and only if for every open set

$V$

$F$

, the preimage,

f^{-1}(V)

, is an open set in

$E$

Proof: $\Rightarrow$ : Let $$V$$ be an open set in $$F$$ . We want to show that $f^{-1}(V)$ is open in $$E$$ . So let $x \in f^{-1}(V)$ . Since $$f$$ is continuous on $$E$$ , it is continuous at $$x$$ in particular, so for the neighbourhood $$V$$ of $$f(x)$$ there exists an open neighbourhood $$U_x$$ of $$x$$ in $$E$$ such that $f(U_x) \subseteq V$ . So:

(1)

\begin{align} \quad f^{-1}(V) = \bigcup_{x \in f^{-1}(V)} U_x \end{align}

Which shows that $f^{-1}(V)$ is an arbitrary union of open sets in $$E$$ , so that $f^{-1}(V)$ is open in $$E$$ .

$\Leftarrow$ Let $x \in E$ and let $V_{f(x)} \subseteq F$ be a neighbourhood of $$f(x)$$ . Then there exists an open set $V'_{f(x)}$ in $$F$$ such that $x \in V'_{f(x)} \subseteq V_{f(x)}$ , and by assumption, $f^{-1}(V'_{f(x)})$ is an open set in $$E$$ which contains $$x$$ . So set $U_x := f^{-1}(V'_{f(x)})$ . Then $$U_x$$ is an (open) neighbourhood of $$x$$ with $f(U_x) \subseteq V'_{f(x)} \subseteq V_{f(x)}$ . So $$f$$ is continuous at $$x$$ and is thus continuous on all of $$E$$ . $\blacksquare$

Proposition 2: Let

$E$

and

$F$

be topological spaces and let

f : E \to F

. Then

$f$

is continuous on

$E$

if and only if for every closed set

$V$

$F$

, the preimage,

f^{-1}(V)

, is a closed set in

$E$

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