Continuity Defined by Closure/Proof 1

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Theorem

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.

Let $f: T_1 \to T_2$ be a mapping.


Then $f$ is continuous if and only if:

$\forall H \subseteq S_1: f \sqbrk {H^-} \subseteq \paren {f \sqbrk H}^-$

where $H^-$ denotes the closure of $H$ in $T_1$.


That is, if and only if the image of the closure is a subset of the closure of the image.


Proof

Necessary Condition

Let $f$ be continuous.

Let $y \in f \sqbrk {\map \cl H}$.

Then:

$\exists x \in \map \cl H: y = \map f x$


Let $U$ be an open set of $T_2$ such that $y \in U$.

Then by definition of continuous mapping:

$f^{-1} \sqbrk U$ is an open set of $T_1$ such that:
$x \in f^{-1} \sqbrk U$

Hence:

$f^{-1} \sqbrk U \cap H \ne \O$

as $x \in \map \cl H$.

Hence:

\(\ds U \cap f \sqbrk H\) \(\supseteq\) \(\ds f \sqbrk {f^{-1} \sqbrk U \cap H}\)
\(\ds \) \(\ne\) \(\ds \O\)

and:

$y \in \map \cl {f \sqbrk H}$

That is:

$f \sqbrk {\map \cl H} \subseteq \map \cl {f \sqbrk H}$

$\Box$


Sufficient Condition

Suppose that for all $H \subseteq S_1$:

$f \sqbrk {\map \cl H} \subseteq \map \cl {f \sqbrk H}$

Let $V \subseteq S_2$ be closed in $T_2$.

Then:

$f \sqbrk {\map \cl {f^{-1} \sqbrk V} } \subseteq \map \cl V = V$

So by Set is Closed iff Equals Topological Closure:

$\map \cl {f^{-1} \sqbrk V} \subseteq f^{-1} \sqbrk V$

and so $f^{-1} \sqbrk V$ is closed in $T_1$.

Hence from Continuity Defined from Closed Sets:

$f$ is continuous.

$\blacksquare$


Sources

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