Continuity Defined by Closure/Proof 1

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:

and:
 * $y \in \map \cl {f \sqbrk H}$

That is:
 * $f \sqbrk {\map \cl H} \subseteq \map \cl {f \sqbrk H}$

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.