Continuity Defined by Closure

Theorem
Let $T_1 = \left({X_1, \vartheta_1}\right)$ and $T_2 = \left({X_2, \vartheta_2}\right)$ be topological spaces.

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

Then $f$ is continuous iff:
 * $\forall H \subseteq X_1: f \left({H^-}\right) \subseteq \left({f \left({H}\right)}\right)^-$

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

That is, iff the image of the closure is a subset of the closure of the image.

Proof
First we establish some details.

Let $H \subseteq X_1$.

Let $\mathbb K_1$ be defined as:
 * $\mathbb K_1 := \left\{{K \subseteq X_1: H \subseteq K, K \text{ closed}}\right\}$

That is, let $\mathbb K_1$ be the set of all closed sets of $T_1$ which contain $H$.

Similarly, let $\mathbb K_2$ be defined as:
 * $\mathbb K_2 := \left\{{K \subseteq X_2: f \left({H}\right) \subseteq K, K \text{ closed}}\right\}$

That is, let $\mathbb K_2$ be the set of all closed sets of $T_2$ which contain $f \left({H}\right)$.

From the definition of closure, we have that:
 * $\displaystyle H^- = \bigcap \mathbb K_1$

That is, the closure of $H$ is the intersection of all the closed sets of $T_1$ which contain $H$.

Similarly:
 * $\displaystyle \left({f \left({H}\right)}\right)^- = \bigcap \mathbb K_2$

That is, the closure of $f \left({H}\right)$ is the intersection of all the closed sets of $T_2$ which contain $f \left({H}\right)$.

We have:

From Subset of Image we have:
 * $H \subseteq K \implies f \left({H}\right) \subseteq f \left({K}\right)$

Suppose $f$ is continuous.

From the above we have that :
 * $\displaystyle \left({f \left({H}\right)}\right)^- := \bigcap \mathbb K_2$

As $f$ is continuous, then:
 * $\forall K \in \mathbb K_2: f^{-1} \left({K}\right)$ is closed in $T_1$

But as $f \left({H}\right) \subseteq K$, it follows from Subset of Image that $H \subseteq f^{-1} \left({K}\right)$.

So:
 * $\mathbb K_3 := \left\{{f^{-1} \left({K}\right): K \text{ closed in } T_2, H \subseteq f^{-1} \left({K}\right)}\right\}$

consists entirely of closed sets in $T_1$ which are supersets of $H$.

That is, $\mathbb K_3 \subseteq \mathbb K_1$.

So:
 * $\displaystyle \bigcap \mathbb K_1 \subseteq \bigcap \mathbb K_3$

and so:
 * $\displaystyle f \left({\bigcap \mathbb K_1}\right) \subseteq f \left({\bigcap \mathbb K_3}\right)$

But from Mapping Image of Intersection we have that:
 * $\displaystyle f \left({\bigcap \mathbb K_3}\right) \subseteq \bigcap_{K \in \mathbb K_3} f \left({K}\right)$

But:
 * $\displaystyle \bigcap_{K \in \mathbb K_3} f \left({K}\right) = \bigcap \mathbb K_2$

and so:
 * $\displaystyle f \left({\bigcap \mathbb K_1}\right) \subseteq \bigcap \mathbb K_2$

which means that:
 * $f \left({H^-}\right) \subseteq \left({f \left({H}\right)}\right)^-$

as we wanted to show.

Suppose $f$ is not continuous.

Then $\exists H \subseteq X_1$ such that $f \left({H}\right)$ is closed in $T_2$, but such that $H$ is not closed in $T_1$.

As $f \left({H}\right)$ is closed in $T_2$, from Closed Set Equals its Closure we have that $f \left({H}\right) = \left({f \left({H}\right)}\right)^-$.

But as $H$ is not closed in $T_1$, we have that $H \subsetneq H^-$.

So from Subset of Image it follows that $f \left({H}\right) \subsetneq f \left({H^-}\right)$.

That is:
 * $\left({f \left({H}\right)}\right)^- \subsetneq f \left({H^-}\right)$

which means:
 * $f \left({H^-}\right) \nsubseteq \left({f \left({H}\right)}\right)^-$

Thus it is not the case that:
 * $f \left({H^-}\right) \subseteq \left({f \left({H}\right)}\right)^-$

and the proof is complete.