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 Image of Subset is Subset of Image/Corollary 2 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 Image of Subset is Subset of Image/Corollary 3 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 $\exists B \subseteq X_2$ which is closed in $T_2$ such that $f^{-1} \left({B}\right)$ is not closed in $T_1$.

We have that $\left(f\left(f^{-1}\left(B\right)\right)\right)^- = B^-$. From Closed Set Equals its Closure we have that $B^- = B$. Transitively, we get $\left(f\left(f^{-1}\left(B\right)\right)\right)^- = B$.

Because $f^{-1}\left(B\right)$ is not closed in $T_1$, we have that $f^{-1}\left(B\right) \subsetneq \left(f^{-1}\left(B\right)\right)^-$. This means there exists an element $x \in \left(f^{-1}\left(B\right)\right)^-$ such that $x \notin f^{-1}\left(B\right)$.

Therefore, $f(x) \notin B$, but $f(x) \in f\left(\left(f^{-1}\left(B\right)\right)^-\right)$. From above, we had $B = \left(f\left(f^{-1}\left(B\right)\right)\right)^-$, so there exists a set $A \subseteq X_1$, namely $A = f^{-1}\left(B\right)$, such that $f\left(A^-\right) \not\subset \left(f\left(A\right)\right)^-$.