Continuity Defined by Closure

Theorem
Let $T_1 = \left({X_1, \tau_1}\right)$ and $T_2 = \left({X_2, \tau_2}\right)$ be topological spaces.

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

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

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

That is, 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 :
 * $H \subseteq K \implies f \left[{H}\right] \subseteq f \left[{K}\right]$

Necessary Condition
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 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 Image of Intersection under Mapping:
 * $\displaystyle f \left[{\bigcap \mathbb K_3}\right] \subseteq \bigcap_{K \mathop \in \mathbb K_3} f \left[{K}\right]$

But:
 * $\displaystyle \bigcap_{K \mathop \in \mathbb K_3} f \left[{K}\right] = \bigcap \mathbb K_2$

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.

Sufficient Condition
Suppose $f$ is not continuous.

From Continuity Defined from Closed Sets, $\exists B \subseteq X_2$ which is closed in $T_2$ such that $f^{-1} \left[{B}\right]$ is not closed in $T_1$.

By Image of Preimage under Mapping, we have that:
 * $f \left[{f^{-1} \left[{B}\right]}\right] \subseteq B$

So from Topological Closure of Subset is Subset of Topological Closure:
 * $\left({f \left[{f^{-1} \left[{B}\right]}\right]}\right)^- \subseteq B^-$

From Closed Set Equals its Closure we have that $B^- = B$.

Transitively, we get:
 * $\left({f \left[{f^{-1} \left[{B}\right]}\right]}\right)^- \subseteq B$

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 $f \left({x}\right) \notin B$, but $f \left({x}\right) \in f \left({\left[{f^{-1} \left[{B}\right]}\right]^-}\right)$.

From above, we had:
 * $\left({f \left[{f^{-1} \left[{B}\right]}\right]}\right)^- \subseteq B$

so there exists a set $A \subseteq X_1$, namely $A = f^{-1} \left[{B}\right]$, such that:
 * $f \left[{A^-}\right] \nsubseteq \left({f \left[{A}\right]}\right)^-$