# Continuity Defined by Closure

## 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 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:

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

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:

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

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

Hence from Continuity Defined from Closed Sets:

$f$ is continuous.

$\blacksquare$

## Proof 2

First we establish some details.

Let $H \subseteq S_1$.

Let $\mathbb K_1$ be defined as:

$\mathbb K_1 := \set {K \subseteq S_1: H \subseteq K, K \text { closed} }$

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:

$\mathbb K_2 := \set {K \subseteq S_2: f \sqbrk H \subseteq K, K \text{ closed} }$

That is, let $\mathbb K_2$ be the set of all closed sets of $T_2$ which contain $f \sqbrk H$.

From the definition of closure, we have that:

$\ds H^- = \bigcap \mathbb K_1$

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

Similarly:

$\ds \paren {f \sqbrk H}^- = \bigcap \mathbb K_2$

That is, the closure of $f \sqbrk H$ is the intersection of all the closed sets of $T_2$ which contain $f \sqbrk H$.

We have:

 $\ds f \sqbrk {H^-}$ $=$ $\ds f \sqbrk {\bigcap \mathbb K_1}$ $\ds$ $\subseteq$ $\ds \bigcap_{K \mathop \in \mathbb K_1} f \sqbrk K$ Image of Intersection under Mapping
$H \subseteq K \implies f \sqbrk H \subseteq f \sqbrk K$

### Necessary Condition

Suppose $f$ is continuous.

From the above we have that :

$\ds \paren {f \sqbrk H}^- := \bigcap \mathbb K_2$

As $f$ is continuous, then:

$\forall K \in \mathbb K_2: f^{-1} \sqbrk K$ is closed in $T_1$

But as $f \sqbrk H \subseteq K$, it follows from Image of Subset under Relation is Subset of Image: Corollary 3 that:

$H \subseteq f^{-1} \sqbrk K$

So:

$\mathbb K_3 := \set {f^{-1} \sqbrk K: K \text { closed in } T_2, H \subseteq f^{-1} \sqbrk K}$

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

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

So:

$\ds \bigcap \mathbb K_1 \subseteq \bigcap \mathbb K_3$

and so:

$\ds f \sqbrk {\bigcap \mathbb K_1} \subseteq f \sqbrk {\bigcap \mathbb K_3}$

But from Image of Intersection under Mapping:

$\ds f \sqbrk {\bigcap \mathbb K_3} \subseteq \bigcap_{K \mathop \in \mathbb K_3} f \sqbrk K$

But:

$\ds \bigcap_{K \mathop \in \mathbb K_3} f \sqbrk K = \bigcap \mathbb K_2$

and so:

$\ds f \sqbrk {\bigcap \mathbb K_1} \subseteq \bigcap \mathbb K_2$

which means that:

$f \sqbrk {H^-} \subseteq \paren {f \sqbrk H}^-$

as we wanted to show.

$\Box$

### Sufficient Condition

Suppose $f$ is not continuous.

From Continuity Defined from Closed Sets, $\exists B \subseteq S_2$ which is closed in $T_2$ such that $f^{-1} \sqbrk B$ is not closed in $T_1$.

By Image of Preimage under Mapping, we have that:

$f \sqbrk {f^{-1} \sqbrk B} \subseteq B$
$\paren {f \sqbrk {f^{-1} \sqbrk B} }^- \subseteq B^-$

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

Transitively, we get:

$\paren {f \sqbrk {f^{-1} \sqbrk B} }^- \subseteq B$

Because $f^{-1} \sqbrk B$ is not closed in $T_1$, we have that:

$f^{-1} \sqbrk B \subsetneq \paren {f^{-1} \sqbrk B}^-$

This means there exists an element $x \in \paren {f^{-1} \sqbrk B}^-$ such that $x \notin f^{-1} \sqbrk B$.

Therefore $\map f x \notin B$, but $\map f x \in f \sqbrk {\sqbrk {f^{-1} \sqbrk B}^-}$.

$\paren {f \sqbrk {f^{-1} \sqbrk B} }^- \subseteq B$

so there exists a set $A \subseteq S_1$, namely $A = f^{-1} \sqbrk B$, such that:

$f \sqbrk {A^-} \nsubseteq \paren {f \sqbrk A}^-$

$\blacksquare$