# Characterization of Closure by Open Sets

## Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.

Let $A$ be a subset of $S$.

Let $x$ be a point of $T$.

Let $A^-$ denote the closure of $A$.

Then $x \in A^-$ if and only if:

for every open set $U$ of $T$:
$x \in U \implies A \cap U \ne \varnothing$

## Proof

### Sufficient Condition

Let $x \in A^-$.

Aiming for a contradiction, suppose there exists an open set $U$ of $T$ such that:

$x \in U$ and $A \cap U = \varnothing$

We have that $U$ is open in $T$.

So by definition of closed set, $\complement_S \left({U}\right)$ is closed in $T$.

Then:

 $\displaystyle A \cap U$ $=$ $\displaystyle \varnothing$ $\displaystyle \implies \ \$ $\displaystyle A$ $\subseteq$ $\displaystyle \complement_S \left({U}\right)$ Empty Intersection iff Subset of Complement $\displaystyle \implies \ \$ $\displaystyle A^-$ $\subseteq$ $\displaystyle \complement_S \left({U}\right)$ Definition of Set Closure: $\complement_S \left({U}\right)$ is closed $\displaystyle \implies \ \$ $\displaystyle A^- \cap U$ $=$ $\displaystyle \varnothing$ Empty Intersection iff Subset of Complement

But we have:

$x \in A^-$

and also:

$x \in U$

and thus by definition of set intersection:

$x \in A^- \cap U$

This contradicts $A^- \cap U = \varnothing$

Hence by Proof by Contradiction the assumption that there exists an open set $U$ of $T$ such that $x \in U$ and $A \cap U = \varnothing$ was false.

So for every open set $U$ of $T$:

$x \in U \implies A \cap U \ne \varnothing$

$\Box$

### Necessary Condition

Let $x$ be such that for every open set $U$ of $T$:

$x \in U \implies A \cap U \ne \varnothing$

Aiming for a contradiction, suppose $x \notin A^-$.

Then:

$x \in \complement_S \left({A^-}\right)$

Then by assumption:

$A \cap \complement_S \left({A^-}\right) \ne \varnothing$

as $\complement_S \left({A^-}\right)$ is open.

By definition of complement:

$A \cap \complement_S \left({A}\right) = \varnothing$
$A \not \subseteq A^-$
$A \subseteq A^-$

But from Set Complement inverts Subsets:

$\complement_S \left({A^-}\right) \subseteq \complement_S \left({A}\right)$

from which by Empty Intersection iff Subset of Complement:

$A \cap \complement_S \left({A^-}\right) = \varnothing$

Hence by Proof by Contradiction the assumption that $x \notin A^-$ was false.

So $x \in A^-$.

$\blacksquare$