Characterization of Closure by Open Sets

Theorem
Let $T = \struct {S, \tau}$ 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 $x \in A^-$ :
 * for every open set $U$ of $T$:
 * $x \in U \implies A \cap U \ne \O$

Sufficient Condition
Let $x \in A^-$.

there exists an open set $U$ of $T$ such that:
 * $x \in U$ and $A \cap U = \O$

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

So by definition of closed set, $\relcomp S U$ is closed in $T$.

Then:

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 $A^- \cap U = \O$

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 $T$:
 * $x \in U \implies A \cap U \ne \O$

Necessary Condition
Let $x$ be such that for every open set $U$ of $T$:
 * $x \in U \implies A \cap U \ne \O$

$x \notin A^-$.

Then:
 * $x \in \relcomp S {A^-}$

Then by assumption:
 * $A \cap \relcomp S {A^-} \ne \O$

as $\relcomp S {A^-}$ is open.

By definition of complement:
 * $A \cap \relcomp S {A^-} = \O$

So by Empty Intersection iff Subset of Complement:
 * $A \not \subseteq A^-$

From Set is Subset of its Topological Closure:
 * $A \subseteq A^-$

But from Set Complement inverts Subsets:
 * $\relcomp S {A^-} \subseteq \relcomp S A$

from which by Empty Intersection iff Subset of Complement:
 * $A \cap \relcomp S {A^-} = \O$

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

So $x \in A^-$.