Characterization of Closure by Open Sets

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

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


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 $A \cap U = \O$


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

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


Then:

\(\ds A \cap U\) \(=\) \(\ds \O\)
\(\ds \leadsto \ \ \) \(\ds A\) \(\subseteq\) \(\ds \relcomp S U\) Empty Intersection iff Subset of Complement
\(\ds \leadsto \ \ \) \(\ds A^-\) \(\subseteq\) \(\ds \relcomp S U\) Definition 3 of Closure (Topology): $\relcomp S U$ is closed
\(\ds \leadsto \ \ \) \(\ds A^- \cap U\) \(=\) \(\ds \O\) 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 $A^- \cap U = \O$

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

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

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

$\Box$


Necessary Condition

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

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


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

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:

$A \nsubseteq A^-$

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 $x \in A^-$.

$\blacksquare$


Sources

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