Condition for Point being in Closure/Proof 2

Theorem

Let $T = \struct {S, \tau}$ be a topological space.

Let $H \subseteq S$.

Let $H^-$ denote the closure of $H$ in $T$.

Let $x \in S$.

Then $x \in H^-$ if and only if every open neighborhood of $x$ contains a point in $H$.

The condition to be proved is equivalent to showing that $x \in H^-$ if and only if, for every open neighborhood $U$ of $x$, the intersection $H \cap U$ is non-empty.

For $U \in \tau$, let $U^{\complement}$ denote the relative complement of $U$ in $S$.

By definition, $U^{\complement}$ is closed in $T$

We have that:

$\ds H \cap U$

$=$

$\ds \O$

$\ds \leadstoandfrom \ \ $

$\ds H$

$\subseteq$

$\ds U^{\complement}$

$\ds \leadstoandfrom \ \ $

$\ds H^-$

$\subseteq$

$\ds U^{\complement}$

$\ds \leadstoandfrom \ \ $

$\ds U$

$\subseteq$

$\ds \paren {H^-}^{\complement}$

$\ds \leadstoandfrom \ \ $

$\ds H^- \cap U$

$=$

$\ds \O$

Thus:

$x \in U \iff x \notin H^-$

The result follows from the Rule of Transposition.

$\blacksquare$