Condition for Point being in Closure/Proof 2

Proof
The condition to be proved is equivalent to showing that $x \in H^-$, 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:

Thus:
 * $x \in U \iff x \notin H^-$

The result follows from the Rule of Transposition.