Condition for Point being in Closure/Proof 1

Proof
From the definition of closure, we have that $H^-$ is the union of $H$ and all the limit points of $H$ in $T$.

By definition, an open neighborhood of $x$ in $T$ is an open set of $T$ which contains $x$.

Necessary Condition
Let $x \in H^-$.

Then either:
 * $(1): \quad x \in H$, in which case every open neighborhood of $x$ in $T$ trivially contains a point in $H$ (that is, $x$ itself);


 * $(2): \quad x$ is a limit point of $H$ in $T$.

Suppose $(2)$ holds.

Then it follows directly from the definition of limit point that every open neighborhood of $x$ in $T$ contains a point in $H$ other than $x$.

Sufficient Condition
Suppose that every open neighborhood of $x$ in $T$ contains a point in $H$.

If $x \in H$, then $x$ is in the union of $H$ and all the limit points of $H$ in $T$.

Hence by definition of closure:
 * $x \in H^-$

If $x \notin H$ then $x$ must be a limit point of $H$ by definition.

So again, $x$ is in the union of $H$ and all the limit points of $H$ in $T$.

Hence by definition of closure:
 * $x \in H^-$