Condition for Point being in Closure

Theorem
Let $T$ be a topological space.

Let $H \subseteq T$.

Let $x \in T$.

Then $x \in H^-$ every open set of $T$ which contains $x$ contains a point in $H$.

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

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

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


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

If the latter is the case, then it follows directly from the definition of limit point that every open set of $T$ which contains $x$ contains a point in $H$ other than $x$.

Sufficient Condition
Suppose that every open set of $T$ which contains $x$ 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 $x \in H^-$.

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 $x \in H^-$.