Condition for Point being in Closure

Theorem
Let $$T$$ be a topological space.

Let $$H \subseteq T$$.

Let $$x \in T$$.

Then $$x \in \operatorname{cl} \left({H}\right)$$ iff every open set of $$T$$ which contains $$x$$ contains a point in $$H$$.

Proof
From the definition of closure, we have that $$\operatorname{cl} \left({H}\right)$$ is the union of $$H$$ and all the limit points of $$H$$ in $$T$$.

Sufficient Condition
Suppose $$x \in \operatorname{cl} \left({H}\right)$$.

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


 * $$x$$ 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 $$x$$.

Necessary 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 $$x \in \operatorname{cl} \left({H}\right)$$.


 * 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 $$x \in \operatorname{cl} \left({H}\right)$$.