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

$\blacksquare$