# Isolated Point of Closure of Subset is Isolated Point of Subset

## Theorem

Let $\struct {T, \tau}$ be a topological space.

Let $H \subseteq T$ be a subset of $T$.

Let $\map \cl H$ denote the closure of $H$.

Let $x \in \map \cl H$ be an isolated point of $\map \cl H$.

Then $x$ is also an isolated point of $H$.

## Proof 1

Let $x \in \map \cl H$ be an isolated point of $\map \cl H$.

Aiming for a contradiction, suppose that $x$ is not an isolated point of $H$.

Then by the definition of a limit point, it follows that $x$ must be a limit point of $H$.

$H \subseteq \map \cl H$

So by Limit Point of Subset is Limit Point of Set it follows that $x$ is a limit point of $\map \cl H$.

But then $x$ cannot be an isolated point of $\map \cl H$.

Therefore every isolated point of $\map \cl H$ is also an isolated point of $H$.

$\blacksquare$

## Proof 2

Let $x \in \map \cl H$ be isolated in $\map \cl H$.

By definition of isolated point, $x$ is not a limit point of $\map \cl H$.

$H \subseteq \map \cl H$

We have that Limit Point of Subset is Limit Point of Set.

But $x$ is not a limit point of $\map \cl H$ .

So by the Rule of Transposition, $x$ cannot be a limit point of $H$.

As $x$ is not a limit point of $H$, it follows that $x$ must be an isolated point of $H$.

$\blacksquare$