Isolated Point of Closure of Subset is Isolated Point of Subset/Proof 1
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
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 $H$.
From Set is Subset of its Topological Closure:
- $H \subseteq \map \cl H$
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 $\map \cl H$.
This is a contradiction.
Therefore every isolated point of $\map \cl H$ is also an isolated point of $H$.
$\blacksquare$