Equivalence of Definitions of Isolated Point

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

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

Definition 1 implies Definition 2
Let $x$ be an isolated point of $H$ by definition 1.

Then by definition:
 * $\exists U \in \tau: U \cap H = \set x$

Thus we have an open set in $T$ such that $x \in U$ contains no other point of $H$ than $x$.

Thus, by definition, $x$ is not a limit point of $H$.

Thus $x$ is an isolated point of $H$ by definition 2.

Definition 2 implies Definition 1
Let $x$ be an isolated point of $H$ by definition 2.

$x$ is a limit point of $H$.

Then by definition every open set $U \in \tau$ such that $x \in U$ contains some point of $H$ other than $x$.

That is:
 * $\forall U \in \tau: x \in U \implies \exists y \in S, y \ne x: y \in U \cap H$

That is:
 * $\not \exists U \in \tau: U \cap H = \set x$

because all $U$ with $x$ in them are such that there is at least one point in $U \cap H$ apart from $x$.

Thus by Proof by Contradiction $x$ is not a limit point of $H$.

That is, $x$ is an isolated point of $H$ by definition 1.