Equivalence of Definitions of Perfect Set

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

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

Proof
Let $T = \left({S, \tau}\right)$ be a topological space and let $H \subseteq S$.

Definition 1 implies Definition 2
Suppose that $H = H'$ where $H'$ is the derived set of $H$.

By definition of derived set, $H'$ is the set of all limit points of $H$.

So $H$ contains all its limit points and so by definition is closed.

Now we also have that any point not in $H'$ is an isolated point.

But there are no points in $H$ which are not in $H'$, so $H$ has no isolated points.

Therefore $H = H'$ implies that $H$ is closed and has no isolated points.

Definition 2 implies Definition 1
Suppose $H$ is closed and has no isolated points.

By Closed Set iff Contains all its Limit Points we have that $H' \subseteq H$ where $H'$ is the derived set of $H$.

As $S$ has no isolated points, all its elements are elements of its derived set.

Therefore $H \subseteq H'$.

So by definition of set equality it follows that $H = H'$.

Definition 2 iff Definition 3
Let $H$ be closed with no isolated points.

By definition, $H$ is closed it contains all its limit points.

Also by definition, $H$ is dense-in-itself it has no isolated points.

Hence the result.