Equivalence of Definitions of Perfect Set

Theorem
The three definitions of a perfect set:


 * $(1): \quad$ A perfect set is a set which equals its derived set
 * $(2): \quad$ A perfect set is a closed set which has no isolated points
 * $(3): \quad$ A perfect set is a set $S$ which is dense-in-itself and which contains all its limit points

are logically equivalent.

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

$(1) \implies (2)$:

Suppose that $S = S\,'$ where $S\,'$ is the derived set of $S$.

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

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

Now we also have that any point not in $S\,'$ is an isolated point.

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

Therefore $S = S\,'$ implies that $S$ is closed and has no isolated points.

$(2) \implies (1)$:

Suppose $S$ is closed and has no isolated points.

By Closed Set contains All its Limit Points we have that $S\,' \subseteq S$ where $S\,'$ is the derived set of $S$.

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

Therefore $S \subseteq S\,'$.

So by Equality of Sets it follows that $S = S\,'$.

$(2) \iff (3)$:

Suppose $S$ is closed and has no isolated points.

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

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

Hence the result.