Equivalence of Definitions of Perfect Set

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Theorem

The following definitions of the concept of Perfect Set are equivalent:


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

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

Definition 1

A perfect set of a topological space $T = \left({S, \tau}\right)$ is a subset $H \subseteq S$ such that:

$H = H'$

where $H'$ is the derived set of $H$.

That is, where:

every point of $H$ is a limit point of $H$

and

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

Definition 2

A perfect set of a topological space $T = \left({S, \tau}\right)$ is a subset $H \subseteq S$ such that:

$H$ is a closed set of $T$
$H$ has no isolated points.

Definition 3

A perfect set of a topological space $T = \struct {S, \tau}$ is a subset $H \subseteq S$ such that:

$H$ is dense-in-itself.
$H$ contains all its limit points.


Proof

Let $T = \struct {S, \tau}$ 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.

$\Box$


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 $H = H'$.

$\Box$


Definition 2 iff Definition 3

Let $H$ be closed with no isolated points.

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

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

$\Box$


Hence the result.

$\blacksquare$


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