# Closure of Set of Condensation Points equals Itself

## Theorem

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

Let $A$ be a subset of $S$.

Then:

$\left({A^0}\right)^- = A^0$

where

$A^0$ denotes the set of condensation points of $A$
$A^-$ denotes the closure of $A$

## Proof

$A^0 \subseteq \left({A^0}\right)^-$

To prove the equality by definition of set equality it suffices to show the inclusion:

$\left({A^0}\right)^- \subseteq A^0$

Let $x \in \left({A^0}\right)^-$.

We will prove that

$(1): \quad \forall U \in \tau: x \in U \implies A \cap U$ is uncountable

Let $U$ be an open subset of $S$ such that

$x \in U$
$A^0 \cap U \ne \varnothing$

By definition of empty set:

$\exists y: y \in A^0 \cap U$

By definition of intersection:

$y \in A^0 \land y \in U$

By definition of set of condensation points:

$y$ is condensation point of $A$

Then by definition of condensation point:

$A \cap U$ is uncountable

We will prove that

$x$ is limit point of $A$

Let $U$ be an open subset of $S$ such that

$x \in U$

By $(1)$:

$A \cap U$ is uncountable
$A \cap \left({U \setminus \left\{{x}\right\}}\right)$ is uncountable

Thus

$A \cap \left({U \setminus \left\{{x}\right\}}\right) \ne \varnothing$

Then by definition and $(1)$:

$x$ is condensation point of $A$

Thus by definition of set of condensation points:

$x \in A^0$

$\blacksquare$