Closure of Set of Condensation Points equals Itself

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

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

Then:
 * $\paren {A^0}^- = A^0$

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

Proof
By Set is Subset of its Topological Closure:
 * $A^0 \subseteq \paren {A^0}^-$

To prove the equality by definition of set equality it suffices to show the inclusion:
 * $\paren {A^0}^- \subseteq A^0$

Let $x \in \paren {A^0}^-$.

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
 * $x \in U$

By Condition for Point being in Closure:
 * $A^0 \cap U \ne \O$

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


 * $A \cap \paren {U \setminus \set x}$ is uncountable

Thus:
 * $A \cap \paren {U \setminus \set x} \ne \O$

Then by definition and $(1)$:
 * $x$ is condensation point of $A$

Thus by definition of set of condensation points:
 * $x \in A^0$