Set of Condensation Points of Countable Set is Empty

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

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

Then:
 * if $A$ is countable,
 * then $A^0 = \varnothing$.

Proof
Assume
 * $A$ is countable.

Aiming for a contradiction suppose that
 * $A^0 \ne \varnothing$

By definition of empty set:
 * $\exists x: x \in A^0$

Then by definition of set of condensation points:
 * $x$ is a condensation point of $A$.

This contradicts Lemma.

Thus the result follows by Proof by Contradiction.