Set of Condensation Points of Countable Set is Empty

Theorem

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

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

Then:

if $A$ is countable,

then $A^0 = \O$.

Lemma

if $A$ is countable,

then there exists no point $x$ of $S$ such that $x$ is a condensation point of $A$.

Proof

Assume

$A$ is countable.

Aiming for a contradiction, suppose

$A^0 \ne \O$

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.

$\blacksquare$

Sources

• 1989: Ryszard Engelking: General Topology (revised and completed ed.)

• Mizar article TOPGEN_4:56