Set of Condensation Points of Union is Union of Sets of Condensation Points/Lemma

Theorem

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

Let $A, B$ be subsets of $S$.

Let $x$ be a point of $S$.

Then:

if $x$ is condensation point of $A \cup B$,

then $x$ is condensation point of $A$ or $x$ is condensation point of $B$.

Proof

Assume $x$ is condensation point of $A \cup B$.

Aiming for a contradiction, suppose

$x$ is not condensation point of $A$ and $x$ is not condensation point of $B$.

By definition of condensation point:

$\exists U_1 \in \tau: x \in U_1 \land A \cap U_1$ is countable

By definition of condensation point:

$\exists U_2 \in \tau: x \in U_2 \land B \cap U_2$ is countable

By definition of intersection:

$x \in U_1 \cap U_2$

By definition of topological space:

$U_1 \cap U_2 \in \tau$

By Intersection is Subset:

$A \cap U_1 \cap U_2 \subseteq A \cap U_1 \land B \cap U_1 \cap U_2 \subseteq B \cap U_2$

By Set Union Preserves Subsets:

$\left({A \cap U_1 \cap U_2}\right) \cup \left({B \cap U_1 \cap U_2}\right) \subseteq \left({A \cap U_1}\right) \cup \left({B \cap U_2}\right)$

By Countable Union of Countable Sets is Countable:

$\left({A \cap U_1}\right) \cup \left({B \cap U_2}\right)$ is countable

Then by Subset of Countable Set is Countable:

$(1): \quad \left({A \cap U_1 \cap U_2}\right) \cup \left({B \cap U_1 \cap U_2}\right)$ is countable

By Intersection Distributes over Union:

$\left({A \cap U_1 \cap U_2}\right) \cup \left({B \cap U_1 \cap U_2}\right) = \left({A \cup B}\right) \cap \left({U_1 \cap U_2}\right)$

By definition of condensation point:

$\left({A \cup B}\right) \cap \left({U_1 \cap U_2}\right)$ is uncountable

This contradicts $(1)$.

Thus the result fallows by Proof by Contradiction.

$\blacksquare$

Sources

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

• Mizar article TOPGEN_4:53