Set of Condensation Points of Union is Union of Sets of Condensation Points

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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


Then:

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

Lemma

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

Set of Condensation Points of Union Subset Union of Sets of Condensation Points

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

By definition of set of condensation points:

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

By Lemma:

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

By definition of set of condensation points:

$x \in A^0$ or $x \in B^0$

Thus by definition of union:

$x \in A^0 \cup B^0$

$\Box$

Union of Sets of Condensation Points Subset Set of Condensation Points of Union

By Set is Subset of Union:

$A \subseteq A \cup B \land B \subseteq A \cup B$

By Set of Condensation Points is Monotone:

$A^0 \subseteq \left({A \cup B}\right)^0 \land B^0 \subseteq \left({A \cup B}\right)^0$

Thus by Union is Smallest Superset:

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

$\Box$

Thus by definition of set equality:

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

$\blacksquare$


Sources