Closure of Set of Condensation Points equals Itself
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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:
- $x \in A^0$
$\blacksquare$
Sources
- Mizar article TOPGEN_4:51