T3 Space with Sigma-Locally Finite Basis is Perfectly T4 Space/Lemma 2
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Theorem
Let $T = \struct {S, \tau}$ be a $T_3$ topological space.
Let $\BB$ be a basis for $T$.
Let $G$ be open in $T$.
Let:
- $\CC = \set{B \in \BB : B^- \subseteq G}$
where $B^-$ denotes the closure of $B$ in $T$.
Then:
- $\CC$ is a is a cover of $G$
Proof
Let $x \in G$.
From Characterization of T3 Space:
- $\exists U \in \tau : x \in U : U^- \subseteq G$
where $U^-$ denotes the closure of $U$ in $T$.
By definition of basis:
- $\exists B \in \BB : x \in B : B \subseteq U$
From Topological Closure of Subset is Subset of Topological Closure:
- $B^- \subseteq U^-$
From Subset Relation is Transitive:
- $B^- \subseteq G$
Hence:
- $B \in \CC$
Since $x$ was arbitrary, we have:
- $\forall x \in G : C \in \CC : x \in C$
Hence $\CC$ is a cover of $G$ by definition.
$\blacksquare$