T3 Space with Sigma-Locally Finite Basis is Perfectly T4 Space/Lemma 2

This article needs proofreading. Please check it for mathematical errors. If you believe there are none, please remove {{Proofread}} from the code. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Proofread}} from the code.

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.

$\blacksquare$