T3 Space with Sigma-Locally Finite Basis is Perfectly T4 Space

Theorem
Let $T = \struct {S, \tau}$ be a $T_3$ topological space.

Let $\BB$ be a $\sigma$-locally finite basis.

Then:
 * $T$ is a perfectly $T_4$ space

Proof
From User:Leigh.Samphier/Topology/T3 Space with Sigma-Locally Finite Basis is T4 Space:
 * $T$ is a $T_4$ space

Let $\BB = \ds \bigcup_{n \mathop \in \N} \BB_n$ be a $\sigma$-locally finite basis where $\BB_n$ is locally finite set of subsets for each $n \in \N$.

Let $G$ be open in $T$.

Let:
 * $\CC = \set{B \in \BB : B^- \subseteq G}$

where $B^-$ denotes the closure of $B$ in $T$.

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:
 * $\forall x \in G : C \in \CC : x \in C : C^- \subseteq G$

For each $n \in \N$, let:
 * $\CC_n = \CC \cap \BB_n$

From Subset of Locally Finite Set of Subsets is Locally Finite:
 * $\CC_n$ is locally finite

For each $n \in \N$, let:
 * $C_n = \bigcup{C^- : C \in \CC_n}$

From Union of Closures of Elements of Locally Finite Set is Closed:
 * For each $n \in \N$, $C_n$ is closed in $T$

From Union of Subsets is Subset:
 * $\forall n \in \N: C_n \subseteq G$

and
 * $\ds \bigcup_{n \mathop \in \N} C_n \subseteq G$

We have:
 * $\forall x \in G : \exists n \in \N : \exists C \in \CC_n : x \in C^-$

From Set is Subset of Union:
 * $\forall n \in \N, C \in \CC_n : C^- \subseteq C_n \subseteq \ds \bigcup_{n \mathop \in \N} C_n$

Hence:
 * $\forall x \in G : x \in \ds \bigcup_{n \mathop \in \N} C_n$

By definition of Definition:Subset:
 * $G \subseteq \ds \bigcup_{n \mathop \in \N} C_n$

By definition of Definition:Set Equality:
 * $G = \ds \bigcup_{n \mathop \in \N} C_n$

Hence $G$ is an $F_\sigma$ set.

Since $G$ was arbitrary, every open set in $T$ is an $F_\sigma$ set.

From Complement of F-Sigma Set is G-Delta Set:
 * every complement of an open set in $T$ is a $G_\delta$ set

Hence every closed set in $T$ is a $G_\delta$ set by definition.

It follows that $T$ is a perfectly $T_4$ space.