T3 Space with Sigma-Locally Finite Basis is T4 Space/Proof 1

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Theorem

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

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

Then:

$T$ is a $T_4$ space

Proof

Let $A$ and $B$ be disjoint closed subsets of $T$.

Lemma 1

Let $F$ be a closed subset of $T$.

Let $X \subseteq S$ be disjoint from $F$.

Then there exists a countable open cover $\WW = \set{W_n : n \in \N}$ of $X$:

$\forall n \in \N : W_n^- \cap F = \O$

$\Box$

From Lemma 1, there exists a countable open cover $\UU = \set{U_n : n \in \N}$ of $A$:

$\forall n \in \N : U_n^- \cap B = \O$

and there exists a countable open cover $\VV = \set{V_n : n \in \N}$ of $B$:

$\forall n \in \N : V_n^- \cap A = \O$

From Countable Open Covers Condition for Separated Sets:

$A$ and $B$ can be separated in $T$

By definition of separated:

there exists $U, V \in \tau$ such that $A \subseteq U, B \subseteq V$ and $U \cap V = \O$

Since $A$ and $B$ were arbitrary disjoint closed subsets of $T$, by definition, $T$ is a $T_4$ space.

$\blacksquare$

Sources

• 1955: John L. Kelley: General Topology: Chapter $4$: Embedding and Metrization: Lemma $19$