T3 Space with Sigma-Locally Finite Basis is T4 Space/Lemma 1
Theorem
Let $T = \struct {S, \tau}$ be a $T_3$ topological space.
Let $\BB$ be a $\sigma$-locally finite basis.
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 $X$:
- $\forall n \in \N : W_n^- \cap F = \O$
Proof
Let $\UU = \set {U \in \BB : U^- \cap F = \O}$.
Lemma 2
Let $x \in S \setminus F$.
Then:
- $\exists U \in \BB : x \in U : U^- \cap F = \O$
$\Box$
From Lemma 2:
- $\forall x \in X : \exists U_x \in \BB : x \in U_x : U_x^- \cap B = \O$
Hence:
- $\forall x \in X : \exists U_x \in \UU$
By definition of open cover:
- $\UU$ is an open cover of $X$
By definition of $\sigma$-locally finite:
- $\BB = \ds \bigcup_{n \in \N} \BB_n$
where $\BB_n$ is a locally finite set of subsets.
For each $n \in \N$, let:
- $\UU_n = \UU \cap \BB_n$
From Subset of Locally Finite Set of Subsets is Locally Finite:
- $\UU_n$ is locally finite
For each $n \in \N$, let:
- $W_n = \ds \bigcup \set{W : W \in \UU_n}$
From Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets:
- $W_n \in \tau$
From Union of Closures of Elements of Locally Finite Set is Closed:
- $W_n^- = \ds \bigcup \set{W^- : W \in \UU_n}$
We have:
| \(\ds W_n^- \cap F\) | \(=\) | \(\ds \bigcup \set{W^- : W \in \UU_n} \cap B\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \bigcup \set{W^- \cap F : W \in \UU_n}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \bigcup \set{\O : W \in \UU_n}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \O\) |
Let $x \in X$.
From Lemma 2:
- $\exists U_x \in \BB : x \in U_x : U_x^- \cap F = \O$
By definition of $\UU$:
- $U_x \in \UU'$
As $\BB = \ds \bigcup_{n \in \N} \BB_n$:
- $\exists n \in \N : U_x \in \BB_n : x \in U_x : U_x^- \cap F = \O$
Hence:
- $U_x \in \UU_n$
From Set is Subset of Union:
- $U_x \subseteq \ds \bigcup \UU_n = W_n$
Hence:
- $x \in W_n$
By definition, $\WW = \set{W_n : n \in \N}$ is a countable open cover of $X$.
$\blacksquare$