User:Leigh.Samphier/Topology/Nagata-Smirnov Metrization Theorem/Sufficient Condition

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

Let $T$ have a basis that is $\sigma$-locally finite

Then:


 * $T$ is metrizable

Proof
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 $n \in \N$.

From T3 Space with Sigma-Locally Finite Basis is Perfectly T4 Space:
 * $T$ is a perfectly $T_4$ space

Let $I = \set{\tuple{B, n} : B \in \BB, B \in \BB_n}$.

By definition of perfectly $T_4$ space:
 * $\forall \tuple{B, n} \in I : \exists f_{\tuple{B, n}} : S \to \closedint 0 1 : B = \set{x \in S : \map {f_{\tuple{B, n}}} x \ne 0}$

Let $\alpha$ be the cardinality of $I$.

Let $H^\alpha$ be the generalized Hilbert sequence space of weight $\alpha$.

We have $H^\alpha$ is a metric space from Generalized Hilbert Sequence Space is Metric Space.

Lemma 1
Let $g_n : S \to \closedint 0 1$ be the mapping defined by:
 * $\map {g_n} x$ is the limit of the generalized sum $\ds \sum_{B \in \BB_n} \map {f_{\tuple{B, n}}^2} x$

From Lemma 1:
 * $\forall n \in \N : g_n : S \to \closedint 0 1$ is well-defined

Lemma 2
From Lemma 2:
 * $\forall x \in S : \ds \family{\dfrac 1 {\paren{\sqrt 2}^n} \dfrac {\map {f_{\tuple{b, n}}} x} {\sqrt {1 + \map {g_n} x}}}_{\tuple{B, n} \in I} \in H^\alpha$

Let $F : S \to H^\alpha$ be the mapping defined by:
 * $\forall x \in S : \ds \map F x = \family{\dfrac 1 {\paren{\sqrt 2}^n} \dfrac {\map {f_{\tuple{b, n}}} x} {\sqrt {1 + \map {g_n} x}}}_{\tuple{B, n} \in I}$

Let $G : S \to \map F S$ be the mapping defined by:
 * $\forall x \in S : \ds \map G x = \map F x = \family{\dfrac 1 {\paren{\sqrt 2}^n} \dfrac {\map {f_{\tuple{b, n}}} x} {\sqrt {1 + \map {g_n} x}}}_{\tuple{B, n} \in I}$

Let $X$ be the $\map F S$ be given the subspace metric.

$G$ is a closed mapping
From :
 * $T$ is homeomorphic to a subspace $X$ of $H^\alpha$

From Subspace of Metric Space is Metric Space:
 * $X$ is a metric space

It follows that $T$ is metrizable by definition.