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:
 * $\forall \tuple{B, n} \in I : \exists$ continuous $f_{\tuple{B, n}} : S \to \closedint 0 1 : B = \set{x \in S : \map {f_{\tuple{B, n}}} x \ne 0}$

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

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

That is, $H^\alpha = \struct {A, d_2}$ where:
 * $A$ is the set of all real-valued functions $x : I \to \R$ such that:
 * $(1)\quad \set{i \in I: x_i \ne 0}$ is countable
 * $(2)\quad$ the generalized sum $\ds \sum_{i \mathop \in I} x_i^2$ is a convergent net.

and
 * $d_2: A \times A: \to \R$ is the real-valued function defined as:
 * $\ds \forall x = \family {x_i}, y = \family {y_i} \in A: \map {d_2} {x, y} := \paren {\sum_{i \mathop \in I} \paren {x_i- y_i}^2}^{\frac 1 2}$

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

For each $\tuple{B, n} \in I$, let:
 * $F_{\tuple{B, n}} : S \to \R$ be the mapping defined by:
 * $\forall x \in S : \map {F_{\tuple{B, n}}} x = \dfrac 1 {\paren{\sqrt 2}^n} \dfrac {\map {f_{\tuple{B, n}}} x} {\sqrt {1 + \map {g_n} x}}$

From Multiple Rule for Continuous Mapping to Normed Division Ring:
 * $\forall \tuple{B, n} \in I : F_{\tuple{B, n}}$ is continuous

Let $F : S \to H^\alpha$ be the mapping defined by:

Let $X$ be $\map F S$ with the subspace metric from the generalized Hilbert sequence space $H^\alpha$:
 * $d_2: \map F S \times \map F S: \to \R$ be the real-valued function defined as:
 * $\ds \forall x = \family {x_i}, y = \family {y_i} \in \map F S: \map {d_2} {x, y} := \paren {\sum_{i \mathop \in I} \paren {x_i- y_i}^2}^{\frac 1 2}$

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

Let $G : T \to X$ be the mapping defined by:

It follows that $G$ is a surjection.

$G$ is an Injection
We show that $G$ is an injection.

Let $x, y \in S : x \ne y$.

From User:Leigh.Samphier/Topology/Characterization of T1 Space using Basis:
 * $\exists n \in \N, B \in \BB_n : x \in B, y \notin B$

Hence:
 * $\map {f_{\tuple{B, n}}} x \ne 0, \map {f_{\tuple{B, n}}} y = 0$

It follows that:
 * $\map {F_{\tuple{B, n}}} x \ne 0, \map {F_{\tuple{B, n}}} y = 0$

So:
 * $\map {F_{\tuple{B, n}}} x \ne \map {F_{\tuple{B, n}}} y$

Hence:
 * $\map G x \ne \map G y$

It follows that $G$ is an injection.

$G$ is a continuous mapping
We show that $G$ is a continuous mapping.

$G$ is a closed mapping
We show that $G$ is a closed mapping.

Let $Z \subseteq S$ be a closed set.

Let $\map G x \notin G \sqbrk Z$.

By definition of image of set under mapping:
 * $x \notin Z$

By definition of regular space:
 * $\exists U, V \in \tau : x \in U, Z \subseteq V : U \cap V = \O$

By definition of basis:
 * $\exists m \in \N : C \in \BB_n : x \in C : C \subseteq U$

From Subsets of Disjoint Sets are Disjoint:
 * $C \cap Z = \O$

By definition of $f_{\tuple{B, n}}$:
 * $\map {f_{\tuple{C, m}}} x \ne 0$

and:
 * $f_{\tuple{C, m}} \sqbrk Z = \set{0}$

Hence:
 * $\map {F_{\tuple{C, m}}} x \ne 0$

and:
 * $F_{\tuple{C, m}} \sqbrk Z = \set{0}$

We have:

We have:

From Subset of Metric Space is Closed iff contains all Zero Distance Points:
 * $G \sqbrk C$ is closed in $X$

Since $C$ was arbitrary, it follows:
 * for all closed $C \subseteq S : G \sqbrk C$ is closed in $X$

It follows that $G$ is a closed mapping by definition.

By definition of homeomorphism:
 * $T$ is homeomorphic to the metric subspace $X$ of $H^\alpha$

It follows that $T$ is metrizable by definition.