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.
==== ====
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
For all $n \in \N$, let:
:$I_n = \set{\tuple{B, k} \in I : k \ge n}$
By definition of $I$ and $I_0$:
:$I = I_0$
==== ====
From :
:$\forall x \in S : \ds \family{\dfrac 1 {\paren{\sqrt 2}^n} \dfrac {\map {f_{\tuple{B, n}}} x} {\sqrt {1 + \map {g_n} x}}}_{\tuple{B, n} \in I} \in H^\alpha$
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 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 everywhere continuous.
Let $x \in S$.
Let $\epsilon \in \R_{\mathop > 0}$.
From Sum of Infinite Geometric Sequence:
:$\ds \sum_{n \mathop = 0}^\infty \dfrac 1 {2^n}$ is convergent
From Tail of Convergent Series tends to Zero:
:$(1) \quad \exists N \in \N : \ds \sum_{n \mathop = N}^\infty \dfrac 1 {2^n} < \dfrac {\epsilon^2} 4$
By definition of locally finite set of subsets:
:$\forall n < N : \exists U_n \in \tau : x \in U_n : \set{B \in \BB_n : B \cap U_n \ne \O}$ is finite
From Set Intersection Preserves Subsets:
:$\forall n < N : \forall B \in \BB_n : B \cap \set x \subseteq B \cap U_n$
From Subset of Empty Set iff Empty:
:$\forall n < N : \set{B \in \BB_n : x \in B} \subseteq \set{B \in \BB_n : B \cap U_n \ne \O}$
From Subset of Finite Set is Finite:
:$\forall n < N : \set{B \in \BB_n : x \in B}$ is finite
From Union of Finite Sets is Finite:
:$\set{B \in \ds \bigcup_{n \mathop < N} \BB_n : x \in B}$ is finite
Let:
:$\set{B_1, B_2, \ldots, B_m} = \set{B \in \ds \bigcup_{n \mathop < N} \BB_n : x \in B}$
Furthermore:
:$\forall B \in \ds \bigcup_{n \mathop < N} \BB_n \setminus \set{B_1, B_2, \ldots, B_m} : x \notin B$
Let $E = \set{\tuple{B_1, n_1}, \tuple{B_2, n_2}, \ldots, \tuple{B_m, n_m}}$ where for each $k \in \closedint 1 m$:
:$n_k < N$
:$B_k \in \BB_{n_k}$
By definition of continuity:
:$\forall k \in \closedint 1 m : \exists V_k \in \tau : x \in V_k \subseteq U_{n_k}$: 
::$\forall y \in V_k : \size{\map {F_{\tuple{B_{n_k}, n_k}}} x - \map {F_{\tuple{B_{n_k}, n_k}}} y } < \dfrac \epsilon {\sqrt {2 m}}$ 
Let $V = \ds \bigcap_{k = 1}^m V_k$.
We have:
:$x \in V$
and
:$(2) \quad \forall y \in V : \forall k \in \closedint 1 m : \size{\map {F_{\tuple{B_{n_k}, n_k}}} x - \map {F_{\tuple{B_{n_k}, n_k}}} y } < \dfrac \epsilon {\sqrt {2 m}}$
Let:
:$I_{N - 1} = \set{\tuple{B, n} \in I : n < N}$
Hence $E$ is a finite subset of $I_{N - 1}$
Let:
:$y \in V$
Then:
:$\forall \tuple{B, n} \in I_{N - 1} \setminus E : \map {F_{\tuple{B, n} } } x = \map {F_{\tuple{B, n_k} } } y = 0$
Hence:
:$\forall \tuple{B, n} \in I_{N - 1} \setminus E : \paren{\map {F_{\tuple{B, n} } } x - \map {F_{\tuple{B, n} } } y }^2 = 0$
We have:
Since $y$ was arbitrary, it follows that:
:$\forall y \in V : \map {d_2} {\map G x, \map G y} < \epsilon$
Since $\epsilon$ was arbitrary, it follows that:
:$\forall \epsilon \in \R_{ > 0} : \exists V \in \tau : x \in V : \forall y \in V : \map {d_2} {\map G x, \map G y} < \epsilon$
It follows that $G$ is continuous at $x$.
Since $x$ was arbitrary, it follows that:
:$G$ is everywhere continuous 
==== $G$ 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:
:$\exists m \in \N : \exists C \in \BB_m : x \in C : C \subseteq U$
From Subsets of Disjoint Sets are Disjoint:
:$C \cap Z = \O$
By definition of $f_{\tuple{C, m}}$:
:$\map {f_{\tuple{C, m}}} x \ne 0$ 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:
Hence:
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.
Category:Nagata-Smirnov Metrization Theorem