# Nagata-Smirnov Metrization Theorem

## Theorem

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

Then:

$T$ is metrizable if and only if $T$ is regular and has a basis that is $\sigma$-locally finite.

## Proof

### Necessary Condition

Let $T$ be metrizable.

#### $T$ is Regular

We show that $T$ is regular.

By definition of metrizable topological space:

there exists a metric $d: S \times S \to \R_{\gt 0}$ on $S$ such that the topology induced by $d$ is $\tau$
$T$ is a regular space

#### Construction of Basis $\VV$

We construct a $\sigma$-locally finite basis $\VV$.

For each $n \in \N$, let:

$\UU_n = \set{\map {B_{1 / 2^n}} x : x \in S}$

That is, $\UU_n$ is the set of all open balls of radius $\dfrac 1 {2^n}$.

$\forall n \in \N: \UU_n$ is an open cover of $T$
$T$ is a paracompact space

By definition of paracompact space:

$\forall n \in \N : \exists$ a open refinement $\VV_n$ of $\UU_n$ which is locally finite

Let $\VV = \ds \bigcup_{n \in \N} \VV_n$.

By definition, $\VV$ is $\sigma$-locally finite.

#### $\VV$ is Basis

We show that $\VV$ is a basis.

Let $U \in \tau$.

Let $x \in U$.

the sequence $\sequence{\dfrac 1 {2^n}}$ is a null sequence
$\exists n \in N : \map {B_{1 / 2^n}} x \subseteq U$

By definition of open refinement:

$\VV_{n + 1}$ is an open cover

By definition of open cover:

$\exists V \in \VV_{n + 1} : x \in V$

By definition of open refinement:

$\exists U \in \UU_{n + 1} : V \subseteq U$

By definition of $\UU_{n + 1}$:

$\exists y \in S : U = \map {B_{1 / 2^{n + 1}}} y$

We have:

$x \in V \subseteq \map {B_{1 / 2^{n + 1}}} y$
$\map {B_{1 / {2^{n + 1}}}} y \subseteq \map {B_{1 / 2^n}} x$

Hence:

$x \in V \subseteq \map {B_{1 / {2^{n + 1}}}} y \subseteq \map {B_{1 / 2^n}} x$

By definition $\VV$ is a basis.

Hence $\VV$ is a basis that is $\sigma$-locally finite.

$\Box$

### Sufficient Condition

Let $T$ be regular.

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

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 $n \in \N$.

$T$ 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 $1$

$\forall x \in S$ and $n \in \N$:
the generalized sum $\ds \sum_{B \mathop \in \BB_n} \map {f_{\tuple {B, n} }^2} x$ converges

$\Box$

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 $I_0$:

$I = I_0$

#### Lemma $2$

for all $n \in \N$ and $x \in S$:
the generalized sum $\ds \sum_{\tuple{B, k} \mathop \in I_n} \sqbrk{\dfrac 1 {\paren{\sqrt 2}^k} \dfrac {\map {f_{\tuple{B, k}}} x} {\sqrt {1 + \map {g_k} x}}}^2$ converges

and:

$\ds \sum_{\tuple{B, k} \mathop \in I_n} \sqbrk{\dfrac 1 {\paren{\sqrt 2}^k} \dfrac {\map {f_{\tuple{B, k}}} x} {\sqrt {1 + \map {g_k} x}}}^2 \le \sum_{m \mathop = n}^\infty \dfrac 1 {2^k}$

$\Box$

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:
$\forall x \in S : \map {F_{\tuple{B, n}}} x = \dfrac 1 {\paren{\sqrt 2}^n} \dfrac {\map {f_{\tuple{B, n}}} x} {\sqrt {1 + \map {g_n} x}}$
$\forall \tuple{B, n} \in I : F_{\tuple{B, n}}$ is continuous

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

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

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:
$\ds \forall x = \family {x_i}, y = \family {y_i} \in \map F S: \map {d_2} {x, y} := \paren {\sum_{i \mathop \in I} \paren {x_i- y_i}^2}^{\frac 1 2}$
$X$ is a metric space

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

 $\ds \forall x \in S: \,$ $\ds \map G x$ $=$ $\ds \map F x$ $\ds$ $=$ $\ds \family{\map {F_{\tuple{B, n} } } x}_{\tuple{B, n} \in I}$

It follows that $G$ is a surjection.

#### $G$ is an Injection

We show that $G$ is an injection.

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

$\exists n \in \N, B \in \BB_n : x \in B, y \notin B$

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.

$\Box$

#### $G$ is a continuous mapping

We show that $G$ is everywhere continuous.

Let $x \in S$.

Let $\epsilon \in \R_{\mathop > 0}$.

$\ds \sum_{n \mathop = 0}^\infty \dfrac 1 {2^n}$ is convergent
$(1) \quad \exists N \in \N : \ds \sum_{n \mathop = N}^\infty \dfrac 1 {2^n} < \dfrac {\epsilon^2} 4$

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
$\forall n < N : \forall B \in \BB_n : B \cap \set x \subseteq B \cap U_n$
$\forall n < N : \set{B \in \BB_n : x \in B} \subseteq \set{B \in \BB_n : B \cap U_n \ne \O}$
$\forall n < N : \set{B \in \BB_n : x \in B}$ 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:

 $\ds \map {d_2} {\map G x, \map G y}^2$ $=$ $\ds \sum_{\tuple{B, n} \mathop \in I} \paren{\map {F_{\tuple{B, n} } } x - \map {F_{\tuple{B, n_k} } } y }^2$ $\ds$ $=$ $\ds \sum_{\tuple{B, n} \mathop \in I \setminus I_N } \paren{\map {F_{\tuple{B, n} } } x - \map {F_{\tuple{B, n} } } y }^2$ $\ds$  $\ds \quad + \sum_{\tuple{B, n} \mathop \in I_N} \paren{\map {F_{\tuple{B, n} } } x - \map {F_{\tuple{B, n} } } y }^2$ Absolutely Convergent Generalized Sum over Union of Disjoint Index Sets $\ds$ $=$ $\ds \sum_{\tuple{B, n} \mathop \in E } \paren{\map {F_{\tuple{B, n} } } x - \map {F_{\tuple{B, n} } } y }^2$ Generalized Sum Restricted to Non-zero Summands $\ds$  $\ds \quad + \sum_{\tuple{B, n} \mathop \in I_N} \paren{\map {F_{\tuple{B, n} } } x - \map {F_{\tuple{B, n} } } y }^2$ $\ds$ $=$ $\ds \sum_{k \mathop = 1}^m \paren{\map {F_{\tuple{B_k, n_k} } } x - \map {F_{\tuple{B_k, n_n} } } y }^2$ Definition of Summation over Finite Index $\ds$  $\ds \quad + \sum_{\tuple{B, n} \mathop \in I_N} \paren{\map {F_{\tuple{B, n} } } x - \map {F_{\tuple{B, n} } } y }^2$ $\ds$ $<$ $\ds \sum_{k \mathop = 1}^m \dfrac {\epsilon^2} {2 m} + \sum_{\tuple{B, n} \mathop \in I_N} \paren{\map {F_{\tuple{B, n} } } x - \map {F_{\tuple{B, n} } } y }^2$ from $(2)$ $\ds$ $=$ $\ds m \dfrac {\epsilon^2} {2 m} + \sum_{\tuple{B, n} \mathop \in I_N} \paren{\map {F_{\tuple{B, n} } } x - \map {F_{\tuple{B, n} } } y }^2$ $\ds$ $=$ $\ds \dfrac {\epsilon^2} 2 + \sum_{\tuple{B, n} \mathop \in I_N} \paren{\map {F_{\tuple{B, n} } } x - \map {F_{\tuple{B, n} } } y }^2$ $\ds$ $=$ $\ds \dfrac {\epsilon^2} 2 + \sum_{\tuple{B, n} \mathop \in I_N} \paren{\map {F_{\tuple{B, n} } } x}^2 - 2 \map {F_{\tuple{B, n} } } x \map {F_{\tuple{B, n} } } y + \paren{\map {F_{\tuple{B, n} } } y }^2$ Square of Difference $\ds$ $\le$ $\ds \dfrac {\epsilon^2} 2 + \sum_{\tuple{B, n} \mathop \in I_N} \paren{\map {F_{\tuple{B, n} } } x}^2 + \paren{\map {F_{\tuple{B, n} } } y }^2$ Inequality Rule for Absolutely Convergent Generalized Sums $\ds$ $=$ $\ds \dfrac {\epsilon^2} 2 + \sum_{\tuple{B, n} \mathop \in I_N } \paren{\map {F_{\tuple{B, n} } } x}^2 + \sum_{\tuple{B, n} \mathop \in I_N } \paren{\map {F_{\tuple{B, n} } } y }^2$ Sum Rule for Convergent Nets $\ds$ $\le$ $\ds \dfrac {\epsilon^2} 2 + \sum_{n \mathop = N}^\infty \dfrac 1 {2^k} + \sum_{n \mathop = N}^\infty \dfrac 1 {2^k}$ Lemma 2 $\ds$ $<$ $\ds \dfrac {\epsilon^2} 2 + \dfrac {\epsilon^2} 4 + \dfrac {\epsilon^2} 4$ from $(1)$ $\ds$ $=$ $\ds \epsilon^2$ $\ds \leadsto \ \$ $\ds \map {d_2} {\map G x, \map G y}$ $<$ $\ds \epsilon$

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

$\Box$

#### $G$ 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:

$\exists m \in \N : \exists C \in \BB_m : x \in C : C \subseteq U$
$C \cap Z = \O$

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:

 $\ds \forall \map G z \in G \sqbrk Z: \,$ $\ds \paren{\map {d_2} {\map G x, \map G z} }^2$ $=$ $\ds \paren{\map {d_2} {\map F x, \map F z} }^2$ definition of $G$ $\ds$ $=$ $\ds \sum_{\tuple{B, n} \in I} \paren{\map {F_\tuple{B, n} } x - \map {F_\tuple{B, n} } z}^2$ definition of $d_2$ $\ds$ $\ge$ $\ds \paren{\map {F_\tuple{C, m} } x - \map {F_\tuple{C, m} } z}^2$ Absolutely Convergent Generalized Sum Converges to Supremum $\ds$ $=$ $\ds \paren{\map {F_\tuple{C, m} } x}^2$ as $F_{\tuple{C, m} } \sqbrk Z = \set{0}$ $\ds \leadsto \ \$ $\ds \forall \map G z \in G \sqbrk Z: \,$ $\ds \map {d_2} {\map G x, \map G z}$ $\ge$ $\ds \map {F_\tuple{C, m} } x$ Power Function is Strictly Increasing on Positive Elements

Hence:

 $\ds \map {d_2} {\map G x, G \sqbrk Z}$ $=$ $\ds \inf \set{\map {d_2} {\map G x, \map G z} : \map G z \in G \sqbrk Z}$ Definition of Distance between Element and Subset of Metric Space $\ds$ $\ge$ $\ds \map {F_\tuple{C, m} } x$ Definition of Infimum $\ds$ $>$ $\ds 0$ as $\map {F_{\tuple{C, m} } } x \ne 0$
$G \sqbrk C$ 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.

$\Box$

By definition of homeomorphism:

$T$ is homeomorphic to the metric subspace $X$ of $H^\alpha$

It follows that $T$ is metrizable by definition.

$\blacksquare$

## Source of Name

This entry was named for Jun-iti Nagata and Yurii Mikhailovich Smirnov.

## Historical Note

The Nagata-Smirnov Metrization Theorem was discovered by Jun-iti Nagata ($1950$) and Yurii Mikhailovich Smirnov ($1951$) independently of Bing's Metrization Theorem discovered in $1951$ by R.H. Bing.

The two theorems Bing's Metrization Theorem and Nagata-Smirnov Metrization Theorem are often merged as the Bing-Nagata-Smirnov Metrization Theorem.