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

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

Then:


 * $T$ is regular and has a basis that is $\sigma$-locally finite.

$T$ 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 $\tau$

From Metric Space is Fully Normal, Fully Normal Space is Normal Space and Normal Space is Regular Space:
 * $T$ is a regular space

Construction of 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 $\dfrac 1 {2^n}$.

From Open Balls of Same Radius form Open Cover:
 * $\forall n \in \N: \UU_n$ is an open cover of $T$

From Metric Space is Paracompact:
 * $T$ 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
Let $U \in \tau$.

Let $x \in U$.

From Sequence of Powers of Number less than One:
 * the sequence $\sequence{\dfrac 1 {2^n}}$ is a null sequence

From Null Sequence induces Local Basis in Metric Space:
 * $\exists n \in N : \map {B_{1 / 2^n}} x \subseteq U$

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

From Open Ball Contains Open Ball Less Than Half Its Radius:
 * $\map {B_{1 / {2^{n + 1}}}} y \subseteq \map {B_{1 / 2^n}} x$

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.