Metrization: Urysohn, Nagata–Smirnov, Bing, Smirnov

For each $x\in X$, discreteness supplies a neighbourhood meeting at most one member of $\mathcal{D}$. That is a finite number, so [L1] makes $\mathcal{D}$ locally finite.

Applying step 1.1 separately to every discrete layer in a $\sigma$-discrete decomposition leaves the same countable decomposition and gives a $\sigma$-locally-finite basis.

Put $V_n=\{(x,y):y\in \mathrm{St}(x,{\mathcal{U}}_n)\}$. Each $V_n$ is symmetric, contains the diagonal, and normality gives $V_{n+1}\circ V_{n+1}\subseteq V_n$: two successive ${\mathcal{U}}_{n+1}$-links lie in the star of one member, which is contained in a member of ${\mathcal{U}}_n$.

Define $d(x,y)$ as the smaller of $1$ and the infimum of ${\sum }_{r=1}^k{2}^{-n_r}$ over finite chains $x=x_0,\dots ,x_k=y$ with $(x_{r-1},x_r)\in V_{n_r}$, taking the infimum of an empty collection to be $+\mathrm{\infty }$. Reversing a chain gives symmetry. Concatenation gives the triangle inequality within a chain-connected component, while points in different components have distance $1$; truncation at $1$ preserves the triangle inequality. The diagonal chains give $d(x,x)=0$.

The containment in step 1.1 lets every chain of total weight below $2^{-n-1}$ be compressed, from its finest links upward, to a $V_n$-link. Hence $d(x,y)<2^{-n-1}$ implies $(x,y)\in V_n$; conversely $(x,y)\in V_n$ gives $d(x,y)\le 2^{-n}$.

Compatibility (i) and step 3.1 show $d(x,y)>0$ when $x\mathrm{\ne }y$. Compatibility (ii), the two bounds in step 3.1, and [L1] show that the $d$-balls and the original neighbourhoods contain one another at every point. Thus $d$ is a metric inducing the given topology.

For each $n\in \mathbb{N}$, let ${\mathcal{C}}_n$ be the cover by balls of radius $2^{-n-3}$. By [L1], choose a locally finite open refining cover ${\mathcal{V}}_n$ of ${\mathcal{C}}_n$.

The family $\mathcal{B}={\bigcup }_n{\mathcal{V}}_n$ is $\sigma$-locally finite. It is a basis: if $x\in O$ with $O$ open, [L2] gives $\epsilon >0$ with $B_d(x,\epsilon )\subseteq O$; choose $n$ with $2^{-n-2}<\epsilon$, and a member $V\in {\mathcal{V}}_n$ containing $x$. As $V$ lies in some $B_d(c,2^{-n-3})$ containing $x$, the triangle inequality gives $V\subseteq B_d(x,2^{-n-2})\subseteq O$.

Thus $\mathcal{B}$ is the asserted $\sigma$-locally-finite basis.

If $X$ is metrizable, it is regular and $T_1$, and [L1] gives the required basis.

Conversely write the supplied basis as $\mathcal{B}={\bigcup }_{n\ge 0}{\mathcal{B}}_n$ with every ${\mathcal{B}}_n$ locally finite. We first prove normality, rather than assuming it. For disjoint closed $F,H$, let $U_n$ be the union of those $B\in {\mathcal{B}}_n$ with $\bar{B}\cap H=\mathrm{\varnothing }$, and define $V_n$ symmetrically for $H$ against $F$. Regularity and the basis property say the $U_n$ cover $F$ and the $V_n$ cover $H$: each point outside a closed set has a basis neighborhood whose closure misses it. By [L2], $\overline{U_n}$ misses $H$ and $\overline{V_n}$ misses $F$. Now put $U={\bigcup }_{n\ge 0}(U_n\setminus {\bigcup }_{k\le n}\overline{V_k}),V={\bigcup }_{n\ge 0}(V_n\setminus {\bigcup }_{k\le n}\overline{U_k})\mathrm{.}$ Both are open, $F\subseteq U$, and $H\subseteq V$. If a point belonged to the $n$th piece of $U$ and the $m$th piece of $V$, then $m\le n$ would contradict its exclusion from $\overline{V_m}$, while $n\le m$ would contradict its exclusion from $\overline{U_n}$. Thus $U\cap V=\mathrm{\varnothing }$ and $X$ is normal.

Every basis member $B$ is the cozero set of a continuous function. For each $n$, let $F_{B,n}$ be the union of $\bar{C}$ over $C\in {\mathcal{B}}_n$ with $\bar{C}\subseteq B$. It is closed and contained in $B$ by [L2]. The union of these $F_{B,n}$ over $n$ is $B$: for $x\in B$, regularity gives an open $O$ with $x\in O\subseteq \bar{O}\subseteq B$, and some basis member $C\ni x$ lies in $O$. By [L3] applied to the disjoint closed sets $F_{B,n}$ and $X\setminus B$, choose $f_{B,n}:X\to [0,1]$ equal to $1$ on $F_{B,n}$ and $0$ outside $B$. AC makes these simultaneous choices for all $(B,n)$. The uniformly convergent sum $u_B={\sum }_{n\ge 0}{2}^{-n-1}{f}_{B,n}$ is continuous, zero outside $B$, and positive at every point of $B$. Hence $B=\{u_B>0\}$.

Index repeated basis members by their layer, writing $i=(n,B)$ with $B\in {\mathcal{B}}_n$. At each $x$, only finitely many members of ${\mathcal{B}}_n$ contain $x$, so the finite sum $S_n(x)={\sum }_{B\in {\mathcal{B}}_n}{u}_B(x{)}^2$ is well defined. It is continuous: around any $x$, local finiteness makes all but finitely many summands identically zero. Define $a_{nB}(x)=2^{-n-1}\frac{u_B(x)}{\sqrt{1+S_n(x)}}\mathrm{.}$ Each coordinate is continuous and ${\sum }_{B\in {\mathcal{B}}_n}{a}_{nB}(x{)}^2\le 4^{-n-1}$ uniformly in $x$. Consequently the nonnegative generalized sum $d(x,y{)}^2={\sum }_{n\ge 0}{\sum }_{B\in {\mathcal{B}}_n}(a_{nB}(x)-a_{nB}(y){)}^2$ is finite. It is the squared ${\mathrm{\ell }}^2$ distance of the two coordinate vectors; the triangle inequality follows from finite-sum Cauchy--Schwarz and passage to the supremum over finite index sets. Symmetry and $d(x,x)=0$ are immediate. If $x\mathrm{\ne }y$, $T_1$ and the basis give a $B$ with $x\in B$, $y\mathrm{\notin }B$; then $a_{nB}(x)>0=a_{nB}(y)$ for a layer containing $B$. Thus $d$ is a metric.

The metric induces the original topology. Given $x\in O$ open, choose a basis member $B$ with $x\in B\subseteq O$ and a layer $n$ containing it. If $d(x,y)<a_{nB}(x)\mathrm{/}2$, then $a_{nB}(y)>0$, so $y\in B\subseteq O$. Conversely fix $x$ and $\epsilon >0$. Choose $N$ so that $4{\sum }_{n>N}{4}^{-n-1}<{\epsilon }^2\mathrm{/}2$. For each of the finitely many layers $n\le N$, take a neighborhood of $x$ meeting only finitely many $B\in {\mathcal{B}}_n$. On the intersection of these neighborhoods, every other coordinate in those layers vanishes both at $x$ and at nearby $y$. Continuity of the remaining finitely many coordinates makes their squared-difference sum $<{\epsilon }^2\mathrm{/}2$ on a smaller neighborhood. The tail estimate $(a-b{)}^2\le 2a^2+2b^2$ and the uniform layer bound make its contribution $<{\epsilon }^2\mathrm{/}2$. Thus $d(x,y)<\epsilon$ there.

Step 1.1 proves the forward implication, and steps 1.2--3.1 construct a compatible metric for the reverse implication under the stated AC.

If $X$ is metrizable, [L1] gives a $\sigma$-discrete basis; metric spaces are regular and $T_1$.

If $X$ is regular, $T_1$, and has a $\sigma$-discrete basis, [L2] converts that basis to a $\sigma$-locally-finite one and then gives a metric.

The cases establish both directions.

First $X$ is normal. Given disjoint closed $A,B$, enumerate the basis members whose closures avoid $B$ as $(U_n)$ and those whose closures avoid $A$ as $(V_n)$, padding by empty sets if necessary. By [L1] and the basis property, the first family covers $A$ and the second covers $B$. Put $U={\bigcup }_n(U_n\setminus {\bigcup }_{i\le n}{\bar{V}}_i)$ and $V={\bigcup }_n(V_n\setminus {\bigcup }_{i\le n}{\bar{U}}_i)$. Each summand is open because only finitely many closed sets are removed. The sets contain $A,B$ respectively. They are disjoint: a point in the $n$th summand of $U$ and $m$th summand of $V$ would contradict the removal of ${\bar{V}}_m$ if $m\le n$, and of ${\bar{U}}_n$ if $n\le m$. This proves normality.

For every pair $(i,j)$ with ${\bar{B}}_i\subseteq B_j$, [L2] gives a continuous $f_{ij}:X\to [0,1]$ equal to $1$ on ${\bar{B}}_i$ and $0$ on $X\setminus B_j$. Use AC to select all these functions, and list them as a sequence $(f_k{)}_{k\ge 1}$, adding zero functions when necessary. For every point $x$ and open neighbourhood $O$, choose $B_j$ with $x\in B_j\subseteq O$, shrink inside $B_j$ using [L1], and choose $B_i$ containing $x$ inside that shrinking. Then ${\bar{B}}_i\subseteq B_j$, so one listed function is $1$ at $x$ and vanishes outside $O$. This also separates distinct points because $T_1$ makes $X\setminus \{y\}$ open.

Define $d(x,y)={\sum }_{k\ge 1}{2}^{-k}\mathrm{\mid }{f}_k(x)-f_k(y)\mathrm{\mid }$. Each term is at most $2^{-k}$, so the sum converges. Symmetry and the triangle inequality follow termwise, and step 2.1 gives $d(x,y)>0$ for $x\mathrm{\ne }y$. Thus $d$ is a metric (including the empty-space case).

For fixed $x$ and $\epsilon >0$, choose $N$ with ${\sum }_{k>N}{2}^{-k}<\epsilon \mathrm{/}2$. Continuity of the first $N$ functions gives an original open neighbourhood $W$ of $x$ where their weighted differences from their values at $x$ sum to less than $\epsilon \mathrm{/}2$. Hence $W\subseteq B_d(x,\epsilon )$. Conversely, for an original open $O$ containing $x$, take the index $k$ supplied by step 2.1. If $d(x,y)<2^{-k}$, then $f_k(y)>0$, so $y\in O$. Thus every original neighbourhood contains a metric ball and every metric ball contains an original neighbourhood at its centre; the topologies coincide. Hence $X$ is metrizable.

Put ${\mathcal{C}}_n=\{B\cap W_s:s\in S,B\in {\mathcal{B}}_{s,n}\}$. Each member is open in $X$, since $W_s$ is open and a set open in the open subspace $U_s$ is ambient open.

Fix $x\in X$. By [L1], choose a neighborhood $N$ meeting only finitely many $W_s$. For each of these indices, if $x\in U_s$, relative local finiteness supplies an ambient neighborhood $N_s$ whose intersection with $U_s$ meets only finitely many $B\in {\mathcal{B}}_{s,n}$. If $x\mathrm{\notin }{U}_s$, then $x\mathrm{\notin }\overline{W_s}$, so choose $N_s$ disjoint from $W_s$. Intersect $N$ with this finite collection of $N_s$. It meets only finitely many members of ${\mathcal{C}}_n$; hence ${\mathcal{C}}_n$ is locally finite.

If $O$ is open and $x\in O$, choose $s$ with $x\in W_s$. A relative basis member $B\in {\mathcal{B}}_{s,n}$ contains $x$ and lies in $O\cap U_s$. Then $x\in B\cap W_s\subseteq O$. Thus ${\bigcup }_n{\mathcal{C}}_n$ is a basis.

Steps 2.1 and 2.2 prove the result.

If $X$ is metrizable, it is Hausdorff and locally metrizable by taking $X$ itself as the open neighbourhood, and it is paracompact by [L1].

Conversely, local metrizability gives an open cover $\mathcal{U}$ by metrizable subspaces. Apply [L3] to obtain a locally finite open cover $\{W_s:s\in S\}$ and assigned $U_s\in \mathcal{U}$ with $\overline{W_s}\subseteq U_s$. Under choice, fix for each $U_s$ a $\sigma$-locally-finite relative open basis ${\bigcup }_{n<\omega }{\mathcal{B}}_{s,n}$ by [L1].

For each $n$, put ${\mathcal{C}}_n=\{B\cap W_s:s\in S,B\in {\mathcal{B}}_{s,n}\}$. Each member is open in $X$, because $U_s$ and $W_s$ are open. The family ${\mathcal{C}}_n$ is locally finite at every $x\in X$: first choose a neighborhood meeting only finitely many $W_s$; for each such $s$, if $x\in U_s$, relative local finiteness of ${\mathcal{B}}_{s,n}$ supplies an ambient neighborhood meeting only finitely many of its members. If $x\mathrm{\notin }{U}_s$, then $x\mathrm{\notin }\overline{W_s}$, so an ambient neighborhood of $x$ misses $W_s$ entirely. Intersect the finitely many chosen neighborhoods.

The union ${\bigcup }_n{\mathcal{C}}_n$ is a basis of $X$: if $x\in O$ with $O$ open, choose $s$ with $x\in W_s$, then choose a relative basis member $B\in {\mathcal{B}}_{s,n}$ with $x\in B\subseteq O\cap U_s$. The open set $B\cap W_s$ contains $x$ and lies in $O$. Thus $X$ has a $\sigma$-locally-finite basis.

Hausdorffness implies $T_1$, and [L2] makes $X$ regular; applying Nagata–Smirnov in [L2] to the basis from step 3.1 yields a metric. Together with step 1.1 this proves the equivalence.