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A Lindelöf space is a topological space in which every open cover has a countable subcover. A closed discrete subset of a Lindelöf space must be countable: cover each of its points by an open set meeting it at only that point and add the open complement. In a metrizable space, the Lindelöf property is equivalent to topological separability.

A metrizable space is separable exactly when it is Lindelöf. A countable dense subset gives a countable topological basis of balls with rational radii, hence the open-cover property. Conversely, choose countable subcovers of all radius-$1/n$ ball covers; their centers form a countable dense subset.

If $\mathcal{B}$ is a countable topological basis of $X$, then $\{B\cap Y:B\in \mathcal{B}\}$ is a countable topological basis of the subspace $Y$. Indeed every subspace-open neighborhood is $Y\cap U$ for an open $U$ of $X$, and a suitable basic set lies between its point and $U$. This proves the first assertion.

Urysohn's metrisation theorem: every second-countable space that is also a T3 space is a metrizable space. Here regularity of a topological space allows a neighborhood of a point to be shrunk to an open neighborhood whose closure is still inside the original one.

We first prove the separation facts needed in a complete proof. Second countability implies the Lindelöf property: for each basic set contained in some member of an open cover choose one such member; these countably many members cover. In particular closed subspaces are Lindelöf. For disjoint closed sets $F,G$, regularity of a topological space and this property give open covers $F\subseteq {\bigcup }_n{U}_n$, $G\subseteq {\bigcup }_n{V}_n$ with ${\overline{U}}_n\cap G=\varnothing$ and ${\overline{V}}_n\cap F=\varnothing$. DefineThese sets are open and contain $F,G$ respectively. They are disjoint: if a point belonged to a term indexed by $n$ in the first union and by $m$ in the second, the term with the larger index would have removed the closure of the other covering set. Thus the space is a normal space.

For completeness, Urysohn's lemma supplies continuous separating functions in a normal topological space. Its construction is as follows. Given disjoint closed $F,G$, choose open sets $O_r$ for dyadic $r\in [0,1]$, with $F\subseteq O_0$, $O_1=X\setminus G$, and ${\overline{O}}_r\subseteq O_s$ whenever $r<s$. Start by shrinking a neighborhood of $F$ inside $X\setminus G$, and insert the successive dyadic sets by the closed-neighborhood shrinking in a normal space. DefineThen $f=0$ on $F$, $f=1$ on $G$, and $0\le f\le 1$. The identities $\{f<a\}={\bigcup }_{r<a}{O}_r$ and $\{f>a\}={\bigcup }_{a<r<1}(X\setminus {\overline{O}}_r)$ prove continuity for $0<a<1$, with the endpoint cases immediate.

Let $\{B_n\}$ be a countable topological basis. For each pair $(n,m)$ with ${\overline{B}}_n\subseteq B_m$, choose a continuous function $f_{nm}:X\to [0,1]$ equal to one on ${\overline{B}}_n$ and zero outside $B_m$. These form a countable family. Given $x\in O$ open, choose $B_m$ with $x\in B_m\subseteq O$, shrink a neighborhood of $x$ inside $B_m$, and then choose $B_n$ inside that smaller neighborhood. ThusConsequently these functions generate the original topology. They also separate distinct points: use the $T_1$ open set $X\setminus \{y\}$ when $x\ne y$. Enumerate them as $f_j$, padding by zero functions if necessary. The map $x\mapsto (f_j(x){)}_j$ is a topological embedding into the countable cube. Pulling back its metric, or directly usinggives a compatible metric. The fact that the functions distinguish distinct points proves positivity for distinct points, the triangle inequality is termwise, and the uniformly bounded tails show that this metric induces exactly the topology generated by the coordinates. Empty and singleton spaces are trivially metrizable spaces. This proves the theorem.

For the application, a separable metric space has a countable topological basis consisting of balls with rational positive radii about a countable dense set, and it is a T3 space. The given product theorems make a countable product of these spaces second countable and a T3 space. Urysohn's theorem makes it a metrizable space. Finally every second countable space is separable: choose a point from each nonempty basic open set; the resulting countable set meets every nonempty open set. Hence the product is separable and a metrizable space.

A topological space is Lindelöf if every open cover has a countable subcover. We also prove the requested equivalence for metrizable spaces here, after the two source definition leaves.

Choose a compatible metric $d$. First suppose that $D$ is a countable dense subset. The balls $B(d_0,q)$ with centers $d_0\in D$ and positive rational radii $q$ form a countable topological basis. Indeed, if $x\in U$ and $B(x,ϵ)\subset U$, choose $d_0\in D$ with $d(x,d_0)<ϵ/3$ and a rational $q$ between $d(x,d_0)$ and $ϵ-d(x,d_0)$. The triangle inequality gives $x\in B(d_0,q)\subset U$.

For any open cover, retain the topological basis elements which are contained in some member of the cover, and choose one covering member for each such topological basis element. There are only countably many choices, and the resulting subfamily covers the topological space, since each point has a topological basis neighborhood inside a covering member. Thus a separable metrizable space is Lindelöf.

Conversely, suppose the metric space is Lindelöf. For each positive integer $n$, the balls $B(x,1/n)$ cover it, so select a countable subcover with centers in a countable set $D_n$. The union $D={\bigcup }_{n\ge 1}{D}_n$ is countable and dense: if $B(x,ϵ)$ is a nonempty open ball, choose $n$ with $1/n<ϵ$. One of the selected radius-$1/n$ balls contains $x$, so its center lies in $D\cap B(x,ϵ)$. ThereforeThe metric hypothesis is used in producing uniform small-ball covers and a countable ball topological basis; the equivalence need not hold for general topological spaces.

For a fixed nonzero second coordinate $c$, the rational expression defining $h(x,c)$ is continuous on the half-line where $xc>0$. On the opposite half-line it is constant. At $x=0$ the ratio tends to $1$, agreeing with the value on the other side. If $c=0$ the entire restriction is constant. By symmetry the restrictions with the first coordinate fixed are continuous too. Hence $h$ is separately continuous, and so is $h_t(x,y)=h(x-t,y-t)$ for every real $t$.

On the diagonal $D$, $h_t(t,t)=1$, while $h_t(s,s)=0$ for $s\ne t$. ThereforeEach diagonal singleton is open in the induced topology; arbitrary unions show that $D$ has the discrete topology.

If $X$ were metrizable, its countable dense subset from the preceding part would make it second countable. Indeed balls with rational radii about dense points contain a countable basis after the usual smaller-ball argument. Every subspace of a second-countable space is second countable. A discrete second-countable space is countable, since each singleton must itself occur as a basis member. But $D$ is uncountable. This contradiction proves $X$ is not metrizable.

A linear map from a metrizable space carrying a vector-space topology is continuous if it takes every null sequence to a null sequence. If continuity fails, choose a point in each shrinking basic neighborhood whose image stays outside a fixed neighborhood of zero. These points converge to zero but their images do not. For a Fréchet space described by increasing seminorms $p_j$, a discontinuous functional admits $p_j(f_j)\le 1/j$ and $|u(f_j)|\ge 1$.

A space is sigma-compact if it is a countable union of compact subsets. A locally compact, sigma-compact metrizable space is second countable: each compact metrizable subset has a countable dense set, their union is countable and dense, and a separable metric space has a countable basis.

A T3 space is a regular topological space which is also a T1 space. Every metric space has this property: singletons are closed, and a sufficiently small ball about a point outside a closed set has closure avoiding that set. A second-countable T3 space is a metrizable space by Urysohn's metrization theorem.

Every second-countable space that is also a T3 space is a metrizable space. First regularity of a topological space and the Lindelöf property yield a normal space: cover each of two disjoint closed sets by countably many open sets whose closures avoid the other, and subtract the first $n$ opposite closures from the $n$th neighborhood. The resulting unions are disjoint open neighborhoods. Urysohn's lemma then gives, for each pair of basic sets ${\overline{B}}_n\subseteq B_m$, a continuous function equal to one on the smaller closure and zero outside the larger set. There are countably many such pairs. The shrinking form of regularity of a topological space shows these functions generate the topology, and $T_1$ makes them separate points. Their coordinate map embeds the space into the countable cube, with compatible metric $d(x,y)={\sum }_j{2}^{-j}|f_j(x)-f_j(y)|$. The bounded tails prove compatibility with the product topology. Thus a countable base converts separation axioms into a numerical distance.