(cross-posted from [this math.SE question](https://math.stackexchange.com/questions/4679118/does-a-completely-metrizable-space-admit-a-compatible-metric-where-all-intersect))


It is well-known that given a metric space $(X,d)$, the metric is complete if and only if every intersection of nested (i.e. decreasing with respect to inclusion) closed balls of vanishing diameters have non-empty intersection. This is usually referred to as Cantor's Intersection Theorem.


*Is it true that every completely **metrizable** topological space **admits one compatible metric** such that every intersection of nested closed balls has a non-empty intersection?*




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Some more details:
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Let's call spherically complete a metric as above where every intersection of nested closed balls has non-empty intersection.


Not every complete metric is spherically complete (see e.g. [this question](https://math.stackexchange.com/questions/3314912/in-a-metric-space-the-intersection-of-nested-closed-balls-is-empty) and the list of its related questions).


However, this question is not about a fixed metric, but about the existence of one metric compatible with the topology.


We know there exist (classes of) completely metrizable spaces that admit compatible spherically complete metrics.


For example:


* All (complete) metrics on a compact space are spherically complete.
* The standard euclidean metric on $\mathbb{R}$ is spherically complete.
* All completely ultrametrizable spaces admit a compatible spherically complete metric.


So to be more precise my question is:


* Do all completely metrizable topological spaces admit a compatible spherically complete metric?
* If not, is it known what is the largest class of (completely metrizable) topological spaces that admit one?


I suspect that every locally compact completely metrizable space admits a spherically complete metric, and probably the same is true for Polish spaces (using that they are homeomorphic to $G\_\delta$ subsets of the Hilbert cube).
But I could not find any reference for this.



Does a completely metrizable space admit a compatible metric where all intersections of nested closed balls are non-empty?
Let us say that a topological space $X$ is *spherically completely metrizable* if the topology of $X$ is generated by a spherically complete metric.



> 
> **Theorem.** Every closed subspace $X$ of the countable product of locally compact metrizable spaces is spherically completely metrizable.
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> 
> 


*Proof.* We lose no generality assuming that $X$ is a closed subspace of the countable power $L^\omega$ of some locally compact metrizable space $L$. By the paracompactness, the locally compact metrizable space $L$ is a topological sum $\bigcup\_{\alpha\in\kappa}L\_\alpha$ of clopen $\sigma$-compact subspaces. Each space $L\_\alpha$ is locally compact and $\sigma$-compact, and hence its topology is generated by a metric $d\_\alpha$ whose closed balls are compact. On the space $L$ consider the metric $d$ defined by
$$d(x,y)=\begin{cases}1&\mbox{if $x\in L\_\alpha$ and $y\in L\_\beta$ for distinct $\alpha,\beta\in\kappa$};\\
\min\{d\_\alpha(x,y),1\}&\mbox{if $x,y\in L\_\alpha$ for some $\alpha\in\kappa$}.
\end{cases}
$$
It follows that for every $c\in L$ and every $r<1$ the closed ball $B\_L(c;r):=\{x\in L:d(x,c)\le r\}$ is compact and for every $r\ge 1$, $B\_L(c;r)=L$.


On the countable power $L^\omega$ consider the complete metric $\rho$ defined by $$\rho((x\_n)\_{n\in\omega},(y\_n)\_{n\in\omega})=\max\_{n\in\omega}\frac{d(x\_n,y\_n)}{2^n}.$$


We claim that the metric $\rho$ induces a spherically complete metric on the closed subspace $X\subseteq L^\omega$.


Let $(B\_n)\_{n\in\omega}$ be a sequence of nested closed balls in $X$. Let $(c\_n)\_{n\in\omega}$ and $(r\_n)\_{n\in\omega}$ be the sequences of the centers and radii of the balls $B\_n$. Since every sequence of positive real numbers contains a monotone subsequence, we lose no generality assuming that the sequence $(r\_n)\_{n\in\omega}$ is monotone.


If the sequence $(r\_n)\_{n\in\omega}$ is increasing (i.e., $r\_n\le r\_{n+1}$ for all $n$), then for every $n\in\omega$ we have $c\_n\in B\_n\subseteq B\_0$ and hence $\rho(c\_n,c\_0)\le r\_0\le r\_n$ and $c\_0\in B(c\_n;r\_n)=B\_n$.


So, we assume that $(r\_n)\_{n\in\omega}$ is strictly decreasing.


If $\inf\_{n\in\omega}r\_n=0$, then $\lim\_{n\to\infty}r\_n=0$ and the intersection $\bigcap\_{n\in\omega}B\_n$ is not empty by the completeness of the metric $\rho$.


So, we assume that $r:=\inf\_{ n\in\omega}r\_n>0$.


If $r\ge 1$, then every ball $B\_n$ coincides with $X$ and hence $\bigcap\_{n\in\omega}B\_n=X\ne\emptyset$.


So, we assume that $r<1$. Let $m\in\omega$ be the largest number such that $2^m r<1$.


Since $\lim\_{n\to\infty}r\_n=r$, we can replace the sequence $(B\_n)\_{n\in\omega}$ by a suitable subsequence, and assume that $2^mr\_0<1$. Then for every $k\in\omega$, the ball $B\_L(c\_0(k),2^mr\_0)$ in $L$ is compact. For every $n\in\omega$, the inclusion $c\_n\in B\_n\subseteq B\_0$ implies $\rho(c\_n,c\_0)\le r\_0$ and hence $d(c\_n(k),c\_0(k))\le 2^kr\_0$. Then for every $k\le m$, the sequence $(c\_n(k))\_{n\in\omega}$ is contained in the compact ball $B\_L(c\_0(k),2^mr\_0)$ and hence has a convergent subsequence.


Replacing $(B\_n)\_{n\in\omega}$ by a suitable subsequence, we can assume that for every $k\le m$ the sequence $(c\_n(k))\_{n\in\omega}$ is convergent in $L$, and moreover $d(c\_i(k),c\_j(k))<r$ for all $i,j\in\omega$.


We claim that $\rho(c\_0,c\_n)\le r\_n$ for every $n\in\omega$. This inequality will follow as soon as we check that $d(c\_0(k),c\_n(k))\le 2^kr\_n$ for all $k\in\omega$.


If $k>m$, then $d(c\_0(k),c\_n(k))\le 1\le 2^{m+1}r<2^kr\_n$ by the definition of the metric $d$.


If $k\le m$, then $d(c\_0(k),c\_n(k))<r\le 2^kr\_n$ by the choice of the (sub)sequence $(B\_i)\_{i\in\omega}$.


Therefore, $c\_0\in \bigcap\_{n\in\omega}B\_n$. $\quad\square$.


Since every Polish space is homeomorphic to a closed subspace of $\mathbb R^\omega$, Theorem implies



> 
> **Corollary.** Every Polish space is spherically completely metrizable.
> 
> 
> 


The Theorem suggests the following



> 
> **Question.** Which metrizable spaces do embed into countable products of locally compact metrizable spaces?
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> 
> 


**Remark 1.** The necessary condition of the embeddability of a topological space $X$ into the countable product of locally compact metrizable spaces is the separability of all quasicomponents of $X$. This condition implies that nonseparable connected metrizable spaces do not embed into countable products of locally compact metrizable spaces.


The following proposition answers the above Question.



> 
> **Proposition.** A topological space $X$ is homeomorphic to a closed subspace of the countable product of locally compact metrizable spaces if and only if $X$ is homeomorphic to a closed subspace of $\mathbb R^\omega\times\kappa^\omega$ for some cardinal $\kappa$ endowed with the discrete topology.
> 
> 
> 


*Proof.* The "if" part of this characterization is trivial. To prove the "only if" part, assume that $X$ is homeomorphic to a closed subspace of the product $\prod\_{n\in\omega}L\_n$ of locally compact metrizable spaces $L\_n$. By the paracompactness, every space $L\_\alpha$ is a topological sum of locally compact $\sigma$-compact metrizable spaces and hence is a topological sum of Polish spaces. Since every Polish space is homeomorphic to a closed subspace of the space $\mathbb R^\omega$, for every $n\in\omega$ the locally compact metrizable space $L\_n$ is homeomorphic to a closed subspace of $\mathbb R^\omega\times\kappa$ for some cardinal $\kappa$. Then $\prod\_{n\in\omega}L\_n$ is homeomorphic to a closed subspace of the space $(\mathbb R^\omega\times\kappa)^\omega$, which is homeomorphic to $\mathbb R^\omega\times\kappa^\omega$. $\quad\square$


Proposition and Theorem imply that every closed subspace of $\mathbb R^\omega\times\kappa^\omega$ is spherically completely metrizable.



> 
> **Problem.** Let $\kappa$ be a cardinal. Is every closed metrizable subspace of the space $[0,1]^\kappa\times\kappa^\omega$ spherically completely metrizable?
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> 
> 


**Remark 2.** For every cardinal $\kappa$, every closed metrizable subspace of the space $[0,1]^\kappa\times\kappa^\omega$ is completely metrizable. On the other hand, closed metrizable subspaces of the space $\mathbb R^\kappa$ are realcompact but needs not be completely metrizable (by Theorem 3.11.12 in Engelking's "General Topology", every Lindelof space is realcompact; in particluar, every metrizable separable space is realcompact).


