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.

*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?

**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?

**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).

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