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(cross-posted from this math.SE question)

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?


Some more details:

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 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.

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    $\begingroup$ A metric space is completely metrizable if and only if it is a $G_\delta$ subspace of its Cauchy completion. The usual proof of the "if" direction gives you an actual formula for a complete metric $\delta$ on $X$ in terms of a complete metric $d$ on its completion: $$\delta(x,y) = d(x,y) + \sum_{n=1}^\infty 2^{-n} \frac{| d(x,U_n)^{-1} - d(y,U_n)^{-1}|}{1+| d(x,U_n)^{-1} - d(y,U_n)^{-1}|},$$ where $U_1,U_2,U_3,\dots$ are open and $X = \bigcap_{n=1}^\infty U_n$. I've stared at this formula for the last few minutes, and I can't decide whether it's spherically complete or not. But maybe? $\endgroup$
    – Will Brian
    May 8, 2023 at 12:04
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    $\begingroup$ It seems to me that another example of “spherically complete” is any Banach space. Indeed, for balls in a normed space, $B(x’,r’)\subset B(x,r)$ implies $\|x-x’\|\le r-r’$ (looking at the line between $x$ and $x’$). So for a nested family of closed balls $\bar B(x_k,r_k)$ the sequence of the centers is a Cauchy sequence. $\endgroup$ May 8, 2023 at 12:28
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    $\begingroup$ Every locally compact separable metrizable space admits a proper metric (i.e. a metric, under which closed balls are compact, sometimes also called Heine-Borel metric.) This one is surely spherically complete. $\endgroup$ May 8, 2023 at 19:40
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    $\begingroup$ Exact; and it seems to me one can make such a metric $d$ even without using paracompactness. Then if one truncates this metric defining $d':=\min\{d, 1\}$, the closed balls of $d'$ are either compact balls of $d$, or the whole $X$, so that $d'$ is spherically complete. $\endgroup$ May 9, 2023 at 19:42
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    $\begingroup$ A condition that implies sph. compl. for a complete metric space $X$ is: for all $x\in X$ and $y\in X$ and $r>0$, there exists $z\in X$ such that $d(y,z)=r$ and $d(x,y)+d(y,z)=d(x,z).$ It's quite strong: it implies that decreasing nested sequences of closed balls have decreasing radii; are a Cauchy sequence of closed sets w.r.to the Hausdorff metric, and their centres are a Cauchy sequence. $\endgroup$ May 13, 2023 at 18:22

1 Answer 1

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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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  • $\begingroup$ Thanks, that gives excellent sufficient conditions. It is maybe worth noticing that a space is completely ultrametrizable if and only if it is homeomorphic to a closed subspace of $\kappa^\omega$ for some $\kappa$, so your proposition gives also a proper extension of what was known for ultrametrizable spaces. $\endgroup$
    – Cla
    May 16, 2023 at 14:03
  • $\begingroup$ Also, the radial metric on the plane is an example of a non-separable connected spherically complete metric space, that thus does not embed into $\mathbb{R}^\omega\times \kappa^\omega$, so this condition is not necessary. $\endgroup$
    – Cla
    May 17, 2023 at 11:25
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    $\begingroup$ @Cla As was observed by Pietro Mayer, Banach spaces are spherically complete and they also do not embed into $\mathbb R^\omega\times\kappa^\omega$. More generally, every complete metric space $(X,d)$ satisfying the Axiom of Segment Construction ($\forall x,y,a,b\in X\; \exists z\in X\; (d(y,z)=d(a,b) \wedge d(x,z)=d(x,y)+d(y,z))$) is spherically complete. $\endgroup$ May 17, 2023 at 12:04

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