# Not-too-non-compact metric spaces

Consider the following two separable metric spaces: Cantor space $$2^\omega$$ and Baire space $$\omega^\omega$$. These spaces give rise to two classes of metric spaces, namely the "big" ones into which Baire space embeds as a closed subset and the "small" ones which are "easily covered" by Cantor space, that is, which are the continuous image of $$2^\omega\times\omega$$. For example, $$\mathbb{R}$$ is small in this sense, since $$[-n,n]$$ is a compact metric space for each $$n\in\mathbb{N}$$ and every compact metric space is the continuous image of Cantor space. In fact, this last fact shows that "small" is just "$$\sigma$$-compact."

Each "natural" separable metric space that I can think of is either big or small in this sense, and this raises the following question:

Suppose $$M$$ is a separable metric space which doesn't contain a closed set homeomorphic to Baire space and is homeomorphic to a Borel subset of Baire space (I'll also accept more complicated examples as long as they are "reasonably natural," e.g. exist in $$L(\mathbb{R})$$ assuming large cardinals). Must $$M$$ be $$\sigma$$-compact?

I suspect the answer is "no," but I haven't been able to whip up a counterexample.

• As Eric Wofsey pointed out, if we drop the "tameness" requirement on $M$ it's easy to whip up a counterexample (e.g. a Bernstein set). This was something I was aware of but forgot to include initially. – Noah Schweber Mar 16 at 20:55
• You probably know this but if $M$ is complete and somewhere nowhere locally compact (i.e. nowhere locally compact in some nonempty open set) it has a closed copy of Baire space. – James Hanson Mar 16 at 22:43
• Yes, but that's worth mentioning (and +1 for "somewhere nowhere"). – Noah Schweber Mar 16 at 22:43

I think the answer is yes for Borel subsets of Polish spaces. I'm not confident enough to say anything about $$L(\mathbb R).$$

Lemma 1. Any second-countable space $$M$$ can be partitioned into an open $$\sigma$$-compact space $$A$$ and a closed subspace $$B$$ such that every non-empty relatively open subset of $$B$$ is non-$$\sigma$$-compact.

Proof: Define $$A$$ to be the union of all $$\sigma$$-compact open sets in $$M.$$ Since $$M$$ is strongly Lindelöf, $$A$$ is $$\sigma$$-compact. Set $$B=M\setminus A.$$ Consider a relatively open $$\sigma$$-compact subset $$U\subseteq B.$$ This is the restriction of some open $$U'\subseteq M.$$ Since $$A\cup U'=A\cup U$$ is open and $$\sigma$$-compact, $$U'\subseteq A.$$ So $$U=\emptyset.$$ $$\Box$$

In fact I'm only going to use a weaker property of $$B$$ as a metric space: closed balls are non-compact. (If a closed ball $$B'(x,r)$$ is compact, then $$B(x,r)$$ is $$\sigma$$-compact.)

Lemma 2. If $$M$$ is a non-empty subspace of a complete metric space $$P,$$ and $$M$$ has the property of Baire, and all closed balls of $$M$$ are non-compact, then there is a closed embedding of $$\omega^\omega$$ in $$M.$$

Proof. Write $$M=U\Delta (\bigcup C_i)$$ where $$U$$ is open and each $$C_i$$ is nowhere dense.

Since $$M$$ is not sequentially compact, it is either not totally bounded or it has a Cauchy sequence that converges to a point in $$P\setminus M.$$ In either case we can find a sequence of disjoint "non-converging" closed balls $$B_n\subseteq U$$ - by this I mean that no sequence with $$x_n\in B_n$$ has limit points in $$M.$$ We can pick the balls $$B_n$$ so that they do not intersect $$C_1$$ and such that they have radius less than $$1$$.

Then for each $$n_1$$ we can pick a sequence of disjoint non-converging closed balls $$B_{n_1,n_2}\subseteq B_{n_1}$$ which do not intersect $$C_2$$ and with radius less than $$1/2.$$ Continue in this manner, defining closed balls $$B_{n_1,\dots,n_{k+1}}\subseteq B_{n_1,\dots,n_k}$$ avoiding $$C_1\cup\cdots\cup C_{k+1}$$ and with radius less than $$1/k.$$

Define an embedding $$f:\omega^\omega\to M$$ by taking $$(n_1,n_2,\cdots)$$ to the unique point in $$\bigcap_{k}B_{n_1,\dots,n_k}.$$ This map is injective because for each $$k,$$ the balls $$B_{n_1,\dots,n_k}$$ are disjoint. For each sequence $$x_1,x_2,\dots\in\omega^\omega,$$ the sequence $$f(x_n)$$ converges if and only if $$x_n$$ converges in $$\omega^\omega.$$ So $$f$$ is a closed embedding. $$\Box$$

Borel subsets $$M$$ of a Polish space certainly have the Baire property. So these two lemmas imply that $$M$$ must be either $$\sigma$$-compact or contain a homeomorphic copy of $$\omega^\omega.$$

• +1 - and since (under large cardinals) all subsets of $\omega^\omega$ have the Baire property, this also goes a way towards answering the more ambitious question. I'm going to hold off on accepting for a bit to see if the more ambitious question gets resolved, but I think this is a very satisfying answer. – Noah Schweber Mar 19 at 19:41
• I just noticed that my previous comment has a horrible typo - of course not all subsets of $\omega^\omega$ have the Baire property, rather all reasonably definable subsets of $\omega^\omega$ have the Baire property with the notion of reasonable definability scaling with the large cardinal hypotheses (including subsuming $\mathcal{P}(\omega^\omega)\cap L(\mathbb{R})$). – Noah Schweber Jul 16 at 19:06