A condition equivalent to $\sigma$-compactness for Polish spaces

Question

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First, happy holidays, everyone!

For context: This question was originally asked by a different user here. However, after I posted my answer, they deleted the question and later deleted their account. The question and the result contained in it are interesting enough that, in my mind, they should be preserved. However, the original question contained several typos, used some imprecise wordings, and was somewhat lacking in context anyway, so I believe it is better to reformulate and re-ask this question with typo fixes and more information than to simply undelete the original question.

Now, onto the question itself:

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Recall that a topological space is called Polish if it is separable and completely metrizable. Up to Borel equivalence, there aren't that many Polish spaces - there is one for each countable cardinality, and then one uncountable one, namely $\mathbb{R}$. Topologically, though, there are a large amount of interesting Polish spaces. For example, every topological manifold (in particular all Euclidean spaces) is Polish, and so is every separable Banach space.

We note that plenty of natural examples of Polish spaces are $\sigma$-compact. Indeed, any topological manifold as well as any closed subset thereof are all $\sigma$-compact. These examples are all $\sigma$-compact for the same reason, namely, they are locally compact, and a locally compact Polish space is $\sigma$-compact. Indeed, let $\mathcal{B}$ be a countable basis of the topology. Then $\mathcal{C}$, the subset of $\mathcal{B}$ containing those elements that are relatively compact, is an open cover of the space (since $X$ is LCH). This means the (countable) union of the closures of elements of $\mathcal{C}$ is $X$, so $X$ is $\sigma$-compact.

Remark: One may observe that the above proof actually shows a second countable LCH space is $\sigma$-compact, which is a priori more general. But a second countable LCH space is Polish anyway - LCH spaces are $T_3$. By Urysohn's metrization theorem, a second countable $T_3$ space is metrizable. A locally compact metrizable space is completely metrizable. And, of course, second countability implies the space is separable.

There are, however, also examples of $\sigma$-compact Polish spaces that are not locally compact. A standard example is the topologist's sine curve. There are also examples of non-$\sigma$-compact Polish spaces. Standard examples include the irrational numbers and the Hilbert space $\ell^2$. In both cases, they are non-$\sigma$-compact for the same reason - any compact subset has empty interior, so the space cannot be a countable union of compact sets per Baire category theorem.

In particular, for a (nonempty) Polish space to be $\sigma$-compact, it has to be the case that there are compact subsets with nonempty interior. This is, of course, not enough - one can simply take the disjoint union of a locally compact Polish space and a non-$\sigma$-compact Polish space. Then the resulting space is Polish and has plenty of relatively compact open sets, but is nevertheless non-$\sigma$-compact. (In fact, one can even get the union of relatively compact open sets to be dense without making the space $\sigma$-compact. An example would be the closed topologist's sine curve minus $\{0\} \times (\mathbb{Q} \cap [-1,1])$.)

The issue with these examples is that there are nonempty closed subspaces $F \subset X$ s.t. all compact subsets of $F$ have no interior in $F$. This cannot happen in a $\sigma$-compact Polish space, since a closed subspace of a $\sigma$-compact Polish space is a $\sigma$-compact Polish space itself. Hence, the following condition is necessary for a Polish space $X$ to be $\sigma$-compact:

$$\text{Every nonempty closed }F \subset X\text{ has a compact }K \subset F\text{ s.t. }K\text{ has nonempty interior in }F\tag{$\ast$}$$

The natural question, of course, is whether this condition is also sufficient:

If a Polish space $X$ satisfies $(\ast)$, is $X$ necessarily $\sigma$-compact?

A search for examples on pi-Base yields no counterexample to the above conjecture.

Answer 1

This is true. Let us begin with a few lemmas:

Lemma 1: Let $U \subset X$ be an open subset of a topological space. Let $V \subset X \setminus U$ be a (relatively) open subset of $X \setminus U$. Then $U \cup V$ is open in $X$.

Proof: Since $V$ is open in $X \setminus U$, there exists open $W \subset X$ s.t. $V = W \setminus U$. Thus, $U \cup V = U \cup W$ is open in $X$. $\square$

Definition: Let $X$ be a topological space. Then we define its locally compact part to be,

$$X^{LC} = \{x \in X: \text{ there exists a compact neighborhood of }x\text{ in }X\}$$

Lemma 2: Let $X$ be a $T_3$ topological space. Then $X^{LC}$ is a locally compact open subspace of $X$.

Proof: We first show $X^{LC} \subset X$ is open. Indeed, for each $x \in X^{LC}$, there is an open subset $U \subset X$ and a compact subset $K \subset X$ s.t. $x \in U \subset K$. But then every point in $U$ has a compact neighborhood, namely $K$, so $U \subset X$.

We now show $X^{LC}$ is locally compact. For any $x \in X^{LC}$, as we have seen, there is an open subset $U \subset X^{LC}$ and a compact subset $K \subset X$ s.t. $x \in U \subset K$. By regularity of $X$, as $X^{LC} \subset X$ is open, there is an open $V \subset X$ s.t. $x \in V \subset \overline{V} \subset X^{LC}$. Thus, $\overline{V} \cap K$ is compact and contained in $X^{LC}$. Furthermore, $x \in V \cap U \subset \overline{V} \cap K$, so $\overline{V} \cap K$ is a compact neighborhood of $x$ in $X^{LC}$. $\square$

Lemma 3: Let $X$ be a second countable topological space. Then any well-ordered strictly increasing chain of open subsets of $X$ must be countable.

Proof: Assume to the contrary that there exists a transfinite sequence of strictly increasing open sets $\{U_\lambda\}_{\lambda < \omega_1}$ of length $\omega_1$. Fix a countable basis $\mathcal{B}$ of $X$. Then for each $U_\lambda$, let $\mathcal{B}_\lambda = \{W \in \mathcal{B}: W \subset U_\lambda\}$. Since $\mathcal{B}$ is a basis, $\{\mathcal{B}_\lambda\}_{\lambda < \omega_1}$ is a transfinite sequence of strictly increasing subsets of the countable set $\mathcal{B}$, which cannot exist. $\square$

Theorem: Let $X$ be a Polish space satisfying the condition,

$$\text{Every nonempty closed }F \subset X\text{ has a compact }K \subset F\text{ s.t. }K\text{ has nonempty interior in }F\tag{$\ast$}$$

Then $X$ is $\sigma$-compact.

Proof: Assume to the contrary that $X$ is not $\sigma$-compact. We shall construct a transfinite sequence of strictly increasing $\sigma$-compact open subsets $\{U_\lambda\}_{\lambda < \omega_1}$ as follows: Set $U_0 = \varnothing$. At any $0 < \lambda < \omega_1$, the inductive assumption shall be $\{U_\kappa\}_{\kappa < \lambda}$ is a transfinite sequence of strictly increasing $\sigma$-compact open subsets. If $\lambda$ is a limit ordinal, let $U_\lambda = \bigcup_{\kappa < \lambda} U_\kappa$. Since $\{U_\kappa\}_{\kappa < \lambda}$ is strictly increasing, $U_\kappa \subsetneq U_\lambda$ for any $\kappa < \lambda$. Since $\lambda < \omega_1$, $U_\lambda$ is a countable union of $\sigma$-compact open subsets, whence it is a $\sigma$-compact open subset as well.

Now, assume $\lambda = \kappa + 1$ is a successor ordinal. Since $U_\kappa$ is $\sigma$-compact and $X$ is not, we have $X \setminus U_\kappa$ is a nonempty closed subset of $X$. As $X$ is Polish, so is $X \setminus U_\kappa$. By $(\ast)$, there exists compact $K \subset X \setminus U_\kappa$ s.t. $K$ has nonempty interior in $X \setminus U_\kappa$. In particular, the interior of $K$ is contained in $(X \setminus U_\kappa)^{LC}$, so the latter space is a nonempty locally compact open subspace of $X \setminus U_\kappa$, by Lemma 2. Being an open subspace of a Polish space, it is Polish itself, whence $(X \setminus U_\kappa)^{LC}$ is locally compact Polish and thus $\sigma$-compact. Now, let $U_\lambda = U_\kappa \cup (X \setminus U_\kappa)^{LC}$. This is open by Lemma 1. It is strictly larger than $U_\kappa$ since $(X \setminus U_\kappa)^{LC}$ is nonempty. It is $\sigma$-compact because it is the union of two $\sigma$-compact spaces.

Polish spaces are second countable, but we have constructed a transfinite sequence of strictly increasing open sets of length $\omega_1$, contradicting Lemma 3. Hence, it must be the case that $X$ is $\sigma$-comapct. $\square$