# A space which is $\sigma$-compact but neither hemicompact nor second countable

The pi-base as of this posting doesn't have a space which is $$\sigma$$-compact but neither second countable nor hemicompact.

As mentioned in A $\sigma$-compact but not hemicompact space?, the set of rationals is a standard example of a $$\sigma$$-compact space which is not hemicompact. However, it is second countable.

What is an example of a space which is $$\sigma$$-compact but neither second countable nor hemicompact?

## 2 Answers

The rational numbers $$\mathbb Q$$ are $$\sigma$$-compact but not hemicompact. The ordinal space $$Y=\omega_1+1=[0,\omega_1]$$ is compact, but not second countable (it's not first countable at $$\omega_1$$ for example).

So their topological sum $$X=\mathbb Q\coprod Y$$ is $$\sigma$$-compact, but not second countable (because of $$Y$$), and not hemicompact (because of its closed subspace $$\mathbb Q$$).

• Thank you for this much more straight-forward example. Commented Aug 2, 2023 at 4:34

We use the idea of Prop. 2.3 from this paper - Islas, Jardon. Vietoris topology on spaces dominated by second countable ones. 2015

Let $$Y$$ be $$[0,\omega_1]$$ with the order topology and $$\mathbf 0$$ be the constant zero function $$\omega \to Y$$. The $$\sigma$$-product on $$Y^\omega$$ is defined to be $$X = \{ f \in Y^\omega : f^{-1}[Y \setminus \{0\}] \in [\omega]^{<\omega} \}$$ with the topology induced as a subspace of $$Y^\omega$$.

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

Since $$X_n = \{ f \in Y^\omega : (\forall x \geq n)\ f(x) = 0 \}$$ is homeomorphic to $$Y^n$$ and $$X = \bigcup \{ X_n : n \in \omega \}$$, $$X$$ is $$\sigma$$-compact.

$$X$$ is not second countable.

Since $$Y$$ can be homeomorphically embedded into $$X$$ and $$Y$$ is not second countable, $$X$$ is also not second countable.

$$X$$ is not hemicompact.

Here, we use Theorem 3.9 of this paper - Cascales, Orihuela, Tkachuk. Domination by second countable spaces and Lindelöf $$\Sigma$$-property. 2011

We will actually show that the closed subspace $$\mathbf F := 2^\omega \cap X$$ is not hemicompact. From there, we can conclude that $$X$$ is not hemicompact since a closed subspace of a hemicompact space must be hemicompact.

Let $$\{ K_n : n \in \omega \}$$ be a collection of compact subsets of $$\mathbf F$$. We will produce a compact subset of $$\mathbf F$$ which is not contained in any $$K_n$$.

First, we note that, for any infinite subset $$\Lambda$$ of $$\omega$$, $$\{ f \in \mathbf F : f^{-1}(1) \subseteq \Lambda \}$$ has limit points outside of $$\mathbf F$$. Indeed, let $$\{ A_n : n \in \omega \}$$ be an ascending sequence of finite subsets of $$\Lambda$$ so that $$\bigcup \{ A_n : n \in \omega \}$$ is infinite. Then the functions $$g_n : \omega \to 2$$ defined by taking value $$1$$ on $$A_n$$ and $$0$$ otherwise have a limit outside of $$\mathbf F$$.

Let $$n \in \omega$$ and suppose we have $$\{ f_\ell : \ell < n \} \subseteq \mathbf F$$ defined so that, for each $$\ell < n$$, $$f_\ell \in \mathbf F \setminus K_\ell$$ and that $$\{ f_\ell^{-1}(1) : \ell < n \}$$ is pairwise disjoint. Note that $$\Lambda := \omega \setminus \bigcup \{ f^{-1}_{\ell}(1) : \ell < n \}$$ is infinite. Then $$A := \{ f \in \mathbf F : f^{-1}(1) \subseteq \Lambda \}$$ has limit points outside of $$\mathbf F$$. So we can choose $$f_n \in A \setminus K_n$$. Observe that $$\{ f^{-1}_\ell(1) : \ell \leq n \}$$ is pairwise disjoint.

Hence, we can construct $$\{ f_n : n \in \omega \} \subseteq \mathbf F$$ to be so that, for each $$n \in \omega$$, $$f_n \in \mathbf F \setminus K_n$$ and that $$\{ f^{-1}_n(1) : n \in \omega \}$$ is pairwise disjoint. We now claim that $$K := \{ f_n : n \in \omega \} \cup \{ \mathbf 0 \} \subseteq \mathbf F$$ is compact. Consider any open set $$U$$ that contains $$\mathbf 0$$. Without loss of generality, we can assume that $$U$$ is a basic open set; that is, that $$U = \{ f \in 2^\omega : (\forall x \in E)\ f(x) = 0 \}$$ for some $$E \in [\omega]^{<\omega}$$. Since $$\{ f_n^{-1}(1) : n \in \omega \}$$ is pairwise disjoint, $$\{ n \in \omega : f_n^{-1}(1) \cap E \neq \emptyset \}$$ is finite. So $$U$$ contains all but finitely many of the $$f_n$$. Hence, $$K$$ is compact.

Therefore, since $$f_n \in K \setminus K_n$$ for every $$n \in \omega$$, we see that $$K \not\subseteq K_n$$ for every $$n \in \omega$$. That is, $$\mathbf F$$ is not hemicompact.

• Why is $A\setminus K_n$ nonempty? Commented Jul 16, 2023 at 1:47
• @StevenClontz my gut reaction was to say because $A$ is not compact and $K_n$ is, but the real answer (I think) is because $A$ has limit points outside of $\mathbf F$. So the closure of $A$ cannot be contained in $K_n$. I'll edit the answer accordingly. Commented Jul 16, 2023 at 1:55
• Sure, (0,1) is not compact and [0,1] is, so we need a bit more about the sets for this to be clear. Commented Jul 16, 2023 at 1:58
• @StevenClontz Ok, I think the limit point comment makes it clearer. If $A \subseteq K_n \subseteq \mathbf F$, then the closure of $A$ would be contained in $\mathbf F$. (Though, my phrasing could probably be improved.) Commented Jul 16, 2023 at 2:08
• Interesting example. Does the next to last paragraph actually show that the sequence of $f_n$ converges to $\mathbf 0$? (Also, minor stuff: "ascending union of ..." was meant as "ascending sequence ..." I suppose?) Commented Aug 1, 2023 at 22:59