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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?

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2 Answers 2

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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 $\mathbb Q$).

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

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    $\begingroup$ Why is $A\setminus K_n$ nonempty? $\endgroup$ Jul 16, 2023 at 1:47
  • $\begingroup$ @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. $\endgroup$ Jul 16, 2023 at 1:55
  • $\begingroup$ Sure, (0,1) is not compact and [0,1] is, so we need a bit more about the sets for this to be clear. $\endgroup$ Jul 16, 2023 at 1:58
  • $\begingroup$ @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.) $\endgroup$ Jul 16, 2023 at 2:08
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    $\begingroup$ 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?) $\endgroup$
    – PatrickR
    Aug 1, 2023 at 22:59

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