I was studying some counterexamples, and found out that, if a space $X$ is Hausdorff and nowhere locally compact, then every continuous function $f : X^* \to \mathbb{R}$ is constant, where $X^* $ is the one-point compactification of $X$. Further, if $V$ is an open set of $X$ which is not dense, this means $\operatorname{cl}_{X^* } (V) = \operatorname{cl}_X (V) \cup \{ \infty \}$ is a proper subset of $X^*$. If $V$ is $\sigma$-compact, this means $V$ is $F_\sigma$ in $X^* $. Therefore, we have an example of an open $F_\sigma$-set whose closure is not the support of a continuous function. This cannot happen for a normal space. I would like to know if the condition $\sigma$-compact for $X$ is sufficient to find such a $V$. Examples of such spaces are: the rational numbers $\mathbb{Q}$, the prime integer topology, the $p$-adic topology, and many others. For countable spaces, the result is trivial, since every singleton is compact.
2026-03-26 06:31:14.1774506674
Is there any Hausdorff, $\sigma$-compact and nowhere locally compact space with every non dense open set not $\sigma$-compact?
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