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Prove: A first countable hemicompact space is locally compact.

A topological space $(X,\tau)$ is said to be hemicompact if it has a sequence of compact subsets $K_n$, $n \in \mathbb{N}$, such that every compact subset $C$ of $(X,\tau)$ satisfies $C \subseteq K_n$, for some $n \in \mathbb{N}$.

In locally compact space each point $x$ has a compact nbhd $C$, s.t. $x \in U \subseteq C$, where $U$ is open.

By first countability, there exists a monotonic decreasing sequence $U_n$ of open nbhds of point $x$. By hemicompactness, there exists an increasing monotonic sequence $K_m$ of compact sets containing $x$. My intuition tells me that $U_n \subseteq K_m$, for some $n,m \in \mathbb{N}$, and $K_m$ is a compact nbhd of $x$, but I can't formalize this. Am I on the right path?

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Andreo
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1 Answers1

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Let $K_n$ be the cofinal sequence of compact subsets and let $p \in X$ and let $U_n$ be a decreasing (WLOG) local base at $p$.

Suppose that for all $n$: $U_n$ is not a subset of $K_n$. Let $x_n \in U_n \setminus K_n$ be a witnessing point of this non-inclusion.

Then $S = \{x_n: n \in \mathbb{N}\} \cup \{p\}$ is a compact subset of $X$ (a sequence together with its limit), and so by assumption on the cofinal sequence, for some $m$, we must have $S \subseteq K_m$. But this cannot be, as $x_m \in S$ and $x_m \notin K_m$. This contradiction shows that the original assumption is false, so that for some $n$, $U_n \subseteq K_n$, which shows that $K_n$ is a compact neighbourhood of $p$.

As this works for all $p$, $X$ is locally compact, as claimed.

Henno Brandsma
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