# A first countable hemicompact space is locally compact

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?

• Please could you state the definitions you're using for hemicompact and locally compact? – Henno Brandsma Sep 15 '18 at 3:10
• @henno-brandsma I've added the definitions of hemicompact and locally compact – Andreo Sep 15 '18 at 17:57
• You're idea is OK, you need a diagonalisation argument to formalise it. See my answer. – Henno Brandsma Sep 16 '18 at 15:34

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.