Is the Arens space hemicompact but not locally compact?

Question

The Arens space $S_2$ is a typical example of a space that is sequential, but not Frechet-Urysohn. As well illustrated in the diagrams here, it can be constructed informally by taking a convergent sequence and then to each isolated point of the first sequence attach another sequence converging to it.

Explicitly, $X=\{\infty\}\cup\mathbb N\cup(\mathbb N\times\mathbb N)$ with $\mathbb N=\{1,2,...\}$. Every point of $(\mathbb N\times\mathbb N)$ is isolated. Every point $n\in\mathbb N$ has a local base consisting of the sets $B_N(n)=\{n\}\cup\{(n,k):k\ge N\}$. So $n$ is the limit of the sequence of points in its column. And a local base at the point $\infty$ consists of the sets equal to $X$ minus a finite number of columns $B_1(n)$ and furthermore omitting a finite number of isolated points $(n,k)$ in each of the remaining columns. So $\infty$ is the limit of the sequence of points in $\mathbb N$.

Can the following be shown?

Show that the Arens space is hemicompact but not locally compact. (Hemicompact means that there is a sequence of compact sets in $X$ such that every compact subset of $X$ is contained in one of those compact sets.)

This is interesting because, as recorded in pi-base:

Every hemicompact first countable space is weakly locally compact.

The Arens space then shows that one cannot relax the hypothesis of first countable to sequential and still expect to keep local compactness.

Answer 1

A subset $A\subseteq X$ that contains isolated points of $X$ (i.e., points from $\mathbb N\times\mathbb N$) from infinitely many columns cannot be compact. Because if we can find a point $(n,k_n)\in A\cap(\mathbb N\times\mathbb N)$ for each $n$ in some infinite set $M\subseteq\mathbb N$, the collection consisting of all the singletons $\{(n,k_n)\}$ and their complement $X\setminus\{(n,k_n):n\in M\}$ is a covering of $A$ by open sets which has no finite subcollection covering $A$.

Since any neighborhood of $\infty$ contains isolated points of $X$ from infinitely many columns, $\infty$ has no compact neighborhood. So the (Hausdorff) space $X$ is not locally compact.

On the other hand, since any compact set has empty intersection with all but finitely many of the columns of $\mathbb N\times\mathbb N$, it must be contained in one of the sets $$K_N=\{\infty\}\cup\mathbb N\cup\{(n,k)\in\mathbb N\times\mathbb N:n\le N\}$$ for some integer $N$. Each of the $K_N$ is compact as a finite union of compact sets (the horizontal converging sequence and some of the vertical ones). Since we have countably many $K_N$, the Arens space is hemicompact.

---

Note (added Sep 2024): From the above, one can also show the following.

Even though the Arens space is not second countable, it has a countable $k$-network.

That is, a countable collection $\mathcal N$ of subsets of $X$ such that for every compact set $K$ and open set $U$ in $X$ with $K\subseteq U$, there is finite subcollection $\mathcal N^*\subseteq\mathcal N$ such that $K \subseteq \bigcup\mathcal{N}^* \subseteq U$.

Spaces that are $T_3$ and have a countable $k$-network are called $\aleph_0$-spaces. Since the Arens space is $T_3$, we can also summarize the above by saying:

The Arens space is an $\aleph_0$-space.

Proof: Each $K_N$ above is the topological disjoint union of finitely many convergent sequences (a tail of the horizontal sequence, plus some vertical sequences); so each $K_N$ is separable and metrizable, hence second countable. Let $\mathcal N_N$ be a countable base for the subspace $K_N$, which will also be a $k$-network for $K_N$. And let $\mathcal N$ be the union of all the $\mathcal N_N$.

Claim: $\mathcal N$ is a countable $k$-network for $X$. It is certainly countable. And given sets $K\subseteq U$ with $K$ compact and $U$ open in $X$, we have $K\subseteq K_N$ for some $N$ by the proof of hemicompactness. So $K$ is contained in $K_N\cap U$, which is open in $K_N$. Since $\mathcal N_N$ is a $k$-network for $K_N$, the set $K$ is contained in the union of finitely many members of $\mathcal N_N\subseteq\mathcal N$, which completes the proof.