Need a unique convergence (UC) space's Alexandrov extension be a UC space?

Question

Background

Say a topological space $X$ is

• a unique convergence (UC) space iff every sequence of points of $X$ converges to at most one point of $X$;
• a unique convergent clustering (UCC) space iff every convergent sequence of points of $X$ has a unique cluster point;
• a KC space iff compact subsets of $X$ are closed.

I've been able to show that Hausdorff $\implies$ KC $\implies$ UCC $\implies$ UC $\implies$ $T_1,$ and that a first-countable UC space is UCC.

Given a space $X$, we define the Alexandrov extension of a space $X$--which I denote $X_\alpha$--by adjoining a point $\infty$ to $X$, and making the neighborhoods of $\infty$ those subsets of $X\cup\{\infty\}$ whose complements are closed compact subsets of $X$. It can be shown that $X_\alpha$ is always compact, $X$ an open subspace of $X_\alpha,$ and that $X$ is dense in $X_\alpha$ if and only if $X$ is non-compact.

It is trivial to show that $X$ is $T_1$ iff $X_\alpha$ is $T_1,$ since finite sets are immediately compact. I'm also aware that $X_\alpha$ is Hausdorff iff $X$ is Hausdorff and locally compact.

Edit: I also "proved" before posting the question that $X$ is KC iff $X_\alpha$ is KC, but this turns out not to be true. When $X$ is the Arens-Fort space, then $X$ is Hausdorff (so KC), but $X_\alpha$ is not even KC, as pointed out in the comments. To see why, note that every neighborhood of the "point at infinity" in $X_\alpha$ is cofinite, so every subset of $X_\alpha$ containing this point is compact. However, the set consisting of "even columns" of $X$ is not open in $X_\alpha$ and doesn't contain the "point at infinity," so its complement is a compact, non-closed subset of $X_\alpha.$ Thanks again to MW for bringing that lovely counterexample to my attention!

Edit: Another "proof" was that $X$ was UCC if and only if $X_\alpha$ was UCC, but again, the Arens-Fort space provides a counterexample (so is an even stronger counterexample to preservation of the KC property by Alexandrov extension), as pointed out in the comments once again by MW.

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It is readily the case that if $X_\alpha$ is a UC space, then so is $X$. However, I've been unable to prove or disprove that if $X$ is UC, then so is $X_\alpha$. I suspect that it isn't true in general, but I haven't found any counterexamples.

If it is true, how might one go about proving it?

If it is not true, can you provide a counterexample?

Answer 1

Here’s a counterexample.

Let $P=\beta\omega\setminus\omega$ be the set of free ultrafilters on $\omega$. Let $X=(\omega+1)\cup P$ be topologized by making $\omega+1$ an open subset of $X$ with its usual order topology and declaring $U\subseteq X$ a nbhd of $p\in P$ iff $p\in U$ and $U\cap\omega\in p$. The only non-trivial sequences in $X$ are those that converge to $\omega$, so $X$ is $US$, but $\langle n:n\in\omega\rangle$ clusters at each point of $P$ (and of course converges to $\omega$), so $X$ is not $UCC$. Note that $\omega$ is dense in $X$, which is not compact, so it’s not the case that the range of a convergent sequence in a $US$ space must have compact closure.

$P$ is an infinite closed, discrete subset of $K$, and $P\cap\operatorname{cl}A$ is infinite for each infinite $A\subseteq\omega$. (In fact $|P\cap\operatorname{cl}A|=|\beta\omega\setminus\omega|=2^{\mathfrak{c}}$ for each infinite $A\subseteq\omega$.) Thus, the only compact, closed subsets of $X$ are the finite sets, and if $q$ is the point at infinity in $X^*$, the sequence $\langle n:n\in\omega\rangle$ converges to both $\omega$ and $q$.