A "simple" space with closed retracts but non-unique sequential limits

Question

This question asking how KC ("Kompacts are Closed") and RC ("Retracts are Closed") are distinct has some good discussion, including a now-published example by Banakh and Stelmakh which is RC but not KC. An earlier revision of their paper had an example which was actually not US ("Unique Sequential limits", weaker than KC), but was revised so the version in print was US. I've read that earlier revision and am convinced their example of an RC-not-US space is valid.

Nonetheless, the example is somewhat involved, as more structure was desired by the authors (this space is semi-Hausdorff, Brown, strongly rigid, etc.). After some attempts to "strip down" the example to get a "clean" example of an RC-not-US space, I came up short.

Can anyone construct an "elementary" example of an RC-not-US space?

Answer 1

Here is a relatively simple example.

Note first that by continuity, if $r\colon X\to A$ is a retraction, and $a_\alpha\in A$ is a net converging to $x$, then $a_\alpha=r(a_\alpha)\to r(x)$, so if $a_\alpha$ has a unique limit then $x=r(x)\in A$.

With this in mind, let $X=[0,\infty)\cup \{\infty_1,\infty_2\}.$

The topology on $[0,\infty)$ is the usual Euclidean topology, a neighborhood base of $\infty_1$ is given by sets of the form $\{\infty_1\}\cup (a,\infty)\backslash 2\mathbb N$, and a base for $\infty_2$ by sets of the form $\{\infty_2\}\cup (a,\infty)\backslash (2\mathbb N +1)$.

Then $X$ is not US, since for example the sequence $.5+n$ converges to $\infty_1$ and $\infty_2$.

On the other hand, if $A$ is a retract of $X$, and $a_\alpha\in A$ is a net converging to $x\in [0,\infty)$, then $x$ is the unique limit (as $[0,\infty)$ is Hausdorff), so $x\in A$. Thus $\overline{A}\cap [0,\infty)=A\cap [0,\infty)$.

Suppose now that $\infty_1\in \overline{A}$. Then either $\infty_1\in A$, or $A$ contains an unbounded subset of $[0,\infty)$.

In the latter case, since $X$ is connected, the retract $A$ is connected, so $A\supseteq [a,\infty)$ for some $a$. Therefore the sequence $2n+1$ eventually lies in $A$, so since $2n+1$ has a unique limit of $\infty_1$, we again have $\infty_1\in A$.

The same argument applies to $\infty_2$ via the sequence $2n$, and so the retract $A$ is closed.

Thus $X$ is RC.