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I want to follow the discussion from here concerning about the strength of the separation "all retract subspaces are closed".

(A retract subspace of a topological space $X$ is a subspace $A$ where there exists continuous $f: X\to A$ such that $f|_A = \mathrm{id}_A$.)

Write $\mathrm{KC}$ as "all compact subsets are closed", and $\mathrm{RC}$ as "all retract subspaces are closed". We have $T_2\Rightarrow \mathrm{KC}$, $T_2\Rightarrow \mathrm{RC}$ and neither of the reversed implications. The cocountable topology over $\mathbb{R}$ is a $\mathrm{KC}$ example not $\mathrm{RC}$: $f(x) = |x|$ is a continuous function from $\mathbb{R}$ with cocountable topology to its subspace $[0,+\infty)$ since the preimages of countable sets are countable, but $[0,+\infty)$ is certainly not closed.

So I would like to ask if $\mathrm{RC}$ implies $\mathrm{KC}$, because the only $\mathrm{RC}$ non-Hausdorff spaces I know are those compact $\mathrm{KC}$ spaces (note that compact $\mathrm{KC}$ implies $\mathrm{RC}$). Any help appreciated.

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RC does not imply KC: in this paper Banakh and Stelmakh construct a semi-Hausdorff countable Brown space $X$ which is strongly rigid (and hence $X$ has $RC$) and contains a non-closed compact subset (so, $X$ fails to have $KC$).

This example also shows that $KC$ does not follow from the semi-Hausdorff property, which is intermediate between $T_1$ and $T_2$.

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  • $\begingroup$ Genius! If possible, please allow me to ask just of curiosity: is that space is $US$ (unique-sequence-limit, every sequence has at most one limit)? If no, is it again possible to construct a strongly rigid space that is $T_1$ but not $US$? $\endgroup$
    – Jianing Song
    Commented Dec 8, 2022 at 20:11
  • $\begingroup$ @JianingSong That space is not US: it has two points $a,b$ which are limits of the same sequence $(s_a(3n))_{n\in\mathbb N}$ (in the notations from that paper). Nonetheless those two points cannot be permuted by a homeomorphism (because of different ultrafilters assigned to those points). $\endgroup$
    – Taras Banakh
    Commented Dec 8, 2022 at 20:22
  • $\begingroup$ @JianingSong To obtain a strongly rigid US-space with non-closed compact subset, one should destroy the convergence of the sequence $(s_a(3n))_{n\in\mathbb N}$ and its subsequences to $b$ and this can be easily done replacing the Frechet filter in the neighborhood base of $b$ by an ultrafilter. $\endgroup$
    – Taras Banakh
    Commented Dec 8, 2022 at 20:33
  • $\begingroup$ @JianingSong Such a spaces can be not only US, but also k-metrizable (= its k-coreflection is metrizable). I have aded this condition to Theorem 1.4 in the paper researchgate.net/publication/… $\endgroup$
    – Taras Banakh
    Commented Dec 10, 2022 at 6:21
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    $\begingroup$ @StevenClontz Indeed, our paper with Stelmakh does not contain such an example. However, I hope it can be easily produced by a suitable modification of the proof of Theorem 1.5. The only differnce is in the definition of neighborhoods of the point $b$ inthe topology $\mathcal T_Y$ on page 188 (in the Topology Proc. paper). Those neighborhoods should be define by analogy with the neighborhoods of the point $a$, so that the sequence $(s_a(3n))_{n=1}^\infty$ will converge to $a$ and $b$, thus witnessing that the space $(Y,\mathcal T_Y)$ is not US. $\endgroup$
    – Taras Banakh
    Commented Aug 23, 2023 at 6:52
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EDIT: This answer relied on an accepted answer elsewhere that has now been updated to remove an oversight. See my note below.

First I need to prove that the Arens-Fort space $X$ is not compactly generated. To do this, I'll show there exists a non-closed set which has closed intersection with every compact. This is very easy: since this space is not discrete, it has an infinite non-closed set $C$. Since this space is anticompact, all compacts are finite, and thus $C$ intersected with every compact is finite. Finally, since the space is Hausdorff, $C$ intersected with every compact is closed.

Theorem 5 of Between $T_1$ and $T_2$ shows us that $X$'s one-point compactification $X^+$ is not $KC$ since $X$ is not compactly generated.

Finally, $X^+$ is $RC$, since it fits the requirements of Paul Fabel's answer here.

Thus $RC$ does not imply $KC$, even for compact spaces.

EDIT: $X^+$ is not $RC$; Fabel's answer has now been updated to require $X$ to be compactly generated, which our $X$ is not.

To see this directly, let $0$ be the non-isolated point of $X$; let $\infty$ be the new point in $X^+$. Then $A=X^+\setminus\{0\}$ is non-closed; we claim it is a retract. Let $f:X^+\to A$ be defined by $f(0)=\infty$ and $f(x)=x$ otherwise. Note $f\upharpoonright A=id_A$ and $A$ is open, so $f$ is continuous at each point of $A$. Let $U$ be a neighborhood of $f(0)=\infty$. Since $X$ is anticompact, $U$ is cofinite. Thus $f^\leftarrow[U]$ is also cofinite, and thus open, proving $f$ is continuous at $0$. This completes the proof of our claim.

And just to round this out, $A$ also provides an explicit example of why $X^+$ is not $KC$: it's compact as every neighborhood of $\infty$ is co-finite.

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    $\begingroup$ Something is fishy here: upon closer inspection, Fabel's answer proves RC by asserting it is compact and KC., which seems to contradict the result I cite from Between T1 and T2. $\endgroup$
    – Steven Clontz
    Commented Nov 25, 2022 at 1:01
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    $\begingroup$ Thanks for your work! What a pity for the "example", of course :) $\endgroup$
    – Jianing Song
    Commented Nov 27, 2022 at 18:55
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    $\begingroup$ As noted, thanks to Steven's inquiry in the comments of the original question, I noticed the mistake in my original answer and corrected it yesterday, and, serendipitously, also constructed a non-example similar to the one at hand. It's nice to revisit the question, thanks to Steven and everyone else for exposing the mistake, and helping to set things right. $\endgroup$
    – Paul Fabel
    Commented Nov 28, 2022 at 15:21
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    $\begingroup$ @StevenClontz: I fooled around with (perhaps) somewhat similar ideas in several versions of my deleted answer below, which you probably can't see: Start out with a space with few or no retracts ("Cook continuum"), add a point $\infty$, and give the extra point only big neighborhoods (an extreme case would be to make the whole space the only neighborhood). Then there should be few retracts, but many compact sets (because just having $\infty$ in the set really helps), so maybe we can get RC, but not KC. But one has to carefully fine tune things at $\infty$, and I gave up after too many mistakes. $\endgroup$
    – Christian Remling
    Commented Nov 28, 2022 at 18:39
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    $\begingroup$ @StevenClontz: In other words, I tried to get a counterexample as a one-point compactification of a space with no nontrivial retracts. If your bet is correct, it would refute this strategy in general (since the first space is metric). $\endgroup$
    – Christian Remling
    Commented Nov 28, 2022 at 18:44

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