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The starting point of this question is the fact that any retract of a $T_2$-space is closed.

Let's say a topological space $(X,\tau)$ is $T_{\textrm{rc}}$ if all retracts of $X$ are closed.

All $T_{\textrm{rc}}$-spaces are $T_1$, because singletons are always retracts, and a space is $T_1$ if and only if all singleton subsets are closed. So we have $T_2 \implies T_{\textrm{rc}}\implies T_1$.

The second implication is not an equivalence: Consider $(\omega,\mathcal{P}_{\text{cf}}(\omega))$ where $\mathcal{P}_{\text{cf}}(\omega) = \{\emptyset\}\cup \{U\subseteq \omega: \omega\setminus U \text{ is finite}\}$. Let $S$ be the set of even numbers and $r:\omega \to S$ be defined by $2n\mapsto 2n$ and $2n+1 \mapsto 2n$ for all $n\in\omega$. Then $S$ is a retract of $(\omega,\mathcal{P}_{\text{cf}}(\omega))$ but it is not closed. So the implication $T_{\textrm{rc}}\implies T_1$ is not an equivalence.

Question: Do we have $T_2 \Leftrightarrow T_{\text{rc}}$?

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No. Let $X$ be a compactly generated Hausdorff space which fails to be locally compact at precisely one point. Now take the Alexandroff compactification of $X$, adding exactly one new point, whose neighborhoods have compact complement in $X$. The new space is not Hausdorff, but has the property that it is compact, and each compact subspace is closed. In general the image of a compact space is compact, and hence each retract of the new space is a closed subspace of itself.

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    $\begingroup$ How do you know each compact subspace is closed without the Hausdorff property? $\endgroup$ Nov 26, 2022 at 17:47
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    $\begingroup$ Thanks Steven, nice catch! We should also assume $X$ is compactly generated, i.e. $A$ is closed in $X$ iff $A \cap C$ is closed in $X$ for each compact $C \subset X$. $\endgroup$
    – Paul Fabel
    Nov 27, 2022 at 16:16
  • $\begingroup$ To see why we must also assume $X$ is compactly generated, let $X$ be the plane, but refine the usual topology so that each countable set is closed. Then no infinite subset of $X$ is compact. However if $Y$ is the Alexandroff compactifictation then each subset of $Y$ which contains $\infty$ is compact. In particular $Y \setminus 0$ is compact but not closed in $Y$. $\endgroup$
    – Paul Fabel
    Nov 27, 2022 at 16:43
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    $\begingroup$ A couple other notes for posterity: local compactness need only fail at at least one point to break T2 in the compactification, not precisely. The proof that compactly generated KC spaces produce KC compactifications in general is Thm 5 of jstor.org/stable/2316017 We investigated some of this at mathoverflow.net/questions/434451 $\endgroup$ Nov 28, 2022 at 17:45
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Let $X$ be the rationals with their subspace topology, and $X^+=X\cup\{\infty\}$ be its one-point compactification.

Because $X$ is not locally compact, $X^+$ is not $T_2$.

The space $X^+$ has the property $T_{kc}$, that is, all compact subsets are closed.

We now show that all compact $T_{kc}$ spaces $K$ are $T_{rc}$. Let $A$ be a retract of $K$. Then $A$ is a continuous image of $K$, and therefore compact. By $T_{kc}$, $A$ is closed.

Therefore $X^+$ is a $T_{rc}$ space which is not $T_2$.

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This is not a complete answer, just a possibly useful observation.

Theorem: If the property $T_\mathrm{rc}$ is preserved by finite products then any $T_\mathrm{rc}$ space is $T_2$.

Proof. Suppose $T_\mathrm{rc}$ is preserved by finite products and $X$ is $T_\mathrm{rc}$. By assumption $X \times X$ is also $T_\mathrm{rc}$, therefore its diagonal $\Delta_X = \{(x,y) \in X \times X \mid x = y\}$ is closed, as it is a retract of $X \times X$ by the map $(x,y) \mapsto (x,x)$. But a space is $T_2$ iff its diagonal is closed, so $X$ is $T_2$.

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    $\begingroup$ You seem to have proved the right-to-left direction twice and not proved the converse. $\endgroup$ Nov 26, 2022 at 22:19
  • $\begingroup$ Oops, sorry about that, I deleted the fake news and alternative facts. $\endgroup$ Nov 27, 2022 at 17:57

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