"All retracts are closed" as separation axiom

Question

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 $T_2 \Leftrightarrow T_{\text{rc}}$?

Answer 1

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.

Answer 2

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 $T_2$.

Answer 3

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 $T_2$.