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 $T\_2 \Leftrightarrow T\_{\text{rc}}$?



"All retracts are closed" as separation axiom
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.


