When are fixed point sets in $T_1$ spaces always closed?

Question

Let $X$ be a topological space, and say that $X$ satisfies the closed fixed point set property if every continuous self-map $f:X\to X$ has fixed point set $\operatorname{Fix}(f)=\{x\in X\mid f(x)=x\}$ which is closed in $X$.

There is a cute result which says that Hausdorff (i.e. $T_2$) spaces satisfy this property. The proof is pretty straight forward (it is an exercise in a topology course I was teaching).

A space $X$ is $T_1$ if for all pairs $x, y\in X$ there exists a neighbourhood $N_x$ of $x$ that does not contain $y$. Hausdorff spaces satisfy this condition. I am currently considering a space which is $T_1$ but not Hausdorff, and would like to prove that this space satisfies the closed fixed point set property. So I'm looking for inspiration:

What properties $P$ are there so that a $T_1$ space which additionally satisfies $P$

• will satisfy the closed fixed point set property, and
• will not necessarily be Hausdorff.

Answer 1

Two simple observations: If $\mathbf{X}$ has the CFPSP, then $\mathbf{X}$ is $T_1$ (the constant functions $x \mapsto a$ show that every singleton is a fixed point set). The space $\mathbf{X} \times \mathbf{X}$ has the CFPSP iff $\mathbf{X}$ is Hausdorff, as witnessed by the maps $(x,y) \mapsto (x,x)$ showing that the diagonal is a fixed point set.

So anything that would ensure that a $T_1$-space has the CFPSP would need to behave a bit odd wrt to products. The pattern is same as the discussed for the $T_{\mathrm{rc}}$-property in this question:

"All retracts are closed" as separation axiom

The $T_{\mathrm{rc}}$-property is that all retracts of the space are closed, and clearly any retract is a fixed point set. Thus, the CFPSP implies being $T_{\mathrm{rc}}$.