Let $(X,\tau)$ be a topological space. Let us call $x,y\in X$ swappable if there is $f:X\to X$ continuous such that $f(x)=y$ and $f(y)=x$. This relation is obviously reflexive and symmetric, but not necessarily transitive.

Moreover, we call $(X,\tau)$ rigid if the identity is the only homeomorphism from $X$ to itself.

Is there a rigid space $(X,\tau)$ with $|X| > \aleph_0$ such that for all $x,y\in X$ we have that $x,y $ are swappable?

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The answer is yes by a 1951 result of Miroslav Katětov, who proved that there is an (uncountable) rigid totally disconnected compact space. Any such space $X$ is rigid and has the property that any two points are swappable by continuous functions, since if $x\neq y$, we can find an open separation $X=U\sqcup V$ with $x\in U$ and $y\in V$, and then let $f$ be constant on each piece with value $y$ or $x$, respectively. This is a continuous function swapping $x$ and $y$.

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  • $\begingroup$ I'm not sure if there would be an easier example or not. $\endgroup$ – Joel David Hamkins Apr 10 '17 at 16:53
  • $\begingroup$ If you don't insist on compactness, then it is relatively easy to show that there are zero-dimensional uncountable rigid spaces (for example, you can use transfinite recursion to construct a rigid subset of the reals with cardinality of the continuum). Your argument still works for such spaces. $\endgroup$ – Will Brian Apr 10 '17 at 17:37
  • $\begingroup$ Thanks! Indeed, any rigid subspace of the reals will be totally swappable, since rigidity ensures that it will be totally disconnected, and then the locally constant maps I mention will perform the desired swaps. $\endgroup$ – Joel David Hamkins Apr 10 '17 at 20:08
  • $\begingroup$ Also every path connected metrizable rigid space is swappable. Such examples are not hard to construct. $\endgroup$ – Moishe Kohan Apr 12 '17 at 23:34

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