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The discrete, indiscrete, cofinite, cocountable, co-(any downwards closed collection of cardinals) topological spaces are part of the unique class of spaces such that all permutations of the space (bijections from the space to itself) are autohomeomorphisms. This is because their topologies may be defined purely in terms of cardinality, so everything is bijection-invariant. If we relax the requirement to just saying that all transpositions (i.e. permutations that just swap two elements) are homeomorphisms, do we get a broader class of spaces, or is this property equivalent to bijection-invariance?

Roselyn van Lauwe
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2 Answers2

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Let's call a topological space $X$ [weakly] anti-rigid, if every permutation [transposition] is a homeomorphism. And filter like, if whenever $U$ is a non-empty open subset of $X$, and $U \subset V \subset X$, then also $V$ is open.

Then we have: $X$ weakly anti-rigid, iff $X$ is filter like and ($X$ is T1 or $X$ is indiscrete).

Proof. $"\Rightarrow"$: If $X$ contains an isolated point, then $X$ is discrete, hence filter like and T1. So let's assume that $X$ contains no isolated points, and $|X| \ge 2$. Let $\emptyset \neq U$ be open, $y \in X$. Case 1: $y \in U$: then $U \cup \{y\} = U$ is open. Case 2: $y \notin U$. Choose an $x \in U$. Then the transposition $f: X \rightarrow X$ swapping $x$ and $y$ shows that $(U \setminus \{x\}) \cup \{y\} = f(U)$ is open. Hence $U \cup \{y\} = U \cup ((U \setminus \{x\}) \cup \{y\})$ is open. Thus, any subset $V$ such that $U \subset V$ is open. Now assume $X$ is not T1: then the intersection of all non-empty open sets is non-empty, hence equals $X$, hence X is indiscrete.
$"\Leftarrow"$: W.l.o.g. $X$ T1. Let $x, y \in X$, $f: X \rightarrow X$ be the transposition swapping $x$ and $y$. We have to show that $f$ is continuous: Let $U$ be open in $X$. Case 1: $x, y \in U$ or $x, y \notin U$: Then $f^{-1} (U) = U$. Case 2: $x \in U, y \notin U$: Then, by T1, $U \setminus \{x\}$ is open, hence $f^{-1}(U) = (U \setminus \{x\}) \cup \{y\}$ is open. Case 3: $x \notin U, y \in U$: analogous.

Now let $\mathfrak{F}$ be a free ultrafilter on $\mathbb N$. Then $\mathfrak{F} \cup \{\emptyset \}$ is filter like topology on $\mathbb N$ and it is $T1$, hence weakly anti-rigid, but not anti-rigid: Let $\mathbb N = A \cup B, A \cap B = \emptyset$, $A, B$ be infinite. W.l.o.g. $A \in \mathfrak{F}$. There is a bijective map $f: \mathbb N \rightarrow \mathbb N$, such that $f(B) = A$. Hence, $f$ is not continuous.

Remark: it is easy to see that weakly anti-rigid Hausdorff spaces are discrete.

Ulli
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  • Wonderful! Thank you for the insightful answer :) – Roselyn van Lauwe Jan 02 '23 at 18:17
  • You can streamline the $\Rightarrow$ argument as follows: Suppose $X$ is weakly anti-rigid. Let $U$ be a non-empty open set, with $x\in U$. Then for any $y\notin U$, swapping $x$ and $y$ and unioning with $U$, we find $U\cup {y}$ is open. By taking unions, it follows that for any $U\subseteq V\subseteq X$, $V$ is open. So $X$ is filter-like. Now suppose $X$ is not indiscrete. Then there is some proper non-empty open set $U$. Let $x\in U$ and $y\notin U$. For any $x'\neq y'$ in $X$, we can swap $x'$ and $x$ and swap $y'$ and $y$ to find a open set containing $x'$ but not $y'$. So $X$ is T1. – Alex Kruckman Jan 02 '23 at 20:02
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The example in Ulli's answer uses a non-principal ultrafilter on $\mathbb{N}$. I'll supplement that nice answer with a more elementary (and choice-free!) example.

Consider the following topology on $\mathbb{Z}$: $U\subseteq \mathbb{Z}$ is open if and only if $U = \varnothing$ or there exists $N\in \mathbb{Z}$ such that $m\in U$ for all $m\geq N$. Any transposition is an autohomeomorphism, but the permutation $n\mapsto -n$ is not, since the preimage of the open set $\{n\in \mathbb{Z}\mid n\geq 0\}$ is the non-open set $\{n\in \mathbb{Z}\mid n \leq 0\}$.

Alex Kruckman
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