Separation of finite sets in homogeneous spaces by homeomorphisms

Question

Call a topological space $X$ flexible, if for each finite set $A \subset X$ there exists a homeomorphism $f: X \rightarrow X$ such that $A \cap f(A) = \emptyset$. (Certainly not a good name, and by far not standard, but for the purpose of this question it might suffice.)

Let $X$ be an infinite, T2, homogeneous topological space (i.e., for all $x, y \in X$ there is a homeomorphism $f: X \rightarrow X$ such that $f(x) = y$).

Is $X$ flexible?

Notes

1. Of course, a flexible, non-empty space is infinite and must provide a certain amount of homeomorphisms. For instance, if it is rigid (i.e, the identity is the only homeomorphism), it can't be flexible. Therefore, it makes sense to restrict to homogeneous spaces.
2. Considering "typical" homogeneous spaces as $\mathbb{R}^n$, the answer seems so obviously to be "yes". However, I couldn't prove the above in general, not even for two-element sets $A$.
3. It is not difficult to prove that $X$ is flexible, if $X$ is infinite and at least one of the following conditions holds:
   a) $X$ is the underlying space of a topological group
   b) $X$ is a product with (at least) one factor flexible (this might indicate how weak flexible is)
   c) $X$ is n-homogeneous for all $n \in \mathbb{N}$
   d) $X$ is strongly locally homogeneous, T2 and contains no isolated points (in particular, if $X$ is a manifold)
   e) $X$ is uniquely homogeneous
   (The notations in c), d) and e) are the standard ones, see for instance here.)
4. My assumption is that the answer is "yes". Perhaps, the proof is more combinatorial (eg. Ramsey theory) rather than topological? Or even with some trivial argument, which I just didn't notice?
5. The pseudo-arc is a standard example of a homogeneous, not strongly locally homogeneous, space. I'm not very familiar with it. Embarrassingly, I don't know, whether it is flexible or not. Perhaps, it provides a counter-example?
6. [edit: I just deleted 6. (and my two related comments below), since after some further consideration it no longer make sense.]
7. Perhaps the T2 requirement in the prerequisite is superfluous? I also don't know of a non-T2 counterexample.

Answer 1

This is in fact automatic for purely combinatorial reasons. By a theorem of Neumann, a group cannot be covered by finitely many cosets of subgroups of infinite index (see https://mathoverflow.net/questions/17396/can-a-group-be-a-finite-union-of-left-cosets-of-infinite-index-subgroups). As a corollary, if a group $G$ acts transitively on an infinite set $X$, then for any finite $A,B\subset X$ there exists $g\in G$ such that $gA\cap B=\emptyset$. Indeed, if no such $g$ existed, that would mean exactly that $G$ is covered by the finitely cosets of the stabilizer subgroups of each element of $A$ which map them to each element of $B$. These stabilizer subgroups all have infinite index because $G$ acts transitively and $X$ is infinite, so this is impossible.

Here is a direct proof of that corollary (this is just what you get by translating Neumann's proof into the language of group actions). We use induction on $|A|$, the base case $|A|=0$ being trivial. Now suppose $|A|>0$ and fix $a\in A$. Pick $h\in G$ such that $h(a)\not\in B$, and also for each $b\in B$ pick $g_b\in G$ such that $g_b(a)=b$. Now apply the induction hypothesis to the sets $A'=A\setminus\{a\}$ and $B'=B\cup\bigcup_{b\in B}g_bh^{-1}B$ to obtain $g\in G$ such that $gA'\cap B'=\emptyset$. If $g(a)\not\in B$ then we have $gA\cap B=\emptyset$ and are done. If $g(a)\in B$, let $b=g(a)$ and observe that $hg_b^{-1}g(a)=h(a)\not\in B$ and also for each $a'\in A'$ we have $hg_b^{-1}g(a')\not\in B$ since $g(a')\not\in g_bh^{-1}B$. Thus $hg_b^{-1}gA\cap B=\emptyset$ and $hg_b^{-1}g$ is our desired element of $G$.

Answer 2

This answer currently has a fatal flaw.

Your supposition 4 is probably correct, in that the proof runs through infinitary combinatorics. I use topology below, but purely instrumentally; with a little more cleverness, it probably could be eliminated. This also implies your supposition 7. Now, on to the proof.

Let $X$ be an set on which $G$ acts transitively and (w/oLoG) faithfully. (For example, let $X$ be an infinite topological space and $G=\mathrm{Homeo}(X)$.) For any finite $A,B\subseteq X$, I claim that either there exists $g\in G$ such that $gA\cap B=\emptyset$, or $X$ is finite. To reduce to your question, take $A=B$ in an infinite $X$.

Fix $x\in X$ and pick any $\{g_a\}_{a\in A\cup B}$ such that $g_ax=a$ (for all $a\in A\cup B$). Note that $$\{g:gA\cap B=\emptyset\}=\bigcap_{a\in A,b\in B}{\{g:ga\neq b\}}=\bigcap_{a\in A,b\in B}{g_b(G\setminus\mathrm{Stab}(x))g_a^{-1}}\tag{1}$$

Now, give $X$ the coarsest topology such that $X$ is $T_1$: the cofinite topology. (If $X$ is finite, this is discrete.) Place on $G$ the initial topology induced by the action on each point of $X$; that is, $$\{\{g:gy\in U\}:y\in X,U\text{ open in }X\}$$ is a basis (flaw: this should be a subbasis) for this topology. $G$ is not necessarily a topological group under this topology, but the following nice properties still hold:

1. For any $l,r\in G$, the maps $p\mapsto lp$ and $p\mapsto pr$ are homeomorphisms. This follows from the explicit $G$-invariance of our basis.
2. The set $\mathrm{Stab}(x)$ is closed. This follows immediately from the definition.
3. If the set $\mathrm{Stab}(x)$ is not nowhere dense in $G$ (i.e. has nontrivial interior), then $X$ is finite.

To see the last claim, let $U$ be open and $y\in X$ such that $$\{g:gy\in U\}\subseteq\mathrm{Stab}(X)$$ Let $K=X\setminus U$, pick any $u\in U$, and, for each $k\in K$, choose $g_k\in G$ such that $g_ku=k$. I claim that $$G=\mathrm{Stab}(x)\cup\bigcup_{k\in K}{g_k\mathrm{Stab}(x)}$$ For, if $g\in G$, then either

1. $gy\in U$, so that $g\in\mathrm{Stab}(x)$, or
2. $gy\in K$, so that $g_{gy}^{-1}gy=u\in U$ and $g_{gy}^{-1}g\in\mathrm{Stab}(x)$.

Thus $$|X|=|G/\mathrm{Stab}(x)|\leq|K|+1<\aleph_0$$ since $U$ is cofinite.

Now, $G$ with this topology need not be Baire. Nevertheless, in any topological space, a finite intersection of dense open sets is dense. Thus (1) is dense, and, in particular, nonempty.

Answer 3

To complete the picture I would like to add the following lemma, which is a trivial consequence of Eric Wofsey's proof:

Let the group $G$ be acting on the set $X$. Then it holds:
For all $A, B$ finite subsets of $X$ there is a $g \in G$ such that $gA \cap B = \emptyset$, if and only if each orbit is infinite.

( "$\Rightarrow$" is obvious. "$\Leftarrow$": Follows by the same proof as above: $h$ can be picked, since the orbit of $a$ is infinite. Instead of $b \in B$ one has to consider $b \in B \cap \mathrm{orbit}(a)$.)

As a corollary we have that a topological space is flexible (as defined at the top), if and only if each homogeneity component is infinite.