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?


  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.
  • $\begingroup$ I added the group-action tag, since the proof turned out to be based upon methods from this topic $\endgroup$
    – Ulli
    Mar 13, 2022 at 9:57

3 Answers 3


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 $G$.

  • 1
    $\begingroup$ Wow, that's really convincing! I didn't know this theorem of Neumann. I think it's a very interesting proof: You apply induction by squaring the size of $B$! Haha! In fact, your proof shows a little bit more, namely that $B$ can also be infinite, as long as $|B| < |X|$. Thank you very much for that! $\endgroup$
    – Ulli
    Mar 8, 2022 at 9:37
  • 1
    $\begingroup$ ... and it's also a striking example for a proof by induction, which goes through only by strengthening the claim, as I originally wanted to show that there exists $g\in G$ such that $gA\cap A = \emptyset$. $\endgroup$
    – Ulli
    Mar 8, 2022 at 9:52
  • $\begingroup$ Yeah, the way it uses induction is indeed striking. Note that the stronger claim is trivially seen to be equivalent to the original claim: if you know $g(A\cup B)\cap (A\cup B)=\emptyset$ then $gA\cap B=\emptyset$. But the stronger claim is nevertheless easier to prove by induction since it uses induction only on $A$ while using different sets $B$ in each step of the induction. $\endgroup$ Mar 8, 2022 at 15:30
  • $\begingroup$ Very nice! This is the sort of argument I was trying to come up with in the first few drafts of my answer, but I couldn't get all the details to work out. A very well-deserved +1 to you. $\endgroup$ Mar 8, 2022 at 23:24

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.

  • $\begingroup$ @Ulli: Yes, it should; I'll fix that momentarily. And no worries about the delay, I myself am very slow to check answers to my questions. $\endgroup$ Mar 4, 2022 at 19:47
  • $\begingroup$ I eventually had the time for a closer look into your proof. Again a minor remark: I think the topology you defined on $G$ is the initial topology rather than the final topology. But now a serious remark: (continuing in next comment due to length restriction) $\endgroup$
    – Ulli
    Mar 6, 2022 at 9:14
  • $\begingroup$ $\{\{g:gy\in U\}:y\in X,U\text{ open in }X\}$ is only a subbase for this topology. This implies that in your central argument 3., you only have $$ \emptyset \neq \bigcap_{i=1}^n \{g:gy_i\in U_i\}\subseteq\mathrm{Stab}(x)$$ for $y_1, \dots , y_n \in X, U_1, \dots , U_n$ non-empty, open in $X$. The $U_i$'s are certainly not the problem, since their intersection is still cofinite, hence non-empty. But I don't see, at least not at first glance, how to repair the problem with multiple $y_i$? $\endgroup$
    – Ulli
    Mar 6, 2022 at 9:15
  • $\begingroup$ Apparently, it doesn't help to consider the initial topology for just one fixed $x$ rather than for all $x \in X$, since then right translations need no longer be continuous, although left translations are still homeomorphic, and $\mathrm{Stab}(x)$ is closed for this fixed $x$. $\endgroup$
    – Ulli
    Mar 7, 2022 at 9:26
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    $\begingroup$ I would still like to thank you for your efforts. Perhaps you will be able to repair it somehow? Anyway, I think it's a very nice application of topological language to combinatorics! $\endgroup$
    – Ulli
    Mar 8, 2022 at 9:41

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.


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