Let $X$ be a metric space and $G$ a group of homeomorphisms. I have to prove the following equivalences:

1) $G$ acts properly discontinuous on $X$ (i.e. any $G$-orbit is discrete and any stabilizer is finite);

2) For any compact $K$ in $X$, $T(K) \cap K \neq \emptyset$ for finitely many $T \in G$;

3) Any point $x \in X$ has a neighborhood $V$ such that $T(V) \cap V \neq \emptyset$ implies $T(x)=x$.

I showed $2) \implies 1)$ as follows: if we assume by contradiction that some orbit is not discrete, the it admits a limit point $z_0$, thus any neighborhood of $z_0$ meets infinitely many of its images under $G$. So, if I take any compact subset $K$ containing $z_0$ and I pass to some open neighborhood contained in $K$, 1) must follow in order to avoid contradictions with the hypotheses in 2).

Moreover, I showed $1) \implies 3)$ in this way: just consider a point $x \in X$. Since its orbit is discrete, we will always find a ball $B_\epsilon(x)$ containing only $x$ (and no other points of the orbit). Just take $V \subseteq B_{\frac{\epsilon}{2}}$. Then the condition $T(V) \cap V \neq \emptyset$ implies necessarily that $T(x)=x$.

I'm trying to prove by contradiction $3) \implies 2)$, but I have no clue. Can you help me, please?


1 Answer 1


It's no wonder you're having trouble proving that (3) implies (2), because it's not true. Here's a counterexample. Let $X=\mathbb R^2\smallsetminus\{(0,0)\}$, and let $G$ be the group of homeomorphisms of $X$ defined by $T_n(x,y) = (2^n x, 2^{-n} y)$ for $n\in\mathbb Z$. This satisfies properties (1) and (3), but not (2).

To see that (2) fails, consider the compact subset $K= \{(x,y): \max(|x|,|y|)=1\} \subseteq X$. (It's a square of side 2 centered at the origin.) For every positive $n$, the element $(1,2^{-n})$ lies in $K$, and it's also equal to $T_n(2^{-n},1)\in T_n(K)$. Thus $T_n(K)\cap K \ne \emptyset$ for infinitely many $n$.

For a discussion of the different definitions of "properly discontinuous," see my answer to this question.


It's also not true that (3) implies (1). A counterexample is given by letting $X = \mathbb R$, and $G$ be the group of homeomorphisms given by $T_n(x) = 2^n x$ for $n\in \mathbb Z$. This satisfies (3), but it doesn't satisfy (1) because the stabilizer of the origin is not finite.

To address your attempted proofs:

There's a flaw in your proof that (2) $\implies$ (1): there might not be any compact set that contains a neighborhood of $z_0$. (This would be true if the metric space were locally compact, but that's not part of the hypothesis.) You have to be more careful about choosing the compact set $K$. Other than that, I think this basic outline can be turned into a proof.

There's also a gap in your proof that (1) $\implies$ (3): I don't see why $T(V)\cap V \ne \emptyset$ implies that $T(x)=x$. If we assume that $T(V)\cap V \ne\emptyset$, this just means there is some $y\in B_{\epsilon/3}(x)$ such that $T^{-1}(y)$ is also in $B_{\epsilon/3}(x)$. You'd like to conclude from this that $T(x) \in B_\epsilon(x)$, but I don't see how to conclude this unless all the homeomorphisms in $G$ preserve the metric, which is not part of the hypothesis. To tell the truth, I'm not sure at the moment whether (1) $\implies$ (3) or not.

Are you sure you copied the question correctly? If you add the hypothesis that $G$ consists entirely of isometries, then I think it's true that all three properties are equivalent.

  • $\begingroup$ What about the proofs I gave? $\endgroup$ Commented Jun 4, 2018 at 23:25
  • $\begingroup$ @TheWanderer: See my addition above. $\endgroup$
    – Jack Lee
    Commented Jun 5, 2018 at 2:37
  • $\begingroup$ I'm doing the proof adding the necessary hypotheses you suggest. Now, with these new hypotheses, I want to show $3) \implies 2)$. I proceed by contradiction. Assume there exists a compact $K$ such that $T(K) \cap K \neq \emptyset$ for infinitely many $T$. Hence I find a sequence $z_k=T_j(x_l) \in K$. I consider $x_l$ converging to $x$ (up to passing to a subsequence). So, any neighborhood of $x$ contains infinitely many $x_l$'s. How may I conclude my argument in order to find a contradiction with 3)? $\endgroup$ Commented Jun 11, 2018 at 10:23

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