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I think there aren't any discrete paradoxical subsets in $\mathbb{R}^2$ (any isometry that mapped a discrete subset into itself would have to either be a glide-reflection, a translation or a rotation by $2\pi/n$, and hence the subgroup of the isometry group on $\mathbb{R}^2$ generated by isometries that map a discrete set into itself would have only translations and glide-reflections as elements of infinite order and thus no free subsemigroup of rank 2, hence the subgroup would be supramenable).

Say that a subset in a metric space is uniformly discrete if there is an $r>0$ such that every pair of points has distance greater than $r$.

Are there any uniformly discrete paradoxical subsets in $\mathbb{R}^3$?

By definition, given a group $G$ acting on a set $X$, a subset $E\subseteq X$ is $G$-paradoxical if $E\neq\emptyset$ and there are $n,m$ and pairwise disjoint subsets $U_1,\dots,U_{n+m}$ of $E$ such that $E=\bigcup_{i=1}^ng_iU_i=\bigcup_{i=n+1}^{n+m}g_iU_i$. That is, one can divide $E$ into pieces that can be reassembled into $E$ twice over. In the above question, $G$ is meant to be the isometry group of the Euclidean space.

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    $\begingroup$ I think it's worth stating the definition of "paradoxical" here. We are talking about the Banach-Tarski sort of thing, right? $\endgroup$
    – Johan Wästlund
    Commented Jul 14, 2013 at 16:22
  • $\begingroup$ Right. $E\subseteq X$ is $G$-paradoxical, where $G$ is a group of actions on $X$, provided that there are pairwise disjoint subsets $U_1,...,U_n,U_{n+1},...,V_{n+m}$ of $E$ and $g_i\in G$ such that $\bigcup_{i=1}^n g_i U_i = \bigcup_{i=n+1}^{n+m} g_i U_i = E$. I.e., you can divide $E$ into pieces that can be reassembled into $E$ twice over. $\endgroup$
    – Alexander Pruss
    Commented Jul 14, 2013 at 18:51
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    $\begingroup$ By the way, the reason we need <i>uniformly</i> discrete is that we can get a discrete set by mimicking the construction of the Sierpinski-Mazurkiewicz set in the xy-coordinates but adding a translation in the z-direction to the rotation. More precisely, write $\mathbb R^3 = \mathbb C \times \mathbb R$. Take the isometries $\rho(z,y) = (e^i z,y+1)$ and $\tau(z,y)=(z+1,y)$. Let $E$ be the orbit of $(0,0)$ under the semigroup generated by $\rho$ and $\tau$. Then $E$ is discrete. Let $U_1 = \rho E$ and $U_2 = \tau E$. Then $E = \rho^{-1} U_1 = \tau^{-1} U_2$ and $U_1$ and $U_2$ are disjoint. $\endgroup$
    – Alexander Pruss
    Commented Jul 14, 2013 at 19:05

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I think it is not possible for a uniformly discrete set $E$ in any ${\bf R}^d$ to be paradoxical, because one can create an invariant (or almost invariant) mean on such a set. Indeed, for any $\varepsilon>0$ and $C>0$, one can use the pigeonhole principle to find a large radius $R$ such that $|E \cap (B(0,R+C) \backslash B(0,R-C))| \leq \varepsilon |E \cap B(0,R)|$, where $|A|$ denotes the cardinality of $A$, because the uniformly discrete nature of $E$ forces $|E \cap B(0,R)|$ to grow at most polynomially. This makes the probability measure $\mu(A) := |A \cap E \cap B(0,R)| / |E \cap B(0,R)|$ approximately invariant (up to error $\varepsilon$) with respect to isometries $T$ that involve a translation by at most $C$, and which map $A$ to a subset of $E$, in the sense that $|\mu(TA)-\mu(A)| \leq \varepsilon$ for such sets. This should be inconsistent with any putative paradoxical decomposition if one chooses $C,\varepsilon$ appropriately with respect to this decomposition. (One could also create a genuinely invariant mean by sending $R \to \infty$ and taking an ultralimit or by using a suitable compactness theorem, although this does not seem necessary for this particular application.)

(There should also be some way to use the Tits alternative or something related to this alternative (e.g. Jordan's theorem, Gromov's theorem on groups of polynomial growth, or the Solovay-Kitaev argument) to show the stronger assertion that the semigroup of isometries that map a uniformly discrete set to a subset should be "virtually nilpotent" (or maybe even virtually abelian) in some sense, although I was not able to make any of these fancier tools work here (largely because one only has a semigroup of isometries to play with rather than a group), whereas the more elementary counting argument above seems to already suffice for the purposes of answering the stated question.)

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    $\begingroup$ @Terry: There is an analogue of Gromov polynomial growth theorem for semigroups with cancelation ($xy=xz\to y=z$, $yx=zx\to y=z$), see Grigorchuk, R. I. Semigroups with cancellations of degree growth. Mat. Zametki 43 (1988), no. 3, 305--319, 428; translation in Math. Notes 43 (1988), no. 3-4, 175–183 $\endgroup$
    – user6976
    Commented Jul 15, 2013 at 6:10
  • $\begingroup$ Great! So one can probably classify all the possible isometry semigroups of uniformly discrete sets using this, though there is still the problem of ensuring that such semigroups are finitely generated... of course to avoid degeneracies one should assume that the uniformly discrete set is not contained in a hyperplane. It might be fun to work out exactly what the answer is here, though I don't think I'll have the time to think about it myself... $\endgroup$
    – Terry Tao
    Commented Jul 15, 2013 at 18:28

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