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

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.

• I think it's worth stating the definition of "paradoxical" here. We are talking about the Banach-Tarski sort of thing, right? Jul 14, 2013 at 16:22
• 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. Jul 14, 2013 at 18:51
• 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. Jul 14, 2013 at 19:05

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.)
• @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