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Consider the finite cross $C$ (=union of line segments $\overline{(0, -1)(0, 1)}$ and $\overline{(-1, 0)(1, 0)}$) and the unit half-circle $H$. It is easy to see that we may pack continuum-many disjoint copies of $H$ into the plane $\mathbb{R}^2$ without overlapping, whereas for $C$ we may never pack uncountably many copies without overlapping. By "copy," here, I mean "copy up to rotation and translation," or more generally "image under an element of the affine special orthogonal group." The specific choice of transformation group to use isn't important, so long as it is reasonable.

For $S\subseteq\mathbb{R}^n$, let $\mathfrak{p}(S)$ be the supremum of the cardinalities of disjoint packings of copies of $S$ inside $\mathbb{R}^n$. In case the continuum hypothesis holds, either $\mathfrak{p}(S)$ is countable or $\mathfrak{p}(S)=2^{\aleph_0}$; however, in general this need not be true. For example, let $G\subseteq (\mathbb{R}, +)$ be a subgroup of index $\aleph_1$ and let $H$ be a coset of $G$; then the copies of $H$ are precisely the cosets of $G$, so $\mathfrak{p}(H)=\aleph_1$.

My question is: suppose $\neg CH$. Under what assumptions - both on the shape $S$ and the ambient axioms of set theory - can we conclude that $\mathfrak{p}(S)$ is either countable or continuum? For example:

Does ZFC prove that, for $S\subseteq\mathbb{R}^n$ compact, $\mathfrak{p}(S)$ is either countable or continuum?

I suspect the answer is "yes," but I don't see how to prove it.

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Suppose $K \subseteq \mathbb{R}^n$ is compact. Let $G$ be the group of isometries of $\mathbb{R^n}$. The compact-open topology on $G$ is defined by declaring the sets $W_{A. U} = \{f \in G : f[A] \subseteq U\}$ open, for each compact $A \subseteq \mathbb{R}^n$ and open $U \subseteq \mathbb{R}^n$. This is the smallest topology that makes the evaluation function $(g, x) \mapsto g(x)$ continuous. It can be shown that this makes $G$ a locally compact, Polish group. Let $d$ be a witnessing metric (Kechris' Descriptive set theory book gives a formula for this metric on page 60).

Let $S = \{f \in G : f[K] \cap K = \phi\}$. Then $S = W_{K, \mathbb{R}^n \backslash K}$ is open in $G$.

Claim: If $id \notin \text{cl}(S)$, then every family of pairwise disjoint copies of $K$ is countable.

Proof: Suppose $X \subseteq G$ is uncountable. Since $G$ is second countable, there exists $g \in X$ such that every neighbourhood of $g$ meets $X$ at uncountably many points. Let $W$ be an open set containing $id$ such that $W \cap \text{cl}(S) = \phi$. Pick $f \in X \cap gW$, $f \neq g$. Then, $(g^{-1}f)[K] \cap K \neq \phi$ hence also $f[K] \cap g[K] \neq \phi$.

Claim: Assume $id \in \text{cl}(S)$. There exists a binary tree $\langle B_{\sigma} : \sigma \in 2^{< \omega} \rangle$ of open sets with compact closure in $G$ such that for all $\sigma$, for $i = 0, 1$, $\text{cl}(B_{\sigma i}) \subseteq B_{\sigma}$ and for every $f_i \in B_{\sigma i}$, $f_0[K] \cap f_1[K] = \phi$. Hence, the family $\{f[K]: (\exists x \in 2^{\omega})(\forall n)(f \in B_{x \upharpoonright n}) \}$ is pairwise disjoint and has size continuum.

Proof: To construct such a tree, choose some $f \in B_{\sigma}$. Using the assumption that the isometries that map $K$ to a set disjoint with $f[K]$ accumulate near $f$, pick $g \in B_{\sigma}$ such that $f[K] \cap g[K] = \phi$. Let $B_{\sigma i}, i = 0, 1$ be open sets such that $f \in B_{\sigma0}, g \in B_{\sigma1}$, $\text{cl}(B_{\sigma i}) \subseteq B_{\sigma}$ and for every $h_i \in B_{\sigma i}$, $h_0[K] \cap h_1[K] = \phi$. This can be done by separating the compact sets $f[K], g[K]$ by open sets $U_f, U_g \subseteq \mathbb{R}^n$ and considering the disjoint neighbourhoods $W_{K, U_f}, W_{K, U_g}$ of $f, g$ respectively.

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    $\begingroup$ +1. Ashutosh, would it be possible for you to augment your solution with a little more explanation? I feel that I would very likely learn something valuable if you did... $\endgroup$ – Joel David Hamkins Oct 20 '14 at 21:05
  • $\begingroup$ I added more details. Please let me know if there are mistakes. $\endgroup$ – Ashutosh Oct 21 '14 at 13:58
  • $\begingroup$ I am very sorry that I didn't get around to accepting this answer until now. Thank you very much! $\endgroup$ – Noah Schweber Jul 29 '16 at 0:46
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Since you ask specifically about compact $S$, it is natural also to consider only very nice packings. So let us consider only Borel packings, and in this simplified case, the answer is yes.

Specifically, we may view a packing as an equivalence relation on the plane, where $x\sim y$ just in case $x$ and $y$ are on the same copy of $S$ (or both are not on any copy of $S$). If this relation is a Borel equivalence relation, then it follows by the Silver dichotomy either that there are only countably many or continuum many equivalence classes.

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  • $\begingroup$ Very nice! I vaguely recall that, assuming large cardinals, Silver's dichotomy can be pushed up the projective hierarchy; is this true? If so, I think this would completely answer my question. $\endgroup$ – Noah Schweber Oct 16 '14 at 20:46
  • $\begingroup$ Good idea! I think something like that is true, but I don't recall exactly how it goes. Please let us know! (But it wouldn't really answer the question completely, would it, since there could be non-projective packings, right?) $\endgroup$ – Joel David Hamkins Oct 16 '14 at 20:56
  • $\begingroup$ It can't be literally like that for projective equivalence relations, since the relation of isomorphism of well-orders has $\omega_1$ many classes, and it is projective. But I think there is something like adding $\omega_1$ as one more possibility... $\endgroup$ – Joel David Hamkins Oct 16 '14 at 20:58
  • $\begingroup$ Would it add anything to consider the case where the equivalence relation determined by the packing is a co-$\kappa$-Suslin equivalence relation (that is, the complement is the projection $p[T]$ of a tree $T$ on $\omega \times \omega \times \kappa$)? A theorem of Harrington and Shelah says the number of equivalence classes is at most $\kappa$ if the complement of $p[T]$ is an equivalence relation in a Cohen generic extension. There is a summary of the relevant results in Schlicht's paper "Thin equivalence relations and inner models", section 4. $\endgroup$ – Avshalom Oct 16 '14 at 23:35
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One simple ZFC-observation about $\mathbb{R}^{2}$, along the lines of the finite cross example, is well-known (see for details e.g. P. Komjath, V. Totik, Problems and Theorems in Classical Set Theory): any disjoint family of $Y$-sets is countable. A set $X$ is a $Y$-set if $X$ is the union of three line segments with a common endpoint. So if $S$ contains a homeomorph of a $Y$-set, the answer is positive in ZFC.

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  • $\begingroup$ In fact, that's true even for homeomorphs of $Y$-sets; only countably many can be packed in the plane. R. L. Moore's triod theorem is a generalization of that fact. $\endgroup$ – bof Oct 16 '14 at 11:07
  • $\begingroup$ Right; I incorporate your observations with thanks. $\endgroup$ – Avshalom Oct 16 '14 at 11:10
  • $\begingroup$ There is a related link about tacks: mathoverflow.net/questions/27244/… $\endgroup$ – Avshalom Oct 16 '14 at 11:59

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