# A set of points congruent to its proper subset

There are sets of points in $\mathbb{R}^n$ congruent to their own proper subsets. A (trivial) example is a ray, or to give a more interesting bounded example, $\{e^{i \cdot n} \mid n\in\mathbb{N}\}$. There are references saying about a proof by Jan Mycielski that there exists an infinite and bounded set in $\mathbb{R}^3$ congruent to every its subset obtained by removing a finite number of points. Is there an explicit construction of such a set, or it is a pure existence theorem? Is there an infinite set congruent to every its subset obtained by removing countable number of points?

-

The finite case is pretty cheap. Take the free group $G$ with countably many generators $R_1,R_2,\dots$. Consider any injection $S:F(G)\to \mathbb N$ where $F(G)$ is the set of finite subsets of $G$ such that $S(W)$ is different from any index of a generator contained in a word $w\in W$ and take $E$ to be all (irreducible) words that do not start with $R_{S(W)}^{-1}w$ for $w\in W$. Then $R_{S(W)}E=E\setminus W$. Now just realize $G$ as a subgroup of the orthogonal group and take a generic point on the sphere.

Whether to call this "constructive" is a matter of taste. Everything can be done explicitly but it is quite a mess.

To do the countable case this way, one would need a free group with continuum generators. If you believe the continuum hypothesis, you still can embed this monster in the orthogonal group. Now I'm not so sure about $R^3$ but $R^9$ will still work (orthogonal matrices acting on themselves). If you believe that the continuum hypothesis fails, I don't know (because then you'll have to add a generator to an uncountable family). However, you can try to construct some explicit family of orthogonal operators without non-trivial relations in some high dimension. It would be interesting to see :).

-

Sorry to put this here, but I'm disreputable and thus I cannot comment on fedja's answer. What does $R_{S(W)}$ mean? It makes an appearance in the answer without being defined. Sorry if it's so standard that even a child should know it. Thanks.

-
Oops, sorry: $R_j$ are the generators of $G$. It is, indeed, impossible to formally derive from what I wrote though it may be possible to guess it from the context. I'll edit them in :) –  fedja Dec 3 '11 at 0:20
D'oh, should've figured that out :) –  Antonio Ramírez Dec 6 '11 at 4:52