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I am asking for a friend who is interested in finding subsets $X$ of $\Bbb R^n$ such that for some rotation $r$ one has $ X \subsetneq r(X)$. Perhaps such things have been looked at and have a name? Any references would be appreciated. Of course, such subsets and rotations exist. But what are they called?

W. Edwin Clark
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  • Just interested in $X$ being a proper subset of $r(X)$. Or $r(X)$ being a proper subset of $X$, if you multiply both sides by $r^{-1}$. – W. Edwin Clark Feb 17 '22 at 18:37
  • That is very weird. My mind doesn't even want to imagine sets/maps like this. The image of a set always has lesser or equal cardinality than the set itself. Some tricks like dividing by two the even numbers in order to get the whole set of integers come to mind. But not for simple maps like rotations. They don't have a name that people know of, that's for sure. – Compacto Feb 17 '22 at 18:52
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    Assuming the Axiom of Choice, the Hausdorff-Banach-Tarski paradox provides lots of such $X, Y$ and $f$. These do not have special names though. – markvs Feb 17 '22 at 19:00
  • I realized later that homotethies in vector spaces provide many such examples. Still doubt that rotations have any such set. – Compacto Feb 18 '22 at 08:27
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    Find a point $x$ and rotation $r_0$ such that the points $r_0^n(x)$, $n \in \Bbb{Z}$, are distinct. Then consider the set $X = { r_0^n(x) \vert n = 0, 1, 2,\cdots }$ . Apply any of the rotations $r=r_0^{-k}$ where $k$ is any positive integer to $X$. Shades of Hilbert's Hotel! – W. Edwin Clark Feb 18 '22 at 19:32
  • So we see it is easy to find such sets $X$. But it is impossible to find a compact set $X$ for this. – GEdgar Feb 20 '22 at 01:06
  • GEdgar, can you indicate a proof that $X$ cannot be compact? – W. Edwin Clark Feb 20 '22 at 01:41

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