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Let A and B be two subsets of R^2. I define the relation T(A,B) to hold between A and B iff there exists a translation f on R^2 such that the image set of A under f is B. It is easy to prove that T is an equivalence relation. In a question on MSE, I asked if the equivalence class under T for every nonempty proper subset of R^2 has cardinality of the continuum. The answer was negative. The answerer gave me an example of a set of points whose associated equivalence class is countably infinite. But the answer to one question gives birth to another question. My question now is, is the equivalence class of every nonempty proper subset of R^2 infinite? Might there be nonempty proper subset P of R^2 such that the equivalence class of P is finite?

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  • $\begingroup$ That doesn't quite answer my question, I think. $\endgroup$
    – user107952
    Dec 21, 2013 at 5:02
  • $\begingroup$ (I deleted a wrong comment.) $\endgroup$ Dec 21, 2013 at 5:44

1 Answer 1

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If the equivalence class of S is finite, then $A=\{a∈R^2:S+a≠S\}$ is finite, pick $r$ such that $r<|a|$ for all $a∈A$. For $x∉S$, $B_r(x)⊂R^2\setminus S$, and similarly with $x∈S$, $B_r(x)⊂S$. So the minimal distance between point in $S$ and not is $S$ at least $2r$, contradiction.

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