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Is there any way to show that the set of disjoint translations of the cantor ternary set is countable?

That is show that there are countably many disjoint sets of the form $\{x+C: x\in \mathbb{R}\}$???

Thanks

Brandon
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  • Don't we have $x + C \cap y + C \ne \emptyset$ iff $x-y \in C- C = [-1,1]$ and hence only countably many disjoint translates? – martini Sep 20 '12 at 08:01

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For any $x,y\in\Bbb R$ we have $(x+C)\cap(y+C)\ne\varnothing$ iff $x-y\in C-C=[-1,1]$ iff $|x-y|\le 1$. Suppose that $A\subseteq R$ and $\{a+C:a\in A\}$ is pairwise disjoint; then for distinct $a,b\in A$ we must have $|a-b|>1$, and $A$ must be countable.

Brian M. Scott
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  • @Ursula: $\Bbb Q-C=\bigcup_{q\in\Bbb Q}(q-C)$ is a countable union of closed nowhere dense subsets of $\Bbb R$, so its complement is dense. $0\in C$, so $\Bbb Q-C\supseteq\Bbb Q$, and we can pick an irrational $x\in\Bbb R\setminus(\Bbb Q-C)$. Then $(x+C)\cap\Bbb Q=\varnothing$. – Brian M. Scott Sep 20 '12 at 08:42