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Feb 2, 2021 at 21:30 comment added Anton Klyachko "...which are not both subsets of $\mathbb Z$." This is not used actually. The argument works for $d=1$ as well, does not it?
Feb 2, 2021 at 17:33 vote accept andres
Feb 1, 2021 at 17:06 history answered Robert Israel CC BY-SA 4.0