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For $t\in \mathbb{Z}\times\mathbb{Z}$ and $A\subseteq\mathbb{Z}\times\mathbb{Z}$ we set $t+A :=\{t+a: a\in A\}$.

Call $A\subseteq\mathbb{Z}\times\mathbb{Z}$ tileable if there is $T\subseteq\mathbb{Z}\times\mathbb{Z}$ such that

  1. $t_1\neq t_2\in T$ implies $(t_1+A)\cap (t_2+A) =\emptyset$;
  2. $\bigcup\{t+A: t\in T\} = \mathbb{Z}\times\mathbb{Z}$.

Is the collection of tileable subsets of $\mathbb{Z}\times\mathbb{Z}$ uncountable?

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    $\begingroup$ So you allow infinite "tiles"; otherwise there are clearly only countably many. Do you require the tile to be connected? $\endgroup$
    – Gabriel Nivasch
    Commented Mar 15, 2016 at 10:38

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Yes. Uncountable. For each set $P\subset \mathbb Z$ take the tile $A = (\{0\}\times \mathbb Z) \cup (\{1\} \times P) \cup (\{-1\} \times (\mathbb Z \setminus P))$.

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  • $\begingroup$ Is the $\{0\}\times\mathbb Z$ part needed? $\endgroup$
    – Anthony Quas
    Commented Mar 15, 2016 at 12:02
  • $\begingroup$ Just to make the tile connected $\endgroup$
    – Gabriel Nivasch
    Commented Mar 15, 2016 at 12:24

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