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?



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


