For any set $S\subseteq \mathbb{Z}\times\mathbb{Z}= \mathbb{Z}^2$ and $a\in \mathbb{Z}^2$, we set $a+S = \{a+s: s\in S\}$, where $+$ is the componentwise addition in $\mathbb{Z}^2$. Moreover, for any collection of subsets ${\frak S}\subseteq {\cal P}(\mathbb{Z}^2)$ we let $a + {\frak S} = \{a+S: S\in{\frak S}\}$.
We say that a partition ${\frak T}$ of $\mathbb{Z}^2$ is a mono-tiling if for any $T_0\neq T_1\in {\frak T}$ there is $a\in\mathbb{Z}^2$ such that $T_0 = a + T_1$. We call the mono-tiling ${\frak T}$ aperiodic if for all $y\in \mathbb{Z}^2$ we have ${\frak T}\neq y + {\frak T}$.
Does $\mathbb{Z}^2$ have an aperiodic mono-tiling?
Best Answer
Yes. A $2$-by-$2$ square $\{0,1\}^2$ can tile $\mathbb{Z}^2$ with just one period. So $\{0,2\}^2$ can tile $2\mathbb{Z}^2 \leq \mathbb{Z}^2$ with just one period. Break other periods in the other cosets.