Intersecting a translation of an integer set with its complement

Question

So I have a quite specific question: Let $A \subset \mathbb{Z}^d$ and suppose that both $A$ and its complement $A^C$ contain infinitely many elements. Now I assume that for every vector $v \in \mathbb{Z}^d$ the set \begin{align*} (A+v) \cap A^C \end{align*} has only finitely many elements. Is this possible for $d>1$?

If I assume $d=1$, I could take for example the natural numbers $\mathbb{N}$, where their translation by any integer number $v$ will only intersect the negative integers in finitely many points. However, in higher dimensions this argument does not seem to hold. Intuitively, I feel like the property stated above should not be possible for $d>1$, i.e. there must be some translation $v$ such that $A+v$ intersects the complement $A^C$ in infinitely many points. Does anyone see a convenient way to prove this, or can I construct a counter example?

Intuitive proof for $d=2$:

I will write here my intuition (for $d=2$), which I think should work for all other $d>1$ as well. Beware that I will be quite sloppy with the mathematical details:

We consider the lines $H_m = \{x \in \mathbb{Z}^2: \, x_2=m\}$. On each line we can only have finitely many transitions $A \to A^C$ (i.e. $x \in A, x+e_1 \in A^C$) and $A^C \to A$. This means that the interval $[c_m, \infty) \subset H_m$ must be fully contained in either $A$ or $A^C$. We assume without loss of generality that it is contained in $A$.

Then the same must hold for all lines $H_m$ (i.e. $[c_m, \infty) \subset A \cap H_m$ for all lines $m$), since otherwise the set $(A+v) \cap A^C$ would have infinitely many elements. In other words, for every line $H_m$ there exists a position $c_m \in H_m$ such that right of $c_m$ everything is in $A$. We can picture this as the right hand side of the plane to be entirely covered by $A$.

Since $A^C$ has infinitely many elements as well, we must have two possibilities: Either there exists a line $H_m$ such that $(-\infty, d_m] \subset H_m$. Then, by the same argument as above, the entire left hand side of the plane is covered by $A^C$. This means that there are infinitely many transitions $A^C \to A$, which is not possible. Or the left hand side is also entirely covered by $A$. Then every line $H_m$ can only contain finitely many elements of $A^C$. Since $A^C$ must still contain infinitely many elements, there are infinitely many lines $H_m$ containing elements of $A^C$. This again leads to infinitely many transitions $A^C \to A$, which is again a contradiction.

Intuitively, the argument above makes sense, and should work for higher dimensions $d>2$ as well. However, the argument is not very elegant, and I have the feeling there should be an easier way to argue this...

Answer 1

Such sets do not exist for $d\ge 2$. For simplicity let me produce a proof for $d=2$.

Assume the contrary that $A\subset\mathbb Z^2$ is infinite, and has an infinite complement $B=A^c$, so that $(v+A)\cap B$ is finite for any $v\in\mathbb Z^2$.

Denote $X_k=\{(x,k): x\in\mathbb Z\}$ as the horizontal line $y=k$, and $A_k,B_k$ as $A\cap X_k$, $B\cap X_k$ respectively.

Look at $k=0$ first. Clearly one of $A_0,B_0$ must be infinite, say $A_0$ w.l.o.g (otherwise swap $A,B$ in the first place).

Take $v=(0,1)$ and consider the set $(v+A_0)\subset X_1$. By assumption $(v+A_0)\cap B$ is finite, thus for the set $B_1$ we have $$B_1\subseteq(v+B_0)\cup\left((v+A_0)\cap B\right).$$

By assumption there shall be only finitely many $k$ such that $v+A\cap B_k$ is non-empty (otherwise $v+A\cap B$ would be infinite), let's denote them as $k_1,\ldots,k_m$. Repeat the argument in the last paragraph on $v+A_1,v+A_2,\ldots$ (where each $A_k$ contains an infinite subset of $v+A_0$), and also on $-v+A_0,-v+A_1,\ldots$, we see that there is some $a$ so that the vertical line $x=a$ is completely contained in $A$, where $a$ can be determined by $B_0,B_{k_1},\ldots, B_{k_m}$.

If $B_0$ is finite, then $B$ is contained in the vertical stripe $S_a=\{(x,y): |x|< a\}$ with a proper choice of $a$. But then the set $(2a,0)+B$ is completely disjoint with $B$, thus its intersection with $A$ is itself, and is therefore infinite, yielding a contradiction.

If $B_0$ is infinite, then we can apply the above arguments again but with $A,B$ swapped to assert the existence of some vertical line $x=b$ that is completely contained in $B$, and then the line $x=a$ (which is inifinite) is in the intersection of $(a-b,0)+B$ and $A$, also yielding a contradiction.