About translating subsets of $\Bbb Z.$

Question

This is a continuation of About translating subsets of R2. Is it possible to find a pair of sets $A,B\subseteq\Bbb Z$ such that

• A is a union of translated (only translations are allowed) copies of B;
• B is a union of translated copies of A;
• A is not a a single translated copy of B (and the other way around, which follows)?

The unions need not be disjoint.

I do not believe it is, but I haven't been able to prove it for a long time.

Just to make things clear: David Moews' answer does not work in this case since $\Bbb Z^\omega\not\hookrightarrow\Bbb Z.$

And to give a bit more background. If there is such pair of sets in $\Bbb Z$ then there is one in any abelian group in which $\Bbb Z$ embeds. And that only leaves torsion groups to check (which might be the next question :) ). If there is no such pair of sets, then I think the question of which infinite abelian groups do have such a pair of sets becomes quite interesting (and that's the question I'm actually interested in).

Added: I've posted the general question on MO: https://mathoverflow.net/questions/187041/sets-which-are-unions-of-translates-of-each-other-but-arent-single-translates

Answer 1

I think it's not possible.

For subsets $X$, $Y$ of $\mathbb{Z}$, I'll denote by $X+Y$ the subset $$\left\{x+y: x\in X, y\in Y\right\}.$$

The assumption is that there are sets $X$, $Y$ with $A=B+X$ and $B=A+Y$, so $A=A+X+Y$ and $B=B+X+Y$.

If $|X+Y|=1$ then $|X|=1$ and $A$ is a translate of $B$. Otherwise, choose $0\ne n\in X+Y$. I'll assume $n>0$; the case $n<0$ is similar.

Then both $A+n\mathbb{N}\subseteq A$ and $B+n\mathbb{N}\subseteq B$, so for each congruence class $i\pmod{n}$ either $A\cap(i+n\mathbb{Z})=i+n\mathbb{Z}$, or $A\cap(i+n\mathbb{Z})=\emptyset$, or there is some $a_i\in A$ such that $A\cap(i+n\mathbb{Z})=a_i+n\mathbb{N}$, and similarly for $B$.

Unless $A$ is a union of congruence classes (i.e., the third possibility above doesn't occur), in which case $B$ is also a union of congruence classes and then it's clear that $A$ is a translate of $B$, this implies that $X+Y$ contains no negative integers (consider where the smallest $a_i$ can be sent by translation by an element of $X+Y$), and (by translating $B$, which doesn't affect the question) we can assume that neither $X$ or $Y$ contain any negative integers. But this is impossible unless $A=B$ since each $a_i$ must be $b_j+x$ for some $b_j$ and some $x\in X$, and each $b_i$ must be $a_j+y$ for some $a_j$ and some $y\in Y$.