Problem 1268

I. Solution. We will show that the questioned conclusion is not true. We will recursively construct the sets $A-t, H-t$, and the $E \neq E'$ translation sets (all non-empty subsets of $\mathbb{R}$), such that

$$H=\bigcup_{e \in E}(A+e)=\bigcup_{e' \in E'}\left(A+e'\right)$$

and all the translations $A+e=\{a+e: a \in A\} (e \in E)$ and $A+e'=\left\{a+e': a \in A\right\} (e' \in E')$ are pairwise disjoint.

First, let $A_1=\{0\}, E_1=\{0\}, E_1'=\{1\}$, and $H_1=\left(A_1+E_1\right) \cup\left(A_1+E_1'\right)=\{0,1\}$. From here, we proceed recursively. Suppose we have already constructed the finite sets $A_n, E_n, E_n'$ such that $E_n \cap E_n'=\emptyset$, and $A_n+E_n$ and $A_n+E_n'$ cover each element they cover exactly once (i.e., the sets $A+e (e \in E_n)$ are pairwise disjoint, and the sets $A+e' (e' \in E_n')$ are also pairwise disjoint). Let $H_n=\left(A_n+E_n\right) \cup\left(A_n+E_n'\right)$. Now, for each $h \in H_n$, we do the following: if $h$ has not yet been in $A_n+E_n$, we add an element $a$ to $A_n$ and an element $e$ to $E_n$ such that their sum is exactly $h$, and to ensure that the previous properties do not break, $a$ should not be of the form $a_1+e_1-e_2$ (where these are previous elements: $a_1 \in A_n, e_1, e_2 \in E_n$), and $e$ should not be of the form $e_1+a_1-a_2$ (where $e_1 \in E_n, a_1, a_2 \in A_n$), and the new $e$ should not be in $E_n'$. Each of these conditions represents a finite number of forbidden elements. We proceed similarly for $A_n+E_n'$. Thus, we obtain the sets $A_{n+1}, E_{n+1}, E_{n+1}'$, clearly $H_n \subseteq A_{n+1}+E_{n+1}$, $H_n \subseteq A_{n+1}+E_{n+1}'$.

It is clear that $A=\bigcup_{n=1}^{\infty} A_n, E=\bigcup_{n=1}^{\infty} E_n, E'=\bigcup_{n=1}^{\infty} E_n', H=\bigcup_{n=1}^{\infty} H_n$ satisfy the conditions, provided that the boundedness also holds. However, we can easily ensure the boundedness by choosing each new element from the interval $(-2,2)$ as follows: in a typical step, we want to write a given $h \in H_n \ni h \in (-2,2)+(-2,2)=(-4,4)$ in the form $a+e$, where $a \in A_{n+1}$, and $e \in E_{n+1}$ (or $E_{n+1}'$). It is clear that since only a finite number of forbidden elements exist, there exist suitable $a, e \in (-2,2)$ numbers.

II. Solution (based on the work of Matolcsi Dávid). Let $H$ be the set of rational numbers with 3-power denominators that fall into the open interval $(-2,2)$. Let $A \subset (-1,1)$ be the set of numbers of the form $\pm(1-3^{-r})$, where $r$ is a non-negative integer. We will show that $H$ can be decomposed into pairwise disjoint translates of $A$ in more than one way.

Let $E$ be the set of rational numbers with 3-power denominators that fall into the closed interval $[-1,1]$. Then $A+E=\{a+e: a \in A, e \in E\} \subseteq H$ (in fact, equality holds). Since $A$ and $A+2/3$ both contain the number $2/3$, these two translates of $A$ are not disjoint. It suffices to show that both can be extended to a decomposition of $H$ into pairwise disjoint translates of $A$. Since $H$ is countable, it suffices to show that if the translates $A+e_i$ for some $e_1, \ldots, e_n \in E$ do not contain the number $h \in H$, then there exists a number $e \in E$ such that the translate $A+e$ contains the number $h$ and is disjoint from each of $A+e_1, \ldots, A+e_n$.

Choose an integer $r \geq 2$ large enough so that $3^{r-2}(h-e_i)$ is an integer for all $i=1, \ldots, n$, and $3^{-r} \leq 2-|h|$. Then $|h|-(1-3^{-r}) \leq 1$, so we can choose a sign such that with $a= \pm(1-3^{-r})$ and $e=h-a$, we have $|e| \leq 1$, i.e., $e \in E$. Then $h=a+e \in A+e$.

We only need to show that for $b, c \in A$ and $1 \leq i \leq n$, $b+e \neq c+e_i$, i.e., $b+h-a \neq c+e_i$, or $a-b+c \neq h-e_i$. Since $h \notin A+e_i$, we have $h-e_i \notin A$, so we are done if $a=b$ or $a=-c$, since in these cases $a-b+c \in A$. (In the latter case, we use the fact that the set $A$ is symmetric about 0.) In other cases, we show that $3^{r-2}(a-b+c)$ is not an integer, from which the desired inequality immediately follows.

Let $b= \pm(1-3^{-s})$ and $c= \pm(1-3^{-t})$, then $3^{r}(a-b+c)= \pm(3^{r}-1) \mp(3^{r}-3^{r-s}) \pm(3^{r}-3^{r-t})$, where $3^{r}$ is an integer divisible by 9, and $\mp 1 \pm 3^{r-s} \mp 3^{r-t}$ is not, because if $\max(s, t)>r$ then it is either $\mp 1$ or not an integer, and if $\max(s, t) \leq r$ then it is either $\mp 3$ or not divisible by 3. Thus, $3^{r}(a-b+c)$ cannot be an integer divisible by 9.