I'm looking for a pair of sets $A,B$ of points in $\Bbb R^2$ 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 are arbitrary, even uncountable, and they need not be disjoint.

I can't even decide whether I believe such $A,B$ should exist or not. Could you help me please?

I've looked at many different sets, and I suspect that if such sets exist, then $A$ and $B$ need to be rather strange, but I can prove nothing interesting other than that $A$ and $B$ cannot be finite. It seems unlikely to me that there could be examples of positive finite measure, but I can't prove it. I've tried with many zero-measure and infinite-measure sets, but I really have no clue. (I should note that if such sets exist but are non-measurable, I'm interested in them too.)

Sorry if the tags are wrong, but I'm just not sure which ones to use.

Added. I think defining $A$ and $B$ recursively could work. If I take two sets of vectors $X$ and $Y$, then a set $A_0$, and then define $B_n=A_{n-1}+X,$ $A_n=B_n+Y$ for $n>0$, then $$\left(\bigcup_{n\geq0}A_n\right)+X=\bigcup_{n\geq0}(A_n+X)=\bigcup_{n>0}B_n$$

and also $$\left(\bigcup_{n>0}B_n\right)+Y=\bigcup_{n>0}(B_n+X)=\bigcup_{n>0}A_n.$$

So if $\displaystyle\bigcup_{n\geq0}A_n=\bigcup_{n>0}A_n$ or, equivalently, $\displaystyle A_0\subseteq\bigcup_{n>0}A_n,$ then the sets $A=\displaystyle\bigcup_{n>0}A_n$ and $B=\displaystyle\bigcup_{n>0}B_n$ are unions of translated copies of each other. One obvious way to achieve that is to have an $x\in X$ such that $-x\in Y.$ Then $$A_0=A_0+x-x\subseteq A_0+X+Y=A_1.$$ But then also $$A+x\subseteq A+X=B$$ and $$B-x\subseteq B+Y=A,$$ whence $B\subseteq A+x$ and so $A+x=B.$ So this doesn't work.

I think that it could be a good idea to make the "comeback to $A_0$" take infinitely many steps, so the vectors that are used to "come back" don't add up to any single vector. Unfortunately this is rather vague. I will think about it, but if you have any ideas on this specifically, please do share.

  • $\begingroup$ A couple clarifying questions: By "union" do you mean "disjoint union" or are you allowing overlap? Do you require the union to be finite? I'm asking because this question has a Banach-Tarski feel, but these are two major differences as written. $\endgroup$ Feb 20 '13 at 23:38
  • $\begingroup$ @MartianInvader The unions are arbitrary, even uncountable, and they need not be disjoint. $\endgroup$
    – Bartek
    Feb 20 '13 at 23:38
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    $\begingroup$ Easy does it, @Asaf. It's probably the kind of question you can only answer with non-measurable sets and Axiom of Choice constructions, and that's close enough to set theory for most of us. $\endgroup$ Feb 21 '13 at 0:08
  • 5
    $\begingroup$ Did you ask about $\mathbb R^2$ because the answer is known for $\mathbb R$? If so, it would be useful to link to or state the result; if not, it would make sense to deal with $\mathbb R$ first. $\endgroup$
    – joriki
    Feb 21 '13 at 2:34
  • $\begingroup$ A few observations: Letting $B = A+X$ and $A = B+Y$ we have $A = A+X+Y$. If $0 \in X+Y$ then $B$ is a translate of $A$. Also, if $A$ is invariant under the group of translations generated by $X+Y$, then $B$ is also a translate of $A$. $\endgroup$ Feb 21 '13 at 3:48

Such a pair of sets does exist.

Let $H={\Bbb Z}^{\omega}$ be the direct sum of countably many copies of the additive group $\Bbb Z$ (indexed by the positive integers), and let $G=H\oplus H$. It's clear that $G$ can be embedded in the additive group ${\Bbb R}^2$, so, it will do to find two subsets $A$ and $B$ of $G$ such that $A$ is the union of translates of $B$, $B$ is the union of translates of $A$, and $A$ and $B$ are not mutual translates.

Given $x=(x_i)\in H$, let $|x|=\sum_i x_i$ be the sum of its coordinates. Also, let $e_1$, $e_2$, $\dots$ be the coordinate basis vectors for $H$, so that $(e_i)_j$ is $1$ if $i=j$ and $0$ otherwise. Then, set $$A:=\{(v,w)\in G\mid |v|=|w|, v_i\ge -1, w_j\ge -1 \text{ for all } i, j\},$$ $$B:=\{(v,w)\in G\mid |v|=|w|+1, v_i\ge -1, w_j\ge -1 \text{ for all } i, j\},$$ $$V:=\{(e_i,0)\mid i=1, 2, 3, \dots\},$$ $$W:=\{(0,e_j)\mid j=1, 2, 3, \dots\}.$$ Then:

  • $B=A+V$. It is clear that $A+V\subseteq B$. For the other direction, if $(v,w)\in B$, there is some $i$ such that $v_i=0$. Then $(v,w)=(e_i,0)+(v-e_i,w)$, and $(v-e_i,w)\in A$.

  • $A=B+W$. It is clear that $B+W\subseteq A$. For the other direction, if $(v,w)\in A$, there is some $j$ such that $w_j=0$. Then $(v,w)=(0,e_j)+(v,w-e_j)$, and $(v,w-e_j)\in B$.

  • $A$ is not a translate of $B$. For, suppose $A=B+(\delta,\epsilon)$ for some $\delta$, $\epsilon\in H$. Then, for all $i$, since $(-e_1-e_2-\cdots-e_i,-e_1-e_2-\cdots-e_{i+1})\in B$, and adding this vector to $(\delta,\epsilon)$ must produce an element of $A$, we must have $\delta_1$, $\delta_2$, $\dots$, $\delta_i$, $\epsilon_1$, $\dots$, $\epsilon_i$ nonnegative. On the other hand, since $(-e_1-e_2-\cdots-e_i, -e_1-e_2-\cdots-e_i)\in A$, and subtracting $(\delta,\epsilon)$ from this vector must produce an element of $B$, we must have $\delta_1$, $\dots$, $\delta_i$, $\epsilon_1$, $\dots$, $\epsilon_i$ nonpositive. Putting these results together, we find that $\delta_i=\epsilon_i=0$ for all $i$. This contradicts $A=B+(\delta,\epsilon)$.

This completes the proof.

  • 1
    $\begingroup$ To point out the obvious, this shows there is an example in $\mathbb{R}^n$ for all $n$ as well -- since $G$ embeds into $\mathbb{R}$ by choosing a countably infinite $\mathbb{Z}$-independent set of real numbers! E.g. powers of a fixed transcendental; or $\log(2),\log(3),\log(5),\log(7),\ldots$ $\endgroup$
    – Mike F
    Feb 21 '13 at 5:51
  • $\begingroup$ Very nice construction, thanks! $\endgroup$
    – joriki
    Feb 21 '13 at 8:31
  • $\begingroup$ Thank you very much. This is a great example, but I don't understand one thing. I agree that you've proved that the images of $A$ and $B$ under the embedding are unions of translated copies of each other. But I'm not sure why the image of $B$ cannot be a single translated copy of the image of $A$. My problem is that I only know that the translations by the vectors in $\mathrm{im}(G)$ don't work. But what about the other vectors in $\Bbb R^2$? $\endgroup$
    – Bartek
    Feb 21 '13 at 11:07
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    $\begingroup$ @Bartek: $\mathrm{im}(G)$ is an additive group so the difference vector of any two points in it is also an element. $\endgroup$ Feb 21 '13 at 14:23
  • $\begingroup$ @Jyrki Thank you, I see! $\endgroup$
    – Bartek
    Feb 21 '13 at 14:29

If I haven't made a silly mistake (I have, see comments), finding such a pair of sets is impossible. Let $A,B \subset \mathbb{R}^n$. Suppose we have found "translation sets" $X,Y$ such that \begin{align*} \bigcup_{x \in X} A + x = B && \bigcup_{y \in Y} B + y = A. \end{align*} We will argue that $A$ and $B$ are translates of one another. The above condition can be rewritten as \begin{align*} A + X = B && B + Y = A. \end{align*} Without loss of generality, $0 \in X$ and $0 \in Y$ or else we can make the replacements \begin{align*} A \to A+x+y && B \to B+x+y && X \to X-x && Y \to Y - y \end{align*} for some points $x \in X$, $y \in Y$ which does not affect the question of whether $A,B$ are translates. But then, we have $A \subset A + X =B$ and $B \subset B+Y = A$ which says $A =B$.

  • $\begingroup$ I don't see how you can assume both $0 \in X$ and $0 \in Y$. Could you explain a bit more? $\endgroup$ Feb 21 '13 at 3:28
  • $\begingroup$ In any case your argument does show that $0 \ne X+Y$, which might be useful. $\endgroup$ Feb 21 '13 at 3:29
  • $\begingroup$ Whoops ok thanks for pointing that out. $\endgroup$
    – Mike F
    Feb 21 '13 at 3:31
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    $\begingroup$ I don't think so. You are shifting $X+Y$ by $-x-y$. To start with, you have $A = A+X+Y$, but now the two sides of this equation get shifted by different amounts. $\endgroup$ Feb 21 '13 at 3:44
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    $\begingroup$ ...oh wait it doesn't preserve those identities. OK I'll stop flogging this dead horse ;) $\endgroup$
    – Mike F
    Feb 21 '13 at 3:52

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