7
$\begingroup$

This question was posed to me by a friend who tells me the answer is yes, but I cannot see why.

Does there exist two nonempty, disjoint sets $A, B \subset \mathbb{R}^2$ such that $A$ and $B$ are both isometric to their union?

I cannot for the life of me come up with a way of constructing both sets - I've noticed that the measure of both has to be $0$, but that's about it.

$\endgroup$
  • $\begingroup$ The measure could also be $\infty$. Or else they could be non-measurable. $\endgroup$ – user357151 Jul 15 '18 at 15:37
  • $\begingroup$ Perhaps A and B are of the same range and 'content' but of opposite sign. $\endgroup$ – poetasis Jul 15 '18 at 15:39
8
$\begingroup$

Let the group with presentation $\langle x,y \mid x^2=e \rangle$ (that is, the free product $C_2 * \mathbb Z$) act on the plane by mapping $x$ to a rotation by $180^\circ$ about $(1,0)$ and $y$ to a rotation about the origin by some angle $\theta$ whose cosine is transcendental.

(Originally I claimed that this action is faithful, but actually it isn't -- the group element $y^{-2}xyxyxy^{-2}xyxyx$ has the identity action no matter what $\theta$ is. Fortunately less can do for this particular purpose).

Let $K$ be the set of group elements that can be written on the form $$ y^{n_m}xy^{n_{m-1}}x\cdots xy^{n_1} x y^{n_0} $$ where $m\ge 0$, and each $n_j\ge 0$ except that $n_0$ may be negative. This set is not a subgroup, but it has two useful properties:

  1. $K$ is the disjoint union of $\{yk\mid k\in K\}$ and $\{xyk\mid k\in K\}$.

  2. Different elements of $K$ map the point $p_0 = (3,0)$ to different points. (This can be seen by going to the complex plane where $y$ is multiplication by $e^{i\theta}$ and $x$ is the map $z\mapsto 2-z$. Then each element of $K$ maps $3$ to a different Laurent polynomial in $e^{i\theta}$ with integer coefficients, and those all have distinct values because $e^{i\theta}$ is transcendental. Phew!)

Now set $$ A = \{ykp_0 \mid k\in K\} \qquad B = \{xykp_0 \mid k\in K \} $$ Then $A$, $B$, and $A\cup B$ are related by rigid motions of the plane, namely: $$ y^{-1}A = (xy)^{-1}B = \{kp_0 \mid k\in K \} = A \cup B $$ so they are isometric.


This construction is inspired by the initial step of the proof for the Banach-Tarski paradox. It doesn't need the axiom of choice because it doesn't need to select a representative of each orbit, because it is not required that $A\cup B$ is an entire pre-existing shape.

It is natural to ask how $A$ looks -- but unfortunately it can't really be seen: it is dense in $\mathbb R^2$.

$\endgroup$
  • $\begingroup$ Thank you, amazing answer! Really basic question, I know, but is the group you described abelian? Also, how do you know $A$ is dense in the plane? $\endgroup$ – NMister Jul 16 '18 at 19:51
  • $\begingroup$ @NMister: No, it is quite non-abelian -- it needs to be; otherwise $yxp_0 \in A$ and $xyp_0 \in B$ would be the same point and the construction wouldn't work. Nontrivial free products are never abelian. $\endgroup$ – Henning Makholm Jul 16 '18 at 20:27
  • $\begingroup$ @NMister: A is dense in the plane because the image of $(3,0)$ under iterations of $y$ is dense in a circle around the origin with radius $3$. Doing an $x$ now moves the circle away from the origin -- so now the distances from the origin are dense in the interval $[1,5]$. Then doing more $y$s will smear the moved circle out to points that lie densely in an annulus with radii from $1$ to $5$, and now the next $x$ will move that annulus again ... as we have more and more $x$s in the word we can get arbitrarily far from the origin. $\endgroup$ – Henning Makholm Jul 16 '18 at 20:33
  • $\begingroup$ Thanks! One more question - How do we know we can't cancel the $e^{i\theta}$s? $\endgroup$ – NMister Jul 19 '18 at 17:17
  • $\begingroup$ @NMister: The Laurent polynomials must be different for different $k\in K$ because (a) the cofficients are always integers, (b) the ones that are not $0$ or $\pm 3$ encode $x$s in the word $k$, and the distances between them count the number of $y$s between successive $x$s, (c) exactly one of the coefficients is odd, and its position relative to rightmost "not $0$ or $\pm 3$" encodes the power of $y$ after the last $x$ (d) the exponent corresponding to the odd coefficient is the total power of $y$s in $k$. $\endgroup$ – Henning Makholm Jul 19 '18 at 18:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.