Decomposable: A set $S \subset \mathbb{R}^n$ is decomposable in $m$ sets $A_1,…,A_m \subset \mathbb{R}^n$ if there exist isometries $\phi_1,…,\phi_m:\mathbb{R}^n \rightarrow \mathbb{R}^n$ such that:

(1) $S=\bigcup_{k=1}^m \phi_k(A_k) $

(2) $\forall i \neq j:\phi_i(A_i) \cap \phi_j(A_j)= \emptyset$

Equidecomposable: Sets $S$ and $T$ are equidecomposable if there exists a set $$X=\{Q_1, ... , Q_m \} \subset \mathcal P ( \mathbb{R}^n )$$ with $\mathcal P ( \mathbb{R} ^n )$ as the power set of $\mathbb{R}^n$ such that both $S$ and $T$ are decomposable into the elements of $X$.

Can someone give me an example of when two sets are equidecomposable with using notations such as of these definitions please.


A simple example: $[0,4)\times[0,1)$ and $[0,2)\times [0,2)$ are equidecomposable in $\Bbb R^2$, because both can be written as four translated copies of $[0,1)\times [0,1)$.

Another example in $\Bbb R^1$: Let $A$ be any subset of $[-1,0)$. Then $S:=A\cup [0,\infty)$ and $T:=[0,\infty)$ are equidecomposable. To see this, let $Q_1=\bigcup_{n\ge 1}(A+n)$ and $Q_2=[0,\infty)\setminus Q_1$. Then $T$ is decomposable in $Q_1,Q_2$ per $\phi_1(x)=\phi_2(x)=x$, and $S$ is decomposable in $Q_1,Q_2$ per $\phi_1(x)=x-1$ and $\phi_2(x)=x$.

Using the axiom of choice, one can show that somewhat surprisingly the sphere $S_2\subset \Bbb R^3$ is equidecomposable with two disjoint copies of it, say $S_2\cup (S_2+(2,0,0))$.

  • $\begingroup$ for your R^2 one, are those cross products simply shapes? Like [0,2)x[0,2) is just like a square of length 2? $\endgroup$ – snowman Mar 3 '16 at 14:11
  • $\begingroup$ what would the R^2 ones look in diagram? would it just be a rectangle length 4 and width 2, and a square of length 4 which then both can be decomposed into squares of length 1??? $\endgroup$ – snowman Mar 3 '16 at 14:24
  • $\begingroup$ rectangle with length 4 and width 1 ***** $\endgroup$ – snowman Mar 3 '16 at 14:34
  • $\begingroup$ @snowman Yes, with the lower and left edge belonging to the square and the top and right edge removed $\endgroup$ – Hagen von Eitzen Mar 3 '16 at 18:46
  • $\begingroup$ would it still work with [0,4]x[0,1] and [0,2]x[0,2]? $\endgroup$ – snowman Mar 3 '16 at 19:52

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