Two sets isometric to their union Does there exist two sets $A, B \subset \mathbb{R}^2$ such that $A$ and $B$ are both isometric to their union $A \cup B$?
I thought of this question as a little exercise and it seems the answer is no, but I can’t prove it, or come up with a counterexample. 
 A: A very trivial example is when $A=B$. Then $A \cup B = A = B$ and the identity function $\mathbb{R}^2 \to \mathbb{R}^2$ is an isometry mapping both $A$ and $B$ onto $A \cup B$. This works for any subset $A \subseteq \mathbb{R}$.
Slightly less trivially, let
$$A = \{ (n,0) \mid n \in \mathbb{Z} \text{ and } n \ge 0 \} \quad \text{and} \quad B = \{ (n,0) \mid n \in \mathbb{Z} \text{ and } n \ge 1 \}$$
Then $A \cup B = A$ so the identity function $\mathbb{R}^2 \to \mathbb{R}^2$ maps $A$ isometrically onto $A \cup B$, and the translation $f : \mathbb{R}^2 \to \mathbb{R}^2$ defined by $f(x,y) = (x-1,y)$ for all $(x,y) \in \mathbb{R}^2$ maps $B$ isometrically onto $A \cup B$.
Lots of other trivial examples like this can be constructed.
A: Here is a solution where $A$ and $B$ are disjoint (and nonempty), which I believe is a hypothesis OP intended to include in the question.
Here we use $\mathbb{N}$ to denote the nonnegative integers. Let $\alpha$ be a transcendental number with absolute value 1; for instance, we may take $\alpha = e^i$. Now set
$$X = \mathbb{N}[\alpha] \subset \mathbb{C}$$
as an extension of the monoid $\mathbb{N}$; that is, $X$ consists of all polynomials in $\alpha$ with nonnegative integer coefficients. Then simply let $A = X + 1$ and $B = \alpha X$, which are disjoint (since $\alpha$ is transcendental) and both isometric to $X = A \cup B$.
Note: I explained this construction in person to OP a few months ago, but I figured I'd post it here in case anyone else is interested. This construction is known as the Sierpinski–Mazurkiewicz paradox; further discussion may be found, for instance, in chapter 1 of Tomokovicz and Wagon's book The Banach-Tarksi Paradox.
