TITLE: Existence of subset of reals such that any real number is unique sum of exactly two elements of the subset
QUESTION [11 upvotes]: It is easy to see (using AC, of course) that there exist two sets $U\subset\mathbb{R}$ and $V\subset\mathbb{R}$ such that any real number $x$ can be represented as unique sum $x=u+v$, where $u\in U$ and $v\in V$.
There are $2^{2^\omega}$ such $(U,V)$ pairs.
The question given by my son: prove existence of pair $(U,V)$ with $U=V$.
In other words, prove the existence of $U\subset\mathbb{R}$ such that for any $x\in\mathbb{R}$ there exists unique pair $\{u,v\}$ with $x=u+v$

REPLY [9 votes]: The usual transfinite construction works. Let $\{r_{\alpha} : \alpha < \mathfrak{c}\}$ list $\mathbb{R}$. Construct $\{U_{\alpha}: \alpha < \mathfrak{c}\}$ by induction on $\alpha$ such that the following hold.
(a) For every $a, b, c, d \in U_{\alpha}$, $a + b = c + d \implies \{a, b\} = \{c, d\}$.
(b) There are $a, b \in U_{\alpha + 1}$ such that $r_{\alpha} = a + b$.
(c) For limit $\alpha$, $U_{\alpha} = \bigcup_{\beta < \alpha} U_{\beta}$.
Then $U = \bigcup_{\alpha < \mathfrak{c}} U_{\alpha}$ is as required. Requirements (a), (c) are trivially satisfied. To ensure (b), at stage $\alpha + 1$, if $r_{\alpha} \notin U + U$, choose $x$ outside the $\mathbb{Q}$-linear span of $U_{\alpha} \cup \{r_{\alpha}\}$ and put $U_{\alpha+1} = U_{\alpha} \cup \{x, r_{\alpha} - x\}$ and note that this does not violate (a).