Existence of subset of reals such that any real number is unique sum of exactly two elements of the subset 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 $x=u+v$
 A: 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).
A: Sets $U$ with this property are sometimes called perfect additive bases. As shown in the paper of Sergei Konyagin and myself "The Erdos-Turan problem in infinite groups", any infinite abelian group $G$ with $|2G|=|G|$ possesses such a basis, unless $G$ is the direct sum of a group of exponent $3$ and the group of order $2$. (The condition $|2G|=|G|$ essentially means that $G$ does not have "too many" involutions.)
