# 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 $x=u+v$

• If $(u,v)$ works, then so does $(v,u)$, so there's hardly ever a unique pair. – Gerry Myerson May 12 '18 at 12:53
• @GerryMyerson: Thank you. I have corrected the error. – ar.grig May 12 '18 at 16:26
• As you've written it as a set, we are insisting $u\ne v$? – Gerry Myerson May 12 '18 at 22:42
• No. They can be equal. Sorry for my inaccuracy. Corrected. – ar.grig May 13 '18 at 2:43
• My point was, if $u=v$, then $\{\,u,v\,\}=\{\,u\,\}$ isn't a pair, it's a singleton. – Gerry Myerson May 13 '18 at 5:08

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).

• This is essentially the argument we use in our paper, specified to the torsion-free case. Things get a little subtler when torsion is present. – Seva May 12 '18 at 15:15
• Thank you. So, the simple proof exists! I'll only add that $\mathbb{Q}$-linear span of $U_\alpha\cap\{r_\alpha\}$ can not be equal to $\mathbb{R}$ because $|\alpha| < |c|$. – ar.grig May 12 '18 at 16:53

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.)

• Thank you for quick and exact answer. I am reading your article right now. But may be there exist some simple answer in case of $\mathbb{R}$, because the problem was given to first-year student of MIPT. – ar.grig May 12 '18 at 10:18
• @ar.grig: The answer is quite simple even in the general case, but as to a simple argument - well, if you manage to find one, I would be very much interested to learn about it. – Seva May 12 '18 at 13:26
• Yes, of course, I mean simple proof. One was given by Mike. Thank you. – ar.grig May 13 '18 at 8:21