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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 $x=u+v$

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  • $\begingroup$ If $(u,v)$ works, then so does $(v,u)$, so there's hardly ever a unique pair. $\endgroup$
    – Gerry Myerson
    Commented May 12, 2018 at 12:53
  • $\begingroup$ @GerryMyerson: Thank you. I have corrected the error. $\endgroup$
    – ar.grig
    Commented May 12, 2018 at 16:26
  • $\begingroup$ As you've written it as a set, we are insisting $u\ne v$? $\endgroup$
    – Gerry Myerson
    Commented May 12, 2018 at 22:42
  • $\begingroup$ No. They can be equal. Sorry for my inaccuracy. Corrected. $\endgroup$
    – ar.grig
    Commented May 13, 2018 at 2:43
  • $\begingroup$ My point was, if $u=v$, then $\{\,u,v\,\}=\{\,u\,\}$ isn't a pair, it's a singleton. $\endgroup$
    – Gerry Myerson
    Commented May 13, 2018 at 5:08

2 Answers 2

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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).

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    $\begingroup$ 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. $\endgroup$
    – Seva
    Commented May 12, 2018 at 15:15
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    $\begingroup$ 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 $|\alpha| < |c|$. $\endgroup$
    – ar.grig
    Commented May 12, 2018 at 16:53
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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.)

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  • $\begingroup$ 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. $\endgroup$
    – ar.grig
    Commented May 12, 2018 at 10:18
  • $\begingroup$ @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. $\endgroup$
    – Seva
    Commented May 12, 2018 at 13:26
  • $\begingroup$ Yes, of course, I mean simple proof. One was given by Mike. Thank you. $\endgroup$
    – ar.grig
    Commented May 13, 2018 at 8:21

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