# Are there unique additive decompositions of the reals?

Given $b\in \mathbb{R}_{>1}$ is there $U\subseteq\mathbb{R}_{\ge 0}$ such that $U+bU=\mathbb{R}_{\ge 0}$ and $(U-U)\cap b(U-U)=\{0\}$ (or equivalently: $u+bv=u'+bv' \implies u=u', v=v'$)?

Here is an example of near miss: if $b=10$ define $U$ as the set of positive reals with $0$ digits in odd places. Then the decomposition fails to be unique, but only for finite decimals, such as $1=1+10\times0=0.0909... +10\times 0.0909...$

This question is a simple case of a larger, still fuzzy problem I'm thinking about, but I don't want to jump into that without knowing what surely must already have been studied in depth.

I welcome suggestions or edits for better title and tags.

• What is the $x$ in $u+xv=u'+av'$? – Gerry Myerson Oct 16 '17 at 22:34
• @GerryMyerson: thank you. I fixed the typo. – Yaakov Baruch Oct 17 '17 at 3:13
• I also changed $a$ to $b$ because I have a hard time distinguishing $a$ from $\alpha$ in Ilya Bogdanov'a answer below (which I'm editing accordingly). – Yaakov Baruch Oct 17 '17 at 3:22
• Not exactly what you are asking, but it is shown in a paper of Sergei Konyagin and myself "The Erdos-Turan problem in infinite groups" that the additive group of real numbers admits a perfect additive basis; that is, there exists a set $B\subset\mathbb R$ such that every real number number is representable as $b'+b''$ with $b',b''\in B$, and the representation is unique up to the order of the summands. (Indeed, we have shown that this is true for almost any infinite abelain group.) The argument is pretty much in the spirit of Ilya Bogdanov's answer below... – Seva Oct 17 '17 at 13:43
• ... (continued) and it should be possible to modify it to get exactly what you want. – Seva Oct 17 '17 at 13:43

Can't this be solved in the following usual manner?

Take the first ordinal $\phi$ of cardinality continuum, and let $\{p_\alpha\colon \alpha<\phi\}$ be an enumeration of points in $\mathbb R_{\geq 0}$, with $0$ being the minimal point. We construct the set $U$ by transfinite recursion on $\alpha$. Initially, we put $0$ into $U$. After performing step $\alpha$, we will have $(U-U)\cap b(U-U)=0$ and $\{p_\beta\colon \beta\leq \alpha\}\subseteq U+bU$.

Assume we have reached step $\alpha$, so now $\{p_\beta\colon \beta<\alpha\}\subseteq U+bU$. If $p_\alpha\in U+bU$ as well, we do nothing on this step. Otherwise, we choose some distinct $x_\alpha$, $y_\alpha$ with $x_{\alpha}+by_\alpha=p_\alpha$ and put $x_\alpha,y_\alpha$ into $U$. There are continuumly many choices for this pair, and less obstructions (each of which involves one, two, or three elements from the recent $U$). Thus such pair can be chosen.

[EDIT] The obstructions mentioned in the previous paragraph are: $$x_\alpha-u_1=\pm b^{\pm1}(u_2-u_3);\\ y_\alpha-u_1=\pm b^{\pm1}(u_2-u_3);\\ x_\alpha-u_1=\pm b^{\pm1}(y_\alpha-u_2);\\ x_\alpha-y_\alpha=\pm b^{\pm1}(u_1-u_2);\\ x_\alpha-y_\alpha=\pm b^{\pm1}(x_\alpha-u_1);\\ x_\alpha-y_\alpha=\pm b^{\pm1}(y_\alpha-u_1).$$ Almost for every choice of the $u_i$, each of the obstructions prohibits a finite set of pairs $(x_\alpha,y_\alpha)$, since we have two linear equations on this pair (along with $x_\alpha+by_\alpha=p_\alpha$). There can be uncountably many such choices of the $u_i$, but their cardinality is still less than continuum --- by the choice of $\phi$; so still there are unobstructed pairs.

Exceptions. It might occasionally happen that the obstruction is linearly dependent with the condition $x_\alpha+by_\alpha=p_\alpha$. These hypothetical occasions are

$\bullet$ in the third equality, read as $x+by=u_1+bu_2$; but then $p_\alpha\in U+bU$ already;

$\bullet$ in the fifth equality, read as $(1-b)x_\alpha-y_\alpha=-bu_1$; this would mean that $\frac{b-1}1=\frac{1}b$, i.e., $b^2=b+1$ --- but then $x_\alpha+by_\alpha=b^2u_1=u_1+bu_1$, so $p_\alpha\in U+bU$ already;

$\bullet$ in the last equality, read as $x_\alpha+(b-1)y_\alpha=bu_1$ --- but the left-hand side is not proportional to $x_\alpha+by_\alpha$.

In all other cases, the signs of coefficients of $x_\alpha$ and $y_\alpha$ in the linear equation provided by the obstruction are not the same. So still the method seems to work.

• I'm trying to understand what you mean by "less obstructions". For example one of the dozen or so obstructions is that $(U-x_{\alpha})\cap b(U-U)\subseteq \{0\}$ and once $U$ has grown beyond countable size, couldn't this potentially block all of the continuumly many choiches for $x_{\alpha}$? – Yaakov Baruch Oct 17 '17 at 6:33
• I've expanded the part about obstructions; I also needed several cases to be treated separately --- but still, everything seems to work. (Sorry, I do not understand the first comment,,,) – Ilya Bogdanov Oct 17 '17 at 9:50
• Ilya, you are right. I deleted now that first comment, which was my misunderstanding. – Yaakov Baruch Oct 17 '17 at 16:15