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.



Are there unique additive decompositions of the reals?
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.


