Are there unique additive decompositions of the reals?

Question

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.

Answer 1

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.