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.

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.

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 $(U-U)\cap a(U-U)=0$ and $\{p_\beta\colon \beta\leq \alpha\}\subseteq U+aU$.

Assume we have reached step $\alpha$, so now $\{p_\beta\colon \beta<\alpha\}\subseteq U+aU$. If $p_\alpha\in U+aU$ as well, we do nothing on this step. Otherwise, we choose some distinct $x_\alpha$, $y_\alpha$ with $x_{\alpha}+ay_\alpha=p_\alpha$ 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.

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.