Consider the (Silver mean) substitution $$\varrho:\begin{aligned}&a\mapsto aba\\&b\mapsto a\end{aligned}.$$ If we take $w^{(1)}=a|a$ and $w^{i+1}=\varrho(w^{(i)})$, then we get: $$a|a\overset\varrho\longmapsto aba|aba\overset\varrho\longmapsto\ldots\overset\varrho\longmapsto w^{(i)}\overset{i\to \infty}\longmapsto w= \varrho(w),$$ where $w$ is a bi-infinite sequence.
By something called the geometric inflation rule, we represent $a$ by an interval of length $1+\sqrt 2$ and $b$ by an interval of length $1$. This allows us to interpret $w$ as a partition of the real number line in intervals of length $1+\sqrt 2$ and $1$, where we identify the reference point of $w$ with $0$.
Now the point sets $\Lambda_a$ and $\Lambda _b$ are the left endpoints of the intervals of type $a$ and $b$ respectively. We should now have $$\Lambda_a=s\Lambda_a\mathbin{\dot\cup} \big(s\Lambda_a+(1+s)\big)\mathbin{\dot\cup}s\Lambda_b,\label{moi}\tag{*}$$ where $s=1+\sqrt2$ and $\mathbin{\dot\cup}$ is the disjoint union operator.
Question: I can't figure out how to derive this. I suspect that the right hand side of \eqref{moi} is a description of the left endpoints of $\varrho(w)$ in terms of $\Lambda_a$ and $\Lambda_b$ somehow. I just can't see how exactly.
N.B. This equation and all terminology come from the book Aperiodic order by Baake and Grimm.
The substitution replaces each $a$ interval, of length $s$, with two $a$ intervals and a $b$ interval, of total length $2s+1=3+2\sqrt2=s^2$. It also replaces each $b$ interval, of length $1$, with an $a$ interval, of length $s$. Thus, it multiplies the total interval length by $s$. In fact, it sends any $x\in\Lambda_a\cup\Lambda_b$ to $sx$, adding two new endpoints between $sx$ and $sx+s^2$ whenever $x\in\Lambda_a$. One of these is $sx+s\in\Lambda_b$, and the other is $sx+s+1\in\Lambda_a$.
The term $s\Lambda_a$ accounts for the ‘old’ members of $\Lambda_a$: each $x\in\Lambda_a$ is sent to $sx\in\Lambda_a$. The term $s\Lambda_b$ accounts for the $a$ intervals derived from $b$ intervals: each $x\in\Lambda_b$ is sent to $sx\in\Lambda_a$. Clearly $\Lambda_a\cap\Lambda_b=\varnothing$, so $s\Lambda_a\cap s\Lambda_b=\varnothing$ as well.
The remaining term, $s\Lambda_a+(1+s)$, must account for the ‘new’ members of $\Lambda_a$ that are filled in when an $a$ interval is expanded to an $a$, a $b$, and another $a$ interval, and indeed it does: we saw above that when $x\in\Lambda_a$, the expansion not only sends $x$ to $sx$ but also adds an endpoint $sx+(1+s)\in\Lambda_a$. Finally, it’s clear from the construction that $s\Lambda_a+(1+s)$ is disjoint from $s\Lambda_a\cup s\Lambda_b$. Alternatively, we can note that $x$ and $x+s$ are consecutive endpoints, and they’re sent to $sx$ and $sx+s^2=sx+1+2s$, so $sx$ and $sx+1+2s$ are consecutive ‘old’ endpoints. Since $sx+(1+s)$ is strictly between them, it cannot be an ‘old’ endpoint, i.e., it cannot belong to $s\Lambda_a\cup s\Lambda_b$.