Based on the answers so far I restate the question: on p. 63 of his volume on Analysis (http://math.univ-lyon1.fr/~okra/2011-MathIV/Zorich1.pdf), Zorich says: “We now answer the question whether some real number $x \in \mathbb{R}$ corresponds to every symbol $\alpha_{p}...\alpha_{0}...$ The answer turns out to be negative.” This seems to mean that there are more sequences of the form $r_n = \alpha_pq^p + \dots + \alpha_{p-n}q^{p-n}$ than real numbers. On the same pages Zorich names those additional sequences: those in which “all these numbers [$\alpha_{p-j}$] from some point on are equal to $q-1$”. What exactly does this mean? Are sequences like 3.999 (with $q=10$) deficient in some sense?
(The old question: One way of dealing with a decimal like $0.\bar{9}$ is simply to equate it with 1, but there is a notion that, if taken at face value, these decimals lead to contradictions and do not approximate real numbers as they should. There is a proof of this fact in Zorich’s volume on analysis, of which I totally fail to understand two critical steps (starting with the third, unnumbered, inequality on page 64 here: http://math.univ-lyon1.fr/~okra/2011-MathIV/Zorich1.pdf), but I would be totally happy to have a rigorous proof of any kind.)
1) there is no special need to deal with 0.999... any differently than any other decimal expansion.
2) Nobody 'simply equates' 0.999.... with 1. The two numbers are precisely equal following the meaning of what a decimal expansion is.
3) .... but there is a notion - which notion???
4) these decimal lead to contradictions - which contradictions?
5) no decimal expansion is an approximation of a real number. Instead it is precisely a real number.
6) I can't quite understand what you think is being proven on page 64.
7) you would be happy with a rigorous proof of what exactly?