Given a real number $x\in\mathbb{R}$ in decimal form $$ x = C_0. C_1 \dotsc, \quad C_0 \in \mathbb{Z}, \quad C_i = 0, \dotsc, 9, \quad i > 0 $$ we may define a sequence of decimals $$ q_n = C_0. C_1 \dotsc C_n = C_0 + \sum_{k = 1}^n \frac{C_k}{10^k} \in \mathbb{Q} $$
Cauchy sequences $(a_n)$ and $(b_n)$ are equivalent iff $\lim_{n \to \infty} a_n - b_n = 0$
It is well known that real numbers $1$ and $0.(9)$ are equal. Are they the only type of equivalent decimal sequences?
By that I mean, is it true that if $$ \lim_{n \to \infty}\sum_{k = 1}^n \frac{C_k}{10^k} = 1, \quad C_k = 0, \dotsc, 9 $$ then $C_k = 9$ for all $k > 0$?
Is it true, that modulo the equivalence relation $$ C_0. C_1 \dotsc C_k (9) \equiv C_0. C_1 \dotsc (C_k + 1) $$ we get unique decimal expansion for every real number?
The answer is yes. Consider the sequence $D_k=9-C_k$ ($D_k$ is also a sequence composed from the numbers $0,...,9$). If $C_k$ is not a constant sequence of $9$, it follows that there is $m\in\mathbb{N}$ such that $D_m>0$, so we would get $\sum_{k=1}^\infty\frac{D_k}{10^k}\geq \frac{D_m}{10^m}>0$. Notice that
$$ \sum_{k=1}^\infty\frac{C_k}{10^k}+\sum_{k=1}^\infty\frac{D_k}{10^k}=\sum_{k=1}^\infty\frac{C_k+D_k}{10^k}=\sum_{k=1}^\infty\frac{9}{10^k}=1 $$ So $\sum_{k=1}^\infty\frac{C_k}{10^k}=1-\sum_{k=1}^\infty\frac{D_k}{10^k}<1$