I believe that $\mathbb{R} \setminus R$ is uncountably infinite. The set $R$ contains $\mathbb{Q}$, and is contained inside the set of real algebraic numbers. But the set of real algebraic numbers is countably infinite. Hence,
- $R$ is countably infinite.
- $\mathbb{R} \setminus R$ is uncountably infinite.
See this post for more details.
$\cal C_k$ is indeed empty for $k > 2$.
Suppose that a number has two distinct decimal expansions $\delta = (\delta_0, \delta_1, \delta_2, \ldots)$ and $\delta' = (\delta_0', \delta'_1, \delta'_2, \ldots)$. Then, we must have
$$0 = \sum_{i \ge 0}\dfrac{\delta_i - \delta'_i}{10^i}.$$
Wlog, we may assume that $\delta_0 \neq \delta'_0$. (There must be some smallest $i$ for which they are unequal. By multiplying by a suitable power of $10$, we may assume that it is $i = 0$.) Furthermore, we may assume that $\delta_0 > \delta'_0$.
Thus, we have
$$1 \le \delta_0 - \delta'_0 = \sum_{i \ge 1}\dfrac{\delta'_i - \delta_i}{10^i}.$$
Taking absolute value on all sides and using triangle inequality for series, we see that
$$1 \le \delta_0 - \delta'_0 \le \sum_{i \ge 1}\dfrac{|\delta'_i - \delta_i|}{10^i} \le \sum_{i \ge 1}\dfrac{9}{10^i} = 1.$$
Thus, we have equality throughout. Note that this means that $|\delta'_i - \delta_i| = 9$ for all $i$. In fact, we see that $\delta'_i - \delta_i$ must have the same sign for all $i$. This sign must, of course, be positive.
Thus, we have $\delta_0 = \delta'_0 + 1$ and $\delta_i = 0$ for all $i \ge 1$, $\delta'_i = 9$ for all $i \ge 1$.
(All the manipulations above were justified as the series converged absolutely.)
The above shows that if a number has two decimal expansions, it must actually just be a variant of $1 = 0.\bar{9}$. In particular, there are at most two decimal expansions.
Best Answer
Your second point is not correct. There are uncountably many strings of infinite length, even though there are only countably many strings of finite but unbounded length. This makes sense because the latter correspond to terminating decimals, which are a subset of the rationals.
You can see by diagonal argument that the set of infinite strings is uncountable: assume it is countable, and order the strings $s_1,s_2,\ldots$, then construct a string which is different to $s_1$ in position $1$, different to $s_2$ in position $2$, and so on.