Consider real numbers $S : x \in [0,1]$ whose decimal expansions are $x = 0.d_1 d_2 d_3 \ldots$. Now institute various exclusions, listed below. I am interested to learn of general principles that will allow me to conclude that $S_X$ is either countable or uncountable.
- $S_{!5}$: The decimal representation excludes all $5$'s.
- $S_{5^{\textrm{th}}}$: In the decimal representation, every $5^{\textrm{th}}$ digit is $5$.
- $S_{\textrm{odd}}$: The decimal representation excludes all even digits: $0,2,4,6,8$.
- $S_{01}$: The decimal representation excludes all but the two digits: $0$ and $1$.
- $S_{1}$: The decimal representation uses (after the $0.$) only the digit $1$: $0.1, 0.11, 0.111, \ldots$.
- $S_{\ge}$: The decimal representation is non-decreasing: successive digits are the same or larger. E.g., $.1144456777777788999\ldots$.
- $S_{k\pm}$: The decimal representation $k$-oscillates: The sequence consists of $k$ or more digits (non-strictly) increasing, followed by a sequence of $k$ or more digits that (non-strictly) decreases, and so on. E.g., for $k=5$, $0.11339 \; 966432 \; 567777 \; \ldots$.
- $S_{\textrm{max/min}}$: The decimal representation is finitely oscillatory: there are only a finite number of digit minima and maxima in the sequence of digits.
Perhaps each case must be handled separately? I am particularly interested in difficult, borderline cases that are not so straightforward to settle, which could be used as good student exercises to distinguish countable from uncountable.
Don't know that I'd call it a general principle, but most of these follow from remapping the restricted set to the reals, perhaps in another basis.
Decrement all digits larger than $5$, then the set becomes the representation of the entire $[0,1]$ in base $9$, thus uncountable.
Drop every $5^{th}$ digit, then the set becomes the decimal representation of the entire $[0,1]$, thus uncountable.
Remap the $5$ remaining digits $1,3,5,7,9$ to $0,1,2,3,4$, then the set becomes the representation of the entire $[0,1]$ in base $5$, thus uncountable.
That gives the representation of the entire $[0,1]$ in base $2$, thus uncountable.
This can be indexed by the number of decimal digits, thus countable.
The sequence of digits will become constant eventually, thus countable.
Drop every other group of $k$ digits, and truncate the remaining ones to the first $k$ digits, which leaves groups of exactly $k$ increasing decimal digits. Consider each such group as one digit in some base made up of all increasing sequences of $k$ decimal digits. Then the set becomes the representation of the entire $[0,1]$ in that base, thus uncountable.
The sequence of digits will become constant eventually, thus countable.
[ EDIT ] Numbered the points to stay in sync with the OP, and made a minor change to #7 which I had first (mis)read as exactly $k$ digits.