The standard Cantor set formed by recursively removing the middle one-thirds, on the interval $[0, 1]$ can be shown to be equal to the uncountable set of the base-3 numbers between $0$ and $1$ with the digits after the decimal point being $0$ or $2$.
But, suppose we remove $1/\pi$-ths. Can we deduce from the above that the Cantor set obtained this way will still be uncountable?
I am aware of the fact (not versed in the proof; it's too complicated yet) that perfect sets are uncountable, and I can show that these generalized Cantor sets removing $r$-th fraction in each step ($0 < r < 1$) are perfect.
I want to know if something like the "decimal representation" proof is possible here, or if the uncountability can be deduced from that in the $r = 1/3$ case.