>>333
>contains countably infinite digits
You're confusing the cardinality of sets with natural numbers.
Let your set of 0.1* numbers be S.
\( S \cong \mathbb{N} \) by the counting the ones in each decimal number.
\( 0.\overline{1} \) has \( \aleph_0 \) digits, and \( \aleph_0 \notin \mathbb{N} \), therefore \( 0.\overline{1} \notin S \).
Just because something is countably infinite, doesn't mean you can reach it by counting.
A more simple refutation would be so say that by nature of the construction of S, every number in it that is not the first has a single unique predecessor that you can follow back to 0.
The predecessor of \( 0.\overline{1} \) would be \( 0.\overline{1} \), which clearly puts it ouside S.
This is a lot more intuitive if you imagine a graph of S with each number connected to the next.
This forms a straight line stretching from 0 outwards.
\( 0.\overline{1} \) is in a single node graph who's only edge is to itself, it clearly can not connect to the line.