Showing $C=\bigg\{ \sum_{n=1}^\infty a_n 3^{-n}: a_n=0,2 \bigg\}$ is uncountable Let us define the set
$$C=\bigg\{ \sum_{n=1}^\infty a_n 3^{-n}: a_n=0,2 \bigg\}$$
This is the Cantor set, could anyone help me prove it is uncountable? I've been trying a couple of approaches, for instance assume it is countable, list the elements of $C$ as decimal expansions, then create a number not in the list, I am having trouble justifying this though.
Secondly i've been trying to create a function $f$ such that $f(C)=[0,1]$.
Many thanks

Why list the numbers as their decimal expansion?
Given a countable list $((a_n^m)_n)_m$ of sequences with values $0$ and $2$,
construct another sequence $(b_n)_n$ that does not appear in the list.
You can do that just as in Cantor's proof that the unit interval is uncountable.
For the surjection from $C$ onto $[0,1]$, notice that every element of $C$ uniquely determines the sequence $a_n$ it comes from.
Now map every sequence $(a_n)_n$ to the sequence $(b_n)_n$ where $b_n=0$ if $a_n=0$ and $b_n=1$ if $a_n=2$.
Now consider $\sum_{n=1}^\infty b_n2^{-n}$.
