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

Note the "obvious" bijection between $C$ and $P(\mathbb N)$ defined as:
$$f(A)= 2\sum_{n=1}^\infty\frac{\chi_A(n)}{3^n}$$
Where $\chi_A$ is the characteristics function of $A$ ($1$ for $n\in A$, $0$ otherwise).
Suppose that $A\neq B$ and $x=\min (A\Delta B)$, wlog $x\in A$ then $f(A)-f(B)\ge\dfrac2{3^x}>0$.
In the other direction, if $x\in C$ then we can write $A=\{n\in\mathbb N\mid a_n=2\}$ and it is quite clear why $f(A)=x$.
Now we use the fact that $P(\mathbb N)$ is uncountable (direct result of Cantor's theorem) and we are done.
