Provide a bijection between power set of natural numbers and the Cantor set in $[0,1]$ 
Question: I am trying to prove that Cantor Middle Third Set $C$ is uncountable, by establishing a bijection $f$ from $C$ onto the power set of $\mathbb{N}.$

My attempt: We know that every element of Cantor set $C$ has a ternary representation.
Let $\mathcal{P}(\mathbb{N})$ be the power set of $\mathbb{N}.$
Define $f:C\to\mathcal{P}(\mathbb{N})$ by
$$f((a_1,a_2,a_3,...,a_n,...)) = \{n\in\mathbb{N}:a_n\neq 0 \},$$
where $a_n\in\{0,2\}$ for all natural number $n.$
Clearly $f$ is well-defined and injective.
To prove that $f$ is surjective, given any subset $X\in \mathcal{P}(\mathbb{N}),$ we pick $x \in C$ such that non-zero entries of $x$ corresponds to numbers in $X.$
For example, if $X=\{1,4,6\},$ then we pick $x = (2,0,0,2,0,2,0,0,0,...).$
Is my proof correct?

Here's another proof:
First, note that we can create a bijection between $C$ and all the decimals written using only 0s and 1s by switching all the 2s to 1s. Now consider all those decimals in base 2. Those are precisely all the rational numbers between 0 and 1, so this is a bijection between $C$ and $[0,1]$, which has cardinality $\aleph_1 = |\scr P(\mathbb N)|$.
