I'm trying to show that the middle third Cantor set expressed in base 3 mapped to the set of real numbers in the interval $[0,1]$ expressed in base 2 according to the rule:
$0.a_1a_2a_3...\rightarrow 0.b_1b_2b_3...$
where $b_i=\frac{a_i}{2}$ is not a bijection.
I know that the map is surjective and not injective because some binary decimal has 2 expansions. However, I don't know how to prove this.
Note that each $x \in C$ has a unique representation $x = 0.a_1a_2a_3...$ with $a_i \in \lbrace 0, 2 \rbrace$ wheras some $y \in I$ have two distinct representations $y = 0.b_1b_2b_3...$ with $b_i \in \lbrace 0, 1 \rbrace$ (see Bungo's comment). In other words you have a bijection $\varphi : \Pi_{i=1}^\infty \lbrace 0, 2 \rbrace \to C$ and a surjection $\psi : \Pi_{i=1}^\infty \lbrace 0, 1 \rbrace \to I$ which is not injective. Your map is the composition $\psi \circ b \circ \varphi^{-1}$, where $b :\Pi_{i=1}^\infty \lbrace 0, 2 \rbrace \to \Pi_{i=1}^\infty \lbrace 0, 1 \rbrace$ is the bijection given by $b((a_i)) = (a_i/2)$.