$\Omega=\{0,1\}^\infty$
$A=\{\omega:\omega=(\omega_n)_{n\ge{1}}\in\Omega\}$
$f:A\longrightarrow(0,1)$ where, $f(\omega)=\underset{n\ge{1}}{\sum}2^{-n}\mathbb{I}_{\{\omega_n=1\}}\in(0,1)$
(doubt) Why not $[0,1]$?
I can take $(0,0,0...)$ for $0$ and $(1,1,1...)$ for $1$.
Show that the map $f$ satisfies $f(\omega_1)=f(\omega_2)$ for $w_1\ne\omega_2$ iff $f(\omega_1)$ is a dyadic rational in $(0,1)$.
My approach:
For the if$(\Leftarrow)$ part, I can think of an example-
$\omega_1=(1,1,0,0,0...$ $_{terminating})$
$\omega_2=(1,0,1,1,1...$ $_{recurring})$
But, how do I prove that for any particular dyadic rational, there are $2$ pre-images, one terminating and another recurring? Will there be exactly $2$ or more could there be more than $2$?
I am unable to think of an argument for the only-if$(\Rightarrow)$ part.
Any dyadic rational is of the form $ \sum\limits_{k=1}^{n} \frac {a_k} {2^{k}}$ with each $a_k \in \{0,1\}$. We may suppose $a_n \neq 0$ which means $a_n=1$. Writing $1=\sum\limits_{k=1}^{\infty} \frac 1 {2^{k}}$ we get a second expansion.Every dyadic rational has exactly two expansions and every other number in $(0,1)$ has a unique expansion. For the uniqueness results assume $ \sum\limits_{k=1}^{\infty} \frac {a_k} {2^{k}}= \sum\limits_{k=1}^{\infty} \frac {b_k} {2^{k}}$ and consider the first $m$ for which $a_m \neq b_m$. Estimate the difference between $ \sum\limits_{k=m+1}^{\infty} \frac {a_k} {2^{k}}$ and $ \sum\limits_{k=m+1}^{\infty} \frac {b_k} {2^{k}}$ using the fact that $|a_k-b_k| \leq 1$. I leave it to you to complete the argument from here.