I want to prove that every real number $x \in [0,1]$ has binary representation in the following way:
Let $B$ denote the set of all sequences $b:\mathbb{N} \rightarrow\{0,1\}$. Consider then $f:B\rightarrow [0,1]$ defined by $$ f(b)=\sum_{k=1}^\infty b_k 2^{-k}. $$ Then I want to prove that $f$ is surjective.
We can note that $f$ is indeed well-defined as by the comparison criterion then $$ 0\leq \sum_{k=1}^\infty b_k 2^{-k} \leq \sum_{k=1}^\infty 2^{-k} = 1. $$ and hence also convergent for all $b \in B$.
I have thought about three ideas for a proof:
(i) If we could show $f$ is continuous, then by the mean value property, this would hold. However this requires some thought into what are open sets in $B$ so I don't think this is the easiest route to take.
(ii) Using a similar idea as in (i), maybe we could somehow take a $b$ such that fraction of $b$'s elements that are 1 is exactly the number $r\in[0,1]$ we are trying to hit. However I can't quite construct this $b$.
(iii) The final and I think perhaps most promising idea is to use that we know every real number has decimal representation in the base-10 system. If we were somehow able to translate this representation into this one, then we would be done. Yet this has only just been trial and error so far for me.
Am I on the right track or is there something I am missing?
The same proof that shows each number has a decimal expansion also works for any other basis.
However, if you don't want to just copy an existing proof, you can define $b_i$ exactly like so:
Let $n\in\mathbb N$ be arbitrary. Now, define $x_{n-1} = \sum_{k=1}^{n-1} b_k2^{-k}$.
Now, define
$$b_n = \begin{cases}0&\text{if}& x-x_{n-1} < 2^{-n}\\ 1 &\text{if}&x-x_{n-1} \geq 2^{-n}\end{cases}$$
and you can then prove that, by construction, $x_n$ converges to $x$.