Prove: $|\mathcal{P}( \{0, 1\}\mbox{*}) |=|\left[0, 1\right]|$

Question

So I need to prove the following:

The cardinality of the real interval from 0 to 1 equals the cardinality of the power set of all finite words over the boolean set. In short:

Prove: $ | \mathcal{P}( \{0, 1\}\mbox{*}) | =|\left[0, 1\right]| $

I understand the proof of $ | \mathcal{P}( \{0, 1\}\mbox{*}) | \geq|\left[0, 1\right]| $ where one creates a set from the real number, and thus can inject it into the power set, but I have no idea on how to prove the inverse. Any ideas appreciated :)

Answer 1

Let the boolean set be $\{1,2\}$ instead, and interpret a word there as a natural number in base three. Then the set of such sequences may be interpreted as a certain subset $B$ of the natural numbers.

We want an injective function $f$ that takes as input a set $N \subseteq B$, and gives us a number $r\in[0,1]$. We define $$ f(N) = \sum_{i = 1}^\infty n_i\cdot 10^{-i} $$ where $n_i = 3$ is $i\notin N$ and $n_i = 7$ if $i \in N$. Since the sequence $(n_i)$ uniquely identifies $N$, we must have that $f$ is injective.

This gives you an injection either way, which means that the sets have the same cardinality.