I need to show that the interval $[0,1[$ is uncountable.
The simple way is to construct a bijection between $\{0,1\}^{\mathbb{N}^*}$ and $[0,1[$.
Basically, using the Cantor's diagonal argument, if there is a bijection between $\{0,1\}^{\mathbb{N}^*}$ and $[0,1[$ then $[0,1[$ is uncountable! So, I need to prove that there is a bijection!
Using the Schröder–Bernstein theorem, if there exist injective functions $f: \{0,1\}^{\mathbb{N}^*} \rightarrow [0,1[~$ and $g: [0,1[ \rightarrow \{0,1\}^{\mathbb{N}^*}$ between the sets $\{0,1\}^{\mathbb{N}^*}$ and $[0,1[$, then there exists a bijective function h.
My problem is that I can't prove the second part! i.e to show that there injective functions satisfy the required relations.
Any help is appreciated.
Best Answer
Take an element in $\{0,1\}^{\Bbb N^*}$, and interpret it as the (base ten) decimals in a number between $0$ and $1$. That gives you an injection $\{0,1\}^{\Bbb N^*}\to [0,1[$. For the other direction, for each number in $[0,1[$, take its binary representation, and there you have an element of $\{0,1\}^{\Bbb N^*}$. This is an injection.
PS. Note that a few elements of $[0,1[$ have two binary representations, analoguous to the well-known $0.999\ldots = 1$ thing. You can decide whether you want to go with the one with infinitely many $0$'s, or the one with infinitely many $1$'s, but in principle you have to pick one and stick with it (or give some rule for determining which cases you will use one and which cases you will use the other). This is the reason you can't just go the easy way and declare that interpreting an element of $\{0,1\}^{\Bbb N^*}$ as a number in binary is a bijection. You have to use a higher base in one direction to avoid this. And then stitch together the two injections to some nasty amalgamation using Schröder-Bernstein.