TITLE: Prove that the open interval $(0, 1)$ contains uncountably infinite numbers.
QUESTION [5 upvotes]: Prove that the open interval $(0, 1)$ contains uncountably infinite numbers.

Apparently, there is a way to prove this proposition using Cantor's diagonalization argument.
How does that work? How does looking at the diagonal of a big grid of numbers help? Is there a more straightforward proof?

REPLY [4 votes]: $\newcommand{\N}{\mathbb N}$
Let $f:\N\to (0,1)$ a function. Given $n\in\N$ consider the infinite representation of $f(n)$, $$f(n)=0.a^n_1a^n_2\ldots a_n^n\ldots$$ in base $10$. This infinite representation is unique.
For each $n\in\N$ define $$b_n=\begin{cases} 5 & \text{if} &a_n^n\neq 5\\ 7 & \text{if} & a_n^n=5\end{cases}$$ we will see that  $y=0.b_1b_2b_3\ldots$ can not be in the image of $f$. 
Suppose that there is an $l\in\N$ such that $$y=f(l)=0.a_1^la_2^l\ldots a_l^l\ldots$$
Then $a_l^l=b_l$, i.e.
$$a_l^l=\begin{cases} 5&\text{if}& a_l^l\neq 5\\ 7&\text{if}&a_l^l=5\end{cases}$$
you can see that such an $l$ is not possible.
Therefore $f$ is not surjective and $(0,1)$ is not numerable.