Proving the open interval $(0,1)$ is uncountable I am currently able to prove this statement using the Cantor diagonalisation argument, my question is whether there is another way (more simple or more complex) to prove this statement, without diagonalisation?
 A: You can do this by showing that there is a bijection between $(0, 1)$ and $\mathbb R$.
Two sets are equivalent (have equal cardinalities) if and only if there exists a bijection between them. $\mathbb R$ is uncountable. So by showing that there exists a bijection from $(0, 1)$ to $\mathbb R$, you thereby show that $(0, 1)$ is uncountable.
However, it is perhaps more common that we first establish the fact that $(0, 1)$ is uncountable (by Cantor's diagonalization argument), and then use the above method (finding a bijection from $(0, 1)$ to $\mathbb R)$ to conclude that $\mathbb R$ itself is uncountable.
A: No one mentioned Baire ?
Suppose for contradiction that $(0,1)$ is countable.
Let $x_n$ be a sequence such that $(0,1)=\{x_n|n\in \mathbb N\}$
Then $(0,1)=\cup_n \{x_n\}$
Each $\{x_n\}$ is a closed, nowhere dense, subset of $(0,1)$ with the usual metric.
Hence, by Baire category theorem, $(0,1)$ is nowhere dense.
This is a contradiction.
A: You can start by showin that $2^{\mathbb N}$ is uncoutable quite easily, using the general proof that there is no bijection between $E$ and $2^E$ for any set $E$.
This is because if such a bijection would exist, the set $M=\{x\in E| x\notin f(x)\}$ could have no antecedent $m$ by $f$, because both $m\in M$ and $m\notin M$ lead to a contradiction.
Then, there is an easy injection $2^{\mathbb N}\to (0,1)$, showing that $(0,1)$ is also uncountable: associate to a set $X\subseteq N$ the number whose $i^{th}$ digit is for instance $2$ if $i\in X$ and $5$ otherwise (to avoid reaching $0$ and $1$).
