Proving the open interval $(0,1)$ is uncountable

Question

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?

Answer 1

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.

Answer 2

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 $1$).

Answer 3

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.