The Set of Real Numbers is Uncountable - Nested Intervals Proof

# The Set of Real Numbers is Uncountable - Nested Intervals Proof

On The Set of Real Numbers is Uncountable page we proved that $\mathbb{R}$ is uncountable using a diagonal argument. We now have the tools to prove that this set is uncountable using the nested intervals theorem.

Theorem 1: The set $\mathbb{R}$ is uncountable. |

**Proof:**Again we show that $[0, 1]$ is uncountable by contradiction.

- Suppose that $[0, 1]$ is countable. Then we can enumerate the set as $[0, 1] = \{ x_1, x_2, ... \}$. Let $I_1 \subset [0, 1]$ be a closed interval such that $x_1 \not \in I_1$. Furthermore, let $I_2 \subset I_1$ be a closed interval such that $x_2 \not \in I_2$. For each $n \in \mathbb{N}$ let $I_n \subseteq I_{n-1}$ be a closed interval such that $x_n \not \in I_n$. Observe that $I_n$ is bounded for each $n \in \mathbb{N}$ since $[0, 1]$ is bounded. Then we obtain the following nesting of closed and bounded intervals:

\begin{align} \quad I_1 \supset I_2 \supset I_3 \supset ... \supset I_n \supset ... \end{align}

- By the nested intervals theorem:

\begin{align} \quad \bigcap_{n=1}^{\infty} I_n \neq \emptyset \end{align}

- So there exists an $m \in \mathbb{N}$ such that $x_m \in I_n$ for every $n \in \mathbb{N}$. But $x_m \not \in I_n$ for all $n \geq m$ which is a contradiction. Therefore the assumption that $[0, 1]$ is countable is false. So $[0, 1]$ is uncountable.

- Since $[0, 1]$ is uncountable and $[0, 1] \subset \mathbb{R}$ we have that $\mathbb{R}$ is uncountable. $\blacksquare$