Countability of Real Numbers

To prove that the interval $[0, 1]$ is not countable, we will use nested intervals. We will demonstrate it by reduction to the absurd, assuming that the interval is countable. So in that case, we can list all the numbers in the interval $[0, 1]$ as $\{x_0, x_1, x_2, \ldots, x_n, \ldots\}$

First of all, let’s divide interval $[0,1]$ in three parts: $[0, 1/3]$, $[1/3, 2/3]$, and $[2/3, 1]$. In that case we can find one of the intervals that does not contain $x_1$, as could happen that $x_1$ is an extreme an interval. Let’s take interval $I_0=[a_0, b_0]$ such that $x_0 \notin I_0$.

Now, let’s divide $I_0$ into three parts: $[a_0, a_0 + \frac{1}{9}]$, $[a_0 + \frac{1}{9}, a_0 + \frac{2}{9}]$, and $[a_0 + \frac{2}{9}, b_0]$. In that case we can find one of the intervals that does not contain $x_1$, as could happen that $x_1$ is an extreme of an interval. Let’s take interval $I_1=[a_1, b_1]$ such that $x_1 \notin I_2$.

Proceeding the same way we can divide $I_n$ into three parts:

$[a_n, a_n + \frac{1}{3^{n+2}}], [a_n + \frac{1}{3^{n+2}}, a_n + \frac{2}{3^{n+2}}], [a_n + \frac{2}{3^{n+2}}, b_n]$

In that case we can find one of the intervals that does not contain $x_{n+1}$, as could happen that $x_{n+1}$ is an extreme of an interval. Let’s take interval $I_{n+1}=[a_{n+1}, b_{n+1}]$ such that $x_{n+1} \notin I_{n+1}$.

We have build a succession of nested intervals:

$[a_0, b_0] \supset [a_1, b_1] \supset [a_2, b_2] \supset \ldots \supset [a_n, b_n] \supset \ldots$

And also, as we said $x_n \notin I_n$ for all $n$.

Using the Nested Interval Theorem, we know that:

So there is a number $x$ that belongs to all the intervals $I_n$. But we know that $x \notin I_n$ for all $n$. So we have reached a contradiction, and the interval $[0, 1]$ is not countable, as we wanted to prove.