Question

Prove [0,1] is not countable without using the outer measure of an interval is its length?

Answer 1

Answer on Question #45826 – Math – Real Analysis

Problem.

Prove [0,1] is not countable without using the outer measure of an interval is its length?

Solution:

Suppose that [0,1] is countable.

If $[0,1]$ is countable, then $[0,1] = (x_n)_{n \geq 0}$. We will construct the decreasing sequence of compact intervals. First we split $[0,1]$ into three equal parts $\left[0, \frac{1}{3}\right], \left[\frac{1}{3}, \frac{2}{3}\right], \left[\frac{2}{3}, 1\right]$. $x_0$ is not in one of this intervals. We denote this interval by $[a_0, b_0]$. Now we split $[a_0, b_0]$ into three equal parts $I_1, I_2, I_3$. $x_1$ is not in one of the given intervals. We denote this interval by $[a_1, b_1]$. In the same way we may construct interval $[a_n, b_n]$ for all nonnegative integer $n$. From the construction of the intervals we may notice that $[a_{n+1}, b_{n+1}] \subset [a_n, b_n]$, $x_{n+1} \notin [a_{n+1}, b_{n+1}]$ and $b_{n+1} - a_{n+1} = \frac{1}{3}(b_n - a_n)$ for all nonnegative integer. $([a_n, b_n])$ is a decreasing sequence of compact intervals with

$b_n - a_n \to 0$ (nested intervals), so by Nested Intervals Theorem they have the common point

$P \in [0,1]$. If $[0,1] = (x_n)_{n \geq 0}$, then there exists $m$ such that $x_m = P$, but $x_m \notin [a_m, b_m]$ and therefore it cannot be in the intersection of all intervals, contradiction. Hence $[0,1]$ isn't countable.

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