08 - Uncountability of ℝ

Uncountability of ℝ

$$\mathbb{R}\text{ is uncountable}$$

Proof

It is sufficient to show that $(0,1]$ is uncountable
We know that $(0,1]$ is uncountable by the archimedean property

Suppose, for a contradiction, that $(0,1]$ is countably infinite so
Let the elements be $x_1,x_2,\cdots$
Each has a non-terminating decimal expression

$$\begin{array}{rl}{x}_1& =0.a_{11}{a}_{12}{a}_{13}{a}_{14}\cdots \\ x_2& =0.a_{21}{a}_{22}{a}_{23}{a}_{24}\cdots \\ x_3& =0.a_{31}{a}_{32}{a}_{33}{a}_{34}\cdots \\ ⋮& \\ x_k& =0.a_{k1}{a}_{k2}{a}_{k3}{a}_{k4}\cdots \\ ⋮& \end{array}$$

Construct a real number $x\in (0,1]$ with decimal expansion $0.b_1{b}_2{b}_3\cdots$
where

$$b_k=\{\begin{array}{ll}5& \text{if }{a}_{kk}=6\\ 6& \text{if }{a}_{kk}\ne 6\end{array}$$

Then $x_k\ne x$ for all $k$ as it differs at the $k^{\text{th}}$ decimal place
Hence $x$ is not in the list so it’s a contradiction
We pick $5$ and $6$ just to avoid $0$ or $9$