Cantor set is uncountable

Definition of the Cantor set: see https://en.wikipedia.org/wiki/Cantor_set

 

Theorem. The Cantor set is uncountable.

proof) Let $\mathcal{C}$ be the Cantor set. Let $J_0=\mathcal{C_0}$. In the process to get the Cantor set, $J_0$ is divided into two line segments($[0,\frac{1}{3}],[\frac{2}{3},1]$). Take one of the two. Let the chosen segment be $J_1$. Of course, $J_1$ also can be divided into two line segments. Again, take one of the two. Let the chosen segment be $J_2$. Repeat this process. Then we get a nested interval $(J_n)_{n=0}^{\infty}$ such that $\bigcap_{n=0}^{\infty}J_n=c$ for some $c\in\mathcal{C}$. Clearly, every element in $\mathcal{C}$ can be represented as in that way by the definition. Now, consider the sequence such that following:

$s_0=0$, $s_n=\begin{cases}
0 \text{ if } J_n=\frac{1}{3}J_{n-1} \\
1 \text{ if } J_n=\frac{1}{3}+\frac{1}{3}J_{n-1}
\end{cases}$.

Then for any $c\in\mathcal{C}$, there exists a nested inverval and for the nested interval, there exists a sequence such that above. And it is easy to know that for distinct $(a_n)$ and $(b_n)$, there exists distinct nested interval $(A_n)_{n=0}^{\infty}$ and $(B_n)_{n=0}^{\infty}$ such that $\bigcap_{n=0}^{\infty}A_n=c_1\in\mathcal{C}$, $\bigcap_{n=0}^{\infty}B_n=c_2\in\mathcal{C}$, and $c_1\neq c_2$. These imply that $\left | \mathcal{C} \right |=\left | 2^{\mathbb{N}} \right |$. $\square $

 

Not a rigorous proof. 맞으려나 모르겠다.