countable sets 2

We accepted that although a set (for example rational numbers) was infinite, it could be countable. But all infinite sets are not countable. We should show that real numbers are uncountable.

At first, we show that $\mathbb{R}=(-\infty,\infty)$ is corresponding to open interval $(0,1)$. It is very simple, because we have only to arrange the function
$$f(x)=\tan \left(x-\frac{1}{2}\right)\pi, \quad x\in (0,1)$$
It may be kind of weird. But the function must have been familiar to you.

Hence, we decide to show that open interval $(0,1)$ is uncountable. The proof is by contradiction. Let the open interval $(0,1)$ be the set $I$. If $I$ is countable, then all elements $x$ of $I$ can be listed in any order.

$x_1=0.c_{11}c_{12}c_{13}\cdots$
$x_2=0.c_{21}c_{22}c_{23}\cdots$
$x_3=0.c_{31}c_{32}c_{33}\cdots$
$\quad \cdots \cdots \cdots$
$x_n=0.c_{n1}c_{n2}c_{n3}\cdots$
$\quad \cdots \cdots \cdots$

In here, $c_{ij}$ is a single digit figure from $0$ to $9$ and suppose that for a $i$, all $c_{ij} (j=0,1,\cdots \infty)$ of $x_i$ are not nine and are not zero, because $0.999\cdots =1$ and $0.000\cdots =0$ are out of interval.

This list is with no limit. However, after listing, we are able to make a new number $y$ by the following procedure.

Let $y$ be $0.d_1d_2d_3\cdots$ where $d_i (i=0,1,\cdots \infty)$ is also a single digit figure. Then, we pick a number different from $c_{11}$ as $d_1$. Next, we pick a number different from $c_{22}$ as $d_2$, and also do a number different from $c_{33}$ as $d_3$. Furthermore we keep carrying on this choice with no limit. As a result, the number $y$ becomes a new number different from every $x_i$. Although all $x_i$ is all elements of $I$, we have gotten a new number $y$. It is a contradiction. Therefore, open interval $I$ is not countable and real numbers become uncountable.

It is the famous Cantor's diagonal argument. How do you feel about two procedures with no end?