Because it's proof is very long and a little bit tedious for blogs.
We will see a brief outline of the construction.
Let us make a subset $\alpha$ in rational numbers $\mathbb{Q}$ as follow.
(1)$\alpha\ne \phi,\mathbb{Q}$ .
(2)$\alpha$ is, for a number $a$, $\left\{x\in\mathbb{Q} : x\lt a \right\}$ .
(3)There is not a maximum number in $\alpha$ .
You must remember that all rational numbers is completely ordered "$\lt$" .
Therefore, if $y\lt a$ ,then $y\in \alpha$ and if $y\ge a$ ,then $y\notin\alpha$ .
As rational numbers are dense, if $y\lt a$ , then there must exists a $z$ such that $y\lt z\lt a$ .
Thus, $\alpha$ has not a maximum number.
By the number $a$ , rational numbers $\mathbb{Q}$ are separated to two sets $\alpha$ and $\alpha^c$ .
(You may prefer to think the both-sides open interval $\alpha=(-\infty, a)$ in the line of $\mathbb{Q}$ .
Please note that $\alpha$ does not have a point $a$ . )
It was called the 'Dedekind cut' $\alpha$ in the preceding post.
When we give many rational numbers to $a$ , there can be many 'Dedekind cuts' $\alpha$ .
We will express the set of all 'Dedekind cuts' $\mathcal{C}$ .
That is, $\mathcal{C}$ has the all intervals in which the left-side is $-\infty$
and the right-side is a rational numbers.
It is an astonishing fact that the element of the set $\mathcal{C}$ is a real number.
(We have to recognize the term "all" is much useful. )
