real number system 8 (nested intervals)

We shall prove that a contracting closed interval becomes a point. It is definitely no doubt. But it is only enabled by infinite operations and is in the base of the real number system.

For any bounded closed interval sequences $I_n=[a_n,b_n], (-\infty<a_n\leq b_n<\infty)$ , if $I_n\supset I_{n+1}, (n\in\mathbb{N})$ and $|a_n-b_n|\rightarrow 0$ as $n\rightarrow \infty$, there exists a real number $c$ such that
\[ \bigcap_{n=1}^{\infty}I_n = c \]
In other words, if the real number sequence $a_1,a_2,\cdots $ is monotonic increasing bounded above, the bounded real number sequence $b_1,b_2,\cdots $ is monotonic decreasing bounded below and $a_n\leq b_n$ for any $n$ , there is a real number $c$ such that $a_n\leq c\leq b_n$ . Furthermore, if $|a_n-b_n|\rightarrow 0$ as $n\rightarrow \infty$, $c$ is a point. That is to say, $\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n=c$ .

In real number system, as a bounded monotone real number sequence has a limit number, we put $\lim_{n\rightarrow\infty}a_n=a,\quad \lim_{n\rightarrow\infty}b_n=b$ . Hence, for any $n$ ,
\[ a_n\leq a\leq b\leq b_n \]
and
\[ I_n\supset [a,b]  \]
It means $\bigcap_{n=1}^{\infty}I_n\ne \phi$ . There is a real number $c\in [a,b]$ .

As $a_n\leq c\leq b_n$ for any $n$ , $|a_n-c|\leq b_n-a_n$ . Therefore $a=c$, because $|a_n-b_n|\rightarrow 0$ . Similarly $b=c$ .

It is called the method of nested intervals. By the method we are able to get a root of some kinds of equations computationally.