Loading web-font TeX/Math/Italic

ページ

2013/11/20

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.



0 件のコメント:

コメントを投稿