Question

Prove that if a sequence {an}∞ n=1 satisfies Cauchy’s criterion, then it is bounded.

Answer 1

Suppose $(a_n)$ is a sequence satisfied Cauchy's criterion .

Claim : $(a_n)$ is bounded i,e, there exit a $C\in \N$ such that $|a_n|\leq C$ for all $n\in \N$ .

Since $(a_n)$ is a Cauchy's Sequence , therefore for any $\epsilon >0$ there exist a $K\in \N$ such that $| a_n-a_m|<\epsilon$ $\forall m,n > K$ .

We know that ,

$|a_n| = | a_n+a_m-a_m| \leq |a_m|+|a_n-a_m|$ by triangular inequality .

Set $\epsilon =1$ , Because this is cauchy ,

There exist $K\in \N$ such that $|a_n-a_m|\leq 1$ $\forall m,n >K$

Set $m=K+1$ , Combined with our initial note , we can re write the following ,

$|a_n|\leq 1+|a_{K+1}|$ and this is true for all $n>K$

This bound all the term beyond the $Kth$ .

Looking at the term before the $Kth$ term , we can find the maximum of them and notes that this bounds that part of the sequence

$|a_n| \leq Max \{ |a_1|,|a_2|,.........,|a_K|\}$

And this is true for all $n\leq K$ .

By choosing maximum of either $1+|a_{K+1}|$ or the maximum of the aforementioned set we can find our $C$

Which bound all the term of the sequence .