Answer to Question #159452 in Calculus for Vishal

Question

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

Answer 1

Let $(a_n)_ {n = 1}^{n = \infty}$ be a Cauchy sequence. This implies that given $\epsilon > 0$ $\exist$ $N \in \mathbb{N}$ $\ni$ $\mid a_n - a_m \mid < \epsilon$ whenever $n,m \geq N$ . When $m = N$ , we have that $\mid a_n - a_N \mid < \epsilon$

But $\mid a_n \mid - \mid a_N \mid \leq \mid \mid a_n \mid - \mid a_N \mid \mid \leq \mid a_n - a_N \mid < \epsilon$

$\mid a_n\mid < \epsilon + \mid a_N \mid$

Let $K := \max (\mid a_0 \mid, \mid a_1 \mid, \mid a_2 \mid, \cdot \cdot \cdot , \mid a_{N-1} \mid, \mid a_N \mid, \epsilon + \mid a_N \mid )$

So we have that $\mid a_n \mid \leq K$ as desired.