# Thread: Cauchy Sequence proof

1. ## Cauchy Sequence proof

I don't understand a part of this proof.

A Cauchy Sequence is bounded.

pf) Let ${a_n}$ be a cauchy sequence

then there is N $\in N$ such that Absolute value ( $a_n - a_m) \leq \epsilon$

Since [tex] abs( $a_n) \leq abs(a_N) +1$

Let M = max {abs (a1), abs(a2) , ... abs(aN), abs(an) +1} <-------------- I don't understand this part.

abs(an) $\leq M \forall n \in N$

Thank you.

2. ## Re: Cauchy Sequence proof

Because this is a Cauchy sequence, you know that for some N, if n,m> N, then $|a_n- a_m|< 1$. In particular, if $a_n> a_{N+1}$, we have $|a_n- a_{N+1}|= a_n- a_{N+1}< 1$ so that $a_n< a_{N+1}+ 1$. So $a_{N+1}+ 1$ is an upper bound for all terms with n> N.

The part you are asking about, where they look at $|a_1|$, up to $|a_N|$ is the easy part- those together with the bound $a_{N+1}+ 1$ form a finite set. And a finite set of numbers always contains a "largest member" which must be an upper bound on the entire sequence.