# Cauchy Sequence proof

• Mar 31st 2013, 04:39 PM
rokman54
Cauchy Sequence proof
I don't understand a part of this proof.

A Cauchy Sequence is bounded.

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

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

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

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

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

Thank you.
• Apr 1st 2013, 05:29 AM
HallsofIvy
Re: Cauchy Sequence proof
Because this is a Cauchy sequence, you know that for some N, if n,m> N, then $\displaystyle |a_n- a_m|< 1$. In particular, if $\displaystyle a_n> a_{N+1}$, we have $\displaystyle |a_n- a_{N+1}|= a_n- a_{N+1}< 1$ so that $\displaystyle a_n< a_{N+1}+ 1$. So $\displaystyle a_{N+1}+ 1$ is an upper bound for all terms with n> N.

The part you are asking about, where they look at $\displaystyle |a_1|$, up to $\displaystyle |a_N|$ is the easy part- those together with the bound $\displaystyle a_{N+1}+ 1$ 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.