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.