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Thread: Cauchy Sequence proof

  1. #1
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    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.
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  2. #2
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    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.
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