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Prove that every Cauchy sequence in $\mathbb{C}$ is bounded.

In $\mathbb{R}$, this is a sketch of the proof that I recall:

Let {${a_k}$} be Cauchy in $\mathbb{R}$, since $1\in\mathbb{R}$, $\exists N$ s.t. $\forall m,n>N$, $|a_n-A_N|<1\rightarrow$|a_n|-|A_N|<|a_n-A_N|<1\iff|a_n|<1+|a_N|,\forall n>N-1. Let $M = \max{|a_1|,|a_2|,\ldots,|a_N-1|,1+|a_N|}$. Then, $M$, $-M$ bound {$a_k$}.

A sequence is bounded in $\mathbb{C}$ if $\exists R\in\mathbb{R}$ and an integer $N$ s.t. $|z_n| $\forall, n>N$. Here's my attempt at the proof at hand then:

Let {${z_n}$} be Cauchy in $\mathbb{C}$. I want to show that there exists an R s.t. that definition above is satisfied. Is this R just the $M$ from the proof in $\mathbb{R}$?

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    Emir, how can possibly «this $R$ be just the $M$ from the proof in $\mathbb R$»?! In the proof in $\mathbb R$ there were $a$s, in the new instances there are $z$s... Can you please write out in detail what you have tried to do?2012-01-25

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To say that $-M$ and $M$ are respectively lower and upper bounds on the sequence $\{a_k\}$ is the same as saying $M$ is an upper bound on the sequence $\{|a_k|\}$. Think about how all that applies to $\mathbb{R}$ and then to $\mathbb{C}$.

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    I don't think you need that property. This proof would work just as well for $\mathbb{Q}$.2012-01-25