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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. $|z_n|<R$ $\forall, n>N$. 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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    $\begingroup$ Does it? Where do you think it might stop working? You have to show us a little bit of what you have tried, for us to be able to help you with it. $\endgroup$ Jan 24, 2012 at 21:39
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    $\begingroup$ See also: math.stackexchange.com/questions/95442/… $\endgroup$
    – user940
    Jan 24, 2012 at 21:46
  • $\begingroup$ What is a bounded sequence in $\mathbb{C}$? $\endgroup$
    – leo
    Jan 24, 2012 at 22:55
  • $\begingroup$ 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? $\endgroup$ Jan 25, 2012 at 3:28

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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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  • $\begingroup$ Doesn't the proof implicitly use the lub property of $\mathbb{R}$? This clearly wouldnt work in $\mathbb{C}$ since it's not an ordered field. $\endgroup$
    – Emir
    Jan 24, 2012 at 22:12
  • $\begingroup$ I don't think you need that property. This proof would work just as well for $\mathbb{Q}$. $\endgroup$ Jan 25, 2012 at 4:31

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