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 $\mathbb{R}$?

  • 1
    $\begingroup$ 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. $\endgroup$ – Mariano Suárez-Álvarez Jan 24 '12 at 21:39
  • 2
    $\begingroup$ See also: math.stackexchange.com/questions/95442/… $\endgroup$ – user940 Jan 24 '12 at 21:46
  • $\begingroup$ What is a bounded sequence in $\mathbb{C}$? $\endgroup$ – leo Jan 24 '12 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? $\endgroup$ – Mariano Suárez-Álvarez Jan 25 '12 at 3:28

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 $\mathbb{C}$.

  • $\begingroup$ 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 '12 at 22:12
  • $\begingroup$ I don't think you need that property. This proof would work just as well for $\mathbb{Q}$. $\endgroup$ – Michael Hardy Jan 25 '12 at 4:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.