Let (An) be a Cauchy sequence such that n is a member of N, and let c be a member of R, prove that (c*An) is also a Cauchy sequence

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    $\begingroup$ In the definition take $N$ s.t. $|A_n-A_m|<\frac{\varepsilon}{|c|}$ when $n,m>N$ and the claim follow. $\endgroup$ – user386627 Nov 3 '17 at 12:27
  • $\begingroup$ And if $c<0$ ??? $\endgroup$ – Fred Nov 3 '17 at 12:29
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    $\begingroup$ @Fred: I just gave a comment, not an answer... The OP can adapt properly I'm sure ;-) By the way, in what your answer give something more than my comment ? $\endgroup$ – user386627 Nov 3 '17 at 12:30

If $c=0$, then we are done.

Let $c \ne 0$. Then

$|cA_n-cA_m|=|c||A_n-A_m| < \epsilon \iff |A_n-A_m|< \frac{\epsilon}{|c|}$.

Can you procced ?

  • $\begingroup$ Does this not prove its a cauchy sequence, doesn't it satisfy the definition of a cauchy sequence? $\endgroup$ – Sam Heslop Nov 3 '17 at 12:45
  • $\begingroup$ If (A_n) is Cauchy and $ \epsilon >0$, then there is $N$ such that $|A_n-A_m|< \frac{\epsilon}{|c|}$ for all $n,m >N$. Hence we have $ |cA_n-cA_m|< \epsilon$ for all $n,m >N$, which shows that $(cA_n)$ is Cauchy. $\endgroup$ – Fred Nov 3 '17 at 12:48
  • $\begingroup$ Thank you very much been struggling to get my head around this definition $\endgroup$ – Sam Heslop Nov 3 '17 at 12:49

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