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Suppose {$x_n$} is a Cauchy sequence and if $c\in \Re$, prove that the sequence {$cx_n$} is Cauchy.

My attempt: Let $\epsilon > 0$. Since {$x_n$} is Cauchy there is some $N$ so that for all $n> m> N$, $|x_n - x_m| < \epsilon$.

So $|cx_n - cx_m|$ = $|c||x_n - x_m|$ < $|c|\epsilon$.

Then I don't know what to do from here.

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Is $\{x_{n}\}$ Cauchy to start? You are going to have a hard time otherwise. Nevermind – TheNumber23 Sep 28 '13 at 22:24
Yes, I've edited it. – user87274 Sep 28 '13 at 22:28

1 Answer 1

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If $c=0$ then $\{cx_{n}\}$ is the zero sequence which is clearly Cauchy. So assume $c\neq 0$. Then for all $\tfrac{\epsilon}{|c|}>0$ there exists $N\in\mathbb{N}$ such that for all $n,m>N$, $|x_{n}-x_{m}|<\tfrac{\epsilon}{|c|}\Rightarrow |cx_{n}-cx_{m}|<\epsilon$.

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I have a dumb question: If $c = 0$, then how can you tell that the {$0$} sequence is a Cauchy? – user87274 Sep 29 '13 at 22:47
Cauchy means for all $\epsilon$ there exists $N$ such that if $n,m>N$ then $|x_{n}-x_{m}|<\epsilon$ Since every term is the same their difference is zero. All constant sequences converge which means they are Cauchy. – TheNumber23 Sep 29 '13 at 23:46

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