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I found a proof on here already. I get the general idea, but I'm not sure I understand the need to say $M=max${$N,K$}. Why wouldn't it be sufficient to just say $n,m>N$ and $n_k>K$?

existing proof: If a subsequence of a Cauchy sequence converges, then the whole sequence converges.

Also, if $n_k$ is strictly increasing, is it possible to conclude either $K>N$ or $N>K$?

Many thanks

Community
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chris
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1 Answers1

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Here in the proof the $M=max\{N,K\}$ is needed because this the domain where both the inequalities hold.

happymath
  • 6,297
  • But since "$n,m>N$ and $n_k>K$" is a conjunction, doesn't that inherently mean we are only talking about where both inequalities are true? Wouldn't "for all $n,m,n_k>M$" amount to the same thing as "for all $n,m>N$ and $n_k>K$"? – chris Mar 08 '14 at 06:13
  • @chris yes we want both inequalities to be true i.e. we want to substitute $n_k$ in the other inequality. So the inequality on $n_k$ as well as $n$ should be valid – happymath Mar 08 '14 at 06:23