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I have been reading the following post:

Prove: If a sequence converges, then every subsequence converges to the same limit.

I understand the idea, but I wonder, does this proof imply that such a subsequence actually exists?

That is, suppose a sequence $s_n$ converges. Then every subsequence $s_{n_k}$ of $s_n$ converges to the same limit. But my question is: does there necessarily exist such a subsequence $s_{n_k}$?

Community
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    Take the subsequence to equal the sequence. So, you have one.. Remove a term from the sequence, then you have another subsequence of the sequence. Remove a finite number of terms to get a third subsequence. – wannadeleteacct Oct 30 '13 at 03:10

2 Answers2

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Take your sequence $s_n$. It is nonempty. Therefore it has a subsequence.

Newb
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  • What do you mean by "[the sequence is] nonempty"? – Pedro Oct 30 '13 at 03:18
  • @PedroTamaroff So long as the $s_n$ is not the empty sequence. – Newb Oct 30 '13 at 03:43
  • I know this is a long time after the fact, but what exactly is "the empty sequence"? – Cameron Buie Oct 15 '15 at 23:09
  • @CameronBuie a sequence in which the set of points is nonempty? To be entirely honest with you, this is terminology I picked up from a class at the time. I could revisit my notes and see exactly what I meant, but the explanation I just offered seems sufficiently plausible... – Newb Oct 16 '15 at 16:17
  • How can a sequence have an empty set of points, though? A sequence is a function whose domain is the natural numbers, effectively, and there is no function from the natural numbers into the empty set. I would be curious what an empty sequence is supposed to be, then. – Cameron Buie Oct 16 '15 at 17:00
  • @CameronBuie Some people call functions of the form $f:{1,...,n} \to X$ "sequences", but then still, the term "empty sequence" doesn't make much sense... – YoTengoUnLCD Apr 03 '16 at 00:07
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Some simple subsequences of $(s_n)_{n=1}^\infty$ would be

  • the original sequence, $(s_n)_{n=1}^\infty$,
  • the subsequences obtained by removing finitely many terms - in particular, all of the tails $(s_n)_{n=N}^\infty$ - are subsequences,
  • the subsequences obtained by taking every $k$-th element, $(s_{kn})_{n=1}^\infty$.
Sig TM
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