# If a sequence converges, then every subsequence converges to the same limit — but how do I know a subsequence exists?

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 $s_{n_k}$?

• 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 3:10

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

• What do you mean by "[the sequence is] nonempty"? – Pedro Tamaroff Oct 30 '13 at 3:18
• @PedroTamaroff So long as the $s_n$ is not the empty sequence. – Newb Oct 30 '13 at 3:43
• 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

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, $(s_{kn})_{n=1}^\infty$.