Every sequence has a convergent subsequence?

[CoachBryan]

I'm not sure if this is true or not. but from what I can gather, If the set of Natural numbers (divergent sequence) {1, 2, 3, 4, 5,...} is broken up to say {1}, is this a subsequence that converges and therefore this statement is true?

 

[Simon Bridge]

You are wondering if it is always possible to find at least one convergent subsequence in any sequence however divergent?
Your specific question amounts to whether a sequence with a finite number of elements is convergent.
To answer that, check the definition of "converge".

Where does the question come up?

 

[CoachBryan]

if it's possible to find at least one convergent subsequence in ANY sequence.

Definition of converge: "A sequence {a(n)} converges to a real number A iff for each epsilon>0 there is a positive integer N such that for all n >= N we have |a(n) - A| < epsilon."

 

[Simon Bridge]

So would {1,-1,1,-1,...} converge?

In the definition you quote - n runs from 0 to infinity, so it won't work for a finite sequence.
http://en.wikipedia.org/wiki/Sequence#Limits_and_convergence

 

[CoachBryan]

Hey, thanks a lot. Now i understand it.

 

[homeomorphic]

Only a bounded sequence has a convergent subsequence. An unbounded one, like 1,2, 3, 4...may not.

 

[WWGD]

Does {1,2,3,..,n,.. } , i.e., a_n:=n have a convergent subsequence? Of course, this depends on your topology, but, as
a subspace of the Reals, does this have a convergent subsequence? This is one of the characterization of compact metric spaces, as every sequence having a convergent subsequence. And, in the subspace topology of the Reals,
{1,2,3,...} is/not compact (find a cover by open sets so that each contains a single number).