Every sequence has a convergent subsequence?

1. Nov 2, 2014

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?

2. Nov 2, 2014

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?

3. Nov 2, 2014

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."

4. Nov 2, 2014

5. Nov 2, 2014

CoachBryan

Hey, thanks a lot. Now i understand it.

6. Nov 4, 2014

homeomorphic

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

7. Nov 6, 2014

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).

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook