# What did I do wrong

1. Jun 8, 2006

### Dragonfall

Suppose (xn) is a bounded sequence that does not converge. Show that (xn) has at least two subsequences that converge to different limits.

By the Bolzano-Weierstrass theorem there exists a subsequence of (xn) that converges. Delete that subsequence from (xn) and form a new sequence (xn1) with the remaining terms. (xn1) is bounded and hence has a convergent subsequence. If that subsequence converges to a different limit then we are done. If it does not, delete that subsequence from (xn1) and form a new sequence (xn2) with the remaining terms; ad infinitum. My argument is that not all of those subsequences can converge to the same limit, because the main sequence (xn) is not convergent.

I dug up this problem while flipping through my old analysis book. What did I do wrong? I feel as though something is not right with this proof.

I found another proof:

There exists a convergent subsequence by B-W, converging to some number x. Since (xn) does not converge, it has a tail that is at least e>0 away from x. From that tail there exists another convergent subsequence which must be at least e away from x, hence there are two convergent subsequences.

Last edited: Jun 8, 2006
2. Jun 8, 2006

### matt grime

The problem is that you don't show that your algorithm ever terminates, indeed it might not: you might keep picking 'bad' subsequences. You would need to invoke some kind of axiom of choice argument to say that there is a maximal set of convergent subsequences converging to a given x.

The second proof is good though I'm not sure that I'd use the word 'tail', that sort of implies that all of the sequence after some point is e away from x. I'd say that there are infinitely many terms at least e away from x (or it'd converge), and these have a convergent subsequence.