# Proving the contrapositive

1. Oct 12, 2005

### Icebreaker

"If X is a bounded sequence that does not converge, prove that there exists at least two subsequences of X that converge to two distinct limits."

There is a what I like to call "mass produced" version of the proof with limsup and liminf (which actually tells you where the two subsequences converge to, but it is not necessary). But I didn't want to use that so I did it another way. Can someone tell me if the following reasoning is right? I won't write out the exact proof because latex would kill me; I'll just briefly explain the logic of my proof:

The Weierstrass Theorem tells us that a bounded sequence has at least one subsequence which is convergent. X has such a subsequence, which we shall denote k. Let k' be the terms that are NOT in k. k' is a bounded subsequence, and therefore is also a sequence. k' therefore has a subsequence which is convergent, which we will denote u. If u converges to some number different from that of k, then the proof is complete. If u does converge to the same number as k, then take the terms in k' that are NOT u, and let i denote that subsequence.

Basically, this process can be repeated until we've exhausted all possible subsequences. The argument now is that they cannot ALL converge to the same limit, because that would contradict the hypothesis that X is divergent. Therefore, at least ONE of those subsequences must converge to some other number than that of k.

There may seem to be some handwaving back there but the jest of it is there.

2. Oct 15, 2005

### Icebreaker

Anyone? A similar process was used to prove the nested interval theorem, if I'm not mistaken.

3. Oct 15, 2005

### fourier jr

i think proving the contrapositive would be easier. use limsup & liminf & get a convergent sequence, which is of course also bounded.