# Thread: Alternative proof of Bolzano-Weierstrass Theorem

1. ## Alternative proof of Bolzano-Weierstrass Theorem

First, use the Monotone Convergence Theorem to show that every bounded sequence with a monotone subsequence has a convergent subsequence.
Then, prove that every bounded sequence has a monotone subsequence.

A hint is given for the second part that says: In a sequence <a>, call an index n a peak if a{m} < a{n} for m > n. Consider two cases, accordingly to whether <a> has infinitely many peaks.

My professor says: How can you get a non-increasing subsequence if you have infinitely many peaks? If you have finitely many peaks, what happens after the last peak. Can you get a non-decreasing subsequence?

2. To remind, there is a Calculus forum on this website...

So I understand you need to prove the second claim, that every bounded sequence has a monotone subsequence. Suppose there are infinitely many peaks. Then corresponding a{n}'s form a non-increasing subsequence. Suppose now that there are no peaks after some point, i.e., each index is not a peak. Write what it means that n is not a peak; this should give you an idea how to form a non-decreasing subsequence.

3. I am usually not allowed to use calculus, so I didn't think I should post this there.

I think I get the infinite peaks one, but I thought it would create a monotone increasing sequence because the a{n}'s keep going up. So if an infinite number of peaks forms a monotone increasing sequence, then by the Monotone Convergence Theorem, this sequence is convergent. This is actually a subsequence of the bounded set, so it converges inside that bounded set.

For the finite number of peaks, it means that there is some maximum n such that x{n}<=x{m} for m>n. I'll make the maximum=M. So for the subsequence x{M+n}, there exists no n such that x{M+n}<x{M+m}. This means that for any point in this subsequence there exists a point less than that point. Does this mean the subsequence is monotone decreasing (I know you said before that it would be non-decreasing, but I thought it would be decreasing because the points keep going down).

Maybe I am misunderstanding how to make decreasing and increasing sequences.

4. I thought it would create a monotone increasing sequence because the a{n}'s keep going up.I thought it would create a monotone increasing sequence because the a{n}'s keep going up.
No, a{n}'s keep going down. A (value at a) peak is greater than all subsequent points, so the next peak is smaller.

For the finite number of peaks, it means that there is some maximum n such that x{n}<=x{m} for m>n.
No. Let M be the maximum peak (index), so each n > M is not a peak, i.e., it is not the case that a{m} < a{n} for all m > n. Write this last statement in a positive form (that does not start with "it is not the case"). Hint: the negation of "for all ..." starts with "there exists".