Proof of existence of "peaks" in The Sequential Compactness Theorem

Question

According to Bolzano–Weierstrass theorem:

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How do we guarantee that EVERY sequence of real numbers has either infinitely or finitely many "peaks"? (Note that a subsequent is ordered in a way of original sequence, and we have limit to define such a "peak").

Here are two examples of sequences with either infinitely or finitely many peaks from 'Hagen von Eitzen'. Not an example, is there a proof of existence of "peaks" such that for EVERY sequence of real numbers we ALWAYS can find '$a_i\le a_j$ for all $i>j$'?

Thank you.

score 2 · Answer 1 · answered Jan 30 '15 at 05:06

2

This is really a question of language and logic. The set of peaks for a given sequence may be finite (including empty) or it may not be. Either the statement it is finite is true or it is false. If it is false then you are in case 2: there are infinitely many peaks. If it is true then you are in case 1.

answered Jan 30 '15 at 05:06

Mark Joshi

5,604

No, please note that a sequence is a set of ordered numbers, as is its subsequence. And, the subsequence is made of those peaks in ordered way. – Jan 30 '15 at 05:08
no - a peak is defined in terms of points after but the set of peaks is just a set – Mark Joshi Jan 30 '15 at 08:31
@L.G. Case 1 in the proof given works if there are no peaks. – Stella Biderman Mar 25 '16 at 14:31
@L.G. Case 1 in the proof given works if there are no peaks. – Stella Biderman Mar 25 '16 at 14:37

Proof of existence of "peaks" in The Sequential Compactness Theorem

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