Prove that every sequence has a monotone subsequence (ILIEKMATHPHYSICS)

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This video references the book "Introduction to Real Analysis" by Bartle and Sherbert (Fourth Edition). The fact we are proving in this video is given by Theorem 3.4.7 and is called the Monotone Subsequence Theorem.

Here is a proof of the result we use in the infinite case:

Thanks and enjoy the video!

  1. ILIEKMATHPHYSICS
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I really like how you make sure these videos are easy to understand for everybody by explaining every step. Which means if it's a really hard problem I can trust that I will understand it with your explanations.

RabinSaidÖsteränggymnasietNAC
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Another way to prove this would be to split the sequence to 2 cases: bounded and unbounded. If it is bounded, it would have a limit point and if it has a limit point, it would be easy to construct a monotone subsequence by adjusting epsilon appropriately. If it is unbounded on any one side, it is trivial to construct a monotone subsequence.

pingdingdongpong
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Something about this feels overly complicated. I'm a bit of an amateur (math-adjacent degree, but didn't get this far into formality). Let me see if I got this:

If I have a sequence of *any* Real numbers (x_n), then I can pick out either a finite or (if available) an infinite number of elements of the sequence, in the order in which they appear in the original sequence, and in doing so, I can ensure that the elements that I choose have at least one of the following patterns:
a) these still sequence-ordered elements are already in nonascending order, or
b) these still sequence-ordered elements are already in nondescending order, or
c) these still sequence-ordered elements are equal to each other (which counts as both nonascending and nondecreasing).

Is that good common-wording understanding of this theorem?

And if that's right, then it sounds like the Empty Sequence and single-element sequences (i.e.: single numbers) may be degenerate/trivial/vacuous cases, but then for any sequence of 2 or more elements, I can pick any 2 different-position elements, and either one must be greater than the other, or they must be equal to each other.

emmeeemm
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Real analysis is tough and I'm not a math major, but I feel like you missed proving that S is not empty.

erfanmohagheghian