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.
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.
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.
your common-wording understanding seems right, and in this theorem, we only care about infinite sequences of real numbers, so the sequence (xn) in this proof means the function x : Z+ --> R whose outputs are x_1, x_2, x_3, ..., x_n, ...; likewise subsequences are infinite
if the subsequence is allowed to be fiinite, it's even simpler than that. Pick the first two elements x1, x2 If x1 > x2, the subsequence {x1, x2} is a monotonic decreasing subsequence. If not, it's monotonic increasing. End of proof. Edit: Oh, I see that you included my example in your comment already.
Sequences and subsequences of real numbers, by definition, are functions from N to R. Thus, they have "infinitely many elements" In other words, tuples are not sequences, they are tuples. Sequences are, by definition, infinite
@@iliekmathphysics Sorry I might be dumb and like I said I'm not an expert in this field, but wouldn't you get into any problem to construct the monotone subsequence if S were empty? 🤔 This was a hard proof for me to follow. I also feel that if Xn is strictly increasing, then S is empty. There would be no Xm that is greater than or equal to all the terms after it. Where am I wrong?
it's all good, in the case where S is infinite clearly S is nonempty so we have nothing to worry about here in the case where S is finite, we are worried about whether or not S is empty. But if S is empty then S is bounded (the empty set is bounded, in fact, every real number is both an upper bound and a lower bound of the empty set); thus, the existence of u also holds for the case that S is the empty set, and the rest of the proof follows from here
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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.
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.
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.
your common-wording understanding seems right, and in this theorem, we only care about infinite sequences of real numbers, so the sequence (xn) in this proof means the function x : Z+ --> R whose outputs are x_1, x_2, x_3, ..., x_n, ...; likewise subsequences are infinite
if the subsequence is allowed to be fiinite, it's even simpler than that. Pick the first two elements x1, x2
If x1 > x2, the subsequence {x1, x2} is a monotonic decreasing subsequence. If not, it's monotonic increasing. End of proof.
Edit: Oh, I see that you included my example in your comment already.
Sequences and subsequences of real numbers, by definition, are functions from N to R. Thus, they have "infinitely many elements"
In other words, tuples are not sequences, they are tuples. Sequences are, by definition, infinite
Real analysis is tough and I'm not a math major, but I feel like you missed proving that S is not empty.
why would we need S to be nonempty?
@@iliekmathphysics Sorry I might be dumb and like I said I'm not an expert in this field, but wouldn't you get into any problem to construct the monotone subsequence if S were empty? 🤔
This was a hard proof for me to follow. I also feel that if Xn is strictly increasing, then S is empty. There would be no Xm that is greater than or equal to all the terms after it. Where am I wrong?
it's all good,
in the case where S is infinite clearly S is nonempty so we have nothing to worry about here
in the case where S is finite, we are worried about whether or not S is empty. But if S is empty then S is bounded (the empty set is bounded, in fact, every real number is both an upper bound and a lower bound of the empty set); thus, the existence of u also holds for the case that S is the empty set, and the rest of the proof follows from here
@@iliekmathphysics ok thanks buddy. Am I right about the emptiness of S if Xn is strictly increasing?
@erfanmohagheghian707 yeah i agree with that