Monotone Subsequence Theorem (Every Sequence has Monotone Subsequence) | Real Analysis
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- Опубліковано 8 лют 2025
- How nice of a subsequence does any given sequence has? We've seen that not every sequence converges, and some don't even have convergent subsequences. But today we'll prove what is sometimes called the Monotone Subsequence theorem, telling us that every sequence has a monotone subsequence. #RealAnalysis
The key idea of this proof is that of a peak, a term of a sequence that is greater than or equal to all following terms. If a sequence has infinitely many peaks, we can construct a decreasing subsequence of peaks. If a sequence has finitely many peaks, we can construct an increasing subsequence of terms after the last peak of the sequence.
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Did you know a monotone sequence converges if it has a single convergent subsequence? Check out the proof! ua-cam.com/video/BZ-LQpz5EBc/v-deo.html
What a beautiful proof!
Very beautiful idea, thank you.
You are welcome!
Beautiful explanation 👌
Thank you so much man! Lifesaver.
Glad to help! Thanks for watching and check out my analysis playlist for more - let me know if you have any questions!
ua-cam.com/play/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli.html
Excellent explanation!
Thank you!
Amazing explanation, ur a legend
You can also apply Ramsey's theorem
Very clear explanation
Glad to hear it, thanks for watching!
Clearly explained 👍
Thank you, glad it helped! Let me know if you have any questions, and if you're looking for more analysis, check out my playlist! ua-cam.com/play/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli.html
Thank you for the video! I just want to confirm, all the work in the proof is specific to infinite sequences and subsequences? Otherwise it would be trivial to make a monotone finite subsequence from any sequence which has more than one term, isn's it? You just have to make a subsequence be two values long by taking a pair of adjacent terms, and the second term will either be larger than the first (monotonic increasing) or less than the first (monotonic decreasing)
That's correct, when I say "sequence" in this course, I am referring only to infinite sequences. A "finite sequence" with n terms I'd refer to as an "n-tuple".
@@WrathofMath Thank you for clarifying. And also, if the sequence has no peaks, then it is monotone increasing so any subsequence will also be monotone increasing right?
@@minamishi just because a sequence has no peaks does not mean that it is increasing nor monotone.
This is gold
Thank you!
Nice! Thanks a lottttt
Glad to help! Check out my analysis playlist for more and let me know if you have any questions! ua-cam.com/play/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli.html
thank you so much
My pleasure, thanks for watching! If you're looking for more real analysis, check out my playlist!
ua-cam.com/play/PLztBpqftvzxWo4HxUYV58ENhxHV32Wxli.html
What if the sequence has 0 peak terms? What is then regarded as the last peak of the sequence? Does a sequence have 0 peak terms iff it is increasing?
Thank you for these videos!
Thanks for watching and great question, I should have mentioned that in the lesson! The last peak in the sequence is not really important, what's important is knowing we have a term with no peaks after it. In the case of 0 peaks, any term can play the role of the term with no peaks after it, and thus from that we can construct an increasing sequence.
And regarding the increasing 0 peaks implication, first consider the forward direction. If a sequence is increasing then it has 0 peaks. This is true except for the case of a constant sequence. I think we defined peaks as terms that are greater than or equal to all of the following terms. In the case of a constant sequence, it is technically increasing and also every term is a peak. I'll leave the proofs to you, they should be straightforward from the definitions!
The other direction is "If a sequence has 0 peaks then it is increasing." This is not true, try to think up an example of an oscillating sequence to show it!
@@WrathofMath Thanks for the clear explanation!
@@WrathofMath Sorry, but I am not able to find any oscillating sequence which has zero peaks, how is that even possible? Please do reply
Is it allowed to use the soft inequalities ( ≥) for monotonic subsequence in the case of a finite number of peaks? I find this more universally understandable
It depends where exactly you mean, but probably. Monotonicity does allow equality.
Can you please clarify for alternative sequence like 1, -1,1,-1.....
Because for alternative sequence all term after 1 is not less than 1 , 1 is coming again and again.
Thanks for watching and good question! This theorem just says every sequence has a monotone subsequence. What terms from 1, -1, 1, -1, ... could we take to get a monotone subsequence? And remember - constant sequences are both increasing and decreasing.
@@WrathofMath {1,1,1,...} and {-1,-1,-1,....} are constant subsequence. Ok I understand
Awesome!
How would this work for a sequence that tends to infinity that is sometimes decreasing? For example (1,2,3,4,2,3,4,5,3,4,5,6...)
It’s still a divergent sequence because it’s limit is infinity
The subsequence (1,2,3,4,5,6,...) is increasing from original sequence (1,2,3,4,2,3,4,5,3,4,5,6,4,5,6,7,5,...)
What happens if no peaks
Then the first term of the sequence will be the first term of the subsequence, let's call it a1. And since we know that there is no peak, there must exist some other term a2 that is greater than a1. Again, since no peak exists, then there must be a term a3 greater than a2, and so on. By taking all of these terms together, we have an increasing, monotone subsequent.