Thank you for this video series. I'm about to start a journey through functional analysis, operator algebras, and TQFTs at my school. I already know algebra and topology (and category theory) decently well, but my analysis isnt strong yet. This is one of the first video series that made analysis super interesting for me. I'll also check out measure theory, and hopefull stop pretending like I aready know the main ideas! (Or learn German and watch the calc videos)
Very good video! I love the proof you gave about sequential compactness implies closeness and boundedness. But what interests me more is the proof of general compactness being equivalent to sequential compactness in metric spaces.
I really want to know more about compactness. Why the cover definition and seqence definition are equivalent? Why bounded close sets are compact only in R^n or C^n? Why open sets are not compact? 7:16 the proof for closedness the sequence x_n could be any sequence, including non-convergent ones. What if x_n is non-convergent?
I was thinking, and what properties would be necessary to make compactness equivalent to closed and bounded? Considering that this equivalence holds in both R^n and C^n; and that if I consider, for example, the open unit interval with the inducted standard metric, this equivalence fails (just considered the subset (0, 1/2], which is closed, bounded, but not compact); my guess is that this happens if the metric space is complete. The way I think you might be able to prove this is to first prove that for all sequences in a bounded subset, you can find some Cauchy subsequence of it (I don't know exactly how to prove this), and then since all Cauchy sequences converge in a complete metric space, and the subset is closed, this subsequence would converge somewhere inside the subset, therefore we can conclude the subset is compact. Is that intuition be correct? And if not, what property/properties would make so that that equivalence holds?
Completeness is surely necessary as you argued. However, the discrete metric space shows that it is not sufficient. We will talk about this in the next video :)
@@brightsideofmaths Ah, that makes sense. Thanks for the reply, and I'm definitely looking forward to the next video, I'm personally finding this series to be really fun to follow through.
9:35 ...you say that the distance d(b, Xnk) is larger or EQUAL, but I think, that it's not quite true, because nk is strictly SMALLER than d(a, b) + d(b, Xnk), so if you substract d(a, b) from both sides you won't get equality. Therefore there should be nk - d(a, b) is smaller (not equal) than d(b, xnk). Great video though!
Of course, you could write < instead of my ≤ but both things are correct here. It is called "smaller OR equal". So for every time there is a strict inequality with
Hello, thank you for all your videos. But I'm in the process of learning English and mathematics, so it would be great if you could turn on the subtitles :) Thank you in advance.
Proving covering compactness implies sequential compactness is strait forward, by contraposition. Given a sequence, if theres no convergent sub sequence, then every element of the space is contained in some ball avoiding all but finitely many sequence elements. These balls form an open cover that has no finite sub cover. The other direction is hard. Pugh has a very nice proof of this fact using the Lebesgue number lemma in his real analysis textbook
This is possibly a stupid question, but why does the definition of sequential compactness start with an arbitrary sequence in A from which we extract a convergent subsequence with limit in A? Can we not just require "all convergent sequences in A have a limit in A", which would mean that A is closed?
In this case we wouldn't need two different notions. Compact is different from closedness. Compact really means that the space is "small" in some sense.
@@brightsideofmaths I have possibly explained myself badly. My point was that it seems to me that with my suggested modification i.e. the simpler statement "all convergent sequences in A have a limit in A" would guarantee both closedness (I think this part is obvious?) and boundedness (this is perhaps not so clear - maybe my error lies here) hence compactness. It's possible that I need to go and look at the proof of the B-W theorem in R^n to see where I'm am making a mistake.
A bounded subsequence with a limit can live inside an unbounded or non converging sequence. So how is it that a converging bounded sib sequence? You say that if the norm/distance goes to infinity you can’t find such a (convergent) subsequence.
@@oldtom541 Yes, at 3:45 I talk about sequentially compact sets which cannot be unbounded. The reason is that then you find sequences that don't have convergent subsequences. Of course, you could still construct examples like you did but the point is that it has to work for every possible chosen sequence.
A bounded subsequence with a limit can live inside an unbounded or non converging sequence. So how is it that a converging bounded subsequence implies boundedness of the sequence. You say that if the norm/distance goes to infinity you cannot find such a (convergent) subsequence. I cannot see why you cannot. Eg x=1,1/2,2,1/3,3,…1/n,n…….
Amazing how you tie and connect all these different ideas together. Simply brilliant and quite compelling.
Ein echter Wunder! :-)
You are a lifesaver. Please keep doing this amazing job. Thank you!
i think im in love with you, thank you very much for your work, it's incredible
Thank you for this video series. I'm about to start a journey through functional analysis, operator algebras, and TQFTs at my school. I already know algebra and topology (and category theory) decently well, but my analysis isnt strong yet.
This is one of the first video series that made analysis super interesting for me. I'll also check out measure theory, and hopefull stop pretending like I aready know the main ideas!
(Or learn German and watch the calc videos)
Thank you very much :)
Hey I just came across your comment and was curious. How have your studies been going?
Very good video! I love the proof you gave about sequential compactness implies closeness and boundedness. But what interests me more is the proof of general compactness being equivalent to sequential compactness in metric spaces.
Thanks! The proof of your last question is not so hard. Just remember that in metric spaces you have epsilon-balls.
I really want to know more about compactness. Why the cover definition and seqence definition are equivalent? Why bounded close sets are compact only in R^n or C^n? Why open sets are not compact? 7:16 the proof for closedness the sequence x_n could be any sequence, including non-convergent ones. What if x_n is non-convergent?
Compact? More like "Compawesome!" Thanks again for making all of these videos.
Thank you so much for your videos!
You are very welcome :)
Absolutely amazing video
Thank you very much! :)
Very well motivated definition.
thx
Are sequentially compact set also set that compact in the topological sense?
Yes, they are, as long as you are working in a metric space.
@@brightsideofmaths i see, thank you for the reply 😊
the explanations are sooooooooooo gooooood
Thank you for this video
excelente video. que plataformas usas para escribir
Thanks so much once more!
Q: Does compactness of a metric space imply its completeness?
Yes, you can show that by using the sequential compactness definition.
I was thinking, and what properties would be necessary to make compactness equivalent to closed and bounded? Considering that this equivalence holds in both R^n and C^n; and that if I consider, for example, the open unit interval with the inducted standard metric, this equivalence fails (just considered the subset (0, 1/2], which is closed, bounded, but not compact); my guess is that this happens if the metric space is complete. The way I think you might be able to prove this is to first prove that for all sequences in a bounded subset, you can find some Cauchy subsequence of it (I don't know exactly how to prove this), and then since all Cauchy sequences converge in a complete metric space, and the subset is closed, this subsequence would converge somewhere inside the subset, therefore we can conclude the subset is compact. Is that intuition be correct? And if not, what property/properties would make so that that equivalence holds?
Completeness is surely necessary as you argued. However, the discrete metric space shows that it is not sufficient. We will talk about this in the next video :)
@@brightsideofmaths Ah, that makes sense. Thanks for the reply, and I'm definitely looking forward to the next video, I'm personally finding this series to be really fun to follow through.
danke für dieses video 😊..
9:35 ...you say that the distance d(b, Xnk) is larger or EQUAL, but I think, that it's not quite true, because nk is strictly SMALLER than d(a, b) + d(b, Xnk), so if you substract d(a, b) from both sides you won't get equality. Therefore there should be nk - d(a, b) is smaller (not equal) than d(b, xnk). Great video though!
Of course, you could write < instead of my ≤ but both things are correct here. It is called "smaller OR equal". So for every time there is a strict inequality with
In the proof of compact implies bounded, why did we need to introduce point b? Isn't it sufficient to just say that d(a, x_nk) > nk ?
We need to show there can't be any limit point for the subsequence. So the fact has to hold for all b in A.
Hello, thank you for all your videos. But I'm in the process of learning English and mathematics, so it would be great if you could turn on the subtitles :) Thank you in advance.
Are you planning on covering Stone-Weierstrass and Arzela-Ascoli eventually?
Of course :)
Great video as always :D
Could you show the equivalence of sequencial compactness and
open-cover-compactness?
Proving covering compactness implies sequential compactness is strait forward, by contraposition. Given a sequence, if theres no convergent sub sequence, then every element of the space is contained in some ball avoiding all but finitely many sequence elements. These balls form an open cover that has no finite sub cover.
The other direction is hard. Pugh has a very nice proof of this fact using the Lebesgue number lemma in his real analysis textbook
@@aleksherstyuk8319 Thanks :)
This is possibly a stupid question, but why does the definition of sequential compactness start with an arbitrary sequence in A from which we extract a convergent subsequence with limit in A? Can we not just require "all convergent sequences in A have a limit in A", which would mean that A is closed?
In this case we wouldn't need two different notions. Compact is different from closedness. Compact really means that the space is "small" in some sense.
@@brightsideofmaths I have possibly explained myself badly.
My point was that it seems to me that with my suggested modification i.e. the simpler statement "all convergent sequences in A have a limit in A" would guarantee both closedness (I think this part is obvious?) and boundedness (this is perhaps not so clear - maybe my error lies here) hence compactness.
It's possible that I need to go and look at the proof of the B-W theorem in R^n to see where I'm am making a mistake.
A bounded subsequence with a limit can live inside an unbounded or non converging sequence. So how is it that a converging bounded sib sequence? You say that if the norm/distance goes to infinity you can’t find such a (convergent) subsequence.
Where do I say this?
@@brightsideofmaths Thank you so much for your reply. I thought you implied this at time 3.45?
@@oldtom541 Yes, at 3:45 I talk about sequentially compact sets which cannot be unbounded. The reason is that then you find sequences that don't have convergent subsequences. Of course, you could still construct examples like you did but the point is that it has to work for every possible chosen sequence.
@@brightsideofmaths Thank you. I will have a big think about your answer! 😊
Just ask again if needed :)
please more videos about functional analysis
A bounded subsequence with a limit can live inside an unbounded or non converging sequence. So how is it that a converging bounded subsequence implies boundedness of the sequence. You say that if the norm/distance goes to infinity you cannot find such a (convergent) subsequence. I cannot see why you cannot. Eg x=1,1/2,2,1/3,3,…1/n,n…….
Where did I say what you said I did? :D
@@brightsideofmaths Thank you so much for your reply. I thought you implied this at time 3.45?
Great videos, but you really need to learn how to draw the number 1. Now it looks like an upside down v.
This is how I draw the number one :)
@@brightsideofmaths I got used to it already but when i first saw that thing i was confused :D
To be fair his 1 aren´t that bad.
I have seen a lot worse examples of the upside down v.
Seems to be a german thing.
Sorry,! It's a German thing for sure. I try to do it less than a upside v next time :D