Functional Analysis 16 | Compact Sets

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This is my video series about Functional Analysis where we start with metric spaces, talk about operators and spectral theory, and end with the famous Spectral Theorem. I hope that it will help everyone who wants to learn about it.

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00:00 Introduction
00:40 Compactness in R^n
01:46 Definition: sequentially compact
04:00 Examples
05:52 Proposition: compact implies closed and bounded
07:14 Proof

#FunctionalAnalysis
#VectorSpaces
#Mathematics
#LearnMath
#calculus

I hope that this helps students, pupils and others. Have fun!

(This explanation fits to lectures for students in their first and second year of study: Mathematics for physicists, Mathematics for the natural science, Mathematics for engineers and so on)

[punditgi]

Amazing how you tie and connect all these different ideas together. Simply brilliant and quite compelling.
Ein echter Wunder! :-)

[davidkwon]

You are a lifesaver. Please keep doing this amazing job. Thank you!

[etabeta]

i think im in love with you, thank you very much for your work, it's incredible

[aleksherstyuk]

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)

[RangQuid]

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.

[PunmasterSTP]

Compact? More like "Compawesome!" Thanks again for making all of these videos.

[xwyl]

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?

[ahmedamr]

Thanks so much once more!
Q: Does compactness of a metric space imply its completeness?

[kingmunch]

Are sequentially compact set also set that compact in the topological sense?

[J.MiguelProf]

excelente video. que plataformas usas para escribir

[tsshamoo]

Are you planning on covering Stone-Weierstrass and Arzela-Ascoli eventually?

[jgy]

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?

[HilalBalci_]

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.

[BedrockBlocker]

Great video as always :D
Could you show the equivalence of sequencial compactness and
open-cover-compactness?

[piy]

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 ?

[felipegomabrockmann]

please more videos about functional analysis

[scollyer.tuition]

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?

[oldtom]

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.

[oldtom]

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…….

[xlfc]

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!