Sequential Compactness

In this video, I discuss the notion of sequential compactness, which is an important concept used in topology and analogy. I also explain the similarities and differences between sequential compactness and covering compactness.

[advaitramesh]

Very nice! The other direction is harder to prove yes, but I don’t think it would take an hour. The way I would do it is through the notion that every infinite subset of X has a limit point => X is compact. Rudin’s way of proving this is very neat, where you first prove that if every infinite subset of X has a limit point, then X is separable, meaning any open cover for X has a countable subcover. And then reducing this countable subcover down to a finite subcover is a bit tricky but very neat in my opinion (we use contradiction).

[RCSmiths]

Very interesting concept man! I look forward to learning it!

[eamon_concannon]

3:14 I guess that to be more careful here in getting the contradiction, we suppose that we have an (infinite) sequence of DISTINCT elements of S that do not have a convergent subsequence. Then the rest of the proof makes complete sense to me as I was confused by the possibility of having an infinite sequence consisting of repeated elements of a finite sequence.
Thanks a lot for this playlist and the sequences and series playlists.

[souvikpramanickofficial]

I didn't knew about all this stuffs, thanks very much for teaching these.

[emmyscamander]

Hi Dr. Peyam! Thank you for the amazing video!! I am a bit confused with the contradiction part. Does S_n have to be infinite?

[xwyl]

Great video, nice proof! I wonder about the proof of sequential->covering. I have a basic idea about it: to show sequential but ~covering. Then there must exist x_n in u_n but not in u_(i<n). Take this x_n as our sequence, which has limit x, then x must be in some u_p, and this contradicts the previous fact about x_n. Therefore sequential->covering.

[demogorgon]

Pretty interesting 😃✌ I am new to this channel and I'm gonna subscribe to it right now.

[toaj]

0:29 Does the sub-sequence have to converge to an element of S? Because otherwise open intervals are sequentially compact too.

[sitienlieng]

Dr. Peyam, I have a stupid question. If every sequence in a set S has a convergent subsequence then will it guarantee that the set S is sequentially compact? I mean for example, we show the “necessary condition” that the interval [0, 1] is sequentially compact by applying Bozanol Weierstrass Thm, but it is not sufficient to conclude that a set is sequentially compact. Would the theorem be an iff statement?

[jonatan]

Does the proof of your claim (about finite elements in neighborhood) needs the axiom of choice? Since we construct a subsequence by picking members from an infinite family of sets.

[dgrandlapinblanc]

Thank you very much even it's hard in english.