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_(icovering.
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.
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?
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.
I don’t think so, we’re given the sequence, not constructing it. For every neighborhood it’s a yes or no question. In any case, impossible to do analysis without the axiom of choice, it’s in our blood
Very interesting concept man! I look forward to learning it!
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_(icovering.
The proof is unfortunately much longer and uses Lebesgue points, check out Pugh for a proof
@@drpeyam what's Pugh?
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?
I didn't knew about all this stuffs, thanks very much for teaching these.
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.
Pretty interesting 😃✌ I am new to this channel and I'm gonna subscribe to it right now.
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?
It’s not stupid :) That is the definition of sequential compactness so it’s an iff statement
@@drpeyam Thank you so much for confirming, Dr. Peyam.
great explanation
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.
I don’t think so, we’re given the sequence, not constructing it. For every neighborhood it’s a yes or no question. In any case, impossible to do analysis without the axiom of choice, it’s in our blood
Thankyou sir 🙏👍
0:29 Does the sub-sequence have to converge to an element of S? Because otherwise open intervals are sequentially compact too.
Yes
Thank you very much even it's hard in english.
Lovely