I don't even need to use this knowledge but I love your videos and these are quiet relaxing and helps me while I'm doing my stuff. Btw I really aprecciate your way of teach us, I'm an student of actuary, thx for help.
I never studied metric spaces as such but have enjoyed your series covering them (pun not intended!). Thank you. Since a set is compact only if *all* open covers of it has a finite covering subset I think the first sentence in the board should replace "an" with "any" so that it becomes: Suppose 𝑢 is *any* open cover of [a,b] ... Or even: Let 𝑢 be any open cover of [a,b] ... This to make it clear the proof handles all open covers of the interval. Or is such phrasing not needed even if being strict?
That proof is more complicated than I expected. It gives me the feeling that compactness is deep.
Yeah, it actually shows how nice abstraction is! This proof for example is more direct, but more complicated
The video is published 1 day ago, how can the comment be 1 month ago??
@@wise_math The date is wrong, somehow. I remember this video from at least three months ago, probably more.
Beautifully done. Thank you for the extensive proof 👏
Beautiful proof, saw this in my class last semester!
thank u for this brilliant proof Mr. Peyam :)
I don't even need to use this knowledge but I love your videos and these are quiet relaxing and helps me while I'm doing my stuff. Btw I really aprecciate your way of teach us, I'm an student of actuary, thx for help.
Thanks so much!!!
Hey a view is a view, money is green, amirite :P
I think you want to say C = {x\in [a,b] | [a,x] CAN be covered by finitely many elements in the open cover}.
cool video, great work!
Yeaaahh..! But is it tight though?
* oh wait..!
Wrong forum..
Dang it!
Hard. Thank you very much.
Do general proofs of compactness from the definition look something like this?
I left a comment few days ago about a possible slightly different proof, but I can't see it anymore. Removed?
I’m not really sure, I don’t think I removed it
I never studied metric spaces as such but have enjoyed your series covering them (pun not intended!). Thank you.
Since a set is compact only if *all* open covers of it has a finite covering subset I think the first sentence in the board should replace "an" with "any" so that it becomes: Suppose 𝑢 is *any* open cover of [a,b] ...
Or even: Let 𝑢 be any open cover of [a,b] ...
This to make it clear the proof handles all open covers of the interval.
Or is such phrasing not needed even if being strict?
Doesn’t matter. Here by an it is implied that it’s an arbitrary sub cover.
What did we do to deserve two videos in one day? 😄
Just for being so awesome :)
@@drpeyam PR: 10000 👌
Sequential compactness seems easier
@Rick Does Math it wouldn't matter for metric spaces either ways. Eh
More Books! I mean More research!