So simple and clean! I don't quite get why it's necessary when you're showing compact => closed, to choose a ball of radius 1/2n_1, rather than just 1/n_1. Wouldn't the ball of radius 1/n_1 be separated too if S is contained in S_{1/n_1}=compliment(closure(B_(1/n_1))), which has to be disjoint to B_(1/n_1)?
Hi Jordan! Two things: 1) That's right, 1/n_1 would have sufficed. I think I naturally go overboard so that I avoid any issues with boundary points, it has become instinct! 2) I still haven't gotten an email from you :D
Thanks for mentioning this wonderful theorem. A couple of things: (1) This is ok in R^n but requires care in general metric space. (2) Even if restricted to R^n, effectively one would need to prove the BW property then use it, which takes extra work. The proofs presented here work directly from the definitions.
@@ProfOmarMath I really meant it as sequential compactness, which, for a metric space, is equivalent to being compact. Of course, you would still need to prove this result if it is not the definition you start with, but I assumed that in most analysis courses, that is what is done before proving properties of compact sets as you did.
@@ProfOmarMath To be fair, I remembered that we took these equivalent statements (I'm including complete and totally bounded) of compactness for granted, and only proved them later since their proofs were quite involved. But then we used these results a lot in our proofs when it made them simpler, so it might not have been the best approach after all.
6:28 is there a way to show the proof directly? I.e. without showing that the complement is open, but rather show the set itself is closed? I am revisiting Real Analysis and while I can follow proofs, I have trouble formulating my own ones.. I have no idea what principle I would use when attempting to prove X in topology. Is showing that the complement of X is open always an easier strategy than showing that X is closed directly?
Hi Kane. I'm going to address the question of how to play with these things. Typically my intuition has been the following: when dealing with wanting to prove something is closed in metric spaces there are two big options: use the fact that their definition is that their complement is open or use the definition that the contain limits of sequences in them. I use the former typically when the question at hand itself involves the use of something involving open sets. For example, in this situation, we were using the definition of compactness that every open cover has a finite subcover. However if there is information given that seems like it is about limits of sequences, I use that definition instead. That's generally what i do at least.
@@ProfOmarMath Beautifully explained thank you @ProfOmarMath. Trying to state simply; if you want to prove something is closed, then if openess is mentioned in the problem statement, you might want to prove that the complement is open. Else if there is a mention of sequences and limits thereof, you might want to prove directly with closedness. Makes sense. I still have the question of whether it is conceptually possible to prove it is closed directly by the definition of a closed set, regardless of the above? Sure, it could be a more roundabout proof (i.e. more difficult), but is that at all possible, or is it only possible one way or another.
Proof is correct but not complete. You need the fact that for given teo points there is a radius such that one contains the other. Then you need the fact that there is always a natural number greater than a positive real number
The techniques used in the proof helped me clarify some other topology properties, thank you so much 👍🏻
why bounded set is less or equal to, but not less than? It is clear that *bounded* is a set {|x+r|
Excellent presentation of this theorem. Great idea to start with a metric space. 👍
Thanks!
If we were in a general topological space and not necessarily a metric space, how would you define the notion of boundedness?
So simple and clean! I don't quite get why it's necessary when you're showing compact => closed, to choose a ball of radius 1/2n_1, rather than just 1/n_1. Wouldn't the ball of radius 1/n_1 be separated too if S is contained in S_{1/n_1}=compliment(closure(B_(1/n_1))), which has to be disjoint to B_(1/n_1)?
Hi Jordan! Two things:
1) That's right, 1/n_1 would have sufficed. I think I naturally go overboard so that I avoid any issues with boundary points, it has become instinct!
2) I still haven't gotten an email from you :D
@@ProfOmarMath I will make sure to send the email later today! I appreciate your follow-up immensely! :)
Well done sir!
I'm hopelessly bad at math, but I love your videos anyway! :)
Thanks for the appreciation, but I am curious about this. What makes you feel that way?
More directly, was it an experience you had when you were young, or a way you were taught, or something someone said or?
Thank you so much Prof.
Definitely!
Amazing video. Thank you so much!
subscribed just from that beautiful "subscribe" pointing
Welcome!
Really, the proof for compact implies closed goes much more smoothly if we use the Bolzano Weierstrass property.
Thanks for mentioning this wonderful theorem. A couple of things:
(1) This is ok in R^n but requires care in general metric space.
(2) Even if restricted to R^n, effectively one would need to prove the BW property then use it, which takes extra work. The proofs presented here work directly from the definitions.
@@ProfOmarMath I really meant it as sequential compactness, which, for a metric space, is equivalent to being compact.
Of course, you would still need to prove this result if it is not the definition you start with, but I assumed that in most analysis courses, that is what is done before proving properties of compact sets as you did.
@@thezombat123 Ah I see what you mean. Interestingly I typically teach sequential compactness later on!
@@ProfOmarMath To be fair, I remembered that we took these equivalent statements (I'm including complete and totally bounded) of compactness for granted, and only proved them later since their proofs were quite involved. But then we used these results a lot in our proofs when it made them simpler, so it might not have been the best approach after all.
@@thezombat123 Makes total sense.
6:28 is there a way to show the proof directly? I.e. without showing that the complement is open, but rather show the set itself is closed?
I am revisiting Real Analysis and while I can follow proofs, I have trouble formulating my own ones.. I have no idea what principle I would use when attempting to prove X in topology. Is showing that the complement of X is open always an easier strategy than showing that X is closed directly?
Hi Kane. I'm going to address the question of how to play with these things.
Typically my intuition has been the following: when dealing with wanting to prove something is closed in metric spaces there are two big options: use the fact that their definition is that their complement is open or use the definition that the contain limits of sequences in them.
I use the former typically when the question at hand itself involves the use of something involving open sets. For example, in this situation, we were using the definition of compactness that every open cover has a finite subcover.
However if there is information given that seems like it is about limits of sequences, I use that definition instead.
That's generally what i do at least.
@@ProfOmarMath Beautifully explained thank you @ProfOmarMath. Trying to state simply; if you want to prove something is closed, then if openess is mentioned in the problem statement, you might want to prove that the complement is open. Else if there is a mention of sequences and limits thereof, you might want to prove directly with closedness. Makes sense. I still have the question of whether it is conceptually possible to prove it is closed directly by the definition of a closed set, regardless of the above? Sure, it could be a more roundabout proof (i.e. more difficult), but is that at all possible, or is it only possible one way or another.
Which definition of a closed set do you view as the direct definition
@@ProfOmarMath The definition laid out in Rudin Definition 2.18 (d): E is closed if every limit point of E is a point of E
Heine Borel Thm bring me here❤️❤️❤️
this is how you prove ?
Proof is correct but not complete.
You need the fact that for given teo points there is a radius such that one contains the other.
Then you need the fact that there is always a natural number greater than a positive real number