A minor mistake at 10:23: I should have said x_n does not belong to the closure of V_(n+1) due to construction, NOT due to the fact x_n does not belong to V_(n+1).
LetsSolveMathProblems it looks like you could have just used closed balls for each V_n in the construction rather than forming open balls for each V_n and then closing them afterward unless there’s a subtle detail that I’ve overlooked. From there, the fact that x_n does not belong to V_(n+1) follows trivially from the construction.
I actually wondered the same question when I first read Rudin's proof, which intentionally avoids closed balls without explicitly providing a reason. I see two potential explanation for this. First, since Rudin introduces a neighborhood as an open ball, it may have seemed logical to him to begin the construction with open balls just to use the terminology he defined. Second, and more likely, it should be pointed out that if we do begin the construction with closed balls, we cannot pick the point on the boundary of a closed ball V during the construction because an open (or closed) ball S centered at the boundary point of V is not a subset of V, nor is it required to contain a point of P that resides in V (a desired point may be inside S and outside V). Of course, we can avoid both difficulties by simply stating that we construct closed balls such that they form a sequence of containment WITHOUT ever intersecting each other (to prevent boundary point intersections), but Rudin may not have wished to introduce a notation for or an explicit mention of this type of containment. Who knows, perhaps there was a subtle point both of us missed. However, it seems to me that the construction you outlined should lead to the same conclusion.
I don't know where country you are from, but it sounds like my country people speak English. and so I felt very friendly :) actually I'm bad at english so I often get in trouble when I watch other channel's video and hear what (s)he says. But I didn't felt any difficulty to hear your saying, and it really helped me, Thank you!
Sorry if this is a stupid question (I'm not in the class or anything), but how can you just assume a neighborhood exists? I guess more specifically, what guarantees that there will be any metric by which you measure distance?
An excellent question. In a metric space, we DEFINE how the distance is measured between two elements of that set such that the distance function follows certain axioms. In R^k, distance is defined by the familiar Euclidean distance formula. Once we decide on the definition of distance function (such that it follows certain axioms), we let a neighborhood of some point x be the set of all points at a distance less than some fixed value from x.
Literally one of the best proofs I have ever seen. Thank you for elaborating on all the steps and recalling all the definitions that we need to know when proving a hard argument!
Ha ha! Perhaps I like them because they are too magical. But you are right--almost all of his proofs are intensely condensed, and many times the motivation for certain proofs is not immediately clear. Then again, his proofs are rigorously tight and illustrate the elegance of logic, which is what real analysis is truly about. =)
I loved Rudin's book (well, maybe more of a love-hate) so I wouldn't steer anyone away from it, but if I had to pick an alternative that covered nearly the same material I'd recommend C.C. Pugh's book on analysis. It's an excellent book with a nice expository tone (and pictures!). And the level of difficulty (including the problems) is roughly the same.
But wait! Haven’t you just constructed a counterexample to the theorem!? Thus disproving the theorem about infinite intersections of nested compact sets? Given: if p then q. Proceed to find: p implies not q Therefore either we don’t have p or p doesn’t imply q. I’m lost.
A minor mistake at 10:23: I should have said x_n does not belong to the closure of V_(n+1) due to construction, NOT due to the fact x_n does not belong to V_(n+1).
LetsSolveMathProblems it looks like you could have just used closed balls for each V_n in the construction rather than forming open balls for each V_n and then closing them afterward unless there’s a subtle detail that I’ve overlooked. From there, the fact that x_n does not belong to V_(n+1) follows trivially from the construction.
I actually wondered the same question when I first read Rudin's proof, which intentionally avoids closed balls without explicitly providing a reason. I see two potential explanation for this. First, since Rudin introduces a neighborhood as an open ball, it may have seemed logical to him to begin the construction with open balls just to use the terminology he defined. Second, and more likely, it should be pointed out that if we do begin the construction with closed balls, we cannot pick the point on the boundary of a closed ball V during the construction because an open (or closed) ball S centered at the boundary point of V is not a subset of V, nor is it required to contain a point of P that resides in V (a desired point may be inside S and outside V). Of course, we can avoid both difficulties by simply stating that we construct closed balls such that they form a sequence of containment WITHOUT ever intersecting each other (to prevent boundary point intersections), but Rudin may not have wished to introduce a notation for or an explicit mention of this type of containment. Who knows, perhaps there was a subtle point both of us missed. However, it seems to me that the construction you outlined should lead to the same conclusion.
I don't know where country you are from, but it sounds like my country people speak English. and so I felt very friendly :) actually I'm bad at english so I often get in trouble when I watch other channel's video and hear what (s)he says. But I didn't felt any difficulty to hear your saying, and it really helped me, Thank you!
wow this was really good lol, was more instructive than my actual real analysis class, def gonna check out that book
Thank you💙
One of the best understanding proof
Good video i love maths and physics
great video and a beutiful proof. i will definitely check out that book! :)
Fantastic explanation, thank you!
Thank you sir. 😊
Wow, that's a great proof. You made it seem really easy
Outstanding explanation
Sorry if this is a stupid question (I'm not in the class or anything), but how can you just assume a neighborhood exists? I guess more specifically, what guarantees that there will be any metric by which you measure distance?
An excellent question. In a metric space, we DEFINE how the distance is measured between two elements of that set such that the distance function follows certain axioms. In R^k, distance is defined by the familiar Euclidean distance formula. Once we decide on the definition of distance function (such that it follows certain axioms), we let a neighborhood of some point x be the set of all points at a distance less than some fixed value from x.
LetsSolveMathProblems ah okay, thanks for the answer! I thought it might have to do with it being in R^n.
Literally one of the best proofs I have ever seen. Thank you for elaborating on all the steps and recalling all the definitions that we need to know when proving a hard argument!
Tnx a lot
How closure of V2 is contained in V1
Nicely ❤❤
good one
Surprised to hear you like Rudin's books, his proofs just seem too magical for me.
Ha ha! Perhaps I like them because they are too magical. But you are right--almost all of his proofs are intensely condensed, and many times the motivation for certain proofs is not immediately clear. Then again, his proofs are rigorously tight and illustrate the elegance of logic, which is what real analysis is truly about. =)
Rudin's is more of a tool for training mathematics students, challenging them to fill in steps, rather than being an expository book
I loved Rudin's book (well, maybe more of a love-hate) so I wouldn't steer anyone away from it, but if I had to pick an alternative that covered nearly the same material I'd recommend C.C. Pugh's book on analysis. It's an excellent book with a nice expository tone (and pictures!). And the level of difficulty (including the problems) is roughly the same.
Is the intersection of all K_n equal to the intersection of all V_n *and* P?
The intersection of all the clausures of all V_n and P.
In general int(K_i int K) where i belongs to a family of indexes I, is equal to int(K_i) int K
My bad, thanks.
But wait! Haven’t you just constructed a counterexample to the theorem!? Thus disproving the theorem about infinite intersections of nested compact sets?
Given: if p then q.
Proceed to find:
p implies not q
Therefore either we don’t have p or
p doesn’t imply q.
I’m lost.
A taste of nice maths
gg men
How did you comment so quickly? I clicked on the video as soon as it was done uploading, and your comment was already here. =)
Oops I know nothing about set theory.
I wanted to listen but this guy can't even speak english...