I really love your enthusiasm and energy. One can see that you really love the things that you teach and you want others to share that same enthusiasm. Currently binge watching all your videos. I am glad you took on this topic as I myself struggled with it a lot when I was doing Analysis. Keep up the good work.
I love how youtube gives people the ability to ask questions (and receive answers from!) of actual mathematicians with doctorates. Thanks for making these! Would love a video on limit point compactness, sequential compactness, compactness, and their equivalence in metric spaces.
Those different forms of compactness I mentioned are topological properties of given spaces. When working in metric spaces, it turns out that these properties are equivalent to each other. Not necessarily so when working in non-metric spaces.
I was lucky to have at a Prof who explained compactness well 48 years ago when I was an undergraduate. It is a tricky topic, but very important. Good examples (& non-examples) are very helpful. A small point in your non-ex 1: you cannot drop ANY of the sets, (n-1,n+1), and still cover U = (0,\inf) since each positive integer, n, belongs to (n-1,n+1) and to no other set in the family. Thus, dropping a set (from the covering) means that one point from U is 'uncovered'.
Liked "You can do things with finite covering that you cannot do with infinite coverings." I was really confused so far about why compactness is important. It's still hard to see an intuitive characterization, especially in R^n, but getting closer! Thanks so much.
I've seen this proof years ago, I've learned it by heart and did not understood it. UNTIL NOW!!! So thankful I can't express even in Italian. Just thanks!
Thank you so so much. You cover these difficult concepts in a very understandable way. Just fantastic. We feel deep admiration for you and BlackpenRedpen.
I have to thank you so very much. This topic (Heine-Borel-Theorem and Heine-Cantor-Theorem) was THE one I really did not get when hearing my Analysis II course the first time and now I'm drastically less confused.
I am sadly not one of your students. But I adore everything about the way you present the material. I understand all of your quirky ideas and examples and they make me smile. This video helped me with my grad analysis class. I am struggling with compactness and the various continuity definitions. You are a beautiful man with incredible intelligence and a magnetic energy. Thank you for the video.
Hello. I'm your new subscriber here. I came here because i really can't figure out the concept of compactness in my topology class today. I was just sitting and seeing my lecture talking about compactness yet i find myself hard to comprehend what this was all about. And you just enlighten me through this very helpful video. Thank you, sir! Keep up the good work 💪
The salesman example was awful! But as someone below said, you've got great enthusiasm. You choose interesting problems that have kept me coming back for weeks now. This is a fantastic channel. Subscribed.
There seems to be a slight inconsistency in notation. We remove the outer braces when we take the big union (or big intersection) of a family (of sets), but not when we use set builder notation. For example we would write U K_i instead of U { K_i} , where i is in some index set I. But in set builder notation we would write U { K_i : i in I }. So it appears that U K_i = U { K_i : i in I }, and that seems slightly contradictory. Big Union is being used differently.
@@drpeyam "What about finite sets. What about (0,1)"? ... lol its not a big issue Dr. Peyam but to someone new to Topology like me, the small errors can be confusing
Very useful, thank you. Hope I clearly assimilate this definitions and properties about these radius definitions for epsilon and delta in the infinity small circles in a U and Ux area
Can you think of compactness kind of like the idea of least upper bound? For example, there may be an infinite amount of numbers that serve as an upper bound for the interval (0,1) but 1 is the smallest one. Likewise, a set that is compact may have an infinite amount of sets that could cover it, but there exists a finite number of sets that covers it.
I think you're right, that definition is (close to) sufficient for intervals in R. From my Analysis course: "An interval I in R of the form I = [a, b] is closed and bounded. Such an interval is also said to be compact."
Since we have a topological space, we can take the union of the Ui,k, so that the space can always be covered by a single set, if it is compact. Finite indeed! Why is finite covering, in the definition, used as superior to just covering by an open set? Just semantics, but it does change how you think about it.
i only start watching but you should have used μ instead of i and set V=I to get u⊆U_(μ∈V)u_μ and have a full family of u's now at 14:48 one of my favorite theorems, with an amazing proof now 19:45 in epsilon-delta you can't choose delta that depends on both x and y, the correct "choice" is something like min{1,epsilon/(1+|2y|)}. i would say that the different between the two continuities is that in Uniform Continuity you have delta that does not depends on the point for which you check Continuity(y) and normal Continuity can depends on that point. which is the same as you said just with different wording(after all we have abs, I just find it easier to understand) also, i always found it ironic that in particular cases we find epsilon(delta) and find the inverse of this to get delta(epsilon)
That was at least 5 free, future journeys home, worth of awesomeness. But, what is the set { Future }, anyway? Definitely finite, very likely UC, but, Compact?? I wish I could say?
Why do you leave out the requirement that the covering be by open sets? Sure would be nice if all subsequent theora apply to open, closed, and mixed coverings, but somehow, I doubt it. What say?
A little confused about the definition at 4:00, if for a set to be compact it must be covered by the union of a finite number of subcoverings in ANY given covering, then for any set we could come up with a covering like {(1/n, 1) | Natural n} [seen later ~12:30] for which we need an infinite amount of subcovers to cover it? This seems like a problem because we can conclude any set S is non-compact by coming up with a similar covering (dividing each bound in (1/n, 1) by the measure of S and translating it by inf(S) - assuming it is bounded below.) . For example, consider the interval [0, 1]. This is closed and bounded, hence compact (according to the definition given for my Analysis course). I could say {U_i} = {(1/n, 1) | Natural n} and so we would have found a case where we'd need an infinite number of subcoverings to cover the set (or perhaps just the one when n goes to infinity, but we probably can't just pick "the one" where n goes to infinity). So we've found a suitable {U_i} which can be used as an example to show that [0, 1] is not compact - which is a contradiction. I'd also be worried about the boundaries -- the interval is closed but our subcoverings are open and so the boundaries will never be covered in this case. Perhaps a better definition would be: A set S is compact if there exists a covering set {U_i} such that you only need the union of a finite number of elements from {U_i} to cover S. For the example at around ~12:30: (0, 1) we have been able to come up with a "bad enough" covering set that we need the union of the whole set to cover the interval, but we could also come up with a set like {(-1/n, 1) | Natural n } where you can make do with just one element (say n=1) so it IS compact according to the proposed definition. In this case it's not the most "efficient" covering but from what I've seen I don't think we care about the efficiency yet... Thanks for these videos, I love your enthusiasm!
Your reasoning that [0,1] is not compact fails since the element 0 is not a member of any Ui. Thus the union of the Ui's fails to be a covering as 0 is not in their union. Your arguement only shows (0,1) is not compact.
Ah, I see your point - The given Ui can not be a covering set to begin with. If you don't mind, what do you think of the definition I've proposed? Thank you :)
@@anitejbanerjee2278 I don't think this proposed definition does anything at all. You can just cover the set with itself so you only need one element and boom you're done. Every set is compact. The standard definition of compact metric spaces is extremely important and useful. And even though it seems quite unintuitive at first, you'll start to see it's benefits as time progresses.
But (0,1) can be covered with finite number of open sets, right? Like, take (-1,1) & (0,2). I know I'm obviously making a mistake. I just didn't understand the concept
Then how the hell am I supposed to understand compactness then? I mean this is literally the first topic in math I can think of where it is damn near impossible to construct examples from the definition. Barring boring ones like finite sets. Like good luck proving say the cantor set is compact from the definition. How am I supposed to understand a definition if I cant construct examples.
12:29 respectfully, not quite. we might indeed need infinitely many of them, however definitely not all of them since there is a hell lot of redundancy in this open covering since the sequence {V(n) = (1/n, 1): n = 2, 3, 4, ...} is ascending, i.e. for every n set V(n) is a subset of V(n+1). which begs my impish mind a question.. what if i device a following fiendish scheme: given open cover V = {V(n) = (1/n, 1): n = 2, 3, 4, ...} of the interval (0, 1) i will build a new open cover U in the following iterative way: U(2) := {V(2)}, U(3) := {V(3)}, ..., U(n) := {V(n)} and foreseeably i will set U := lim(n -> oo) of U(n). now since for every n we have #U(n) = 1, then it's tempting to conclude (possibly wrongly), that #U = 1 as well.. your move doctor! ];->
michał botor Yeah I would rather use a covering V(n)=(1/n,2/n): n=2, 3, 4... because than you need the union of all the coverings and can easily see that you indeed need infinity many V’s
Does this work similarly in 3-space but with open spheres instead of circles? We typically represent sets as 2-D regions (venn diagrams, partitions etc.) But is this entirely necessary? Aren't sets fundamentally 1-D? And we abstract them to higher dimensions or how should I be thinking about the dimension of a set?
In practice you solve for |x-y| in the equation |f(x) - f(y)| < epsilon. For example, for f = 2x + 3, the latter equation becomes 2|x - y| < epsilon, so delta = epsilon/2. Of course in harder examples it’s more complicated than that; in this video you construct delta using the info that’s given to you in the problem!
@@drpeyam i dont get it actually....is there a possibility of you making a video to show how to find such epsilon in practise, with more complex examples?
We can find finitely many Uk from the covering. We can’t explicitly choose them, but we know they are there. Basically infinitely many of them are useless
Watching up to 10:00, I am a bit confused because why can’t we choose a “patch” that is infinite in size to cover up the domain. Or is it that the definition of patch Ui allowed is only finite in size?
The (0,1) non-example is even more confusing. It is (one of the) “bad” covering I get it. But why can’t I pick the covering {Un=(-N,N)} again? Then I can easily find one finite patch such as U2=(-2,2) it will cover (0,-1)
You don’t get to choose the covering. What I’m saying is that *no matter what* covering I give you, you can always find a sub covering that covers the whole set. So the covering will always be given to you, and you’ll have to pick the sub covering. Think of it a bit like Jenga :P
I'm not sure this answered the question. If you have the space on (0,1) then (-2, 2) **IS** a sub-covering 'patch' of sorts. Is the problem that all the 'patches' need to be within the initial space? For example, you show (-inf, inf) to not be compact because you can't make patches U1=(0,2); U2=(1,3); ... without remaining ω-incomplete. However, the patch U1=(0,2) completely covers the space (0,1). So why not pick that and be done with it? Thanks! I love your videos.
JPK314 The problem is that of course you can find a covering that has a finite sub-covering which will suffice to cover your domain. But this property needs to hold for any arbitrary covering one can think of. So giving an example of a covering set which does not have a finite sub-covering shows that the given domain is not compact. Think about the interval [0,1] which is the same as (0,1) but including the endpoints. This set in fact is compact. The example he is showing in the video doesn’t work for this one, because even with the whole set wouldn’t cover the interval. Inevitably you would need to add sets containing 0 and 1. But now you will be able to choose those 2 added sets and an finite sup-set of the given covering. Thus [0,1] is compact
This is maybe too naive and simplistic, but I think that it might be better to think of it as being that you can rule out compactness as soon as you can find an example of coverings which need to be infinite? Might need to be corrected on that
I really love your enthusiasm and energy. One can see that you really love the things that you teach and you want others to share that same enthusiasm. Currently binge watching all your videos. I am glad you took on this topic as I myself struggled with it a lot when I was doing Analysis. Keep up the good work.
I love how youtube gives people the ability to ask questions (and receive answers from!) of actual mathematicians with doctorates. Thanks for making these! Would love a video on limit point compactness, sequential compactness, compactness, and their equivalence in metric spaces.
John Wroblewski Do those all have equivalents in metric spaces? There must be some types that have no counterpart at all.
Those different forms of compactness I mentioned are topological properties of given spaces. When working in metric spaces, it turns out that these properties are equivalent to each other. Not necessarily so when working in non-metric spaces.
Thanks, that's most interesting. I'm tempted to learn more.
I am trying. Lol
And paracompactness.
Thank God for you and thank you for being so humble and not acting like others are stupid for struggling with these concepts.
I was lucky to have at a Prof who explained compactness well 48 years ago when I was an undergraduate. It is a tricky topic, but very important. Good examples (& non-examples) are very helpful. A small point in your non-ex 1: you cannot drop ANY of the sets, (n-1,n+1), and still cover U = (0,\inf) since each positive integer, n, belongs to (n-1,n+1) and to no other set in the family. Thus, dropping a set (from the covering) means that one point from U is 'uncovered'.
Your enthusiasm makes this dense material so much "absorbable" lol. Thank you!
Liked "You can do things with finite covering that you cannot do with infinite coverings." I was really confused so far about why compactness is important. It's still hard to see an intuitive characterization, especially in R^n, but getting closer! Thanks so much.
Thank you for explicitly stating the difference between continuous and uniform continuous. That makes sense...finally.
I've seen this proof years ago, I've learned it by heart and did not understood it. UNTIL NOW!!!
So thankful I can't express even in Italian. Just thanks!
mathematics is so beautiful
Thank you so so much. You cover these difficult concepts in a very understandable way. Just fantastic. We feel deep admiration for you and BlackpenRedpen.
I have to thank you so very much. This topic (Heine-Borel-Theorem and Heine-Cantor-Theorem) was THE one I really did not get when hearing my Analysis II course the first time and now I'm drastically less confused.
Well done explanation. This is the best lecture about covering I’ve ever have. !!
I am sadly not one of your students. But I adore everything about the way you present the material. I understand all of your quirky ideas and examples and they make me smile. This video helped me with my grad analysis class. I am struggling with compactness and the various continuity definitions. You are a beautiful man with incredible intelligence and a magnetic energy. Thank you for the video.
Thank you so much, this seriously made my day!!!!
Much better explained than at my uni, well done mate.
Awesome lecture! I tried many CZcams videos and this is the only one that worked for me. Keep up the good work!
So grateful for this straightforward explanation. Thank you Dr. Peyam!
This is a really good lesson. Thank you so much I finally understand what compactness means!
Thank you!!! 🙂
Compact explanations are your specialty!
Hahaha
@@drpeyam 😁
Thank you Sir for your great video .In fact your explanation is so clear that an average student like me can understand the very abstract concepts !
A compact set which has a cover of open sets, has a finite subcover of those open sets. An amazing and non obvious theorem.
Hello. I'm your new subscriber here. I came here because i really can't figure out the concept of compactness in my topology class today. I was just sitting and seeing my lecture talking about compactness yet i find myself hard to comprehend what this was all about. And you just enlighten me through this very helpful video. Thank you, sir! Keep up the good work 💪
Thank you for ending my misery by explaining this concept in English.
The salesman example was awful! But as someone below said, you've got great enthusiasm. You choose interesting problems that have kept me coming back for weeks now. This is a fantastic channel. Subscribed.
I love it when, "BAM some magic happens".
Great lecture!
your videos of intuition are unique in youtube
Great video, thank you!
Gotta say. I love your taste in maths.
There seems to be a slight inconsistency in notation.
We remove the outer braces when we take the big union (or big intersection) of a family (of sets),
but not when we use set builder notation.
For example we would write U K_i instead of U { K_i} , where i is in some index set I.
But in set builder notation we would write U { K_i : i in I }.
So it appears that U K_i = U { K_i : i in I }, and that seems slightly contradictory.
Big Union is being used differently.
Really useful, thanks a lot!!
wow! supercalifragilisticexpialidocious!
Another great lesson!....
You are the best sir. Thank you so muchhhh... ❤️❤️❤️
Thanks for this very helpful video Prof! Also to point out an error-- (0,1) is an uncountably infinite set with a bijection to R
What does that have to do with the video?
@@drpeyam 10:30 ... not a big issue but hopefully it helps someone else who may have been slightly confused there
It’s not a mistake? (0,1) may have as many elements as R but it does not cover R.
@@drpeyam i think there you said the interval (0,1) is a finite set... that is all... nothing to do with (0,1) as a cover
@@drpeyam "What about finite sets. What about (0,1)"? ... lol its not a big issue Dr. Peyam but to someone new to Topology like me, the small errors can be confusing
Love it.... 👍
Finally I got it! Yoy are marvellous. Would you think to come my university METU????? pleaaassseeeeee
Very useful, thank you. Hope I clearly assimilate this definitions and properties about these radius definitions for epsilon and delta in the infinity small circles in a U and Ux area
Can you think of compactness kind of like the idea of least upper bound? For example, there may be an infinite amount of numbers that serve as an upper bound for the interval (0,1) but 1 is the smallest one. Likewise, a set that is compact may have an infinite amount of sets that could cover it, but there exists a finite number of sets that covers it.
I think you're right, that definition is (close to) sufficient for intervals in R.
From my Analysis course:
"An interval I in R of the form I = [a, b] is closed and bounded. Such an interval is also said to be compact."
at 12:17 why can't you just cover (0,1) with a bigger interval (-1,2) so you don't need infinitely many intervals to cover it.
The point is that *every* cover needs to have a finite sub-cover!
Thank you! Very well explained!
Since we have a topological space, we can take the union of the Ui,k, so that the space can always be covered by a single set, if it is compact. Finite indeed! Why is finite covering, in the definition, used as superior to just covering by an open set? Just semantics, but it does change how you think about it.
You’re misunderstanding it, we want *every* open cover to have a finite sub cover, not just that there is a finite covering of a set
i m still at home but fortunatly much excited after this proof
thank YOU!!!!!!!
i only start watching but you should have used μ instead of i and set V=I to get u⊆U_(μ∈V)u_μ and have a full family of u's
now at 14:48 one of my favorite theorems, with an amazing proof
now 19:45 in epsilon-delta you can't choose delta that depends on both x and y, the correct "choice" is something like min{1,epsilon/(1+|2y|)}. i would say that the different between the two continuities is that in Uniform Continuity you have delta that does not depends on the point for which you check Continuity(y) and normal Continuity can depends on that point. which is the same as you said just with different wording(after all we have abs, I just find it easier to understand)
also, i always found it ironic that in particular cases we find epsilon(delta) and find the inverse of this to get delta(epsilon)
Love this video. I was wondering if it is better to say "bounded set" as opposed to finite at 10:55?
Agreed!
Really good video, but I wish you would show the work for how you solved for delta around 20 min in :/
keep up the good work
Great videos!!!! Greetings to you and Oreo
Please keep uploading videos I feel empty if you dont so thanks
That was at least 5 free, future journeys home, worth of awesomeness.
But, what is the set { Future }, anyway?
Definitely finite, very likely UC, but, Compact??
I wish I could say?
Even stronger, all opens intervals in R are NOT compact. Take (a,b). However [a,b] is! This uses LUP of R right ? If you don’t invoke H-B.
Great video. Thanks.
What is the tangible characteristic of a compact and connectedness?
How do you compact 3.141414…..
Thank you sir was really confused
Wonderful !
I think that it would be better to use d(x,y) instead of absolute value at continuity proof ? Would not it ?
No, we’re talking about R^n
Why can't you use two sets like (0, 1/2) and (1/2, 1) to cover (0,1)?
*every* cover needs to have a finite sum cover
Also your cover excludes 1/2
What are the books/sources used in this video?
Hey sir
Please calculate the integral of
1/(×+e^×) and (cos(×^2))/(×^2)
Amazing
6:32 Who? Me? Why?
Hey, Dr. Peyam, can you make it an video on Perron-Frobenius Theorem in Linear Algebra?
Why wont (0, inf) cover (0, inf) ?
Yesss new video!
I love you Peyam
Locally compact Hausdorff space says hi.
Could you elaborate on this problem: Lim x-->0 [sin(tan(x)) - tan(sin(x))]/x^7?
Why do you leave out the requirement that the covering be by open sets? Sure would be nice if all subsequent theora apply to open, closed, and mixed coverings, but somehow, I doubt it. What say?
The sets are open by assumption
This guy as a strange personality, but nobody else seems to be able to explain it in a video on youtube so im watching.
😞
A little confused about the definition at 4:00, if for a set to be compact it must be covered by the union of a finite number of subcoverings in ANY given covering, then for any set we could come up with a covering like {(1/n, 1) | Natural n} [seen later ~12:30] for which we need an infinite amount of subcovers to cover it? This seems like a problem because we can conclude any set S is non-compact by coming up with a similar covering (dividing each bound in (1/n, 1) by the measure of S and translating it by inf(S) - assuming it is bounded below.) .
For example, consider the interval [0, 1]. This is closed and bounded, hence compact (according to the definition given for my Analysis course). I could say {U_i} = {(1/n, 1) | Natural n} and so we would have found a case where we'd need an infinite number of subcoverings to cover the set (or perhaps just the one when n goes to infinity, but we probably can't just pick "the one" where n goes to infinity). So we've found a suitable {U_i} which can be used as an example to show that [0, 1] is not compact - which is a contradiction. I'd also be worried about the boundaries -- the interval is closed but our subcoverings are open and so the boundaries will never be covered in this case.
Perhaps a better definition would be: A set S is compact if there exists a covering set {U_i} such that you only need the union of a finite number of elements from {U_i} to cover S.
For the example at around ~12:30: (0, 1) we have been able to come up with a "bad enough" covering set that we need the union of the whole set to cover the interval, but we could also come up with a set like {(-1/n, 1) | Natural n } where you can make do with just one element (say n=1) so it IS compact according to the proposed definition. In this case it's not the most "efficient" covering but from what I've seen I don't think we care about the efficiency yet...
Thanks for these videos, I love your enthusiasm!
Your reasoning that [0,1] is not compact fails since the element 0 is not a member of any Ui. Thus the union of the Ui's fails to be a covering as 0 is not in their union. Your arguement only shows (0,1) is not compact.
Ah, I see your point - The given Ui can not be a covering set to begin with. If you don't mind, what do you think of the definition I've proposed? Thank you :)
@@anitejbanerjee2278 I don't think this proposed definition does anything at all. You can just cover the set with itself so you only need one element and boom you're done. Every set is compact.
The standard definition of compact metric spaces is extremely important and useful. And even though it seems quite unintuitive at first, you'll start to see it's benefits as time progresses.
@@gdsfish3214 thanks a lot for your reply! I'll keep this in mind for when I read more!
But (0,1) can be covered with finite number of open sets, right? Like, take (-1,1) & (0,2).
I know I'm obviously making a mistake. I just didn't understand the concept
Every cover has a finite sub cover, not some cover has a finite sub cover
@@drpeyam ahh, now it makes sense. Thank you☺️
finally! thank you!
Isn't UC the same as linearity? What's the difference?
no not at all. they are totally different
One question is. How do you construct a finite subcover from an open cover? Is this something thats hard to do in general Dr. Peyam?
Exactly
Then how the hell am I supposed to understand compactness then? I mean this is literally the first topic in math I can think of where it is damn near impossible to construct examples from the definition. Barring boring ones like finite sets. Like good luck proving say the cantor set is compact from the definition. How am I supposed to understand a definition if I cant construct examples.
Thanks Peyam's ! Holidays ?
12:29 respectfully, not quite. we might indeed need infinitely many of them, however definitely not all of them since there is a hell lot of redundancy in this open covering since the sequence {V(n) = (1/n, 1): n = 2, 3, 4, ...} is ascending, i.e. for every n set V(n) is a subset of V(n+1). which begs my impish mind a question.. what if i device a following fiendish scheme: given open cover V = {V(n) = (1/n, 1): n = 2, 3, 4, ...} of the interval (0, 1) i will build a new open cover U in the following iterative way: U(2) := {V(2)}, U(3) := {V(3)}, ..., U(n) := {V(n)} and foreseeably i will set U := lim(n -> oo) of U(n). now since for every n we have #U(n) = 1, then it's tempting to conclude (possibly wrongly), that #U = 1 as well..
your move doctor! ];->
michał botor Yeah I would rather use a covering V(n)=(1/n,2/n): n=2, 3, 4... because than you need the union of all the coverings and can easily see that you indeed need infinity many V’s
that was dense, I mean compacted :D
Does this work similarly in 3-space but with open spheres instead of circles?
We typically represent sets as 2-D regions (venn diagrams, partitions etc.) But is this entirely necessary?
Aren't sets fundamentally 1-D? And we abstract them to higher dimensions or how should I be thinking about the dimension of a set?
Of course!
Nice video Peyam! Why did you write |y-x| but then |f(x)-f(y)|
Doesn't U have to be bounded?
How would one find that delta in practice?
In practice you solve for |x-y| in the equation |f(x) - f(y)| < epsilon. For example, for f = 2x + 3, the latter equation becomes 2|x - y| < epsilon, so delta = epsilon/2.
Of course in harder examples it’s more complicated than that; in this video you construct delta using the info that’s given to you in the problem!
@@drpeyam i dont get it actually....is there a possibility of you making a video to show how to find such epsilon in practise, with more complex examples?
If you are trying to cover (0, infinity), are you allowed to use a patch like (10, infinity)? And if not, what is the rule that prevents that?
(10,infinity) works since it’s open. Patches can be unbounded it’s ok. But this one alone wouldn’t cover (0,infty), you need more than that
Does it means we “choose” finitely many Uk from the covering, or we can find a finitely amount of Uk( not necessarily same as Ui) that cover U?
We can find finitely many Uk from the covering. We can’t explicitly choose them, but we know they are there. Basically infinitely many of them are useless
Dr Peyam Thank you, Peyam. You clear my doubts!!!
but R+ covers R+
So? :) You need to show every cover has a finite sub cover (or one cover that doesn’t have a finite subcover)
Yes!!! YES YES YES!!!!
Watching up to 10:00, I am a bit confused because why can’t we choose a “patch” that is infinite in size to cover up the domain. Or is it that the definition of patch Ui allowed is only finite in size?
The (0,1) non-example is even more confusing. It is (one of the) “bad” covering I get it. But why can’t I pick the covering {Un=(-N,N)} again? Then I can easily find one finite patch such as U2=(-2,2) it will cover (0,-1)
You don’t get to choose the covering. What I’m saying is that *no matter what* covering I give you, you can always find a sub covering that covers the whole set. So the covering will always be given to you, and you’ll have to pick the sub covering. Think of it a bit like Jenga :P
I'm not sure this answered the question. If you have the space on (0,1) then (-2, 2) **IS** a sub-covering 'patch' of sorts. Is the problem that all the 'patches' need to be within the initial space?
For example, you show (-inf, inf) to not be compact because you can't make patches U1=(0,2); U2=(1,3); ... without remaining ω-incomplete.
However, the patch U1=(0,2) completely covers the space (0,1). So why not pick that and be done with it?
Thanks! I love your videos.
JPK314 The problem is that of course you can find a covering that has a finite sub-covering which will suffice to cover your domain. But this property needs to hold for any arbitrary covering one can think of. So giving an example of a covering set which does not have a finite sub-covering shows that the given domain is not compact. Think about the interval [0,1] which is the same as (0,1) but including the endpoints. This set in fact is compact. The example he is showing in the video doesn’t work for this one, because even with the whole set wouldn’t cover the interval. Inevitably you would need to add sets containing 0 and 1. But now you will be able to choose those 2 added sets and an finite sup-set of the given covering. Thus [0,1] is compact
This is maybe too naive and simplistic, but I think that it might be better to think of it as being that you can rule out compactness as soon as you can find an example of coverings which need to be infinite? Might need to be corrected on that
Like 333 is more than 313
Peyam, you're supposed to wash and iron new dress shirts, not wear them straight out of the package. lol
It’s a non-iron shirt! 😛
That's what they tell you...
first once again
ههههههههههه 50K times
U cute