2:20 sequentially compact set 1:07 empty set is compact 1:17 {5} is compact 1:50 R is not compact 2:10 prove [a,b] is compact 3:15 Heine Borel theorem 3:58 Proof
Hello, Professor. Does 3:02 mean: for any sequence {a_n} with a_n in the closed set A, all of its accumulation values (if it has) are actually in A. Compare to 'closed', the condition of 'compact' only adds a requirement that the sequence must have at least one accumulation value?
Hello, Professor. Greetings from Brazil! You mentioned that the set {5} has a constant sequence (an) = 5, 5, 5, 5..., therefore the accumulation value is 5, but in that is different from what I have learned. I've learned in Jay Cumming's Real Analysis the term limit points. I'm assuming that is the same as an accumulation value. According to what I've learned there and in other sources, {5} contains no accumulation values/limit points, since in this set there is only one number therefore there is no sequence approaching any limit. What is your view on this alternate explanation. Can it coexist with your version or are they opposing views? Thank you for your great work!
Thanks and greetings to the Brazilian summer :) This is exactly why I chose the name "acc. VALUE" and not "acc. POINT". If you see the sequence as a set, you only find one point, so no accumulation. But if you see it as how often a value is hit, we have an accumulation. That is the whole thing :)
The proof of the last statemenr that was left as an excersise is by taking the limit sup and as n goes to infinity an also goes to infinity so it has no accumulation value?
My proof for 5:50 Assume a sub n has an accumulation value, Then there is some subsequence a sub k sub n such that the limit of the subsequence as n goes to infinity exists. For any limit a, a choice of epsilon value smaller than one will force that for all n>a+1, the sequence takes on values such that the absolute value of a sub k sub n minus a is greater than one and thus the limit does not exist thus there are no accumulation values.
4:38 how can a sequence be subset of any set since its a list, i know list are sets too but they're defined differently. Isnt belongs to symbol appropriate.
It's just a short notation for saying that the elements for the list come from the set. You are right, the subset symbol is abused in this sense but it's a common notation. One usually reads it like: "This is a sequence and if I wrote {a_n}, then I would have a subset relation."
Sir, while proving the statement : "A compact set is closed and bounded", we penultimately come across proving the contraposition : "If A is unbounded, then its not compact." Here, in the implication that just follows the statement that "A is an unbounded set", are we just considering (rather precisely, framing such a "possible" sequence) a "particular"(?) sequence whose all members are in A but their absolute values exceed n, n being the natural numbers; (which then implies inexistence of any accumulation values). I'm focussing my query upon the meaning of that immediate implication, if it is : "there exists such a sequence...."? Please see sir. Thank you.
Hi, i think im missing something. Why does the set being bounded imply there is no accumulation value? The "fact" you glossed over doesnt,seem inherently,obvious to me. Im picturing the compactification of R where you "include" infinity as a point. The accumulation point for such a sequence would just be infinity?
I guess, you have a typo there and wanted to say "unbounded". By the definition of accumulation value in R (not the compactification!), the sequence (n) has no accumulation value. That is what you can try to prove by just using the definition.
In the proof of C compact then C closed and bounded. From (1) --> (2) you startthe argument by "Let (an) a convergent sequence...." then you used that C is compact so there is a convergent subsequence and the limit it's the same. BUT why have you started from "Let (an) be a convergente sequence" when that is not the definition of compactness.
@@brightsideofmaths I had the same issue with this part, the question is how do we know that there exists a sequence with a limit a \tilde, since compactness makes a claim only about subsequences, but every subsequence is a sequence so we just skipped over the part that a_n is a subsequence of some other sequence. Does this make sense? Thanks
I see your confusion. A closed set does not have to be bounded. See [0, inf) for example, its complement (-inf, 0) is open as you can always find a neighbourhood around any point in it. Thus the former set is closed even though it’s unbounded.
The set of all real numbers R is closed (every accumalation point of R is within R), but obviously not bounded. Closed also isn't the opposite of open. R is also open as any p of R has a Ball which is a subset of R. R is open, closed, but not bounded. Its helpful to *not* think of an interval [a,b] in R for closed, but of the actual definition that is: Set E closed : p is accumulation point of E => p is within E. For [0,1) 1 is an accumulation point as within any ball of 1 you'll find some p element of [0,1). Thus such an intervall isn't closed. The set of all sequence members of a convergent sequence is bounded yet not closed, but if you unify this set with the accumulation/ limit point of this sequence it is closed as well (thus compact).
2:20 sequentially compact set
1:07 empty set is compact
1:17 {5} is compact
1:50 R is not compact
2:10 prove [a,b] is compact
3:15 Heine Borel theorem
3:58 Proof
Bright and clear. Thanks!
You are amazing! I have really enjoyed all your series, especially the real analysis and measure theory ones though. VG aus Chicago :)
Wow, thank you! :)
The proof was so clear, thank you!
You should be my uni mathmatics professor
This is great... Helped me a lot for the exam...Thank youuu and keep this up ❤️
Most welcome 😊 And thanks for the support!
does "any sequence" here solely mean complete sequence or extend to "every subsequence"?
Any sequence means any possible sequence :)
Hello, Professor. Does 3:02 mean: for any sequence {a_n} with a_n in the closed set A, all of its accumulation values (if it has) are actually in A. Compare to 'closed', the condition of 'compact' only adds a requirement that the sequence must have at least one accumulation value?
For "closed", we consider *convergent* sequences. For compact we consider all possible sequences.
@@brightsideofmaths But the sequence {a_n} at 3:02 may not necessarily converge. How is 'closed' used?
@@konataizumi6358 Just take the convergent subsequence :)
@@brightsideofmaths So, for any accumulation value of {a_n}, we can assert that it's in A by taking the convergent subsequence, right?
@@konataizumi6358 It's the definition of an accumulation value :)
Ausgezeichnet!
Hello, Professor. Greetings from Brazil!
You mentioned that the set {5} has a constant sequence (an) = 5, 5, 5, 5..., therefore the accumulation value is 5, but in that is different from what I have learned. I've learned in Jay Cumming's Real Analysis the term limit points. I'm assuming that is the same as an accumulation value. According to what I've learned there and in other sources, {5} contains no accumulation values/limit points, since in this set there is only one number therefore there is no sequence approaching any limit.
What is your view on this alternate explanation. Can it coexist with your version or are they opposing views?
Thank you for your great work!
Thanks and greetings to the Brazilian summer :)
This is exactly why I chose the name "acc. VALUE" and not "acc. POINT". If you see the sequence as a set, you only find one point, so no accumulation. But if you see it as how often a value is hit, we have an accumulation. That is the whole thing :)
@@brightsideofmaths Thank you for the answer!
The proof of the last statemenr that was left as an excersise is by taking the limit sup and as n goes to infinity an also goes to infinity so it has no accumulation value?
Sir eagerly waiting for part 15 ❤️❤️
When will it be uploaded ?
Danke 🙏🙏
It will arrive today or tomorrow :)
My proof for 5:50
Assume a sub n has an accumulation value,
Then there is some subsequence a sub k sub n such that the limit of the subsequence as n goes to infinity exists.
For any limit a, a choice of epsilon value smaller than one will force that for all n>a+1, the sequence takes on values such that the absolute value of a sub k sub n minus a is greater than one and thus the limit does not exist thus there are no accumulation values.
It holds as long as the space is hausdorf, correct?
No, the discrete metric is a counterexample.
4:38 how can a sequence be subset of any set since its a list, i know list are sets too but they're defined differently. Isnt belongs to symbol appropriate.
It's just a short notation for saying that the elements for the list come from the set. You are right, the subset symbol is abused in this sense but it's a common notation. One usually reads it like: "This is a sequence and if I wrote {a_n}, then I would have a subset relation."
Sir, while proving the statement : "A compact set is closed and bounded", we penultimately come across proving the contraposition : "If A is unbounded, then its not compact."
Here, in the implication that just follows the statement that "A is an unbounded set", are we just considering (rather precisely, framing such a "possible" sequence) a "particular"(?) sequence whose all members are in A but their absolute values exceed n, n being the natural numbers; (which then implies inexistence of any accumulation values).
I'm focussing my query upon the meaning of that immediate implication, if it is : "there exists such a sequence...."?
Please see sir. Thank you.
Oh, I don't really understand your problem. Maybe you can rephrase it.
Hi, i think im missing something. Why does the set being bounded imply there is no accumulation value? The "fact" you glossed over doesnt,seem inherently,obvious to me. Im picturing the compactification of R where you "include" infinity as a point. The accumulation point for such a sequence would just be infinity?
I guess, you have a typo there and wanted to say "unbounded". By the definition of accumulation value in R (not the compactification!), the sequence (n) has no accumulation value. That is what you can try to prove by just using the definition.
@@brightsideofmaths I did mean unboounded thank you. So the definition is different under a compactification. That makes sense, thanks!
So the null set is an example of an open compact set, cause its clopen.
You mean the empty set?
@@brightsideofmaths yeah
I hope you make analysis in n-dim'l spaces in the near future.
It's already out there: tbsom.de/s/mc
@@brightsideofmaths
Thanks a lot!
But I meant to extend the theorems such as Bolzano-Weirstraß or Heine-Borel in higher dimensions.
@@hassansameh3404 Yeah, but they are completely the same there.
In the proof of C compact then C closed and bounded. From (1) --> (2) you startthe argument by "Let (an) a convergent sequence...." then you used that C is compact so there is a convergent subsequence and the limit it's the same. BUT why have you started from "Let (an) be a convergente sequence" when that is not the definition of compactness.
Why shouldn't we start with that? We have to show that C is closed. So it's the correct starting for that :)
@@brightsideofmaths I had the same issue with this part, the question is how do we know that there exists a sequence with a limit a \tilde, since compactness makes a claim only about subsequences, but every subsequence is a sequence so we just skipped over the part that a_n is a subsequence of some other sequence. Does this make sense? Thanks
A closed set is always bounded??? Why the distinction is made as bounded and closed???
A *compact* set is always bounded. Do I say otherwise?
I see your confusion. A closed set does not have to be bounded. See [0, inf) for example, its complement (-inf, 0) is open as you can always find a neighbourhood around any point in it. Thus the former set is closed even though it’s unbounded.
The set of all real numbers R is closed (every accumalation point of R is within R), but obviously not bounded. Closed also isn't the opposite of open. R is also open as any p of R has a Ball which is a subset of R. R is open, closed, but not bounded. Its helpful to *not* think of an interval [a,b] in R for closed, but of the actual definition that is:
Set E closed : p is accumulation point of E => p is within E.
For [0,1) 1 is an accumulation point as within any ball of 1 you'll find some p element of [0,1). Thus such an intervall isn't closed. The set of all sequence members of a convergent sequence is bounded yet not closed, but if you unify this set with the accumulation/ limit point of this sequence it is closed as well (thus compact).
❤
accumulation value?
Is that a question?
@@brightsideofmaths yes
@@faizajabeen1738 Then, my answer will be: "Yes" :)