Mr. Michael I am a Ugandan lady ,I have a test soon ,I would like to say thank you for teaching me coz reading this stuff on my own is tough and I end up taking alot of time , God bless you for real coz eeeh..
Examples of sets in which cauchy sequences which do not always converge: The rational numbers Polynomials with any Lp norm Smooth solutions of a PDE (If I remember correctly, they're dense in Sobolev space)
As someone who labored mightily in the SDSU math dept. tutoring room to acquaint struggling students with limit theory, the hardest idea in calculus, I have to say that the need for a SCRATCHWORK area to work backwards in, is underemphasize in this video. I had to tell students: imagine you're walking backwards down a staircase into a dark cellar. That's how it feels to work with limit theory . . . . until you get a feel for coming up with the mechanisms that ensure the operation of the Squeeze-Pipe.
The SCRATCH is back-asswards to the proof; the proof is incomprehensible on first reading, before you've worked through the scratch. Pain in the neck!!!
@@a.nelprober4971 Limit theory tends to make strong men cry, yes. It was figured out fairly LATE in the history of math, 19th c. You didn't see Pythagoras or Newton coming up with rigorous limit-theory proofs.
@@a.nelprober4971 It's a perpetual pedagogical challenge to get kids to think in infinitesimals, when they're more comfortable with CONCRETE stuff. For instance, take a long strip of adding machine tape, like 3' to 6'. Have the student fold in half, then fold one piece back to make a quarter, then keep working with the end zig-zag fashion to make 1/8, 1/16, etc, as far as they can go. When then open the strip out again, they'll have a CONCRETE representation of the geometric series 1/2 + 1/4 + 1/8 . .. = 1. And of course when it's folded zig-zag style, they'll have the alternating series . . .I'll let you calculate what that adds up to!
and hence, we prove that the space of rationals is not a complete metric space. since the sequence (1+1/n)^n is a cauchy sequence in Q but does not converges in Q. R is a complete space which means there's equivalence between cauchy's sequences and convergent sequences
All convergent sequences in any metric space are Cauchy. The converse is not always true. For example the sequence (1+1/n)^n does not converge in the rational numbers (in the real numbers it converges to e, but e is not rational). A metric space where every Cauchy sequence is convergent is known as a complete metric space. In complete metric spaces a sequence is convergent if and only if it is Cauchy. The real numbers with the absolute value metric is complete. If we have a space that is not complete, we can make it complete, or more precisely, we can construct a complete metric space with a subspace that is isometric to the original metric space. This complete metric space is called a completion of our original not complete space. Importantly, the completion of a space is unique up to isometry. With this notion of the completion of a metric space, one was we can construct the real numbers from the rationals is by defining the the real numbers as the completion of the rationals under the absolute value metric.
The fact that every Cauchy sequence in R converges depends on the axiom of completeness. The axiom of completeness implies the fact that every monotone bounded sequence in R converges. The fact that every monotone bounded sequence in R converges implies the fact that every bounded sequence in R has a convergent subsequence. Using this fact the convergence of Cauchy sequence is proved.
Can you explain how a convergent subsequence is related to the Cauchy-ness of a sequence? Also, can you explain how a Cauchy sequence requires the axiom of completeness?
this has been very helpful tbh. i’m 17 years old and in uni (ik shocking) and my class has been struggling to understand what a Cauchy sequence is. luckily, i stumbled across your video and it makes perfect sense now! since we’re all from romania, i’ll try and translate everything for those who need extra help learning and credit you for your really good work!!
There's an equivalent statement: For all epsilon > 0 : there exists an index N : so that for all indices n >= N : | a_n - a_N | < epsilon You can check the equivalence by yourself. One direction is trivial, for the other direction you can use epsilon' := 2*epsilon. I find this alternative definition much better, because it uses less indices (only n >= N instead of n,m >= N), and it's in my opinion also more intuitive. Is there a reasonable reason :-) why the original definition is used?
Love it. This video makes as much (or as little) sense as when I was first taught these thing by the brilliant Sandra Pott in my first year at York uni. I gradually went on to understand. But it takes me right back to those lectures and initially not having a clue what was going on as it was an entirely new topic and idea. In the end took loads of optional real analysis courses right in to my fourth year.
For the second proof, just take the rationals, then clearly if a sequence is Cauchy but in the reals converges to an irrational, then it does not converge to something in the rationals.
@@leif1075 you can look up the specifics of the property but roughly speaking he's using it here to say that the number n/(n+1) is a number which actually exists (since in a Proof it would not be reasonable to pull a number out of thin air without justifying it). As for why it is less than 1; consider that to be equal to 1 the fraction would need to be n/n, if I increase the denominater by 1 I decrease the overall value of the fraction. For instance 2/2-->2/3, 2/3
@@SpartaSpartan117 Thanks yea ofcourse I know why it's less than one i understand basic fractions but i was asking why he knows it's less than EPSILON not one..also later how can he use the triangle inequality if we don't have a triangle here?
@@leif1075 The Archimidean principle states in brief that for any ε>0 - Arbitrarily - small there is a number N (big enough depending on ε) such that Nε >1 (This is a special case. It states that for any y,x>0 nx>y if n is big enough) That answers your question Nε>1 nε > Nε >1 for every n>N nε > 1 1/n < ε for every n>N (N depends on the ε value of course) Considering the ε-definition of a limit this leads to lim(1/n)=0 as n is getting bigger an bigger (as n goes to infinity)
I did not understand one thing. Let An = sum[k=1,n](1/k). An is cauchy, since the difference of their therms get as small as you want, but it diverges. How can that be possible?
"...the difference of their terms get as small as you want" - incorrect. In fact, the difference S_m-S_n where m,n >=N can be as big as you want, cause the initial series diverges, so you can enlarge the difference by choosing big enough m, keeping n fixed.
Is there a sequance that has a limit but is not Cauchy?? It is easy to find example of the converse, a sequance that is Cauchy but has no limit, for instance ia rational sequance converging to an itrational has no limit in the rationals. But having a limit and not being Cauchy?
at 9:28 , when we say that a series is convergent, shouldn't An approach 0 always? Is that 'L' there simply to count for all other possible emmm spaces (like not real numbers?)
Could you always attach a pdf of a screenshot of all the board at the end, or just move out of the picture at the end. You know like teachers/professors do in actual school. great explanation btw. Thanks.
Also weierstrass theorem uses the fact that the set containing an is compact, which uses the fact that R is complete, so i feel like there is some circular logic going on
12:40 When you say you take the absolute value of each side, do you mean all three parts of the compound inequality? It looks like you are only considering the middle and right parts. Why are you allowed to take the absolute value of both sides? If a_n was negative, but with greater absolute value than a_N1 + 1, wouldnt this be illegal?
@@Pklrs you have a mistake (why don't people ever edit their work :/) By substitution and triangle inequality we have |An | = | An + 0 | = | An - Am + Am |
This lecture is mostly good, but I must object to the use of the Bolzano-Weierstrass theorem to prove that every Cauchy sequence of reals converges. Any proof of BW will use the completeness of R or some equivalent property, so the proof is circular. You have to use the Cauchy sequence construction to exhibit a Cauchy sequence of rationals to which a Cauchy sequence of reals necessarily converges. The key step is to show there is a rational between any two reals.
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Mr. Michael I am a Ugandan lady ,I have a test soon ,I would like to say thank you for teaching me coz reading this stuff on my own is tough and I end up taking alot of time , God bless you for real coz eeeh..
Examples of sets in which cauchy sequences which do not always converge:
The rational numbers
Polynomials with any Lp norm
Smooth solutions of a PDE (If I remember correctly, they're dense in Sobolev space)
Real analysis? Pshaw! We must delve into the "polynomial analysis"...
Well, it's enough to have any pre-Hilbert space which is not Hilbert (supposing we have norm induced by the inner product).
As someone who labored mightily in the SDSU math dept. tutoring room to acquaint struggling students with limit theory, the hardest idea in calculus, I have to say that the need for a SCRATCHWORK area to work backwards in, is underemphasize in this video. I had to tell students: imagine you're walking backwards down a staircase into a dark cellar. That's how it feels to work with limit theory . . . . until you get a feel for coming up with the mechanisms that ensure the operation of the Squeeze-Pipe.
The SCRATCH is back-asswards to the proof; the proof is incomprehensible on first reading, before you've worked through the scratch. Pain in the neck!!!
I am a first year student. Is it normal i find this shit hard?
@@a.nelprober4971 Limit theory tends to make strong men cry, yes. It was figured out fairly LATE in the history of math, 19th c. You didn't see Pythagoras or Newton coming up with rigorous limit-theory proofs.
@@mrminer071166 thanks. Makes me feel better. These are the inequalities they should be talking about in school
@@a.nelprober4971 It's a perpetual pedagogical challenge to get kids to think in infinitesimals, when they're more comfortable with CONCRETE stuff. For instance, take a long strip of adding machine tape, like 3' to 6'. Have the student fold in half, then fold one piece back to make a quarter, then keep working with the end zig-zag fashion to make 1/8, 1/16, etc, as far as they can go. When then open the strip out again, they'll have a CONCRETE representation of the geometric series 1/2 + 1/4 + 1/8 . .. = 1. And of course when it's folded zig-zag style, they'll have the alternating series . . .I'll let you calculate what that adds up to!
Thousand thanks for sharing this video. It really helps me figure out the basic idea of couchy sequence.
and hence, we prove that the space of rationals is not a complete metric space. since the sequence (1+1/n)^n is a cauchy sequence in Q but does not converges in Q. R is a complete space which means there's equivalence between cauchy's sequences and convergent sequences
We can extand Q , R as it is for now, is problematic
Plz explain the complex analysis
He did though..
@@literallynull 3 years ago I need that :) thanks
All convergent sequences in any metric space are Cauchy. The converse is not always true. For example the sequence (1+1/n)^n does not converge in the rational numbers (in the real numbers it converges to e, but e is not rational).
A metric space where every Cauchy sequence is convergent is known as a complete metric space. In complete metric spaces a sequence is convergent if and only if it is Cauchy. The real numbers with the absolute value metric is complete.
If we have a space that is not complete, we can make it complete, or more precisely, we can construct a complete metric space with a subspace that is isometric to the original metric space. This complete metric space is called a completion of our original not complete space. Importantly, the completion of a space is unique up to isometry.
With this notion of the completion of a metric space, one was we can construct the real numbers from the rationals is by defining the the real numbers as the completion of the rationals under the absolute value metric.
The fact that every Cauchy sequence in R converges depends on the axiom of completeness.
The axiom of completeness implies the fact that every monotone bounded sequence in R converges.
The fact that every monotone bounded sequence in R converges implies the fact that every bounded sequence in R has a convergent subsequence.
Using this fact the convergence of Cauchy sequence is proved.
Can you explain how a convergent subsequence is related to the Cauchy-ness of a sequence? Also, can you explain how a Cauchy sequence requires the axiom of completeness?
thank you for your good effort
this has been very helpful tbh. i’m 17 years old and in uni (ik shocking) and my class has been struggling to understand what a Cauchy sequence is. luckily, i stumbled across your video and it makes perfect sense now! since we’re all from romania, i’ll try and translate everything for those who need extra help learning and credit you for your really good work!!
There's an equivalent statement:
For all epsilon > 0 : there exists an index N : so that for all indices n >= N : | a_n - a_N | < epsilon
You can check the equivalence by yourself. One direction is trivial, for the other direction you can use epsilon' := 2*epsilon.
I find this alternative definition much better, because it uses less indices (only n >= N instead of n,m >= N), and it's in my opinion also more intuitive. Is there a reasonable reason :-) why the original definition is used?
Very helpful video👍..please make videos on group theory ☺️
I found this helpful, thank you sir.
We want the backfliiiiipsss!!! Haha nice videos sir!
Very easy to understand
Love it. This video makes as much (or as little) sense as when I was first taught these thing by the brilliant Sandra Pott in my first year at York uni. I gradually went on to understand. But it takes me right back to those lectures and initially not having a clue what was going on as it was an entirely new topic and idea. In the end took loads of optional real analysis courses right in to my fourth year.
what part doesnt make sense. it helps to look at concrete examples of sequences
19:13
0:00 : good place to start.
Very nice. I'm watching from Brazil.
Thanks for the explanation
Completely awesome.
This is the most headache part in my entire RA course
this has been so helpful. thank you for this!
Thank you so much. Please make an ultimate Real Analysis playlist covering Rudin proofs.
Well done 👍
I dey vex o @Michael Penn
Nice explained but we need more clarify concept
For the second proof, just take the rationals, then clearly if a sequence is Cauchy but in the reals converges to an irrational, then it does not converge to something in the rationals.
What is archimedian pricniple and why is 1/n plus 1 leas than epsilon.
@@leif1075 you can look up the specifics of the property but roughly speaking he's using it here to say that the number n/(n+1) is a number which actually exists (since in a Proof it would not be reasonable to pull a number out of thin air without justifying it). As for why it is less than 1; consider that to be equal to 1 the fraction would need to be n/n, if I increase the denominater by 1 I decrease the overall value of the fraction. For instance 2/2-->2/3, 2/3
Vice versa should work too? Take the sequence a[n] = sin(n)/n, then for all n≥1, a[n] is irrational, but it converges to a rational number.
@@SpartaSpartan117 Thanks yea ofcourse I know why it's less than one i understand basic fractions but i was asking why he knows it's less than EPSILON not one..also later how can he use the triangle inequality if we don't have a triangle here?
@@leif1075 The Archimidean principle states in brief that for any ε>0 - Arbitrarily - small there is a number N (big enough depending on ε) such that Nε >1 (This is a special case. It states that for any y,x>0 nx>y if n is big enough)
That answers your question
Nε>1
nε > Nε >1 for every n>N
nε > 1
1/n < ε for every n>N (N depends on the ε value of course)
Considering the ε-definition of a limit this leads to lim(1/n)=0 as n is getting bigger an bigger (as n goes to infinity)
Great teacher
Good job
Indian btech student here ❤
Thank you
Can you push ahead on solving Math Olimpiad problems?
I like this
Sir .can you please explain why you cancel that (m+1) and (n+1)
dude's jacked and smart
The theorem would fail using a p-adic metric right?
I did not understand one thing. Let An = sum[k=1,n](1/k). An is cauchy, since the difference of their therms get as small as you want, but it diverges. How can that be possible?
"...the difference of their terms get as small as you want" - incorrect. In fact, the difference S_m-S_n where m,n >=N can be as big as you want, cause the initial series diverges, so you can enlarge the difference by choosing big enough m, keeping n fixed.
15:09 what about the sequence a_n = (-1)^n? It's bounded, but it doesn't have a convergent subsequence.
it does, a_n with n odd or n even would both be convergent subsequences. {1,1,1,1,1...}, and {-1,-1,-1,...} respectively converge to 1 and -1.
@@xibbit6322 I figured it out later lol. Thank you.
I can't really see why you can cancel n/n+1 just because it's less than 1. Can someone explain to me, pls?
Ok if we take n=2, we get 2/3 which is 0.66.. And if we round off this value is equal to 1.
Similarly for 10/11 = 0.9 something on round off it is 1
3:22
Wasn't that supposed to be abs value
Is there a sequance that has a limit but is not Cauchy??
It is easy to find example of the converse, a sequance that is Cauchy but has no limit, for instance ia rational sequance converging to an itrational has no limit in the rationals.
But having a limit and not being Cauchy?
Is it true that if I skip ads or use an adblocker, you don't get paid? I feel like that is not common knowledge.
There is a small typo at 17:31; you should take n >= N = max{n_K, N_2} because you defined K such that |a_{n_k} - L| < eps when k>=K (so n_k >= n_K)
at 9:28 , when we say that a series is convergent, shouldn't An approach 0 always? Is that 'L' there simply to count for all other possible emmm spaces (like not real numbers?)
it doesn't always approach 0 right? The first example, A_n = n/(n+1) approaches 1 as n goes to infinity
5:33 it should be m/(m+1) and not n/(m+1), right?
yea
Could you always attach a pdf of a screenshot of all the board at the end, or just move out of the picture at the end. You know like teachers/professors do in actual school. great explanation btw. Thanks.
Do you work out by any chance sir
Binod
why is n/n+1 smaller than 1?
Because n+1 is always bigger than n
Usually R is defined as the rationals completed so you shouldn t have to prove that R is complete, right? It s a property of construction.
Also weierstrass theorem uses the fact that the set containing an is compact, which uses the fact that R is complete, so i feel like there is some circular logic going on
@@Joffrerap Not really, you can construct R with Dedekind cuts and prove that R is complete as a theorem.
12:40 When you say you take the absolute value of each side, do you mean all three parts of the compound inequality? It looks like you are only considering the middle and right parts.
Why are you allowed to take the absolute value of both sides? If a_n was negative, but with greater absolute value than a_N1 + 1, wouldnt this be illegal?
Suppose |An-Am|
@@Pklrs you have a mistake (why don't people ever edit their work :/)
By substitution and triangle inequality we have
|An | = | An + 0 | = | An - Am + Am |
Wait is Cauchy pronounced co-she? Always pronounced it as cow-chi😳😳😳
Lol same. Fortunately, I corrected it within a week as I immediately started looking for videos online as my college course began
4:30
Can anybody state for me why 1/N+1
By Archimedean property, for any positive number r there is a natural number N such that 1/N
This lecture is mostly good, but I must object to the use of the Bolzano-Weierstrass theorem to prove that every Cauchy sequence of reals converges. Any proof of BW will use the completeness of R or some equivalent property, so the proof is circular. You have to use the Cauchy sequence construction to exhibit a Cauchy sequence of rationals to which a Cauchy sequence of reals necessarily converges. The key step is to show there is a rational between any two reals.
Cool❤
That's so cauchy
lord have mercy, i am failing Analysis 1
sounds cauchy
🙂👍
I didn't like it I was expecting something else from you sir 🙏
How much meth did it take