Real Analysis 3 | Bounded Sequences and Unique Limits

[@chilledvibes99]

i like how you describe everything so simply. my uni lecturers usually just rush through the small details so u never really grasp the full thing

[@PunmasterSTP]

How have your classes been going?

[@punditgi]

Another excellent video. This series is truly a gem! 😃

[@suryanarayanachebolu7870]

Thanks for the Valuable Lecture. Entire Math from Degree to P.G has a lot to do with "Real Analysis" .

[@covariance5446]

Hey! Just wanted to say thanks so much for these excellently made videos! I'm, uh, kind of a screw up in life (did a bachelor's and master's in psych; dropped out of a phd psych program after many years... returning to school as a mature student) and am planning on taking Real Analysis I in January 2024 (*knocks on wood that nothing unexpected comes up to stop me from doing that*) and I've heard lots of stories about how difficult this course/topic tends to be. Thus, I am very grateful to be able to see your videos! I don't have a lot of money at the moment, but if things go swimmingly, I will definitely say thanks with the "Tip" button on youtube. Or actually, I see the Paypal link in the youtube description; perhaps I'll use that instead! :) Anyway, I hope to continue watching this playlist and making notes. In the meantime, I just wanted to express my gratitude in case I may forget later. On topic: I definitely would not have thought to use the triangle inequality to prove the diveregence of (-1)^n; I never even thought to formally prove it, before! I've only had a little experience with epsilon-delta proofs in Calc II (but obviously not enough). Will be taking Calc III and a discrete math (they call the course mathematical structures at my school) in the fall prior to real analysis in the winter, so fingers crossed I'll be a little less inept by then :)

[@brightsideofmaths]

Great :) I wish you good luck with studying. I hope my videos will help you :)

[@ChrisOffner]

I don't understand why you choose *epsilon = 1* here, and why then *|a_N - a| < epsilon* and *|a_(N+1) - a| < epsilon* are supposed to hold. If *a = 0,* then neither of those hold, because the distance between *a_N* and *a = 0,* as well as the distance between *a_(N+1)* and *a = 0,* are both equal to *epsilon = 1,* not less than *epsilon = 1.* What am I missing here? My confusion starts basically at ~1:42 when you say "We have to choose *epsilon* so small that in the *epsilon* neighbourhood around *a,* we don't have *-1* and *1* at the same time. This means we need a number smaller or equal than *1.* " Why a number *equal* to *1* ? Doesn't this only work with numbers *smaller* than *1* (see case *a = 0* )?

[@Zumerjud]

The definition uses less than. So it will be

[@doublecheeseburgirl-]

yeah im having the exact same issue. shouldn't we choose some epsilon in (0,1)? Unless a = 1 or -1, but from what he said it seems like you could choose any value for a since such an a doesn't exist

[@doublecheeseburgirl-]

@@Zumerjud yes, exactly. if a=0, then |1-0|=1, not

[@MrCryzs123]

My prof used epsilon = 1/2, that way there is no issue @@doublecheeseburgirl-

[@luissalasar2129]

Congrats on the excellent lecture! Could u tell me what software you use to write on the screen?

[@brightsideofmaths]

Xournal :)

[@luisernestobuenosalasar5738]

Thanks!

[@atlanta2203]

Thank you for the videos:) These help me understand topics so much better than my uni lectures

[@brightsideofmaths]

Thank you for the support and watching the videos :)

[@jaydensidabutar4629]

hello, thank you for putting out these fantastic mathematics series! these help me understand topics like analysis and calculus so much better, and just a quick question though which sort of went over my head, why is it that the part in 8:40 can't happen?

[@brightsideofmaths]

Glad you like them! We have a positive number A that fulfills A < 1/2 A, meaning A < 0. Hard to satisfy for a positive number.

[@jaydensidabutar4629]

​@@brightsideofmathsOHH RIGHT! thank you so much, i tried reading the proof from left to right and understanding it but that was the only thing i missed.. such a simple thing 😅, thank you again! and keep up the good math content

[@priyamhazarika2744]

Very helpful . Understanding everything vry clearly . Thank you

[@brightsideofmaths]

Glad it was helpful!

[@triton62674]

Do we have to use the TI in the end there or is it enough to say mod(a-ā)

[@PunmasterSTP]

I’m not sure, but I think you have to. I think the main idea is to use the triangle inequality to show that the sum of the distances of a_n to each separate limit is an upper bound, and then 2e = |a - a~|/2 is strictly greater than |a - a~|, which is a contradiction. As a side note, it seems like this proof would have worked just as well by choosing epsilon = |a - a~|/2.

[@IITIAN_ROHIT616]

well understood and I'm in IIT DELHI the professor here is not understanding me . anything if it is sufficient for engineering maths then I can follow your lecture what is proper please tell me SIR 🙏🙏🙏

[@hyperduality2838]

Convergence (syntropy, homology) is dual to divergence (entropy, co-homology) -- the 4th law of thermodynamics! Bounded is dual to unbounded, closed is dual to open. Elliptic curves are dual to modular forms. "Always two there are" -- Yoda.

[@jorgequispe8304]

beautiful example

[@tatasharada]

Thanks for this excellent course material. Many many thanks for all the efforts. I have a question. If I have a sequence an = 1/ (n-1); clearly this sequence is convergent because as n goes to infinity the sequence goes to zero. But will it be considered bounded? Because at n = 1, it starts off at 1/0 ~ infinity? How would that be different from a sequence say an = n^2 which is divergent. So in that vein is the sentence all convergent sequences are bounded valid?

[@brightsideofmaths]

"infinity" is not a real number and cannot be a member of the sequence. Hence, for an = 1/ (n-1), the sequence should start with index n = 2.

[@tatasharada]

@@brightsideofmaths OK. Thanks for the explanation!

[@sarvesh_soni]

@@tatasharada ​ infinity is a Hyperreal number i heard sometime (read on wikipedia)

[@jamesyeung3286]

what's the difference between ε-N proofs and infinitesimal proofs? isn't ε also getting "infinitely close to zero"?

[@brightsideofmaths]

ε-N proofs are the rigorous formulation for some infinitesimal proofs. If you do standard analysis, the ε-N is the correct way to go for describing things like "infinitely close to zero".

[@ThreeMiningHD]

Hi, du machst echt coole Videos! Ich wollte Mal fragen, ob du auch über Übungszettel zu deinen verschiedenen Videoreihen verfügst? LG

[@brightsideofmaths]

Yes, I do! They are in production for the videos :)

[@ThreeMiningHD]

@@brightsideofmaths Cool! Do I have to be a member to have access to the sheets? And will they come in the Future or do they already exist? :)

[@brightsideofmaths]

@@ThreeMiningHD When I publish them, I will talk about it, here on the channel :)

[@brightsideofmaths]

@@ThreeMiningHD Moreover, being a member is not really about getting stuff for the videos but about supporting me for making videos :)

[@ThreeMiningHD]

@@brightsideofmaths okay nice! I am studying physics in the second semester and next to "Mathe für Physiker 2" i try to educate myself in further mathematical topics, but i guess Your awesome Videos arent enogh, practice is important too :D I am looking forward to Your Future Videos :)

[@maxpercer7119]

For the proof of ' (a_n) convergent → (a_n) is bounded ' at 6 minutes, I would just add, you have to pick a specific epsilon. A convenient one is, e = 1. Or if you prefer to leave it as epsilon, without specifying its length, call it epsilon* , or epsilon_0.

[@brightsideofmaths]

Thank you for your comment :) Yes, it's a fixed epsilon. The name is not so important when you know the meaning. Therefore, I prefer it to keep it simple. Of course, one should add more details to my short sketch of the proof.

[@angusclark6170]

Hello! I am slightly confused about convergence implies bounded. In your example, a1 and a2 are both outside the epsilon neighbourhood you drew. How do we know that those elements outside the epsilon neighbourhood do not ruin the boundedness? Is it because there are only finitely many of them?

[@brightsideofmaths]

Yes, for finitely many, there is always a largest element.

[@sarvesh_soni]

i wanted to ask a conjecture i have, I used calculator, and whatever number i start with, If i take the cos of that number infinitely many times , it converges to the same number like if we define a sequence (An) : An = cos(cos(cos... n times (n)...)) then We have to prove An is convergent!, Is this a well known convergence? Thank you for the quality content!, i will be joining UG course in some months! this will help

[@brightsideofmaths]

Yes, it's a convergent sequence :) A typical application of a fixed point theorem.

[@sarvesh_soni]

@@brightsideofmaths ohk, Thank you sir for quick reply

[@johnsu9949]

do you have a playlist for different proofs such as contradiction and contraposition, also do these methods come from the deduction rules stated in your "start learning mathematics" playlist?

[@brightsideofmaths]

Yes, they come from the deduction rules of my "start learning mathematics" playlist.

[@johnsu9949]

@@brightsideofmaths I see, also I'm still stuck at the first proof, at the part where u mentioned that epsilon

[@brightsideofmaths]

@@johnsu9949 We choose epsilon small enough such that our argument works. Don't forget that the convergence definitions states that it has to work for every epsilon, not matter how small.

[@johnsu9949]

@@brightsideofmaths hmm that still doesn't answer why we don't want -1 and 1 at the same time😢

[@brightsideofmaths]

@@johnsu9949 Having two different limits is simply not possible.

[@DISHANTJHA]

it feels illegal for this video to be so good. thanks from the bottom of my heart

[@brightsideofmaths]

Haha, thanks :)

[@rajinfootonchuriquen]

You are great :D

[@xoppa09]

Interesting, for your proof at 3 minutes that the sequence { (-1)^n } diverges, any epsilon less than 1 will produce a similar contradiction. The epsilon value of 1 is the largest of such epsilons that lead to an absurd result (e.g. a false inequality like 2 < 2).

[@brightsideofmaths]

All correct :)

[@PunmasterSTP]

Bounded sequences? More like “Boredom is left in pieces!” Thanks for another stimulating lecture.

[@oxygen2671]

These ideas are very easy but he makes them seem very difficult

[@brightsideofmaths]

Who? :)

[@oxygen2671]

My professor

[@guru0503p]

Why shouldn't the epsilon neighbourhood of ‘a’ contain -1 and 1 simultaneously?

[@brightsideofmaths]

Because it is too small for this :)

[@guru0503p]

I get it, sir. But, if you don't mind, I have one other question. How big can a neighbourhood of a point be?

[@brightsideofmaths]

@@guru0503p As large as possible. The whole space (the whole real number line) is always a neighbourhood.

[@dziugaschvoinikov4440]

5:56 why is |a_N| not included in the definition of C?

[@brightsideofmaths]

Because that is how we define C :)

[@quads4777]

Badass video very well done

[@brightsideofmaths]

Thanks a ton! :)

[@nickl.2565]

hey, I don't get it why it is |a-ã| at 8:12?

[@brightsideofmaths]

That is the quantity we want to calculate or rather estimate.

[@ramyramy9620]

| 1 - a | + | (-1) - a | < 2 we have -a and another -a how are they gonna cancel each other using only | x | + | y | >= | x+ y | 6:25

[@timothypulliam2177]

I love you

[@keptyouwaiting]

...

[@brightsideofmaths]

Question?

[@keptyouwaiting]

@@brightsideofmaths I have no question. I just commented for comment.

[@sandfordstebbings125]

Unfortunately I find this very hard to follow. There is no intuition here, its like your just reading it.

[@brightsideofmaths]

Sorry for that.

[@Stoic_Persistence]

Deutscher

[@brightsideofmaths]

No way!