Real Analysis 7 | Cauchy Sequences and Completeness

[@lucvansprang357]

imagine taking real analysis before this series, i love you man.

[@brightsideofmaths]

thanks a lot :D

[@michaeltamajong2988]

If it is not an application, this man has the most organized handwriting I've ever seen!

[@brightsideofmaths]

Oh, my handwriting is much worse as you can see in the Functional Analysis series: tbsom.de/s/fa

[@qiaohuizhou6960]

1:31 idea of Cauchy sequence , the sequence eventually lay arbitrarily close to each other 2:24 fact: Cauchy sequence is equivalent to convergent sequence when dealing with reals 3:15 dedekind completeness and the properties for subset of real numbers 3:51 proof ( read again ) 7:40 criterion for convergence of sequence

[@qiaohuizhou6960]

Hi, I am revisiting the materials and am a bit confused. I think from Cauchy to the convergent sequence we only need the triangular inequality and I am not sure if the completeness axiom is needed? I think the criterion for convergence of sequence needs the completeness axiom as shown the monotone convergence theorem of the Wikipedia page.

[@Algebraictivist]

Thank you for the amazing explanation, but how do i get the (1/2)^n-1*|b_1-a_1| part at 6:58

[@suryanarayanachebolu7870]

Super explanation sir.

[@NotFound-bg4sr]

This is interesting coming from your presumably newer videos on introductory math concepts, where you motivate convergent sequences using Cauchy sequences, the other way around. I found this other approach helpful too

[@brightsideofmaths]

I don't know what you exactly mean. Convergent sequences and Cauchy sequences are related by the completeness.

[@NotFound-bg4sr]

@@brightsideofmaths I found it interesting that you introduced the definitions in different orders in the two series. In the learning reals series, it made sense to introduce Cauchy sequences first to demonstrate that they don’t always converge for rational sequences. In this series, you started with convergent sequences and worked with them a lot before introducing the Cauchy definition, which made sense to me because intuitively completeness was built into my idea of what the number line should be, from my previous math courses. Idk, I just found it an interesting pedagogical choice I suppose? I think it helped me better understand the relationship between the two sequence types and the real numbers. Both sides of the logical equivalence within the reals, as you highlighted. Sorry if that doesn’t make sense or if I am messing something up with the understanding, it just felt like a aha moment to me watching that part of the video, relating the new to the old. Love these videos, hoping to learn a lot of interesting math working through these playlists!

[@enriquecorimayo2395]

I have a question... in the last part of the video, shouldn't be "the existence of the infimum of this set?" instead of supremun ...minute 8:15. Thanks

[@brightsideofmaths]

Oh yeah, you might be correct there. Sorry for the confusion.

[@danielyoo828]

@@brightsideofmaths Yeah, it should be the infimum since "bounded from below" ==> there's a lower bound, which is one of the criterion for an infimum. It would be helpful if you could add an annotation in the video! Thanks for the lecture series; they are very helpful :)

[@xoppa09]

Dear brightside, excellent video. One issue i am doubtful about. The definition of a cauchy sequence states that for any epsilon > 0 there is an N such that for any m,n >= N (note the comma between m and n, not just m > n) then |a_n - a_m| < epsilon. However in your proof at 6:33 you showed that m > n . Do you have to do a seperate proof for n > m ? In the expression |b_n - b_m| it looks like b_m is varying while b_n is fixed, but you could also vary b_n and fix b_m, or does it make no difference because of absolute value, i.e. without loss of generality.

[@brightsideofmaths]

yeah, it makes no difference because of absolute value/ norm :)

[@chair547]

That convergence application was not trivial and I wish you would have gone over proving it.

[@brightsideofmaths]

Since we have proven Dedekind's completeness, this convergence application follows in a direct way. Just try writing it down :)

[@PunmasterSTP]

Cauchy? More like "Cool and catchy!" These videos are very nice; thank you for making and sharing them.

[@monadic_monastic69]

I had no clue this was also called the join btw, I've been learning some geometric algebra and wonder if there's a relation to its non-linear join and meet operators.

[@Study-lx9lt]

Thanks for the quality content, I'm just wondering what can I do after watching the videos to improve?

[@brightsideofmaths]

Thanks! First, after each video, you should do the quiz about the video: tbsom.de/s/ra And then, you can apply your knowledge to some more exercises :)

[@michaelnicodemus9109]

I don't understand why you define a_n+1 as x in the second case. Why is it not c_n?... You say there exists an element x in M such that x > c_n, but how do you pick an x when there are multiple choices (possibly infinitely many)?

[@brightsideofmaths]

It is not important to pick an x. We only need to know that there is at least one x with the property. (Maybe you have already heard of the axiom of choice which we could apply here)

[@filmmyduniya-mf1hq]

Thank you sir for this wonderful lecture but i did not understand why the limit is SupM and also in case of (1+1/n)^n the limit is e how do we now this by conv. Criteria

[@brightsideofmaths]

Thanks a lot! Limit of (1+1/n)^n is e by definition of Euler's number.

[@user-ij2ue1cg9i]

THANK YOU

[@Leslie.Green_CEng_MIEE]

At 5:36 and 6:00 there is an typo with the iteration values in the *case (2)* parts. (I am using an underscore for the subscripts) In the *case (2)* assignments, when we don’t have a closer upper bound, it is written as a_2 := x and a_(n+1) := x These should read *a_2 := c_1* and *a_(n+1) := c_n* *Discussion* x is a value which exists, showing that c_1 is not an upper bound. Agreed. However, x is not a specific iterated value. Certainly there is a value x > c_1, we just don’t know what x is. If you like, the iteration has left values (a’s) and right values (b’s). We pick which ‘side’ is changed to progress the iteration. The calculated c value goes either to the left or to the right.

[@brightsideofmaths]

Yes, thanks!

[@brightsideofmaths]

However, of course, the whole procedure also works with the chosen x here. It's just that, in practice, we would rather choose the already known c_2 than the value x. And moreover, in the end, we are only interested in the sequence of b_n.

[@yaoweizhang8354]

Dear The Bright Side of Mathematics, Last video on your list is still set as private video. I guess it is meant to be a video talking about sup and inf. Regards.

[@hn97754]

Thank you you save my life!

[@brightsideofmaths]

Happy to help!

[@maiSenpaiDaisuki]

7:01 hello, I don’t understand why did you put (1/2)^n-1, I think this power will be increasing instead of decreasing. I think it will be more reasonable to put n+1 instead, so that the denominator will get larger and larger, then it will actually be decreasing. Can someone explain?

[@brightsideofmaths]

Please note that (1/2)^{n-1} *(1/4) is equal to (1/2)^{n+1}.

[@gingervacation]

the proof of completeness reminds me of binary search

[@elidoz7449]

there is one thing I don't understand for example with the sequence where a_n=1/1+1/2+1/3+...+1/n the terms of the sequence get arbitrarily close to each other, but at the limit it is infinity and doesn't converge

[@brightsideofmaths]

I would suggest that you watch my videos about series :)

[@pencil033]

That is because a_n is not a Cauchy sequence. Notice that the definition is: Given eps > 0, there exists an N such that FOR ALL m, n > N, |a_n - a_m| < esp, not just n and n+1. In your example, say esp = 0.1, you might want to argue that N = 9. Indeed |a_11 - a_10| = 1/11 < 0.1, however, |a_12 - a_10| = 1.742 > 0.1 In fact, no matter how large the N is, if you take the nth item and the (n+ big enough number)th item, their distance can always surpass your chosen esp. Details are left as an exercise :)

[@ezranathanael9566]

@@pencil033 It is a Cauchy Sequence actually.

[@pencil033]

@@ezranathanael9566 a_n is divergent so it cannot be Cauchy sequence. I think I even explained how it is not. Maybe share your thoughts? I'm more than happy to discuss more.

[@cadenrowe738]

@@pencil033 so this sequence does converge to 0, however the corresponding series does not converge.

[@brockobama257]

At 6:30 you meant n>m right?

[@brightsideofmaths]

No, I guess, I really meant m > n. But with the absolute value, this is not so important. Why do you think otherwise?

[@ashwnicoer]

The sequence is monotonically decreasing hence there should be no equality sign(in Important application line)

[@brightsideofmaths]

There is no equality sign there. It is "less or equal", which is the definition of mon. decreasing.

[@ashwnicoer]

@@brightsideofmaths Yes ,I am speaking of it only (. less than or equal to),I have read till now that a monotonically decreasing function decreases throughout i.e. it doesnot becomes constant anywhere . So ,if it doesnot becomes constant then there should be only less than sign in place of less than or equal to.

[@brightsideofmaths]

@@ashwnicoer Of course, there might be other definitions, but for me mon. decreasing functions could be constant. Otherwise, I would call them strictly (monotonically) decreasing.

[@ashwnicoer]

@@brightsideofmaths I read in books that strictly increasing or decreasing functions are called Monotonic else simply writing increasing or decreasing would suffice . Thanks a lot for clarification and good content that you are providing !

[@VaheTildian]

​@@ashwnicoer For this definition, I think it's fine either way (strictly decreasing or just decreasing) as in both case, it's either constant (so it converges) or it's decreasing until reaching the lower bound (so it converges). You're question was very legitimate though! And it's good to be precise in definitions.

[@ffar2981]

Why is it not a strict inequality at 6:45? I.e., can b_m really be a_n?

[@brightsideofmaths]

I get this question a lot. Wherever there is a strict inequality, you can use ≤ as well. This is never wrong :)

[@monadic_monastic69]

Gotta love those or statements.

[@claudiamaquedadiaz4082]

I don't understand the bit |bn-bm|

[@claudiamaquedadiaz4082]

Like: i understand that as n increases, diff between an+1 and an is smaller. And i get that diff between 2 upper bounds("original al better one")is smaller than diff between "original upper bound" and a value in the sequences (an). Then, I understand that that's smaller than the diff between "first bound b1 and first value of the sequence a1" but, why do we need the (1/2)^(n-1)

[@nm-de3bw]

@@claudiamaquedadiaz4082 if x is the minimum value above Cn, then the distance between the two is halved when each an or bn substitution happens. It's halved because Cn is the average/middlepoint between the anterior an and bn.

[@herp_derpingson]

5:12 thats looks like binary search

[@brightsideofmaths]

It is :D

[@oxyrox7194]

Not really happy with completeness axiom. This is extremely non-trivial property to just present as an "axiom". (A bit like introducing complex numbers as: let i be square root of -1). Isn't the proper way to develop the theory is via Dedekind cuts?

[@brightsideofmaths]

Yes, that is one possibility. Another, you can see in my Start Learning Mathematics series.

[@brightsideofmaths]

Here you can find the part: kzbin.info/www/bejne/f2LJeZxvp710d6s

[@fullfungo]

Dedekind cuts are used to construct a MODEL of real numbers, not the THEORY of real numbers. These are two different things.

[@qianliu7471]

The proof of completeness seems not complete, you have proved “bn” is convergent, and the limit of “bn” is a upper bound of M, but it can’t show it’s the “least” one.

[@brightsideofmaths]

Why shouldn't it be the least one?

[@qianliu7471]

sup(M) is the least upper bound of M, the lim(bn) is smallest in the sequence bn, and indeed it’s the supremum, but it seems you don’t give the details from my perspective.

[@brightsideofmaths]

Yes, you are right. I always skip some details if I think it's appropriate and that one can fill them in easily. Do you think it's too hard in this case?

[@qianliu7471]

For this case, it skips a bit too much for me, at first glance, I thought I have missed some points.😅

[@brightsideofmaths]

@@qianliu7471 That's okay. It's good that you noted which gaps need to be filled. Please try to prove it and let me know if you have problems with it.