Real Analysis | The density of Q and other consequences of the Axiom of Completeness.

[@00tact]

Finally, a channel with some real mathematics.

[@augustbergquist5649]

real in two ways lol

[@vaibhavdev974]

yessssssssss

[@harryh5666]

I know right - fantastic

[@DarkMonolth]

You can't just release 900 videos a day on complex math topics! Michael: Hold my beer

[@parthanaved3866]

Um, 900 videos, 24 hours=86400 seconds, every video would take 96 seconds, which is about 1 and 1/2 minutes. Would that be possible? I think maximum 20 videos. I get the point your trying to say 'a lot'.

[@rusejames7242]

@@parthanaved3866 mafs

[@duckymomo7935]

how does this guy do it?

[@jb_1971]

@@parthanaved3866 no, he just talks really fast and then slows it down in post-production. Easy

[@deuteriumtritium9700]

@@parthanaved3866 I work 40 hrs per day

[@Bermatematika]

For the density of the rationals, my favorite argument is as follows: Let says you take your n such that 1/n

[@adviolin]

This is my favorite proof too!

[@anomalous5048]

wtf ? didnt get

[@khbye2411]

Is it right to think of it like this? we want to design a rational number q = m/n = m * (1/n) we have a little rabbit positioned at 0 on the number line the rabbit hops to the right with each step size = 1/n the rabbit takes m many hops to the right so, the total distance travelled by the rabbit is q=m * (1/n) to ensure the rabbit ends up in the trench (a,b)=(left endpoint, right endpoint) we need: -the left endpoint < total distance travelled by the rabbit (so the rabbit cannot be at 'a' itself since the trench is an open interval, the rabbit must move a bit more to the right of 'a')...this means a < m/n -m to be the smallest possible number of steps...this means (m-1)/n

[@Bermatematika]

@@khbye2411 Yes, excellent!

[@schmud68]

i just like to think of rational sequences built by matching the first n decimal places of the desired number. Gets you in range of 10^-n of your desired number. This is the sequential definition of dense, but you could use similar arguments to prove the statement in the video. The sequential statement generalises better to arbitrary metric spaces.

[@jaredvv86]

This was explained extremely well. Keep them coming please

[@jonaskoelker]

Theorem (used at 14:50): for all x in R there's a unique m in Z such that m -n and -n is in Z. Let m = sup(A). Clearly m

[@Aman_iitbh]

thnx

[@catherineferris4379]

thank you for your analysis videos! finally, a professor who EXPLAINS their steps! please continue making!

[@adarshkumar7238]

Your channel is best for real analysis. Keep upgrading

[@atomic_soup_juice]

I wish I had some of this content when I was talking my real analysis classes in school

[@thedoublehelix5661]

This is like the book I'm reading rn called "Understanding Analysis"

[@EdgarCamacho11729]

By Abbott?

[@thedoublehelix5661]

@@EdgarCamacho11729 yes

[@EdgarCamacho11729]

@@thedoublehelix5661 I enjoyed that book, specially in the chapter about continous functions. I hope you enjoy it as well

[@danimillson923]

Like Edgar said, it's a great book. Excellent textbook for introductory real analysis.

[@JasonOvalles]

That's actually the book that Michael Penn uses to teach the Real Analysis course for which he's making these videos. These videos came just at the right time for me too since I just picked up Abbott's book about a month ago! It's ALMOST like I'm taking the course with him! Minus the office hours, groupwork, individual and group feedback, in-person lectures, graded assessments, student presentations, strict schedule, clear syllabus, midterm and final exams.... okay, its still nothing like taking a real college class. But still great! Thanks @Michael Penn!

[@CallMeIshmael999]

I'm getting very impressed by how wide a variety of topics you're putting videos up about. Are you teaching classes on all these things, or is this just a fun thing you're doing in your free time?

[@karthikthiagarajan2331]

Wonderful flow. You explain some subtle points concerning the theorems as well. Are you planning to teach all of real analysis and upload them?

[@joshbolton2782]

Great video. Amazing example for students on how to use sup and inf effectively.

[@quantabot1165]

Playlist length so far is 12hours...Thank you for your time Professor!

[@user-uw1ut4ss2q]

Monotone convergence theorem is also an important consequence of the axiom of completeness from which we can derive a lot of essntial theorems in real analysis.

[@alex_8704]

9:59 There is an opinion that *1/n < y* violates the quantification domain n ∈ N, because natural numbers (unlike the rational numbers) have no division operation, and that it would be better to write *1 < ny* instead.

[@PunmasterSTP]

The density of Q? More like "Terrific information for you!" Thanks again so much for putting all these wonderful videos together.

[@GoutamDAS-ls1wb]

Thank you very much sir for posting these lectures. I have started learning Real Analysis from your website.

[@davidjflorezr3099]

very glad to find this, will thoroughly enjoy my winter break now

[@Nilius01]

Great courses in mathematics! I enjoy showing your videos and I look forward with great interest to further helpful topics.

[@moshadj]

They also all have to be non-empty right. Which I guess is covered if you don't consider the empty set an interval (as it is closed)

[@erikb3799]

You should put links to these videos in your competition videos. I'm interested in these videos but didn't know they existed until you mentioned them in your Q&A video.

[@roberttelarket4934]

Michael you are SUPREMUM!

[@quantabot1165]

Beautiful!, Some REAL analysis Finally

[@oida10000]

What about half open intervals: [a, b)={r in R: a

[@mohamedaminehadji6415]

Still not true, because you can have I_n = [0,1/n), and the intersection of all I_n is empty

[@lemons107]

Brilliant. Simply brilliant.

[@parthanaved3866]

Hey Michael, try these ones: Go to the link and download the file. I swear it isn't something else. It's the questions of the Bangladesh Mathematical Olympiad 2019, Classes 11-12 (basically high school but we call it 'Higher Secondary' here). The questions are hard, especially the function one. So, can you help me?

[@shurjoaunibar]

True! I also had a hard time.

[@sunnyday1239]

OOOH! Fellow Bangladeshi here! This comment is so recent!

[@parthanaved3866]

Um, I wasn't expecting Bangladeshis... hey when did you start watching him?

[@maxpercer7119]

michael penn makes real analysis look easy ;o (just kidding, but he does make it understandable and intuitive)

[@sunnykim7246]

i love how every maths person writes suppose as sps lmaoo - great lecture!!! thanks for uploading

[@maxpercer7119]

11:30 you might add a few details. Letting a = sup(N) we have that a - 1 is not an upper bound of N. Why isn't it an upper bound. Because if a -1 was an upper bound, by the definition of supremum it would be the case that a 0 , then is an axiom, then -1 < 0 , then adding 'a' to both sides, -1 + a < 0 + a.

[@evenaicantfigurethisout]

aren't real numbers also unbounded? i.e. given any n in N there exists an x in R s.t. x>n?

[@2kreskimatmy]

i would never come up with these proofs on my own, although they're pretty straightforward, cool also

[@robertkrohn9385]

For: an < m < an +1 < bn in the density of rationals proof, should it have: an < m

[@zcombe2]

Your proof of 'why not in Q' doesn't seem to work as {x in Q; pi - 1/n

[@senhueichen3062]

Super clear!!!!!

[@tgx3529]

The sequence bn-an has guaranteed not negative limit.

[@evenaicantfigurethisout]

i read that the archimedean property has something to do with infinitesimals and convergence, but it went over my head. can you explain their relationship in layman's terms?

[@briansmith9053]

My only concern is, in the nested interval property proof: are the nested intervals you've constructed truly arbitrary? I understand that A and B are arbitrary to begin with, but then it seems that we impose the extra condition bn >= an for all natural numbers n, in order to construct our intervals. But if they were truly general, then we could have some bn < an, which it seems would violate the requirement that the intervals always be nested. This is the point that makes me worried that these nested intervals aren't perfectly arbitrary. Is this a "WLOG" moment that I'm missing somehow? Clarification: the specific claim that might be problematic is "For all natural numbers n, bn is an upper bound for A." I'm just wondering how we can make that claim, given that the nested intervals must be arbitrary.

[@ekeebobs7520]

Any closed interval with n€|R, [a_n,b_n] = {x€|R: a_n

[@jorgearturoquirozcabrera3372]

Can we see some kind of explanation around Souslin Trees?

[@joshhickman77]

It took me a couple seconds to see the proof that it doesn't work for open intervals -- my proof ending up being the 2nd part of the archemedian principle thing. Also, more backflips haha :)

[@TheMauror22]

What about inf(B) in the proof of the NIP? Can we say that sup(A)=inf(B) or just sup(A)

[@mohamedaminehadji6415]

We cannot say that sup(A) = inf(B). You can for instance have a_n = 1-1/n and b_n= 2+1/n, you'll have sup(A)=1 and inf(B) = 2

[@sonnyjordan332]

I have a question on one of the proofs. The set of natural numbers is discrete. If we let alpha = Sup(N), then we know alpha - 1 is not an upper bound. But since the set of N is discrete, ie there should be no element b/w alpha and alpha - 1, how do we assume there's an element?(I know we assume that there is an element greater than alpha - 1 because it's not an upper bound but this discreteness confuses me). Thank you, sir.

[@ronzang]

At first we did not know anything about alpha other than alpha is a real number that acts as a supremum of natural number. We only knew that alpha is an element of natural number after we found that alpha is equal to n+1.

[@JasonOvalles]

What Michael is proving in the video is that R does not contain an upper bound for N. Essentially, that when we added the AoC, we didn't create some magic number that is larger than all N. Because he's verifying this within R, then there are elements between alpha - 1 and alpha. Hope that helps!

[@doctorwonderful]

How can infinity (or negative infinity ) have an upper bound (or lower)? For example if you have the closed interval [1, infinity], wouldn't that make the axiom of completeness not complete because the upper bound would have to be in the set itself and s = a and not a < (or equal). Or is it okay for the least upper bound to be contained in the set?

[@2funky4u88]

The Axiom of completeness only applies for sets who already have an upper bound. [1,infinity] does not have an upper bound. But for example [1,40] would have upper bounds [40;40,2;43;100;...], where 40 is the least upper bound. So you only consider sets who have an upper bound and then notice that if the set has an upper bound, there has to be a least upper bound. As stated in the video, the least upper bound can be contained in the set, since s>=a.

[@JasonOvalles]

To add to what 2funky4u mentioned, writing infinity as a "bound" is really just a shorthand for saying "unbounded." So when we write [1, inf) we are saying that the interval is unbounded above. The Axiom of Completeness (AoC) states that IF a set has an upper bound, then it has a least upper bound. Since any interval like [1, inf) is unbounded, then AoC doesn't apply. In addition, we never write [1, inf] if we're talking about the real numbers. This is because inf is not a real number so it can't be included in an interval consisting of real numbers. We always write it as [1, inf). Lastly, the supremum of a set can definitely be contained within a set. In fact, any closed interval like [1, 4] will contain its supremum. In this case, the supremum is 4. This is because 4 is an upper bound for [1,4] and 4 is smaller than all other upper bounds.

[@mrsv5287]

I have a question concerning the first result. I understand the example of why open sets doesn't work. However, where in the proof we use the fact that our intervals I_n are closed? The proof seems to work for open sets as well no?

[@mrsv5287]

@Mathew Briguglio I understand that, however b_n would still be an upper bound if it was open. There is no contadiction for the bounds if it was the case because a_n would still be less then b_n for all n therefore by the axiom we can find a=sup(A) and the reste follow...

[@mrsv5287]

@Mathew Briguglio I found another proof of this result and it use the fact that the intervals are closes so I got it. Thanks to take the time to respond to my comment tho!

[@johnwroblewski6458]

@@mrsv5287 He fails to mention that because the intervals are closed, they contain all their limit points. Thus sup(A) in A. To use his counter example with his proof scaffolding, see that A = {0}. Thus sup(A) = sup({0}) = 0 which is not in I_n for any n much less some n. Thus the argument would fail here. The fact that sup(I_n) is in I_n along with the fact that b_n >= supA >= a_n allows us to state that supA is in [a_n, b_n] for all n.

[@joostouweneel7184]

Hey, does anyone have some book recommendations for an introduction to analysis? (For a high school student). Thanks!

[@stephenhu2000]

That said, PDFs are always available online if you know where to look

[@beatoriche7301]

Analysis is a very broad subject, and it really depends on what you want to learn about. I’ve mostly read German textbooks on analysis so far, but there are a few English ones I can recommend. First, it is important to understand that there are two equally valid approaches to analysis - _standard_ analysis, as developed by Cauchy, Bolzano and Weierstrass (among others) in the 19th century, which defines limit processes via epsilon and delta (if you’re not familiar with the idea, you might want to watch 3Blue1Brown’s calculus series and chapter 7 in particular), and _nonstandard_ analysis, which uses _infinitely small quantities_ to define derivatives, integrals, etc. - it’s basically the intuition the founders of calculus had in mind, only on a mathematically rigorous footing. For the former, I recommend Abbott’s “Understanding Analysis,” and for the latter, I recommend “Elementary Calculus: An Infinitesimal Approach” by Keisler (which, as a side note, is also available for free online). The latter requires no prerequisite knowledge of calculus whilst, in the case of the former, you might want to already be somewhat familiar with the ideas. An excellent book I would highly recommend to anyone with even a passing interest in the subject is Tristan Needham’s “Visual Complex Analysis” - it’s an amazing piece of work conveying the intuition behind elementary complex analysis remarkably well, with all sorts of interesting detours into connections with other areas of math. It’s often not particularly rigorous, but it’s not too hard to adapt most of the arguments into rigorous proofs if you’ve studied some analysis before. However, real analysis isn’t really a prerequisite, and many central notions you might come across in a real analysis course (such as limits via epsilon-delta definitions) are also covered, albeit with a focus on complex functions. If you’re interested in integrals and the like, I recommend “Measures, Integrals and Martingales” by René Schilling, which is a pretty dense but still really good textbook. However, it does assume familiarity with basic epsilon-delta analysis.

[@joostouweneel7184]

Beatorīche Thanks for your help! Really appreciate it :)

[@JojiThomas7431]

@@beatoriche7301 Thanks

[@Jim-be8sj]

I think I once had to prove the nested interval thing but it was, nested balls in a general metric space. I probably couldn't prove it now.

[@ndelo4947]

That's a good place to stop... 😁 new favourite phrase...

[@liyuan-chuanli8468]

Here, the axiom of completeness = the least upper bound property of the real numberes.

[@GusTheWolfgang]

Wasn't what you call the "Archimedes property" another axiom necessary for the construction of the real numbers?

[@Bignic2008]

No. I believe it’s equivalent to the completeness axiom (although only one implication was shown here), so you only need one or the other.

[@EdgarCamacho11729]

If that were the case, how would do you state it? since it begins with "for every real number ..." I think you are confusing it with the so called "Dedekind cuts". As far I know, there are only two ways of getting the reals from the rationals; one is via Cauchy sequences, and the other is with the Dedekind cuts

[@beatoriche7301]

There are multiple valid formulations of the completeness of the real numbers - you may also assume the Archimedian property as an axiom; along with the axiom that every Cauchy sequence of real numbers converges, this is equivalent to assuming the least-upper-bound property.

[@beatoriche7301]

@@EdgarCamacho11729 There are actually more than just two - it depends on your analysis textbook. I'm not sure if I understood your question, but if I did, it's actually not that difficult, since the rationals are also an Archimedian field. The real numbers are the rational numbers, an Archimedian field, extended using the limits of all Cauchy sequences of rationals - that's basically it.

[@EdgarCamacho11729]

@@beatoriche7301 Oooops, I got confused, since I continued following the explanation of why the nested interval property fails in the rationals, so I was with the idea of "we ALWAYS need to work over the reals", my bad. Thanks for you response. Also, what are those constructions? I mean, I know there are many axiomatic choices for defining the reals, as many as equivalences of the least upper bound property. In my Analysis course, we took the archimedean property and the nested interval property

[@mazc8958]

Love you

[@Absilicon]

Seconded

[@BookofYAH777]

I disagree with your proof/disproof for nested interval property. If you used a number outside of Q to prove that nested interval property is inapplicable to Q, then you can also prove that nested interval property is inapplicable to R by defining a_n as complex number: a_n = c_n + d_n * i. If you nest the complex number along the real axis, you still end up having a complex number that's outside of real numbers, thus the nested interval property is inapplicable

[@user-kp1ws9ql1v]

He only gave the simple and intuitive proof, and the rigorous proof I saw on the textbook is based on Weirstrass thereom.

[@BookofYAH777]

@@user-kp1ws9ql1v thanks! That's pretty advanced stuff for me haha, but maybe I'll check it out sometime in the future

[@fatih7809]

hey there. i have a question and hope you could help me (it is actually pretty easy): i don't understand why the nested interval property doesn't work with open intervals. (0, 1/n) is bounded with (0, 1) if im not mistaken. and every (nth) subset of (0, 1) such as (0, 1/2), (0, 1/3)... contains the consecutive ((n+1)th) subset. so how is their intersection an empty set?

[@BookofYAH777]

@@fatih7809 Your question seems to be based on the false assumption that their intersection is an empty set. I don't get it: if it's easy, then why do you need help with it?

[@fatih7809]

@@BookofYAH777 sorry i mean it is basic. and michael showed that their intersection is an empty set.

[@yowut8075]

What

[@gustavosouza5600]

This guy basically hates the set of the rational numbers. None of the theorems work for them lmao

[@sharpnova2]

it's super unintuitive to me that there is a rational between every two reals yet rationals have lebesgue measure zero

[@HPTopoG]

That's precisely why they have Lebesgue measure zero. Any open cover of the rationals can be shrunk arbitrarily small by choosing smaller intervals as a refinement. The outer measure of that cover tends to zero.

[@abnerandreymartinezzamudio3366]

Thanks for being so smart and hot. It's nice to learn from a handsome teacher.