You're a legend, dude. I'm relearning this subject right now on my own.

@JasonOvalles

4 жыл бұрын

Ditto. I started going through this on my own about two weeks ago. These videos came in at just the right time. Best math teacher ever!

@kevinb.3541

3 жыл бұрын

@@JasonOvalles Hey, may I ask how is it going? And what textbook are you using for self-study? I'm also going through some introductory analysis (using Stephen Abbott's wonderful book).

@JasonOvalles

3 жыл бұрын

@@kevinb.3541 Using that same book. It's going really slowly, but I don't mind. I'm taking my time to make sure I feel comfortable with each section before moving on.

@tomatrix7525

3 жыл бұрын

@@JasonOvalles was thinking of picking up that book, have you finished it by now? Any thoughts or recommendations would be appreciated, thanks

@JasonOvalles

3 жыл бұрын

@@tomatrix7525 it's going very slowly. But not because of the book, I'm just pretty busy. The book itself is pretty well written. Professor Penn's videos help a lot. It's also such a popular book, that it's pretty easy to find solutions online to check my work. Here's what I usually do: one week, I'll read a section. During this week, I try to do the proofs myself. Often times, I can't and I'll go to the book or these videos for a hint. Or two. Or three. The next week, I'll do some of the exercises in that section. I choose the exercises based on what I can find solutions for. That way, if I'm stuck I can get hints and at the end, I can check my work. Then, if I feel comfortable, the next week, I move on to the next section.

Supremum, or sometimes we call it *soup*

@Sorya-gf7qw

2 жыл бұрын

😂😂

@movierecaps4856

Жыл бұрын

***sup

@mgmartin51

9 ай бұрын

No soup for you.

@ronycb7168

9 күн бұрын

@@mgmartin51 That means I shall grow without bounds toward +infinity for sure

I'm learning real analysis on my own, your videos are incredibly helpful!

@chessematics

Жыл бұрын

Me too

I've just found your channel, and man, it is a blessing! Keep up this amazing work!

This channel is pure gold!!! Thank you prof. Penn

@abdulrahmansaber1768

4 жыл бұрын

منور يسطا واضح اني مش انا لوحدي هنا xD

@newkid9807

4 жыл бұрын

John Wick Egypt numbah one!

@hassan.aladawy

3 жыл бұрын

منورين يا رجالة

Awesome! Please also cover Axiom of Choice and ZFC Set Theory if possible.

You’re just in time. I’ve just learned that thing on calculus. Thank you so much for the proof.

Wow thank you! Currently taking a real analysis class and this was the clearest and most concise explanation that I've heard on this topic. Cheers.

@PunmasterSTP

2 жыл бұрын

How did your real analysis class go?

A useful theorem: For a non empty set S, given any epsilon greater than zero, if S is bounded above there exists an A in S such that sup(S) - e and if S is bounded below there exists a B in S such that inf(S) ≤ B < inf(S) + e.

I am learning Real Analysis on my own and this is really good. Thanks so much because otherwise I’d be sort of astray with no clear path

Very good detailed and to the point. Thank you so much. Saved me hours of non-productive studying.

This is great, I've been looking for something like this and bam here you are .

this is amazing, this is by far the most underrated math channel on youtube, i am still pretty young but i am passionate about mathematics, and this channel has helped me alot, thank you so much for this.

Fantastic, but at the very end you forgot to put a line through 'belongs to' sign as you say pie doesn't belong to rational numbers.

loved the video. Keep doing a great job!

Wow great video, I think I'll need to watch it a second time to fully digest it, but very clear, thank you!

Will we see a follow up video on the point set topology of the real number line now? Sure hope so!

god bless you man. thanks for doing these lectures!

Wow!! You're great, best math channel on KZread! The very best was the last example, with the sequence of digits of pi. It shows that even the algebraic numbers are not complete, as pi is transcendental. And the completeness is an axiom of the Real numbers. But there are countable sets that are complete too. So it is not this axiom that requires the Reals to be uncountable. Must be other axioms...

Love your video!!! Keep doing it!

Amazing lecture as always.. V.helpful👏

Great explanation of sup and inf! :D

Thanks man, all the way from South Africa

Thank you! Please make more videos on topics of Real Analysis! This video was really helpful!

@MichaelPennMath

4 жыл бұрын

I am teaching Real Analysis this Fall and I will be making videos to support the whole course.

@5180073a

4 жыл бұрын

@@MichaelPennMath Great!! Thank you!!

@RhynoBytes857

Жыл бұрын

@@MichaelPennMath What textbook does this course use?

Completeness? More like "You need this"...if you're taking a real analysis course! Thanks again for making and sharing all these wonderful videos.

This really help me understand!

I would like more to this series

Really good video.

Great teacher 🔥🔥🔥🔥

so blessed to find you at the start of my real analysis course

@PunmasterSTP

2 жыл бұрын

I came across your comment and I'm just curious; how did your real analysis course go?

@silversky216

2 жыл бұрын

@@PunmasterSTP Overall good. The examination was online so I managed to score really good. At the end of the day your course will highly depend on your Professor...So that's that.

@PunmasterSTP

2 жыл бұрын

@@silversky216 Yeah I totally understand. I'm glad it went well!

Studied this in first year real analysis Polytech yaounde. Really interesting topic.

Great stuff Michael! Just a mistake in the notation at the end. You wrote pi as the element of rational numbers (Q).

Much love from Cameroon 🇨🇲. Great content sir.

@newkid9807

4 жыл бұрын

Ghislain Leonel how is it in Cameroon?

@directordissy2858

4 жыл бұрын

@@newkid9807 most likely civil war

These are great, more analysis videos would be awesome.

Can you make a video on the cut property of real numbers?

Man, where was this when I took real analysis?!

do you have some tutorials for further real analysis like lebesgue integral?

What are the pre-requisite to follow this real analysis playlist?

Sir make more videos on Olympiad calliber questions

Perfect!!

Good work

Thanks a lot

Wouldn't it work to just let a=sup(A) then s-epsilon

please suggest a textbook to follow along with these videos on real analysis

Finally got it❤

Thank you so much =)

Awesome 👍

Could you please add Indian math olympiad 2020 problem 2 in your list

wo i love ur teaching

thank you

Dear Micheal, Thank you for this amazing video. I have two doubts 1) Shouldn't the set builder form be written in a way to show its elements, therefore can we include the equality sign as x will never take value of √2 and thereby no situation of x^2 =2 arises ? 2) can we say the second sequence has supremum as it cannot be clearly defined since the decimal points do not end and as pie is not a number(in the sense, it is not defined to the exact value with regard to decimal points)?

@cardinalityofaset4992

2 жыл бұрын

1) The set notation he uses stems from a ZFC axiom called Axiom schema of specification which states that from any arbitrary set, you can choose a subset. The way we do it is that we write the squirly brakets, on the left we write elements of a set that we choose from, and on the right we specify property of those elements. This property ultimately tells which elements we think and so the new set is well defined. So if you wrote equality sign instead of inequality the set would be empty since there are no rational numbers whose square is equal to 2 (as Michael eplained). 2) The second sequence must have supremum according to the axiom of completeness, since 4 is clearly an upper bound and it is not empty because 3 is element of the set of sequence values. However, the supremum is real. Now, there is nothing wrong if supremum of a set is a natural number, or rational number, but it is NOT guaranteed. In this case the supremum is pie, which btw is a number. It is an irrational number which means it is a real number that can be represented as an infinite decimal (number with infinitely many decimal points that never repeat)

@johanroypaul2816

2 жыл бұрын

@@cardinalityofaset4992 Thank you for the reply. I wrote the first the qts wrong. 1) My doubt was that shouldn't we write it as 'x^2 < 2' instead of 'x^2

@cardinalityofaset4992

2 жыл бұрын

@@johanroypaul2816 No, x

@johanroypaul2816

2 жыл бұрын

@@cardinalityofaset4992 Here , since x is an element of Q, x cannot be ✓2 , then why are we writing x^2

that epsilon replacement was kinda cool

In all of these "Sup[aSet]=u", there's an implied "Universe". Where "aSet" is a subset of some "Universe" and "u" is an element in some "Universe", but not necessarily is "u" in "aSet". I mention this because I've seen what looks like "Sup[Real]=w", and I don't know in what "Universe" "w" is in.

@theemperor-wh40k18

8 ай бұрын

What are you on about.

5:40 you used the archimedian principle, so R must be an archimedian set ;o

Hi, sorry I may have missed some basic idea, but isn't the sup(B) @7:50 equal to 1/2?

@JoaoVictor-gy3bk

3 жыл бұрын

Nop. Notice that the numbers on that first roll have the form n/(n+1), so they look like 2/3, 3/4, ..., 1999/2000 etc etc. They get closer and closer to 1.

14:45 I did not understand what you meant by 'If Q is our universe bcz the sqrt(2) is not in Q'. Does that mean that for a set to be complete, for all subsets A that member of S with upperbounds, the SUP(A) must also be a member of S?

@baljeetgurnasinghani6563

3 жыл бұрын

No it's not necessary. If this set was defined in R, we would have sqrt(2) as the sup. However, since this set is defined in Q, we can't use sqrt(2) since it's not in Q. But that leads to a problem: u don't have a sup for this set now If u consider a rational number r So, a rational number less than sqrt(2) can never be the sup. Similarly, if u consider a rational number r > sqrt(2) as sup, you can always find another rational number which is an upper bound and also less than r. Since the sup can't be greater than or less than sqrt(2), and there is no number in Q whose square is 2, it basically leads us to the conclusion that this set has no sup. sry if I over complicated the explanation

@SartajKhan-jg3nz

3 жыл бұрын

@@baljeetgurnasinghani6563 Actually no! I understood it completely! Thanks a bunch man for taking your time

@baljeetgurnasinghani6563

3 жыл бұрын

@@SartajKhan-jg3nz My pleasure 😊

Note that he meant pi is not an element of the rationals at the end, but wrote the opposite.

perfect

Nice.

Thx

Teacher: Time for some Real Analysis Kid named Ysis: 😮

@jamesyeung3286

3 жыл бұрын

Complex and functional AnalYsis must be interesting

@henk7747

3 жыл бұрын

WTF

I have a doubt: given supremum of some set is "not in that set" but how this possible? I mean isn't it we can choose the largest number in the set... and take it the supremum ??!?!?

How do you get zero down the column

It's a little confusing how you worded the axiom of completeness. Is it right to say that both sets A have least upper bounds and supremums as opposed to saying their lub and supremum exist but are not in A? Or is it more correct to say that A has no least upper bound but has a supremum?

@michaelbillman4789

2 жыл бұрын

@@angelmendez-rivera351 thanks for that explanation.

@mbrusyda9437

2 жыл бұрын

@@angelmendez-rivera351 greetings! Sorry for replying after more than a year, but I was just recommended to this video. "every nonempty set A which is a subset of R, if it has an upper bound, then it has a supremum that is an element of A. This is true for finite and infinite sets alike." But what about the set A=[0,1)? If I understood correctly, sup(A)=1, but 1 is not in A?

The Lemma 8:10 looks kinda hairy and counterintuitive. If {U} are upper bounds of A , than by common sense , u≥a for ∀ a∈A, u∈U. That's neat. Now where it boggles me is the sup(A). That s is sup(A) if for every ε>0, ∃ a∈A, s.t s-ε

9:10 We can choose a = (s + s - Ɛ)/2 = (2s - Ɛ)/2

@henk7747

3 жыл бұрын

We don't know that (2s-e)/2 is necessarily in the set A. Very tricky detail.

Lovely

10/10

As my doubt, at 4:43 proffessor why 16 cannot be sup(A) ?!?

Doesn't that theorem and proof only work for a continuous set, how would it work for a discrete set. Do the concepts of continuity even apply to sets?

@kehrierg

3 жыл бұрын

i wondered the same. i think the condition for truth of the lemma that wasn't written down might be that the set A must be infinite (not contain finitely many elements). for instance with the A = {1/n : n a natural number} example, the lemma seems true, even though it's still true that A is a discrete set (each element of A has a neighborhood containing no other elements of A).

@kevinb.3541

3 жыл бұрын

@@kehrierg I think it works with a set that has finitely many elements. Let A be a set with n elements, and we claim that max{A} = sup A (max{A} is just the largest element of A which is well defined because A is finite, notice that it is in A). Thus to prove the equivalence we use the theorem; Let e > 0 be given, notice that max{A} > max{A} - e and since max{A} is in A, then the proof is done i.e max{A} = sup A.

can you do a video one day on your workout routine. you look jacked ;o

Does this course happen to follow a textbook?

16:05

Yo the amount of people self-learning real analysis in this comment section is just wonderful. Thought I was alone honestly.

u replaced s-u with epsilon since the both are higher than zero. Idk it's a little bit strange to me. Care to explain . TIA

Can you give subtitle Indonesia?

I can't get the last example 🥺

Sir We want more IMO problem,,,pigeon hole problem,,,love from Bangladesh 🇧🇩🇧🇩🇧🇩🇧🇩🇧🇩🇧🇩

I didn't know this was a thing

What's the completeness property of lR?

@user-uw1ut4ss2q

2 жыл бұрын

We say an order field F is complete if Every nonempty subset of F which is bounded above has the least upper bound. Since Q does not satisfy the axiom of completeness, this axiom is the characteristic that distinguishes R with Q.

life savior

Anyone heard Nintendo switch sound effect? 10:14

I always pronounced it like ‘sup lol

lit

What is the supremum of a bull set

@terfatyokula1761

2 жыл бұрын

Null

@terfatyokula1761

2 жыл бұрын

Null

Dedekind cuts, nice

gotta love the *soup*

Nice and beautiful Oh I meant the guy, the math is too

This cannot be true for a finite set.

How is the inf 0 it makes no sense 😭

@MohamedBenamer940

Ай бұрын

Inf(A) = 0, Sup(A) = 1: Because: 1/n = 1 for n = 1, and lim(n->infinity) 1/n = 0 And the sequence of numbers 1/n is strictly decreasing from 1 to 0 (0 not included, 0 not in A) as n goes from 1 to infinity. Inf(B) = 0, Sup(B) = 1: For a fixed m: lim(n->infinity) m/(n+m) = 0 and the sequence of numbers m/(m+n) is strictly decreasing from m/(m+1) from 1 to infinity. For fixed n: lim(m->infinity) m/(n+m) = 1 and the sequence of numbers m/(m+n) is strictly increasing from 1/(1+n) infinity) m/(m+n) = n/(2n) = 1/2

@MohamedBenamer940

Ай бұрын

1 and 0 not in B also, because we get it from taking n or m approaching infinity, which is a limit not exactly equal to infinity (because it's not possible) و الله اعلم

He moves kind of fast

Man's fuckin' YOKED

Guys that know math are hot hehe.

No

عراقي بجامعة بغداد مر من هنا ❤

@user-bb8fo6gb9e

11 ай бұрын

السلام علیکم و رحمة الله وبركاته أهلا حياك الله شاهد قناة نواف يوسف محمد الزهراني ستجد فيها ما قد يساعدك.

soup

Dude, take a breath. You speak too fast for instruction.

@tomatrix7525

3 жыл бұрын

Maybe he doesn’t suit your pace. That’s fine, find another video or something. The majority really find it great.