@@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).
@@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 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.
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 < A ≤ sup(S), and if S is bounded below there exists a B in S such that inf(S) ≤ B < inf(S) + e.
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.
@@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.
Wow!! You're great, best math channel on TH-cam! 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...
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)?
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)
@@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
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.
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
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?
@@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?
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?
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 < sqrt(2), u can always find a number greater than r and less than sqrt(2). 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
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).
@@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.
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.
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) < 1 to 0 as n goes 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
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
You're a legend, dude. I'm relearning this subject right now on my own.
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!
@@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).
@@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.
@@JasonOvalles was thinking of picking up that book, have you finished it by now? Any thoughts or recommendations would be appreciated, thanks
@@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*
😂😂
***sup
No soup for you.
@@mgmartin51 That means I shall grow without bounds toward +infinity for sure
This channel is pure gold!!! Thank you prof. Penn
منور يسطا واضح اني مش انا لوحدي هنا xD
John Wick Egypt numbah one!
منورين يا رجالة
I'm learning real analysis on my own, your videos are incredibly helpful!
Me too
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.
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 < A ≤ sup(S),
and if S is bounded below there exists a B in S such that inf(S) ≤ B < inf(S) + e.
Awesome!
Please also cover Axiom of Choice and ZFC Set Theory if possible.
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.
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.
You’re just in time. I’ve just learned that thing on calculus. Thank you so much for the proof.
I've just found your channel, and man, it is a blessing! Keep up this amazing work!
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.
so blessed to find you at the start of my real analysis course
I came across your comment and I'm just curious; how did your real analysis course go?
@@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.
@@silversky216 Yeah I totally understand. I'm glad it went well!
Wow!! You're great, best math channel on TH-cam! 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...
Thank you! Please make more videos on topics of Real Analysis!
This video was really helpful!
I am teaching Real Analysis this Fall and I will be making videos to support the whole course.
@@MichaelPennMath Great!! Thank you!!
@@MichaelPennMath What textbook does this course use?
This is great, I've been looking for something like this and bam here you are .
Completeness? More like "You need this"...if you're taking a real analysis course! Thanks again for making and sharing all these wonderful videos.
Wow great video, I think I'll need to watch it a second time to fully digest it, but very clear, thank you!
god bless you man. thanks for doing these lectures!
Great explanation of sup and inf! :D
Will we see a follow up video on the point set topology of the real number line now? Sure hope so!
Thanks man, all the way from South Africa
I would like more to this series
loved the video. Keep doing a great job!
Great teacher 🔥🔥🔥🔥
Love your video!!! Keep doing it!
Sir make more videos on Olympiad calliber questions
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.
Ghislain Leonel how is it in Cameroon?
@@newkid9807 most likely civil war
I appreciate this video.
Thank you
Sare IIT’s me same h kya MA 101 ka syllabus
Btw IIT BHU
@@Lakshya_Raj07 Maybe. But humara Calculus course hai and it is MA 105 instead
Amazing lecture as always..
V.helpful👏
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)?
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)
@@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
@@johanroypaul2816 No, x
@@cardinalityofaset4992 Here , since x is an element of Q, x cannot be ✓2 , then why are we writing x^2
This really help me understand!
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.
What are you on about.
Really good video.
Man, where was this when I took real analysis?!
please suggest a textbook to follow along with these videos on real analysis
Studied this in first year real analysis Polytech yaounde. Really interesting topic.
that epsilon replacement was kinda cool
Does this course happen to follow a textbook?
Teacher: Time for some Real Analysis
Kid named Ysis: 😮
Complex and functional AnalYsis must be interesting
WTF
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
Can you make a video on the cut property of real numbers?
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
Note that he meant pi is not an element of the rationals at the end, but wrote the opposite.
wo i love ur teaching
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?
@@angelmendez-rivera351 thanks for that explanation.
@@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?
How do you get zero down the column
16:05
Yo the amount of people self-learning real analysis in this comment section is just wonderful. Thought I was alone honestly.
5:40 you used the archimedian principle, so R must be an archimedian set ;o
Good work
What are the pre-requisite to follow this real analysis playlist?
do you have some tutorials for further real analysis like lebesgue integral?
Wouldn't it work to just let a=sup(A) then s-epsilon
Could you please add Indian math olympiad 2020 problem 2 in your list
Thank you so much =)
9:10 We can choose a = (s + s - Ɛ)/2 = (2s - Ɛ)/2
We don't know that (2s-e)/2 is necessarily in the set A.
Very tricky detail.
These are great, more analysis videos would be awesome.
thank you
Thanks a lot
Finally got it❤
Perfect!!
Hi, sorry I may have missed some basic idea, but isn't the sup(B) @7:50 equal to 1/2?
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.
Awesome 👍
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?
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 < sqrt(2), u can always find a number greater than r and less than sqrt(2).
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
@@baljeetgurnasinghani6563 Actually no! I understood it completely! Thanks a bunch man for taking your time
@@SartajKhan-jg3nz My pleasure 😊
Thx
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?
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).
@@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.
I can't get the last example 🥺
What is the supremum of a bull set
Null
Null
Sir We want more IMO problem,,,pigeon hole problem,,,love from Bangladesh 🇧🇩🇧🇩🇧🇩🇧🇩🇧🇩🇧🇩
perfect
What's the completeness property of lR?
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.
10/10
Nice.
Can you give subtitle Indonesia?
Lovely
can you do a video one day on your workout routine. you look jacked ;o
life savior
Anyone heard Nintendo switch sound effect? 10:14
How is the inf 0 it makes no sense 😭
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) < 1 to 0 as n goes
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
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
I didn't know this was a thing
He moves kind of fast
No
This cannot be true for a finite set.
Dedekind cuts, nice
Man's fuckin' YOKED
lit
Nice and beautiful
Oh I meant the guy, the math is too
I always pronounced it like ‘sup lol
gotta love the *soup*
عراقي بجامعة بغداد مر من هنا ❤
السلام علیکم و رحمة الله وبركاته
أهلا حياك الله
شاهد قناة نواف يوسف محمد الزهراني ستجد فيها ما قد يساعدك.
Guys that know math are hot hehe.
soup
Dude, take a breath. You speak too fast for instruction.
Maybe he doesn’t suit your pace. That’s fine, find another video or something. The majority really find it great.