Real Analysis, Lecture 4: The Least Upper Bound Property

[@tanaynarshana4056]

The class has a mathematically sound audience. Amazing questions from their end. Kudos to the prof. who takes these tricky questions so well.

[@emilyla6415]

Francis Su is my hero...He makes it incredibly simple to follow along.

[@sdoken]

if anyone else is confused the way i was for a second, a lot of times he means "less than or equal to" when he says "less than". these lectures are GREAT! Praying that he will put videos of the next analysis course after this since thenlectures here cover only half of undergrad analysis.

[@uzername90]

If dank means what I think it means, then it's worth mentioning that these lectures are awesome when you're high.

[@Das0s]

I got all the way to the end of this. Don't waste your time here folks. There is no dank engine.

[@kazifamily8042]

43:00 Consequences of LUB property. The Archimedean property 43:00 to 52:00) Q has the property but does not share this proof. 55:17 Q is dense in R. take a distance between an x and y. Multiply this by a discrete number till it exceeds x and another while it is less. 1:00:00. Properties of the Sup. The Supremum is less than or equal to any upper bound. Then gamma is the infimum of the set of upper bounds. Any upper bound is less than or equal to gamma. If for all a in A, a

[@jamesh625]

44:55 "If y is negative then n = 1 will suffice." Technically, n = 0 will also suffice if x > 0, since by hypothesis y < 0 = nx. However, the statement does assert existence only, so there are actually infinitely many n that satisfy the property.

[@JaspreetSingh-zp2nm]

No not because 0 is not in set of positive integers(natural numbers)

[@DeadFfitchy]

This right here is the real Dank Engine

[@QuantaLumin]

Did you film this with a toaster?

[@alexhenson]

Welcome to 2010

[@greg55666]

THE thing that makes math hard is when there is some reasoning assumed but not stated in a proof. At 51:00, he says that a - x < mx for some m. That is true, but it is not OBVIOUS why it's true. The point is, there has to be some mx s.t. a-x < mx < a. One of the nx has to land in that interval because the length of the interval is x. That's how my brain works.

[@SequinBrain]

sry I'm 11 years late, but that's what I notice from most "proofs" regardless of the class. They fail my definition of what a proof is sometimes. The ones that work well, usually the axioms, are beyond incredible. The most common thing I find missing from proofs is to prove there's any reason for them to exist. If they just did that, the proof would make 1000% more sense to everyone. Not being sarcastic. If we have anything, we should know why we have or need it.

[@enkii82]

It's clearer for me at 56:03, if he says x

[@amnassimi]

I really wish if I could see this video with a higher resoluytion.

[@theunknownscientist3249]

The proof of the density of the rationals in the reals that ends at 1:00:00 is actually incorrect because it bases itself on the well-ordering principle, and m there could be 0, in which case, m-1 would not make sense so you cannot claim that inequality, further more, even if you define the naturals starting with 1, 0 then would not be in the set so it is not correct either. It can be seen easily by choosing x and y to be negative, the proof would then entail that there is a positive rational between two negative reals. Still, the correction of it can be done by considering first x and y both positive, then one negative, one positive, and finally both negative, in which case the inequality -y

[@genghis_cohen]

Wait a second, why on earth is this in a Thomas the Dank engine playlist?

[@imnimbusy2885]

Don’t let Sir Toppemhat know about this, Thomas! You know how he is… when he knows… Thomas, where were you on the morning of September 11th 2009? Thomas..

[@JaspreetSingh-zp2nm]

Please if possible re-upload these lectures with better video quality.

[@trukkstop1]

At 30 minutes, the lecture states the uniqueness of an ordered field having the least upper bound property (the Reals, R). It's interesting to note that a definition which defines R uniquely does not exist in "First Order Logic", due to the "Lowenheim-Skolem" theorem. Any attempt to define R using only "First Order Logic" allows models of R having different cardinality, some models being countable. Thus defining R uniquely requires a stronger logical formalism, typically "Second Order Logic", which does not behave as well as "First Order Logic". Troubling freshman Analysis students with this would only cause confusion, I suppose. But I was astounded to learn this, years after my freshman Calculus class.

[@bluestarfractal5434]

Even if this is so, so what? What is your point?

[@bluestarfractal5434]

As a student of analysis, I couldn't care less.Yes, I know that there is this hair brain crackpot on the net named Wildburger leading a lot of mathematical simpletons down the mud path with his "Real Analysis is incorrect" parade. But I choose to follow my own mind based on the evidence that I see in my studies at the University of Michigan rather than follow a self appointed "expert" who obviously has an ax to grind.

[@bluestarfractal5434]

Quite correct. My post was mistaken. I meant to direct a correction to the poster who claimed that Dr Su said that the rational numbers have the least upper bound property when in fact he stated quite clearly that they do not.Sorry about that.

[@maxpercer7119]

@@bluestarfractal5434 axe*. totally agree , wildburger is a crackpot. he is sending simpletons into the colloquial 'blind leading the blind' ditch with his infinity skepticism. Though i do think it should bear mentioning there isn't much use for cantor's transfinite cardinal (or ordinal) theory, outside of mathematical logic and set theory foundations, and we could recast cantor's work as 'what we can do with mappings and set builder notation'. Practically speaking cantor's legacy will probably be 'naive set theory' as Paul Halmos calls it, because after all sets are so damn useful in stating mathematical results , they are like the concept of folders or directories in operating systems. Similarly with kurt godels work, his legacy might be 'godel numbering' which is a useful device.

[@ldb579932]

The Archimedian property for rationals was asked (46:00) and caught the prof off guard. For the rationals it can be proven easily, the proof in this video proves something more powerful - true for all reals. To prove Arch. property for rationals, suppose a/b, c/d are any rationals and assume a,b,c,d>0. To find an int n, s.t. n*(a/b) > c/d note that a/b * bc/ad = c/d. So any integer n > bc/ad works, say n = bc+1.

[@cognificent]

1, 1.4, 1.41, etc. are all terminating decimals. Any terminating decimal is in Q: e.g. 1.4 = 14/10, 1.41 = 141/100, etc.

[@roly4301]

At 59:16, should the last two strict inequalities be loose inequalities? That is, should (m-1)/n < x instead be (m-1)/n ≤ x, and should m/n > y be m/n ≥ y? I think the first one is not strict, since we only required that m/n to be least such that m/n > x, so the previous fraction can in fact coincide with x. The second inequality (m/n ≥ y) is simply the contrary of m/n < y, what we're trying to show. But the rest of proof would work anyway. Assume for contradiction that m/n ≥ y. Subtract x from both sides: m/n - x ≥ y -x > 1/n. So m/n - x > 1/n. (The strict inequality is preserved). This implies that (m-1)/n > x, a contradiction to the fact that (m-1)/n ≤ x.

[@BladtMartin]

An upper bound can be an element of the set, namely the largest one and in that case it is the supremum. Yes, the property you mention is a property of cuts, not of sets of cuts like A. Gamma is the union of alphas, thus can't be strictly larger than all alphas.

[@greg55666]

"Yes, Emil?" Class 4 and he knows your name? Someone's been hanging out at office hours I would say!

[@zzasdadsdk]

Hi Greg, It might not be a great observation but it's worth pondering. I would suggest you to try to prove archimedean property natural number i.e for given p,q elements of natural numbers, one can always find a natural number r, such that r*p > q. But only use the well ordered principle of natural numbers (i.e. every non-empty set of natural numbers contains a least element).

[@thatcat6025]

how does this definition work for uncomputable numbers; how would you define the cut?

[@peki_ooooooo]

Wow, good question!

[@Shotblur]

How did I get here from this playlist

[@jiggasnap]

I meant to say "If gamma is equal to an element in the set of cuts, then gamma is not an upper bound." The inequality in the left hand side of the second property must be strict in order for the property to hold. If little a is equal to gamma then gamma is an element in the set of cuts and therefore gamma is NOT greater than or equal to supA. This is so by the properties of D cuts (no maximal element). If you just make that inequality into a strict inequality then the statement works.

[@greg55666]

But the question I am asking is, in doing the proof, WHY look at a - x? That was the point of my post, to explain what he did not: WHY choose a - x. It is because, in the interval between a - x and a, it is guaranteed that EXACTLY ONE mx will land in that interval. If we took a - 0.5x, then there might be NO mx in the interval. If we took a - 2x, there would be TWO mxs in the interval. a - x to a is guaranteed to have exactly one mx in the interval. (So that (m+1)x is greater than a.

[@peki_ooooooo]

Well, I post something that I struggled with that may help someone. The biggest problem is not understanding what a "CUT" is. Problem is we have alread been tought what is square root 2. When compare a "CUT" and square root 2, you will find they are different in some sense hence you can not figure out how q^2

[@findclue]

@tubedudelive I don't think he needed to define an injection mapping the cuts to the real numbers since after we understood the injection between the rationals and the cuts that he defined, we can treat cuts as numbers instead of the numbers that we were using, and he and the book showed that the set of every cut is field with addition and multiplication, so now we have this perfect algebraic structure which we tempted to call R. I think he is good at explaining things in easier language.

[@BladtMartin]

no, he meant the irrational number (√2) which he claims is the supremum of the set of all x in Q such that x^2

[@jiggasnap]

That 2nd property of least upper bounds that he listed is wrong. Dedekind cuts have no maximal element so having an upper bound which is not a strict inequality is impossible to begin with. If gamma is equal to an element in the cut, then gamma is not an upper bound bc by the properties of cuts there exists a gamma1>gamma in the cut.

[@researchacademia556]

These lectures are so helpfull...

[@greg55666]

I think the question was missing the point. The LUB can be used to prove the Archimedean property. That is, the Arch. Prop. follows from the LUB. That doesn't mean the converse, that the LUB follows from the Arch. Prop. Rationals have the Arch. Prop., but not the LUB, but that is not the brilliant observation Emil thought it was.

[@TehFingergunz]

40:31 I don't understand how this construction of the reals allows for a greatest lower bound. How can there ever be a lower bound for a set of reals, if each real extends infinitely downwards (is a set of rationals extending infinitely downwards)? I think a student asks this question when he addresses this issue, but it gets brushed to the side because "from now on, we'll think of R as just the reals". Can someone explain what is going on here?

[@milksushi6640]

Think about how order is defined in this definition of the reals. For example think about the set of all positive reals. 0* is less than all of them. How? Because the set of all rationals less than zero is contained in the set of all rationals less than any positive value ("value" being either a limit like root(2) or a straight rational), or in non-mathematical terms: any fraction less than zero is also less than anything positive. Is there any real greater than 0* but still less than all positive reals? No, because you will always be able to construct a positive real (through arithmetic) less than that number but greater than 0*. The trick is that in this definition he kind of messes with how you think of the number line. So when you ask for a bound of the reals (upper or lower) you are really asking for a bound of a set of sets, specifically considering the order that was defined for the outer set. I guess I'm a little late trying to answer this now, but hopefully it still helps.

[@jonhillery7736]

Remember that the order is defined in terms of CONTAINMENT

[@dartme18]

When he writes out "1, 1.4, 1.41", etc., what are those "1.4", etc.? Are they elements of Q? But then, what is "1.4142135..."? (by which he seems to mean x where x*x = 2, which is not in Q.)

[@greg55666]

In fact, his example, when he drew a - x on the number line, was wrong--a - x CAN'T be two multiples of x to the left of a. That is impossible. Not only is there one mx between a - x and a, there is ONLY ONE mx between a - x and a.

[@Leo.majorr]

Why the hell is highest quality available is 240p? I can barely see anything.

[@maxpercer7119]

more like 40p

[@dartme18]

You didn't quite answer the whole question :) I understand that a terminating decimal is in Q. Perhaps by "1.4142135..." he meant "the terminating decimal arbitrarily close to 2^(.5)"? It looks like he meant 2^(.5) which isn't in Q.

[@tubedudelive]

Well, I now see that at 16:42 he corrected himself by adding an '*' to the '2' and therefore making it a cut.

[@greg55666]

But what's the difference then, regarding the Arch. Prop. for rationals and reals. Why can't we use your proof for reals? (I mean, is it only that x might be irrational, or is there some deeper reason?)

[@christineohanyan6216]

could anyone please let me know if professor su goes over lim infs and lim sups? and if so, which lecture is this in?

[@JaspreetSingh-zp2nm]

yes,may be after lecture 15 or so not sure but yes he covers it.

[@SequinBrain]

Why wouldn't the contradiction from 52:38 be the same problem for y? Change alpha to y, x to alpha, then y isn't an ub for A either. how is this wrong? So using precisely the same proof method, y - alpha < m*alpha for some other n ∈ N. So, y < (m*alpha +1)alpha, so y is NOT an ub for A. I find things like this frequently in RA & topology, which is why I don't get it.

[@andrewstallard6927]

His proof of the density of rational numbers in the reals appears to implicitly assume the least upper bound property for the rationals, which they don't have, when he considers a smallest rational m/n that is greater than a real x?

[@tajmahall]

That's for a fixed denominator n, in which case there is a smallest m/n larger than a given number.

[@JaspreetSingh-zp2nm]

Smallest rational m/n greater than X came from Archimedean property used twice.

[@zzasdadsdk]

The professor has clearly dodged the question on Archimedean property for rational numbers asked @ 53:00. I cannot see any any straight forward way to replace lub argument for rationals. One has to use Euclidean division theorem (which relys on well order property of natural numbers) to prove the Archimedean property for rationals.

[@MadhusudanSinha]

"See you next time" sounds so familiar :D :D

[@kylepoe5139]

I am disappointed that there is no allusion in this lecture to the existence of roots of unity when discussing the validity of the definition $a^{1/n} = \{ r \in \mathbb{Q}: r^n < a \}$.

[@greg55666]

I said there is SOME m such that a - x lt mx lt a. You are right, of course, there is SOME OTHER m such that mx gt a. It's not important; it's not the point. As for the proof itself, you CANNOT "take mx greater than a." a is an upper bound of A, NO mx gt a, by definition. That is the contradiction. We started by assuming A had an upper bound; that assumption led to a contradiction.

[@peki_ooooooo]

Aha, finally I figure out what a cut is, and how to construct reals.🤟✊👏🤷‍♀✨🎉🎊🔆🔅

[@greg55666]

But this is the mistake that Emil was making as well. We're getting far far from the point of this lecture, which is about the LUB. Arch. Prop. may or may not be easy or difficult to prove for some other set of numbers, and the reasons for those difficulties might or might not be interesting. It's easy to find mathematical difficulties, but they are not necessarily INTERESTING difficulties. This class is about the LUB and real numbers.

[@greg55666]

This guy's a good teacher. It's weird that this is taught in a big lecture hall. At Berkeley, these classes are taught in small classrooms.

[@dartme18]

He never did define integers, did he? Curious; it seems that it would be worthwhile to spend some time defining them first.

[@BladtMartin]

gamma is not an element in the cut gamma is a cut

[@xfluffypredatorx]

why is this in the thomas the dank engine playlist?

[@findclue]

Typo: field -> a field.

[@zzasdadsdk]

You can not use the same argument, because you can not say much about the ratio of two irrationals (for rationals it's relatively easy). Why don't you try to write a formal proof using Larry's method, I am sure you will realize where are you making mistake.

[@jiggasnap]

Yeah you're right. Thanks.

[@Denesweetz]

@Anirban Pal

[@CrimsonFlameRTR]

'there exists a positive integer,' otherwise known as a natural number lol.

[@maxpercer7119]

7:30 dab nation

[@maxpercer7119]

i cant hear this guy and i cant see this guy . wtf

[@tubedudelive]

He is at times sloppy and consequently confusing. Based on his definition of cut, at 15:38 the equality should be "gamma squared = 2*", i.e. the right hand side should be a cut. The root of the problem is that he defined a cut as the subset, but he never fully developed the injection that maps the cuts to the corresponding real numbers.

[@hejsan7846]

He claims that an injection is one to one, but that's false isn't it? A one to one map is a bijection; injection simply means the map is invertible.

[@AndyTutify]

A function can be one-to-one without the image of the function being equal to the codomain. An injective function says that every element in the domain is mapped to one and only one element in the codomain, hence one-to-one. A bijection just further asserts that the codomain is equal to the image. This, as you know, is a surjection. So a bijection means that every element in the domain is mapped to one and only one element in the codomain (same as injection) AND there is no element in the codomain that hasn't been mapped to (surjection). Hope that makes sense :)

[@x0cx102]

Wtf are these comments