Real Analysis Ep 3: The Axiom of Completeness

[@lucassaito1791]

Your videos are really helpful, I don't know why so few views, thanks for your work/time!

[@coolrunnings617]

Unfortunately, most of the world is more interested in Kim K's butt.

[@passedexams7217]

This is great! You are so good at visualizing what the statements mean. Super helpful!

[@briannewman9285]

I find your videos so helpful. Thank you for all your hard work.

[@vincenzo6412]

For the => proof of s being supA, can you find an a by saying a = s - ep/b, where b > 1, therefore s-ep < a < s. There exists a number in A, s-ep/b less than s and in A, less than s but greater than s-ep meaing s-ep can't be an upper bound ? This would also mean that there exists an infinite number of a's in A between s-ep and s.

[@ChrisStaecker]

I don't think this will work- in general, it may not be possible to find a number a with s-ep < a < s. Sometimes you must take a=s. For example if A = {0,1}, then s=1 but for small epsilon, there is no a with s-ep < a < s. (In particular, s-ep/b will not be an element of A for any b>1, and there will not be infinitely many a's between s-ep and s.)

[@vincenzo6412]

Ok thank you for the response

[@69imbatman]

Hi professor, don’t know if you’re still reading these comments but was hoping for a bit of guidance. I took a stab at the proof for nested intervals being non empty (the one you give your class a chance to complete towards the end of the video). In this proof, we needed only to proof that for x = supremum of An, x is greater than or equal to all An and less than or equal to all Bn. I proved the first exactly the same as you, but I proved the second part different. FSOC, I assumed there exists some Bn such that x is greater than Bn. Then by the nature of x being supremum for An, there exists An < x - epsilon for all epsilon > 0. If we let epsilon = Bn - An, then there exists An > x - Bn + An, or simply x < Bn. This contradicts our assumption that there exists Bn such that x > Bn so we conclude that x is less than or equal to Bn. Is this a valid proof? Thanks for the videos.

[@ChrisStaecker]

This looks great! The idea is right on- there are two little mistakes: when you say "there exists An < x - epsilon", the < should be > by the definition of the sup. Also you cannot assume that n in this formula is the same as n above. In the previous sentence you say there is some Bn. Then in this sentence you say there is some An, but really it should be something like Ak- there is no reason for the two n's to be the same. Neither of these mistakes really effect the end of the proof- everything else goes just like you said. (let epsilon= Bn - Ak)

[@69imbatman]

@@ChrisStaecker Thank you professor!

[@蔺美云]

Thank you so much for the free lectures! Love……

[@lucassobolev]

Hi Chris, I was just considering the lemma proposed in this video for the supremum of a given set, “Assume s is a real number where s is an upper bound for a set A. Then, s = sup(A) if and only if for all ε > 0, there exists an element a in A satisfying s - ε < a.”. Notice however, the lemma is not satisfied in the backward direction, namely, “Assume s is a real number where s is an upper bound for a set A. If for all ε > 0, there exists an element a in A satisfying s - ε < a, then s = sup(A).” However if this statement is false, then the lemma cannot have if and only if in it. Take this example: Take A = {1, 2, 3}. Notice that A is bounded, hence the supremum exists. Take s = 5 (notice this is an upper bound of A by assumption) and ε = 3. Then, s - ε < a is satisfied because we can find an a, namely a = 3. But note that s is not the supremum of A. So just because this inequality is true does NOT imply that s is the supremum of the given set. I am a bit sceptical about this so please do correct me wherever you see the gap in my understanding! Thank you!

[@ChrisStaecker]

Thanks for thinking carefully about this! I believe the "if and only if" is correct in this case. Your example is totally right, but the lemma says "for all ε" This means that a true counterexample would be one in which s - ε < a can hold for EVERY ε (maybe using different a's). In your example, we can use ε=3 and a=3 we do get s-ε

[@lucassobolev]

@@ChrisStaecker Thanks for the reply! I believed that the, “for all ε > 0” statement was referring to all epsilon’s that would satisfy the inequality or is it that the choice of epsilon shouldn’t matter? Because if s = sup(A) then the choice of epsilon wouldn’t matter, as the inequality will always hold. Thanks for your time.

[@ChrisStaecker]

I think you have the right idea with "the choice of ε shouldn’t matter", though I would rephrase it to say: any ε will work. Depending on which ε you use, you may need to use a different value for a. So in that sense, the choice of ε does matter for something. But any ε will work. Thanks for watching!

[@lucassobolev]

@@ChrisStaecker Thank you so much for your help! I really enjoy your videos alongside the book, Understanding Analysis by Abbott!

[@ruipenghan535]

Great lectures! Much appreciation from U of I!

[@kiara4345]

Thankss! You are such a great teacher

[@grantmoore8790]

@24:00 ish. I don't know how to make the fancy epsilon symbol so let e = epsilon. s - e < s, but how do you know s - e is an element of A? It makes sense intuitively, but I don't see how it follows necessarily from the assumptions. x = sup A means x>=a for all a in A, but how do you know subtracting e from x doesn't "skip" over A and leave you on the other side of A, if that makes sense?

[@ChrisStaecker]

This is a good point- I agree that s-ε may not be an element of A. (I always google "epsilon symbol", then copy/paste ε.) But the proof does not require that s-ε is in A. All we use is that s-ε < supA, which means that s-ε is not an upper bound of A.

[@hayallerimverenklerim9733]

Teacher! I have an important question. In the nested interval example was I_n=[-1/n,3] in this example right bound i.e. 3 is a constant but left bound i.e. -1/n is a sequence of rationals {-1,-0.5,-0.333...,-0.25,...} and 3 is a constant {3,3,3,3,...} by the definition of the nested intervals is this a nested interval: I_n=[-1/n,1/n] for n=1 I_1=[-1,1] and n=2 I_2=[-0.5,0.5] and so on... the intersection one to infinity of this intervals is the singleton {0} but in the proof of theorem we choose x is supremum which is for this example is 1. But x is not an element of {0}. Please can describe is if this a counterexample? If so, then is this proof incorrect? Thanks for great video-lecture!

[@ChrisStaecker]

The definition of x is that it is the sup of the left-hand endpoints of the various intervals. So in your example it would be the sup of {-1/n}, which is 0.

[@hayallerimverenklerim9733]

Thank you so much

[@ms_21_levitating23]

Thank you Sir🥺💖

[@ashishrajyaguru5761]

Hello Chris, Do you have similar lecture recordings for measure theory?

[@ChrisStaecker]

Sorry I don’t! I’ve never taught measure theory. Almost all my courses are undergraduate so it doesn’t come up.

[@algorithmo134]

@@ChrisStaecker Hi Chris, at 26:07, you want to prove by contradiction. But i have a question, don't you need to negate the original statement first and assume it is true and find a contradiction. It looks like you only negate the last part only.

[@ChrisStaecker]

@@algorithmo134 Either strategy could lead to a correct proof. As a simpler example, say I want to prove: Say I want to prove something like "∀x, x*x≥0" by contradiction. One strategy, like you suggest, would be to negate the whole thing, and assume FSOC that "∃x, x*x < 0", and then derive a contradiction. Another strategy (like I'm doing in the video) would be to say "Let x be any number, then assume FSOC that x*x

[@algorithmo134]

@@ChrisStaecker how would you finish off your approach? Mine would be proof by cases 1) let x be negative 2) let x be positive 3) let x be 0 and show that all three gives a result >= 0 hence contradict the fact there exist x such that x *x < 0. Hence the original statement ie, ∀x, x*x≥0 is true.

[@algorithmo134]

​@@ChrisStaecker I guess the reason you negate the last part only is because it is an if else statement. So ~(p=>q) is p and not q.​

[@lonnybulldozer8426]

Chris, the class homepage is no longer working.

[@ChrisStaecker]

Sorry- this is a recent problem. Something messed up with networking on campus, out of my control unfortunately. Hopefully it'll be up again soon!

[@TheTacticalDood]

Can you check my reasoning for exercise 1.3.5. It asks us to show that sup(c + A) = c + sup(A), given that A is bounded above, c is some real number, and c + A = {c + a : a in A}. By the definition of the supremum of a set, we have sup A >= a for all a in A. Since A is a subset of R and R is an ordered field, then c + sup A >= c + a for all a in A, hence c + sup A is an upper bound for the set c + A. Moreover, by lemma 1.3.7., we know that there exist some a in A such that for any eps > 0, sup A - eps < a, thus (c + sup A) - eps < c + a for all a in A, hence c + sup A is indeed the supremum of the set c + A. Am I correct?

[@ChrisStaecker]

Looks good to me!

[@iremiposiajayi122]

Hi. The Axiom of Completeness is true for all Real numbers but isn’t true for the set of Rational Numbers. Why? Isn’t the set of Rational Numbers a subset of R? Thus, shouldn’t AoC be true for Q since Q is a subset of R?

[@ChrisStaecker]

The AoC is true for any subset of A⊆R, but it just says that any bounded subset of A has a sup in R. And this is true also for Q: any bounded subset of Q has a sup in R. I think you're thinking of a statement like "any bounded subset of Q has a sup in Q", which is false. The AoC is always about "sup in R", not sup in any other set.

[@steviesdigitdungeon7757]

Hi there! Is there anyway you could post the homework questions that go along with this course? Or some other set of questions? I checked on the course website, but I could not access any of the questions. I am not enrolled at Fairfield University. (Also really great videos btw)

[@ChrisStaecker]

Hi Steven- thanks for watching. I don't post the textbook questions because that would be a copyright violation- the book is Abbott, "Understanding Analysis", 2nd edition. If you don't care about copyright violations, you can easily find a bootleg PDF of the textbook online with a search engine. Good luck!

[@steviesdigitdungeon7757]

@@ChrisStaecker Thank you so so much. Keep making awesome vids!

[@quant-prep2843]

hey awesome class !!!! Could you please confirm whether you covers these topics in Real analysis. Real Analysis: Interior points, limit points, open sets, closed sets, bounded sets, connected sets, compact sets, completeness of R. Power series (of real variable), Taylor’s series, radius and interval of convergence, term-wise differentiation and integration of power series. what are the stuffs you are not covered in here ? and again this is the best Real analysis course on whole planet, cheers for that!

[@ChrisStaecker]

I do all that stuff (not sure if I mention "interior points" by name) except nothing about series. Thanks for watching!

[@quant-prep2843]

@@ChrisStaecker thanks man.

[@TheTacticalDood]

In the proof for the second theorem about the supremum, can we use contradiction? Given s = sup A, for contradiction assume that s - eps >= a for some positive number eps and for all elements in A. This means that s - eps is an upper bound for A for some eps, but this is not possible since s is the least upper bound and no other upper bound can be less than it, hence we reach a contradiction. Is my reasoning correct?

[@ChrisStaecker]

Yes that would be half of the proof of the theorem at 16:45. That theorem is "if and only if", so you need two parts to the proof. Your reasoning would prove the "⇒" part.

[@khushiagarwal13]

In the nested intervals property where you prove the theorem, you take x as the sup(a_n). Will we be able to solve it if we take inf(I_n) as x and then continue?

[@khushiagarwal13]

Ohh, never mind, I got confused. It just goes in the opposite direction. Since we want the nested interval, taking inf will contain every interval's a_n , so we take sup.

[@ChrisStaecker]

@@khushiagarwal13 You can make the same argument work if you let x=inf b_n, then reverse the inequalities.

[@khushiagarwal13]

@@ChrisStaecker Yeah understood, thank you so much!😄

[@066_vikash_kumar5]

In the second sup's proof, shouldn't it be s-ε ≤ a ?, instead of s-ε < a, CZ sup can be outside of set A (excluded from), and when we move toward left, it just stops just at the very boundary of set A, and then there exist a∈A, which is equal to s-ε

[@ChrisStaecker]

I think you mean the theorem I write starting at 16:44. In that theorem, I believe that < could be changed to ≤, and the theorem would still be true. But the situation you describe, where the ≤ would be necessary, will never happen. For what you describe (if I understand correctly) we would need for s to be outside of A, and some values to the left of s are also outside of A, until at a certain point we have some s-ε equal to an element of A. But this cannot happen: this requires that there are some values less than s but also greater than all elements of A. Such values would be upper bounds for A which are less than s, and there are no such values because s is the sup of A. (I hope that's what you meant!)

[@066_vikash_kumar5]

@@ChrisStaecker yes, this is exactly what my doubt is, and I understand what are u saying, I thought about it already. Now, my doubt is that then what is sup for open sets, and how can it be outside of the set. For example, let's say set A= (-2,2), so according to what I have learned, SUP would be 2, but still, it is not the maximum of set A. if we go just left of 2 that would be the maximum value within set A. So, how come SUP(A) is 2. And surely, we cannot find what will be the maximum of set A actually, cz it will be an irrational and also we cannot say which number will be nearest to 2, but we can still write its an equivalent definition if we put '≤' this in the definition. Put it simply, my doubt is how are we finding the SUP of open bounded sets

[@ChrisStaecker]

@@066_vikash_kumar5 The open interval (-2,2) does not have a maximum element. That's because there is no "biggest number less than 2". Many bounded sets do not have a maximum or minimum element, and this is why the concept of sup and inf were created- we would like to talk about something which behaves similarly to a maximum or minimum element, even for sets (like open sets) which have none. As for "how we are finding the sup of open bounded sets", this can be difficult depending on the set. If the set is an open interval (A,B), then the sup is B. Calculating the sup for a different type of open set might be very difficult.

[@066_vikash_kumar5]

​@@ChrisStaecker Oh, okay, then it means that max. value doesn't exist for all types of sets. and at 16:44 , so for every positive number, if we took it away from S, we would end up with something, that is less than some elements of Set A or we can also say that there is an infinite number of elements that is greater than S-ε in case of open sets, so it can't be sup. I think I get it now. Thank you so much for taking the time to clear my doubt.

[@youtubeiscruel3946]

@@ChrisStaecker why isn't 1.999... with the 9 repeating to infinitely many decimal places the maximum and supremum of (-2,2)? There is no a in A larger than it, right?

[@Naoseinaosei213]

Nice. Thanks.

[@anonymoususer2271]

Is x the only element of the intersection of intervals?

[@ChrisStaecker]

Not necessarily- for example if the intervals are [1-1/n, 2+1/n], then the intersection is [1,2], which contains many points. (In this case the x is 1, which is not the only point in the intersection.)

[@anonymoususer2271]

@@ChrisStaecker Ahhh thank you so much!

[@luispinzon1660]

What analysis book do you recommend for an undergrad looking for rigorous proofs or detailed proofs?

[@ChrisStaecker]

I actually don't know the area too well- this was my first time teaching the course, and I used the book that was suggested by a friend. I liked the one I used: "Understanding Analysis" by Abbott.

[@pramodsheokand3962]

Sir please tell me if Walter Rudin is the right book to start real analysis

[@ChrisStaecker]

It’s a popular choice! But I don’t really have strong opinions about textbooks. If you like it and find it easy to read and understand, then it’s the right book!

[@pramodsheokand3962]

@@ChrisStaecker thank you sir

[@biquinary]

Doesn't the axiom of completeness also apply to the integers?

[@ChrisStaecker]

Yes, but for the integers it is uninteresting as an axiom. Every bounded set of integers is finite, and therefore has a maximum element, and this maximum element is its least upper bound. So for integers, we can actually prove the completeness axiom using other basic facts. But for the real numbers, the completeness axiom is an essential ingredient in the definition of what a real number is. Unlike for integers, there is no easy way to identify what the least upper bound of a set of reals is. And there is no way to prove that a bounded set of reals has a least upper bound- it must be simply taken as an axiom. (Or you must assume some other axiom that ends up being equivalent to AOC.)

[@biquinary]

Thanks for the answer! If I understand correctly, the reason we care about the AOC for reals is because it provides for the existence of sup (inf) on any upper (lower)-bounded set of reals. It also does this for the integers but it's already trivial that a bounded set of integers had a sup. Did I understand properly?

[@ChrisStaecker]

Yes you got it- in the case of the integers, it’s more like a theorem rather than an axiom.

[@cameronnelson602]

Hi Chris, Thanks very much for posting the excellent videos! Question related to: “Assume s is a real number where s is an upper bound for a set A. Then, s = sup(A) if and only if for all ε > 0, there exists an element a in A satisfying s - ε < a.”. In the first proof, you said, "Assume s = sup(A), Let ε > 0 be given, we will find a∈A s.t. s - ε < a". Does it also have to be explicitly mentioned that this must hold for ALL ε (or was this implicitly assumed)? I'm saying this because if you choose ε arbitrarily, there may not be an a that exists in the set to satisfy a > s-ε. Thanks!

[@ChrisStaecker]

This thing does indeed need to hold for all ε>0. But this is built in to "Let ε>0 be given..." This assumes that ε is any positive real number, and then whatever we prove about this ε must be true for any positive real.

[@xoppa09]

it is shewn!

[@aaa.o.r7987]

😍😍😍

[@xoppa09]

lub = best upper bound

[@ChrisStaecker]

The only good upper bound is a least upper bound

[@stumbling]

S H E W N S H E W N S H E W N S H E W N

[@anonymoususer2271]

Shewn!

[@Pravuz]

S H E W N