Real Analysis Ep 3: The Axiom of Completeness

Episode 3 of my videos for my undergraduate Real Analysis course at Fairfield University. This is a recording of a live class.

This episode is about the completeness axiom for the real numbers.

[lucassaito]

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

[passedexams]

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

[briannewman]

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

[vincenzo]

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.

[蔺美云]

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

[ruipenghan]

Great lectures! Much appreciation from U of I!

[imbatman]

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.

[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!

[grantmoore]

@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?

[lonnybulldozer]

Chris, the class homepage is no longer working.

[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?

[hayallerimverenklerim]

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!

[iremiposiajayi]

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?

[ashishrajyaguru]

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

[khushiagarwal]

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?

[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?

[_vikash_kumar]

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

[anonymoususer]

Is x the only element of the intersection of intervals?

[pramodsheokand]

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

[steviesdigitdungeon]

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)