Real Analysis | The countability of the rational numbers.

[@littlenarwhal3914]

I think my brain died watching this. So many subscripts make me turn off im gonna have to watch this multiple times...

[@sadececns00]

I did not understand anything. What is the meaning of the piecewise function? how did he find f_2(2)+1=3? What is g function? 😭😭😭😭😭😭😭😭

[@akshatb]

​@@sadececns00 1. Piecewise function is basically a function defined differently on different parts of the domain, for example if domain of a funciton is 0,1 and it has different definitions on 0,0.5 and 0.5,1 it'll be called piecewise 2. f_2(2) is the second element of f evaluated at 2 f2 = {1,2} the second element is 2 thus f_2(2)+1 = 2+1 = 3 hope that helps

[@KostasOreopoulos]

For those that code too. The recursive function defined is a very nice mathematical way to define a double for loop. The f1 is the outer loop counting the set ordinal number, while f2 loops over the cardinality of the set for i = 1 to n for j = 1 to cardinality(i) I have never thought of a recursive function to simulate that! :)

[@DarkCloud7]

Thanks I needed that hint to get the recursion, it didn't come through in the video imo. And people wonder why functional programming isn't more popular. :-P

[@debendragurung3033]

6:49 what does k stand for here in k_f_1(n)

[@izumiasmr]

@@debendragurung3033 f_1(n) will be the index of a "hosting" set let's say f_1(n) = 3, it means we're referring to an element from the set A_3. Then k of that thing = k_f_1(n) = k_3 would be the number of elements in that set

[@pkmath12345]

This reminds me of this one simple proof of making a waiting line to order rational number for one to one correspondence.

[@lionking2192]

It helped me to prove that a continuous function cannot map rational to irrational numbers and vice-versa... So thank you very much... Your contents are very nice... Keep it up...

[@johnwoolley2198]

Thanks for producing these videos!

[@fracaralho]

That is, indeed, a good place to sto--.

[@tomatrix7525]

Holy moley, this was some brain frying stuff with the indexing and inductions but worked out nicely

[@lemons107]

you can say that again brother

[@izumiasmr]

I think it should be: Case 1. f(m-1) + 1 ≤ k_{f_1(m-1)} From which the similar statement written about n in red in parentheses is not instantaneously obvious, however still is a consequence (to show it we need to observe that f2(n+1)=f2(m)=f2(m-1)+1>1 which implies that n also falls into case 1). It's also worth to mention that in order to start talking about m-1 we need to observe that one of the f(m) components is greater than 1 (as f(m)=f(n+1)), so m ≠ 1. If these corrections are valid, then i have an impression that all the proof steps are almost obvious, except for the key moments when we analyse that something is greater than 1 and make important decisions based on that.

[@ferranroigtio5184]

Hi, Michael! I've recently discovered your channel and it's so good! I'm not sure whether there's a better path for suggestions, but I'd really like you to elaborate on the Navier-Stokes equations. Cheers!

[@izumiasmr]

I am proving this once again, I believe f_1 and f_2 are sort of excessive & a bit unnatural notations, I think it'd be easier to index applications of f like: f(1)_1 f(1)_2 perhaps an extra set of parenthesis might be used as well: (f(1))_1 (f(1))_2 or maybe just introduce notation for elements of a tuple as it occurs as in: "let (s, t) = f(n)". Actually I think it's perfectly alright to work with f_1 and f_2 as functions, but as they work together it's a bit counterintuitive to instill them with too much of independency.

[@izumiasmr]

update: whilst in the process of writing a proof, I feel that f_1 and f_2 are not such bad as separate entities, however still I would introduce them somewhat more accentuated, and probably still would like to think of them as a tuple as an option😀

[@davidgillies620]

Incidentally, the cardinality of the A_n's is the Euler totient function phi(n) for n = 1 and 2 phi(n) for n > 1 (sort of by definition). A_1 = {0}, A_2 = {1/1, -1/1}}, A_3= {1/2, 2/1, -1/2, -2/1}, A_4 = {1/3, 3/1, -1/3, -3/1}, ... phi(1) = 1, phi(2) =1, phi(3) = 2, phi(4) =2, ...

[@thesecondderivative8967]

15:55 It feels like we're still using the assumption to prove the inductive step. It feels like we're just taking extra steps to say that since g(m) = g(n) implies m = n. Then, g(m-1) = g(n) implies that m-1 = n. But then I have a question, our function stated that if g(m) =g(n) then m = n. Can we not assign n+1 to a variable p. Therefore we have g(m) = g(p) implies m = p. m = n+1 For the assignment; f_1(m-1) = f_1(n) for similar reasons proven. Also, f_2(m) = f_2(n +1) = 1 It's trivial to see that if f_2(m) is 1, then the element assigned to m is the first element of the current set. Thus, f_2(m - 1) will show the cardinality of the previous element in the previous set. The cardinality of the last element of the set is the amount of elements in that set. Thus f_2(m -1) = k_(f_1(m-1)) = k_(f_1(n)) = f_2(n). The second and third parts of the equality statement are equal due to the fact that f_1 shows the sets they are in. And if the sets are the same, from the first part of the argument, then the cardinality of the sets are the same. Therefore, the cardinality of the sets they are in are the same . It follows that f_2(m -1) = f_2(n). f(m-1) = f(n), g(m-1) =g(n). Proof complete. Edit: I feel like there's an easier way to do this ...

[@AlexBesogonov]

I've been using a much simpler proof by construction, just construct a "snake" that starts at (0, 0) and then goes in a spiral pattern through all natural numbers on x and y axes. x/y will cover all rational numbers.

[@josvanderveeken8373]

Hi Michael, I have seen a number of your presentations on KZbin recently, and find them both very educational and entertaining. Math is fun! However for this particular one, I am a bit puzzled by the consequences of the Lemma : “The countable union of (disjoint) finite sets is countable”. Obviously the Lemma serves its purpose for proving that Q is countable. But if the Lemma is correct, it looks like it is also possible to prove that the power set of the natural numbers P(N) is countable. Let A0 = Φ, and An = P({1,…,n})\An-1 for n ≥ 1. Then the An are all disjoint, |An| = 2^(n-1) which is finite for all n, and UAn = P(N), which according to the Lemma is therefore countable. However it is well known and easy to prove (Cantor) that P(N) is not countable. So somewhere something must be wrong here. Apart from a few minor inaccuracies without any consequence (eg in the proof that g is surjective the expression should be g(k1 + ….. + kn-1 + m) = am(n) ) the proof of the Lemma looks rock solid. So where is the problem? I noted that for the Q case, |An|≤ n for all n, whereas for the power set |An|, while finite for all n, is increasing beyond all bounds. That is probably the reason that Q is countable, but P(N) is not, but that still suggests that the Lemma (and the proof thereof) should be qualified, but I am not sure. Any answer to this conundrum? (Apologies for the fact that the sub- and superscripts do not come out, but hopefully it is clear what is meant)

[@conorbrennan5838]

That union you have won't give you P (N) it will only give finite subsets of N from how you construced it. So you will not get subsets such as N\{1} or the subset of all even numbers etc and thats where the problem is.

[@josvanderveeken8373]

Yep, in the mean time I kind of have arrived at the same conclusion myself, however large n, An only contains finite subsets of N, so we never get to P(N) this way. It shows how tricky dealing with infinities can get. Thanks!

[@hgnb1001]

Very nice constructive demonstration.

[@sergiogiudici6976]

consider all the powers of all prime numbers p(k)^n meaning the n-power of the k-th prime number. You can make a one-to-one correspondence between p(k)^n and the element a(k,n) of the Union. p(k)^n are countable because they form a subset of N, then also all the a(k,n) are countable. Is It correct?

[@paretodeficiente9586]

Some hints for proving the function is surjective via induction?

[@oxygen2671]

When he defined the first function at 8:00 when he evaluated 3 in f(n)=(f1(n),f2(n)) . According to what he chose f1(n) and f2(n) when n=1 he chose f1(n)=1 and when n=2 f1(n)=1 it seems arbitrary to me .

[@maxmegel8861]

5:12 Go ahead, clean this ketchup

[@jkqa911]

Thank you for your video. I have two questions: 1) at 14:03 I don't understand why f(m)=(f1(m-1), f2(m-1)+1) insetad of f(m+1)=(f1(m), f2(m)+1) 2) to choose with rigour a^(n)_k for every set An is it necessary to use the axiom of choice?

[@izumiasmr]

I know the question is old, but... To address the first question please check the top level comment that i just wrote, i think there's a mistake in the proof and your question stems from exactly the problematic place whereas the rest of proof is correct

[@izumiasmr]

The second question seems interesting! I realize the intuition: there are infinite amount of choices we need to do, but I'm definitely not familiar with the AC enough to support our contradict it. Have you found out now?

[@jkqa911]

@@izumiasmr I think there's no contradiction only an implicit use of AC. Axiom AC - For any set X of nonempty sets, there exists a choice function f that is defined on X and maps each set of X to an element of the set. en.wikipedia.org/wiki/Axiom_of_choice Thanks for you answers.

[@user-fq7kv9ik8x]

i don't sure that your prove is correct (13:52) You do the prove to show that hyp is correct using hyp. what if hyp is not correct in fact?? then your prove goes to wrong. let me know what am i missing

[@borisbryant.007]

muscles well defined just as the theorems. you look tougher than your questions....... what do you do aside math????

[@vtvtify]

He does rock climbing.

[@nativechatter999]

Hi, What book would you recommend to learn Real Analysis? Thanks.

[@pkmath12345]

People seem to use Rudin thesedays very much

[@stephenbeck7222]

We used baby Rudin in my undergrad analysis class. Library had a copy of papa Rudin which actually had some proofs in the body of the text that were assigned problems in baby Rudin (and likely much harder assigned problems).

[@mushroomsteve]

Folland is another good one. Also, Taylor & Mann "Advanced Calculus" if you want something more at the undergrad level. Folland covers measure theory, point-set topology and functional analysis. Taylor & Mann covers topics like epsilon-delta proofs, differentiability, pointwise and uniform continuity, the Heine-Borel Theorem, etc.

[@pkmath12345]

Stephen Beck yeah I used baby Rudin in my undergrad real analysis class. I used a different one in my grad school class, but they were based on professors personal notes most likely~

[@kabirsingh4230]

If u r beginner then Kenneth Ross analysis is good

[@aritradey7465]

Can you pls upload a video explaining putnam 1995 B1

[@cariboubearmalachy1174]

But each An isn't finite! There are an infinite number of natural numbers p and q such that p+q=n! What am I missing?

[@cariboubearmalachy1174]

Natural numbers are positive... oh

[@laplacesdemon82]

A1 = {+- p/q s.t p+q =1} A2 = {+- p/q s.t. p+q =2} A10 = {+- p/q s.t. p+q= 10} for p,q of Natural numbers, only few of them satisfy p+q = 10, therefore, A10 is finite. Similarly for each n.

[@laplacesdemon82]

or should it be the absolute value of p + absolute value of q = n?

[@sword7163]

can we establish induction on Q by showing that if a proposition P holds for a rational number then it holds for all elements of a certain sequence ?

[@habibullah-ki7ok]

Induction can be applied whenever there is a notion of "next element". But then you end up with natural elements (only that they hace a different name). So, the answer will be: yes, you can apply induction on the set of rationals. And: No, you can not apply inducton rationals seong them a a field or ring.

[@zmaj12321]

At 16:02 I do not see how to prove for the second case. Since f2 is not in the definition I do not see how to show that f2(m-1)=f2(n).

[@zmaj12321]

Intuitively, they are both equal to k_f1(n) but how do you prove this?

[@user-A168]

Good

[@janshepard55]

Why does the intersection of these finite sets give you the empty set?

[@thesecondderivative8967]

The question is old. But if two or more sets are disjoint, their intersection is zero (by definition). Disjoint sets share no common elements.

[@cariboubearmalachy1174]

Is the assumption that the rational number is in lowest terms necessary to the proof?

[@tiripoulain]

No, in the sense that the union would still be Q, but there would be non trivial intersections. Keeping to the spirit or extreme rigor of this proof, you would probably then want to show that the lemma implies that countable unions of finite sets are countable, regardless of them being disjoint.

[@jaca2899]

my boyfriend just called you hot and im here to prove him wrong

[@urieldaboamorte]

son are you blind

[@andywright8803]

Rather confusing jargon. I would only consider the rationals to be countable once someone has completed the task. I could say the moon is jump able and the Pacific is swimmable.

[@morganhopkins204]

Lol. "Countable," or more precisely for this video, "countably infinite" is not "confusing jargon." It has a very precise definition in Math. Namely, a set is countably infinite iff it has the same cardinality (number of elements) as the natural numbers, which is shown by establishing the existence of a bijection between the set and the natural numbers. This is exactly what Michael did in the video.

[@andywright8803]

@@morganhopkins204 so you are defining the word countable in maths to be something that is clearly not countable, and you say that's not confusing? There are two groups of people who make no sense; pure mathematicians and flat earthers. I am a physicist with feet definitely in the real world. I use maths, but I find so much of the jargon to be almost deliberately confusing. "Real" numbers that can't be written down, "countable" sets that aren't countable at all. The universe is quantised. Why then do you still insist in the continuum of 'real' numbers?

[@andywright8803]

@Mathew Briguglio I fully understand the meaning of countable in mathematical terms. I am a physicist and work with mathematicians. I find the term itself confusing. It is patently impossible to count from 1 to infinity, yet in maths, the set of Natural numbers is called "countable". Even after a couple of decades this term still grates, and I think surely a better term could have been used

[@morganhopkins204]

@@andywright8803 I can see where you are coming from. It may seem that mathematical language is needlessly obtuse, to the point of being exclusionary. And I concede that sometimes the jargon/definitions can be unintuitive. However, this is for good reason! This jargon is quite necessary, unless you want to have some very wordy mathematicians. Put simply, it's often prudent to sacrifice some clarity for brevity and precision. Mathematics is built upon the rules of logic, which I'm not experienced enough to go into great detail on beyond the basics, and a goal: to rigorously and unambiguously prove logical statements using the rules of logic based on as few assumptions as possible. Such basic assumptions are known as "axioms," and typically describe some sort of definition(s). For example, sets are characterized by set axioms, the most popular collection of which is the basis of Zermelo-Fraenkel set theory (ZF). Natural numbers are characterized by the Peano axioms. Often times, the implications of a collection of axioms go far beyond the basic statements of the axioms themselves, with many logical steps according to the rules of logic along the way. It's useful to introduce new definitions, built from previous ones, to organize these steps. Did you know you can define the concept of a function completely in terms sets? That is, the axioms of set theory, which define what a set is and say basic things like, "the union of two sets exists," (I'm referring to ZF, in this case), plus the rules of logic imply what we know as functions. You could talk about functions using entirely the language of set theory, but then a simple statement like "f(x) = y," would become more like "There exists a subset f of the power set of the power set of the union of the sets X and Y, consisting of pairs of elements of the power set of the union of X and Y where the first element of the pair is a set containing a single element of X and no others and the second element of the pair is a pair of elements of the union of X and Y where the first element of the pair is the same element of X and the second is an element of Y, such that if f contains a pair where the first element is a set containing a and the second element is a pair where the second element is b and f also contains a pair where the first element is a set containing a and the second element is a pair where the second element is c, then b equals c, and f contains a pair where the first element is a set containing x and the second element is a pair where the second element is y." Obviously, that is unacceptable and wholly confusing, so instead mathematicians create new definitions and notation based on the axioms and build up to the definition of a function. By using more abstract language, they can take advantage of the fact that, given the axioms of set theory, the rules of logic imply functions without any additional assumptions (axioms), and they can focus on the properties that make functions interesting and useful, rather than having to deal with the details of how they are ultimately defined in terms of sets.

[@pecfexfextus4437]

@@andywright8803 wanna talk about the color charge in qcd? its also unrelated to the everyday meaning of color. isnt it confusing

[@heewahhin7470]

At 17:14, I think what we should check is g(k_1 + ... + k_n + m) = a_(n+1)_m Where m ∈ {1, ..., k_n+1}