Real Analysis | Perfect Sets

[@Theguyfromvoid]

These videos are really helpful, especially how much monotone most professors teach math. I would like really you to generalise it for R^k but I guess it's a nice little exercise.

[@datsmydab-minecraft-and-mo5666]

Thanks a lot for these videos. It's rare to see a channel post videos that aren't long lectures on these sorts of topics, keep it up!

[@youtubeuser8232]

What if p_n=x_{n+1}? Shouldn't you impose also x_{n+1} eq p_n? Anyway it shouldn't be a problem, since the points of P \setminus {p_n} are obviously not isolated and you can choose x_n eq x_{n+1} \in [a_n, b_n] \cap ( P \setminus {p_n} ) .

[@TheQEDRoom]

This is the kind of content I am looking for! Thanks!

[@IsaacBroudy]

Will you be continuing the representation theory video series?

[@tomkerruish2982]

I believe this is how Cantor originally proved that the reals are uncountable.

[@johnluin6793]

0:43 That’s not correct. There must exist ε>o so that infinitely many elements are within B_ε’(x) for every 0

[@simoanwar490]

Thanks a lot ,This is the kind of content I am looking for

[@shimon6689]

Dear Professor Penn, Can you please solve the famous question number 6 from International Math Olimpiad of 1988? I'm very curious to see your approach to that problem. Thanks in advance, Shimon

[@Wurfenkopf]

Isn't there just one way to solve that problem?

[@shimon6689]

As far as I recall I've seen at least a 2 different approaches to that problem.

[@aadfg0]

I found a proof without contradiction: Let p_(1/2) ∈ P and set ε_1 = 1. Pick p_(1/4), p_(3/4) ∈ [p_(1/2) - ε_1, p_(1/2) + ε_1]. To construct p_(1/2^n), p_(3/2^n), ..., p_((2^n-1)/2^n), pick p_((2a-1)/2^n), p_((2a+1)/2^n) ∈ [p_(a/2^(n-1)) - ε_(n-1), p_(a/2^(n-1)) + ε_(n-1)] where ε_(n-1)>0 is chosen so that all 2^(n-2) intervals involved are disjoint (*) and ε_(n-1) ≤ ε_(n-2)/2 (**). For any x ∈ [0,1], define p_x as the limit of the sequence {p_(a_n)} where a_n is the number corresponding to the first n digits of x in binary; the limit exists since (**) ensures the sequence is Cauchy. Since P is closed, p_x ∈ P, so P contains {p_x : x ∈ [0,1]}, which naturally bijects to [0,1] since (*) ensures all p_x's are distinct. Thus, |P| ≥ |[0,1]| = R and P is uncountable.

[@soloanch]

How is this possible : The intersection btw the subset of P and P is an empty set? Please 🙏 reply

[@soloanch]

@ゴゴ Joji Joestar ゴゴ alright thanks

[@johnluin6793]

You could just use stereographic projection to construct a bijection between the open neighborhood around a and R. Done.. Or just notice also that f(x)= 1/x - 1 is an isomorphism between (0,1) and the real line and extend that.

[@CM63_France]

Hi, Mickael, I see a problem: let us take a closed intevalle in Q, say [a,b], it does not have any isolated point, nevertheless it is countable. Where is my mistake?

[@tracyh5751]

Your example is closed in the rational numbers, but as a subset of the real numbers it has limit points that are not included in the set and so your example is not a closed subset of the real numbers.

[@CM63_France]

@@tracyh5751 Ok, thanks.

[@rodrigoromero537]

Hello everyone, can someone give me a hint on how to prove that a topological space is contractible iff every pair of curves with the same endpoints are homotopic? thank you.

[@________6295]

14:26

[@tobiasgorgen7592]

What

[@metakaolin]

@@tobiasgorgen7592 a perfect place to stop

[@tobiasgorgen7592]

@@metakaolin You got to be kidding me xDDD

[@biggerthaninfinity7604]

Mathematics is beautiful, but easy to hide. Who in the world proved it at first?

[@drewkavi6327]

I don’t really enjoy real analysis, but I am applying to uni for maths, should I be worried will I likely dislike my degree if I don’t find this stuff very interesting?

[@matthewryan4844]

I didn't either but got through to analysis 2 just following the rules they lay out for you, then you can choose whatever else you like (my 3rd year was things like group theory, number theory, non-linear dynamics). If you really dislike it but want to do maths, you can probably steer away in year 2; it does provide a good basis for general mathematical thinking though.

[@tonysplodge44]

I agree with Matt, but I would go a little further to say that in general over any degree course, most people find bits they like and bits they dislike - but you have to put in the work, even on the bits you don't like. I'm a massive fan of Michael Penn, but I also find this stuff hard going. I'm a big fan of Normal Wildberger's videos where he argues that much of Reals, Set Theory, Analysis, the Axiom of Choice, yada yada, is all based on very hooky foundations - but you can only justify your point of view if you've worked hard and understood the concepts.

[@goodplacetostop2973]

14:25