Real Analysis | Perfect Sets
ฝัง
- เผยแพร่เมื่อ 22 ธ.ค. 2024
- We define the notion of a perfect set of real numbers and prove that all perfect sets are uncountable.
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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.
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!
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} ) .
This is the kind of content I am looking for! Thanks!
Will you be continuing the representation theory video series?
0:43 That’s not correct. There must exist ε>o so that infinitely many elements are within B_ε’(x) for every 0
I believe this is how Cantor originally proved that the reals are uncountable.
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
Isn't there just one way to solve that problem?
As far as I recall I've seen at least a 2 different approaches to that problem.
Thanks a lot ,This is the kind of content I am looking for
How is this possible :
The intersection btw the subset of P and P is an empty set?
Please 🙏 reply
@ゴゴ Joji Joestar ゴゴ alright thanks
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.
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?
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.
@@tracyh5751 Ok, thanks.
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.
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.
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?
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.
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.
Mathematics is beautiful, but easy to hide. Who in the world proved it at first?
14:26
What
@@tobiasgorgen7592 a perfect place to stop
@@metakaolin You got to be kidding me xDDD
14:25