Real Analysis | Perfect Sets

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  • 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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ความคิดเห็น • 28

  • @Theguyfromvoid
    @Theguyfromvoid 3 ปีที่แล้ว +1

    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
    @datsmydab-minecraft-and-mo5666 4 ปีที่แล้ว

    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
    @youtubeuser8232 4 ปีที่แล้ว +1

    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
    @TheQEDRoom 4 ปีที่แล้ว +2

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

  • @IsaacBroudy
    @IsaacBroudy 4 ปีที่แล้ว +3

    Will you be continuing the representation theory video series?

  • @johnluin6793
    @johnluin6793 ปีที่แล้ว

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

  • @tomkerruish2982
    @tomkerruish2982 4 ปีที่แล้ว +3

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

  • @shimon6689
    @shimon6689 4 ปีที่แล้ว +2

    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
      @Wurfenkopf 4 ปีที่แล้ว

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

    • @shimon6689
      @shimon6689 4 ปีที่แล้ว

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

  • @simoanwar490
    @simoanwar490 3 ปีที่แล้ว

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

  • @soloanch
    @soloanch 4 ปีที่แล้ว +3

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

    • @soloanch
      @soloanch 4 ปีที่แล้ว

      @ゴゴ Joji Joestar ゴゴ alright thanks

  • @johnluin6793
    @johnluin6793 ปีที่แล้ว

    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
    @CM63_France 4 ปีที่แล้ว

    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
      @tracyh5751 4 ปีที่แล้ว

      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
      @CM63_France 4 ปีที่แล้ว +1

      @@tracyh5751 Ok, thanks.

  • @rodrigoromero537
    @rodrigoromero537 4 ปีที่แล้ว

    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.

  • @aadfg0
    @aadfg0 4 ปีที่แล้ว +1

    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.

  • @drewkavi6327
    @drewkavi6327 4 ปีที่แล้ว

    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
      @matthewryan4844 4 ปีที่แล้ว

      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
      @tonysplodge44 4 ปีที่แล้ว +2

      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.

  • @biggerthaninfinity7604
    @biggerthaninfinity7604 3 ปีที่แล้ว

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

  • @________6295
    @________6295 4 ปีที่แล้ว +1

    14:26

    • @tobiasgorgen7592
      @tobiasgorgen7592 4 ปีที่แล้ว

      What

    • @metakaolin
      @metakaolin 4 ปีที่แล้ว +1

      @@tobiasgorgen7592 a perfect place to stop

    • @tobiasgorgen7592
      @tobiasgorgen7592 4 ปีที่แล้ว

      @@metakaolin You got to be kidding me xDDD

  • @goodplacetostop2973
    @goodplacetostop2973 4 ปีที่แล้ว +10

    14:25