Real Analysis | The Heine-Borel Theorem

Поділитися
Вставка
  • Опубліковано 16 гру 2024

КОМЕНТАРІ • 37

  • @MichaelPennMath
    @MichaelPennMath  4 роки тому +32

    I just woke up to notice this video cuts off in the last step. Sorry everyone! My files have been strangely exporting recently... I will make a video with the last step and post it unlisted later today -- linked at the end of this video.

    • @MichaelPennMath
      @MichaelPennMath  4 роки тому +6

      Here is the new video: ua-cam.com/video/p9nKuqz6D9w/v-deo.html

  • @goodplacetostop2973
    @goodplacetostop2973 4 роки тому +67

    17:54 “So this is...” not a good place to stop

  • @Jim-be8sj
    @Jim-be8sj 4 роки тому +13

    One of two things happened here: 1) Cauchy sequences do not necessarily converge in the UA-cam space. 2) Gödel's famous theorem was being demonstrated in some way.

  • @bencheesecake
    @bencheesecake 4 роки тому +6

    One minor note is that we must show that S (from 12:08) is nonempty before determining the existence of its supremum. This is done by taking some U_i containing a, and finding some x_i such that [a,x_i] is contained within U_i, then this single U_i provides a finite (containing only one element) subcover of [a,x_i], so x_i is in S. For those of you preparing for exams, this is the kind of detail that I have gotten dinged on in my oral exams.

  • @mohithraju2629
    @mohithraju2629 4 роки тому +33

    The rest is left as an exercise to the viewer.

    • @raphaelkelly861
      @raphaelkelly861 4 роки тому +2

      The rest of the solution is trivial and left as an elementary exercise

    • @ian731
      @ian731 3 роки тому +2

      @@raphaelkelly861 a tensor is something that transforms like a tensor.

  • @aristo7051
    @aristo7051 Рік тому +2

    Can someone help explain why c < b is a contradiction of s @16:15 thank you

    • @alegal695
      @alegal695 7 місяців тому

      Case c' != b. Since c' = min{ s + eps, b } --> c' = s + eps --> s = c' - eps --> s < c' which is in S --> s is not an upper bound of S --> s != sup S, which contradict that s = sup S.

  • @get2113
    @get2113 4 роки тому +6

    Nice treatment of tougher proof, but there is an editing problem near end .

  • @philperfect8800
    @philperfect8800 3 роки тому +3

    Lemma: the proof look like an induction. But don't you have to prove first that S is not empty. In that way, could we say that [a,a] is obvious part of S.

  • @sitienlieng
    @sitienlieng Рік тому

    At 10:26, we contradicted to the hypothesis by showing y must be in K, but I could not see how y must be in K. In fact, we actually contradicted to the hypothesis that some terms of {y sub n} is not in K, and it makes more sense to me. Is there a mistake?

    • @heewahhin7470
      @heewahhin7470 Рік тому

      We started off with a K that satisfies the property that every open cover of it has a finite subcover. We also assumed that the K has a limit point 'y'. If K does not have a limit point then K is simply closed. From the proof, we know that if such K exists, i.e. satisfying the two conditions then y not being an element of K is not possible. Therefore, if this K exists then we must have all of its limit points to be in K, which makes K a closed set.

  • @joaomegazen
    @joaomegazen 4 роки тому

    You are saving me...my master exam is next month and it's all real analysis! Thank you for your classes!

  • @PowerhouseCell
    @PowerhouseCell 4 роки тому +12

    *Imagine watching the whole video waiting to see how it ends 😢*

  • @tomkerruish2982
    @tomkerruish2982 4 роки тому +2

    Well, this answers my question from yesterday. Apparently I just need patience.

  • @KurdaHussein
    @KurdaHussein 7 місяців тому

    where can I find that proof for Rⁿ ?

  • @Ahmed2-hz7yz
    @Ahmed2-hz7yz Рік тому

    At 9:58,
    Why is it true that yN doesn't belong to any of U sub xi?

    • @fahadqureshi1226
      @fahadqureshi1226 11 місяців тому

      By definition, the limit (introduced around 7:18) must be within an infinitely small distance from the terms in the sequence. However, U sub xi caps this distance at 1/2 min {(x1-y), ...., (xn-y)}.

  • @AmanGupta-sj1rx
    @AmanGupta-sj1rx 4 роки тому +5

    Video just stopped 😩😩
    Eagerly waiting for the ending 😩😩

  • @caladbolg8666
    @caladbolg8666 4 роки тому

    My favorite theorem

  • @CM63_France
    @CM63_France 4 роки тому

    "Ok, great". I think we'll call you the Hitchcock of the maths :)

  • @pandas896
    @pandas896 4 роки тому +3

    I couldn't understand anything. And doesn't makes sense , how is it useful?? Can someone explain in what are the uses of these things? Any use in physics?

    • @DarGViD
      @DarGViD 4 роки тому +2

      That's basically just a part of the common language of advanced analysis. (Physics loves it)

    • @duckymomo7935
      @duckymomo7935 4 роки тому +2

      The result isn’t applied directly but is important to use in physics sure
      Complex numbers is a nice field with compactness property
      A notion of finiteness is nice
      Many theorems use compactness (see compact support)

  • @valeriobertoncello1809
    @valeriobertoncello1809 4 роки тому

    I have only one concern: when we define epsilon using the minimum function, aren't we assuming that a minimum exists? Why should this be true in general?

    • @AP0PT0SIS
      @AP0PT0SIS 3 роки тому +2

      The set is finite by the finite subcover condition, so finite sets always admit a minimum element by the well ordering principle.

  • @user-A168
    @user-A168 4 роки тому

    Good

  • @GrandMoffTarkinsTeaDispenser
    @GrandMoffTarkinsTeaDispenser 4 роки тому +2

    UA-cam: *This is a good place to stop*

  • @philperfect8800
    @philperfect8800 3 роки тому

    I do not understand why c'= min(s+epsi,b). The sets of finite cover Ui are not constraint to be in S, nor in [a,b], so the upper bound of the cover could be anything bigger than c.
    Your demo is a special case wher [a,c] = Union of Uin.

    • @carl3260
      @carl3260 3 роки тому +1

      The aim is to prove finite cover of [a,b], by proving a finite cover of [a, s+e] where either s+e=b and so the required finite cover exists (that it may cover more than [a,b] is irrelevant)

  • @CDChester
    @CDChester 4 роки тому

    ahhh more analysis

  • @Wultuswuffel
    @Wultuswuffel 4 роки тому

    youtube thinks Im smart enough to understand any of this

  • @lyjv
    @lyjv 2 роки тому

    this looks so useless