How many functions are there?

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  • เผยแพร่เมื่อ 7 ก.พ. 2025
  • The answer is: a lot of them! In this video, I show that F(R), the set of functions from R to R, has the same cardinality as P(R), the set of subsets of the real numbers, which, in a previous video, I’ve shown to be much bigger than R. This is set theory at its finest :)

ความคิดเห็น • 45

  • @turtle1907
    @turtle1907 7 ปีที่แล้ว +17

    Oh man, I just took a set theory exam today. Peyam, you should have made all these videos a couple months ago xD.
    Also the proof that |ℝ| = |ℝ^2| that I received brought attention to the fact that most proofs online of this fail to form a direct bijection because of base funkiness, or at least the proofs that attempt to even form such a bijection. Of course, unleashing schroeder bernstein makes it a delight.
    Also I've been pronouncing schroeder bernstein horrendously wrong :c.
    Also what's perhaps more interesting is that |continuous functions from ℝ to ℝ| = |ℝ|. Continuity is quite the restriction!

    • @RJYounglingTricking
      @RJYounglingTricking 6 ปีที่แล้ว +2

      Lol’d at when u said you’ve been pronouncing it wrong🤣
      I have no idea how to pronounce it either cuz I’ve only read it!
      (Also, sometimes online, I’ve seen it called Cantor-Schröder-Bernstein, which always makes me 😆 like.. can you make it longer?!)
      I have that w symbols sometimes cuz in the book I use they don’t always tell the reader what they’re called and I’m doing this solo.

  • @fullfungo
    @fullfungo 7 ปีที่แล้ว +4

    I think that Hilbert curve is a good explanation why |R|=|R^n|.
    It's not only a curve but most importantly it's a function F:R^n->R or F:R->R^n.

  • @tomknapton1461
    @tomknapton1461 7 ปีที่แล้ว +2

    Fantastic!! Absolutely love your videos. Particularly this one!

  • @RJYounglingTricking
    @RJYounglingTricking 6 ปีที่แล้ว

    Thank you so much brother.. again ur saving my behind! I’m working through a math book and the author often doesn’t give the solution to problems cuz he probably assumes you’ll have a teacher or smth (which I do not, cuz I just got a book and started working through it.)
    If it weren’t for you, I’d be so stuck! Really, really appreciate it!
    Now just gotta keep playing w these proofs until they’re part of my ‘spleen’.
    You da man Doc🙏🏼🤟🏼

  • @estuardodiaz2720
    @estuardodiaz2720 7 ปีที่แล้ว

    I really liked it!! Please keep making videos of set theory :D

  • @chessislife3429
    @chessislife3429 6 ปีที่แล้ว

    I took a course that introduced the concept of different sizes of infinity in my year 1 sem 1, everything seems blur. Now I am year 3, finally can appreciate the beauty of this. It's the most beautiful concept I have learned so far in uni.

  • @Zonnymaka
    @Zonnymaka 7 ปีที่แล้ว +3

    Gotta love these videos about cardinalities!

  • @xy9439
    @xy9439 7 ปีที่แล้ว +3

    Awesome video! Best content on TH-cam

    • @cycklist
      @cycklist 7 ปีที่แล้ว

      Adàlia Ramon I agree!

  • @Lklibertad
    @Lklibertad 7 ปีที่แล้ว +4

    more videos of general topology :) its getting interesting

  • @leonardromano1491
    @leonardromano1491 7 ปีที่แล้ว +1

    Is it really necessary to find a bijection between the two sets? Wouldn't it be sufficient to find a surjection s:X->Y and another surjection s':Y->X, to show that they have the same cardinality?
    At least for finite X and Y I think it would make sense, are there any trivial counterexamples for the infinite case?

    • @drpeyam
      @drpeyam  7 ปีที่แล้ว +3

      It’s true but nontrivial, your argument relies on a variant of the Schroeder Bernstein theorem!

  • @davidwright8432
    @davidwright8432 7 ปีที่แล้ว

    Dr. Payam - it might be helpful to number the videos dealing with a particular topic. This would make dependencies clear, and let people search for other videos they need to understand a particular video
    That aside, thanks again. My Brain feels good after a workout like this! (It insists on uppercase 'B'. I have nothing to do with it.)

  • @qcard76
    @qcard76 7 ปีที่แล้ว

    If one has a one to one relation in |R s.t. f(x) = y where f varies, does this mean that there is a countably infinite amount of possible functions to describe it (i.e. the cardinality is the same as the natural numbers)?

  • @TrackopGaming
    @TrackopGaming 4 ปีที่แล้ว

    My god! That was beautiful.

  • @cedricp.4941
    @cedricp.4941 7 ปีที่แล้ว +10

    Great !! Can you please make a video for |R×R| = |R| ? 😃

    • @drpeyam
      @drpeyam  7 ปีที่แล้ว +7

      I’ll do once I find a simple enough proof! The one I’ve seen so far is very complicated :O

    • @stevethecatcouch6532
      @stevethecatcouch6532 7 ปีที่แล้ว +7

      I can do half a proof. |R x R| >= |R x {1}| = |R|. I'll let you take the other half, Dr. Peyam.

    • @giladreti
      @giladreti 7 ปีที่แล้ว

      Take the arctan(x)/pi + 0.5 of each number of each pair of real numbers, then construct a number 0.x1 y1 x2 y2... where xn is the nth digit of the first number, and yn is of the second. then multiply by pi, and subtract pi/2. You will get a number between -pi/2 and pi/2. Take the tan of that number, and you're done. proving the bijectiveness of each part is easy, hence the function is bijective.

    • @drpeyam
      @drpeyam  7 ปีที่แล้ว +1

      Gilad Reti Yes, except there’s this tricky issue that 0.99999 = 1, which may pose problems to the bijectiveness!

    • @giladreti
      @giladreti 7 ปีที่แล้ว

      Hmm... where? The first and the last part are just the standard bijective from R to (0,1) and vice versa, and I the middle we have only numbers from (0,1). Am I missing something?

  • @emmanuelcelio6166
    @emmanuelcelio6166 7 ปีที่แล้ว

    Great video, man!!! :)

  • @ekadria-bo4962
    @ekadria-bo4962 7 ปีที่แล้ว +1

    wonder, how about cardinality of function to R^n to R ? .its same or not?

    • @drpeyam
      @drpeyam  7 ปีที่แล้ว +1

      Yes, since Rn and R have the same cardinality

    • @drpeyam
      @drpeyam  7 ปีที่แล้ว +3

      Fun fact: Because of this, strictly speaking, multivariable calculus is as hard as single variable calculus :P

  • @MathIguess
    @MathIguess 5 ปีที่แล้ว

    No idea what any of that meant.
    I wrote a calculus exam today and played games for the rest of the day. My mental energy is depleted haha.

  • @vs-cw1wc
    @vs-cw1wc 7 ปีที่แล้ว

    Is it possible to explicitly construct a 1-1 onto map from P(R) to F(R)?

    • @drpeyam
      @drpeyam  7 ปีที่แล้ว

      Perhaps, but I’m not 100% sure how!

  • @dougr.2398
    @dougr.2398 5 ปีที่แล้ว

    Ah, three videos to watch in sequence about sequences (or sets!).... this is a linear regression!! :)

    • @drpeyam
      @drpeyam  5 ปีที่แล้ว +1

      Hahaha

  • @tracyh5751
    @tracyh5751 7 ปีที่แล้ว +1

    sin(1/x) is not a function from R to R.

    • @drpeyam
      @drpeyam  7 ปีที่แล้ว +3

      Tracy H If you define it to be 0 at 0 (for example) it becomes one :)

    • @drpeyam
      @drpeyam  7 ปีที่แล้ว +2

      Also the cool thing is that even if your functions are undefined at some (even all) points, there are still as many such functions as P(R). :)

  • @Shadow4707
    @Shadow4707 7 ปีที่แล้ว

    I'd like to see why |RxR| = |R|. I can't think of a bijective function.

    • @arongil
      @arongil 6 ปีที่แล้ว +2

      Wouldn't a space-filling curve like a Hilbert or Peano curve be a bijective function linking RxR and R?

  • @TheRealSamSpedding
    @TheRealSamSpedding 7 ปีที่แล้ว

    Cool.