"infinite" arithmetic is strange

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  • เผยแพร่เมื่อ 11 ม.ค. 2025

ความคิดเห็น • 99

  • @tylerduncan5908
    @tylerduncan5908 3 หลายเดือนก่อน +11

    Not sure why, but this is the first video I've seen (and i watch a LOT of math videos) on set theory that not only made it make sense, but also gave me somewhat of an idea of why set theory is so important for defining basic algebraic operations, and also why it's required to analyze all types of post-finite numbers in a meaningful way.

    • @dannyyeung8237
      @dannyyeung8237 3 หลายเดือนก่อน +1

      When are we reaching ε₀, ζ₀, η₀, φ(ω, 0) and Γ₀ for our set? And what does each of the set mentioned above contains?

    • @tylerduncan5908
      @tylerduncan5908 3 หลายเดือนก่อน

      @@dannyyeung8237 you're asking the wrong guy. I have no idea.

  • @luizmenezes9971
    @luizmenezes9971 3 หลายเดือนก่อน +5

    You got me with "that never ends, but at the end..."

  • @worldnotworld
    @worldnotworld 3 หลายเดือนก่อน +11

    Superb. There are some small and hardly noticeable errors (the kind I often make!): Right at the beginning, n should be defined as {0, 1, 2, ..., (n-1)}; i.e. including zero. Also, at 2:04 we should have "the elements of one, *and one itself."* At 12:56 and thereafter we also have omega plus one and the omega inside it missing zero. Same at 24:26.

    • @ingiford175
      @ingiford175 3 หลายเดือนก่อน

      Was looking at it and stopped it to make sure someone commented that n was misdefined as it did not have 0 element

  • @Happy_Abe
    @Happy_Abe 3 หลายเดือนก่อน +9

    @13:01 “that never ends”
    @13:03 “but, at the end”
    😂

    • @retrogamingfun4thelife
      @retrogamingfun4thelife 3 หลายเดือนก่อน

      I suppose you can do it just with sets, since you can always add an element ( imagine at the beginning) to a set, then do an imaginary sorting of the elements just to present the set in a better way.
      With real numbers - for instance - you cannot say “take all the digits of PI then add the digit 1 after the end”.

    • @Happy_Abe
      @Happy_Abe 3 หลายเดือนก่อน

      @@retrogamingfun4thelife as a sequence I suppose you can’t do that but with sets we can always union to add another element. There’s no real ordering to elements of a set, a set is just defined as the elements in them, not how they’re put in.

  • @xinpingdonohoe3978
    @xinpingdonohoe3978 3 หลายเดือนก่อน +7

    This makes more sense than the other ordinal videos I've seen. I think quite a few pulled ω like a rabbit out of a hat, simply saying "this is after the natural numbers" and not elaborating.

    • @dannyyeung8237
      @dannyyeung8237 3 หลายเดือนก่อน

      What aboutε₀, ζ₀, η₀, φ(ω, 0) and Γ₀ for our set? And what does each of the set mentioned above contains?

  • @supratimsantra5413
    @supratimsantra5413 3 หลายเดือนก่อน +6

    What a tremendous beautiful discussion sir!
    Thanks would be very tiny to appreciate your activities sir!.
    So I want to pay my gratitude only towards your unparalleled vidioes

  • @Bodyknock
    @Bodyknock 3 หลายเดือนก่อน +43

    Nice video, only minor thing being he keeps accidentally omitting 0 when he writes {1,2,3,…} . (Which I’m guessing is because he does so much work in number theory he keeps subconsciously wanting to exclude 0 as a Natural Number. 🤷‍♂️ He did catch himself on it once or twice though.)

    • @Udidda
      @Udidda 3 หลายเดือนก่อน +3

      I’ve initially assumed that it is because 0 is an empty set which is a member of any set by default… but that doesn’t seem to excuse {1,2,3,…,1,2,3,…} so you’re probably right.

    • @dannyyeung8237
      @dannyyeung8237 3 หลายเดือนก่อน

      By the way what branch of mathematics is this. Is this something you can study as a math degree in uni?

    • @darksecret965
      @darksecret965 3 หลายเดือนก่อน +1

      ​@@dannyyeung8237 Infinite set theory I'm guessing

    • @okoyoso
      @okoyoso 3 หลายเดือนก่อน

      ​@@dannyyeung8237Ordinals are a topic in set theory. If you want to learn more, I recommend Goldrei's for a beginner friendly book

    • @Alan-zf2tt
      @Alan-zf2tt 3 หลายเดือนก่อน

      I am not suggesting there is a connection b-u-t the average current (and average voltage(?) any physicists out there?) of an alternating current is: zero.
      In a full cycle starting at zero going, wlog, to max(positive) down to zero down to max(negative) then back up to zero with symmetry cancelling of positive and negative values the only non-zeroed values are 0, 0 and 0. And as everyone can see the average of 0, 0 and 0 is zero making four zeroes in total.
      And so the average current in an ac circuit is zero BUT DO BELIEVE THAT AC CURRENT IS DANGEROUS!
      Whether something similar happens with voltage is for an expert to comment.
      Electrical engineers get around it by using root mean square (RMS) values and they certainly do not equate to zero in this case.
      So, the moral of this comment is: not everything works out nicely in reality as it does on paper?
      I think the math area in discussion here may be number theory but partitioning of math into one things or other things is not really a precise science or so it seems to me. Partitioning of math on pedagogical basis has to happen because of pedagogical practices rather than math awareness - in my opinion

  • @AleksyGrabovski
    @AleksyGrabovski 3 หลายเดือนก่อน

    I think explanation starting from the natural number construction as sets is really helpful. I gave more solid ground to my understanding of ordinals. Thanks

  • @jneal4154
    @jneal4154 3 หลายเดือนก่อน

    Thank you for not pay-walling your educational content. 🙏 You're a real one.

  • @stephenhamer8192
    @stephenhamer8192 3 หลายเดือนก่อน +10

    I didn't quite get the definition of omega (W, the set of all finite ordinals, a.k.a "the natural numbers") as the union of all ordinals less than W. Isn't this circular?

    • @bartekabuz855
      @bartekabuz855 3 หลายเดือนก่อน +1

      It can also be definie as the smalest inductive set that is: x is inductive iff 0 is in x and for all y in x s(y) is in x

    • @ingiford175
      @ingiford175 3 หลายเดือนก่อน

      in this case for one set to be less then another, the one set must be contained by the other

    • @dannyyeung8237
      @dannyyeung8237 3 หลายเดือนก่อน

      When are we reaching ε₀, ζ₀, η₀, φ(ω, 0) and Γ₀ for our set? And what does each of the set mentioned above contains?

    • @funktorial
      @funktorial 3 หลายเดือนก่อน +1

      it might be a little confusing because it's not a definition, it's just a condition that omega satisfies. (in fact, for any ordinal alpha, we have that alpha is the union of all ordinals strictly less than alpha.) but here's where your suspicion is kinda correct: to prove that omega exists, we really do need to use something beyond taking unions. that's the axiom of infinity: some inductive set exists (where an inductive set contains 0 and is closed under taking successors). omega is then *defined* as the smallest inductive set. and that really is what michael did here, but without so much bookkeeping

  • @PPolycephalum
    @PPolycephalum 3 หลายเดือนก่อน

    Looking forward to the followup covering uncountable ordinals!

  • @funktorial
    @funktorial 3 หลายเดือนก่อน

    the condition that every element of an ordinal is a subset is called being "transitive." it's so-called because the element-of relation becomes transitive. so, x is transitive means: if y is an element of x, and z is an element of y, then z is an element of x. neat fact: ordinals can be defined as the transitive sets of transitive sets!

  • @roberttelarket4934
    @roberttelarket4934 3 หลายเดือนก่อน +9

    It’s a good place not to stop but go on and on and on to infinity!!!!!!! . . .

    • @allanjmcpherson
      @allanjmcpherson 3 หลายเดือนก่อน +2

      And beyond!

    • @roberttelarket4934
      @roberttelarket4934 3 หลายเดือนก่อน

      @@allanjmcpherson: Beyond infinity is a contradiction of infinity!

    • @allanjmcpherson
      @allanjmcpherson 3 หลายเดือนก่อน

      @@roberttelarket4934 and yet we did it

  • @IngvarLind
    @IngvarLind 3 หลายเดือนก่อน

    I believe it would help if you compared the ordinal operations to hyperreals, since they are now used in analysis with big advantages, and hyperreals simply follow the standard arithmetical rules used with the standard real numbers, despite dealing with infinite, finite, and infinitesimal numbers.

  • @Calcprof
    @Calcprof 3 หลายเดือนก่อน +1

    It gets even weirder if you go to the surreal numbers, which contain both the Reals and all the ordinal numbers, You can picture some of these by infinite Hackenbush stalks. See Winning Ways, Volume 2, by Berlekamp, Conway and Guy. Or "On numbers and Games" by Conway.

    • @tomkerruish2982
      @tomkerruish2982 3 หลายเดือนก่อน

      There's also _Surreal Numbers_ by Donald Knuth.

  • @tom-e1w
    @tom-e1w 3 หลายเดือนก่อน +2

    It seems odd that left addition and right addition are defined differently for infinite ordinals, does anyone know why it was defined this way??

    • @thatoneguy9582
      @thatoneguy9582 3 หลายเดือนก่อน +3

      they're not defined _differently_ it's just that the way addition itself is defined and the way infinite ordinals are defined combine to mean that it just isnt commutative

    • @radadadadee
      @radadadadee 3 หลายเดือนก่อน

      @@thatoneguy9582 they are MOST DEFINITIVELY defined differently as you can see with your eyes right there on the board.

    • @BridgeBum
      @BridgeBum 3 หลายเดือนก่อน +4

      ​@@radadadadeeSemantics perhaps but it isn't the operator that is defined different, they are both defined in terms of "tail recursion" of the second parameter. How that operator acts on finite parameters is different than how it acts on infinite parameters. Since the recursion only operates on the second parameter, it makes a difference whether or not the infinite parameter appears first or second.
      This feels different from "standard" addition since if both parameters are finite then the position doesn't matter. (Commutativity).

    • @tomkerruish2982
      @tomkerruish2982 3 หลายเดือนก่อน +5

      ​@radadadadee There is no left addition and right addition; there's simply ordinal addition, which happens to be noncommutative in general.

    • @tom-e1w
      @tom-e1w 3 หลายเดือนก่อน +2

      @@BridgeBum Yhea but we could have defined it so that if at least one of the ordinals is infinite we have the union of sets definition so it's non commutative nature seems like a choice.

  • @arijin
    @arijin 3 หลายเดือนก่อน +8

    I rather like the surreal numbers as a model to address the noncommunitivity of addition of transfinite ordinals. It feels much nicer to me. (I’m a bit biased because I wrote my graduate thesis on it, though, so I spent a LOT of time in that alternate universe.)
    Both models are perfectly valid alternatives, but I think Conway really figured some stuff out in inventing the surreals to address certain limitations of this model.
    I expect you’ve gotten into that at some point as well. What do you think?

    • @radadadadee
      @radadadadee 3 หลายเดือนก่อน +1

      I think he has a video on that as well

    • @gwalla
      @gwalla 3 หลายเดือนก่อน +1

      I'm curious, are surreal numbers ever applied outside of their original context of game theory? For example, would a complex number a+bi where a and b are surreal make sense?

    • @arijin
      @arijin 3 หลายเดือนก่อน

      @@gwalla it would definitely make sense . I’ve seen Surcomplex numbers written up here and there.
      I didn’t get deeply into that personally, but I was struck by a line from On Numbers and Games that said “every game has a square root that is a game”. That would imply to me there is a direct way to extend things in that way, but I never looked into it. I don’t believe that was the approach taken in the surcomplex explorations out there though. Worth looking into and exploring, I’d say.

    • @gwalla
      @gwalla 3 หลายเดือนก่อน

      @@arijin Wait, if every game has a square root that is a game, and games can be negative, then are the complex numbers *already* in the surreals?

    • @arijin
      @arijin 2 หลายเดือนก่อน +1

      @@gwalla no. Because the restrictions that qualify a game as a number involve every element of the right set not being less than or equal to any member of the left set, so the square root of the game that is negative doesn’t qualify as a number.
      I always thought perhaps that we could make a modification to the requirements that would include it, but I never explored how that would work exactly or whether or not such a modification. Oils hold up without introducing a whole bunch of other undesirable stuff into the mix.

  • @nerdatmath
    @nerdatmath 3 หลายเดือนก่อน +4

    One should really point out that we're transitioning to proper classes, not sets, when talking about transfinite numbers. Otherwise one runs afoul of Russell's Paradox.

    • @jcsahnwaldt
      @jcsahnwaldt 3 หลายเดือนก่อน +3

      No, the transfinite numbers in this video don't have anything to do with Russell's paradox. And we don't need classes to handle them. More precisely: That depends on the underlying set theory. In ZF(C), all the things mentioned in the video are ordinary sets. I'm pretty sure they are sets (not proper classes) in all common set theories. I guess one could define a set theory in which even ω is not a set, but it would be rather restrictive.
      (Edited to clarify that this doesn't have anything to do with Russell's paradox.)

    • @MartinClark-e1w
      @MartinClark-e1w 3 หลายเดือนก่อน +2

      @@jcsahnwaldt finitists would indeed reject the axiom of infinity in ZF without which there is indeed no set of all the Natural numbers

    • @jcsahnwaldt
      @jcsahnwaldt 3 หลายเดือนก่อน

      ​@@MartinClark-e1wYes, but they would also reject the existence of a class of all natural numbers. I guess they'd reject the existence of anything that's a proper class in ZF(C) and related set theories, because all proper classes are infinite. But my main point was that the comment I responded to is confused about classes and Russell's paradox. This stuff about other set theories is just a tangent. :-)

    • @Smitology
      @Smitology 3 หลายเดือนก่อน +3

      transfinite numbers as a whole form a proper class, but each transfinite number is a set

  • @dronevluchten
    @dronevluchten 3 หลายเดือนก่อน

    In your video omega + 1 1 + omega. What however if you take the Euclidean axioms for number, but only exclude the axiom of induction. That introduces as well a concept for Omega, but with the extra functionalit that by axiom omega +1 = 1 + omega and that even omega -1 exists (= -1 + omega).

    • @dannyyeung8237
      @dannyyeung8237 3 หลายเดือนก่อน

      By the way do you know that 2^ω is also equal to ω?

  • @Alan-zf2tt
    @Alan-zf2tt 3 หลายเดือนก่อน

    Hmmm ... maybe the sums of split infinities differ from the linear sum to infinity? Now if that is true - well there ought to be a research grant for that

  • @MarcinSzyniszewski
    @MarcinSzyniszewski 3 หลายเดือนก่อน

    But 1+omega = union_{n\in N} (n+1) seems to be 1 u 2 u 3..., which is omega \ {0}? So it shouldn't be equal to omega? Or am I missing something?

    • @funktorial
      @funktorial 3 หลายเดือนก่อน +1

      1 u 2 u 3 ... = 0 u 1 u 2..., because 0 is the empty set! taking unions with 0 does nothing :)

  • @jackdeegan3141
    @jackdeegan3141 3 หลายเดือนก่อน +1

    can this be combined with a wheel structure in any sort of meaningful way

  • @Debg91
    @Debg91 3 หลายเดือนก่อน

    Question: don't we need to include a recursion (or induction) axiom (which wasn't mentioned in the previous videos) to justify we can define ω in the first place?

    • @MartinClark-e1w
      @MartinClark-e1w 3 หลายเดือนก่อน +1

      You just need the axiom of infinity from ZF

    • @Debg91
      @Debg91 3 หลายเดือนก่อน

      @@MartinClark-e1w yep that's the axiom I'm referring to

  • @radadadadee
    @radadadadee 3 หลายเดือนก่อน

    15:06 How can you add the empty set just because it's empty? Clearly 1 U 2... is not the same as 0 U 1 U 2... I don't see how they could be the same

    • @radadadadee
      @radadadadee 3 หลายเดือนก่อน +2

      22:58 Also the union of all 2*n is the set of all EVEN numbers {0, 2, 4, 6...} which may well be a copy of omega but it's not exactly omega. Is this sloppy math or am I wrong?

    • @radadadadee
      @radadadadee 3 หลายเดือนก่อน +1

      @@portmanteau7885 I see. You understand the union to be "up to 2n" but he says "the union over all the natural numbers of 2n" which I don't know exactly what it means. Usually the upper limit of a sum or a union goes on top of the symbol. Anyway, it's probably along the lines of what you said, but I don't think it was clear enough (to me).

  • @jan.kowalski
    @jan.kowalski 3 หลายเดือนก่อน

    If you say that "one" is a set with "one element", you didn't defined "one", you just defined "a set". You still need to define "one" as a counting of "number of elements".

  • @quietcollector
    @quietcollector 3 หลายเดือนก่อน +1

    So I guess when kids said “infinity plus 1” they were right

  • @baxtermullins1842
    @baxtermullins1842 3 หลายเดือนก่อน

    It is how you define infinity and it seems to be different based on who you ask!

  • @jesusthroughmary
    @jesusthroughmary 3 หลายเดือนก่อน +7

    The set defining n should also contain 0

    • @dannyyeung8237
      @dannyyeung8237 3 หลายเดือนก่อน

      By the way what branch of mathematics is this. Is this something you can study as a math degree in uni?

  • @johnpaterson6112
    @johnpaterson6112 2 หลายเดือนก่อน

    I can see that nerdy types might find this stuff amusing, but does it serve any practical purpose not more easily accessed by other means?

  • @cmilkau
    @cmilkau 2 หลายเดือนก่อน

    ω contains 0.

  • @dannyyeung8237
    @dannyyeung8237 3 หลายเดือนก่อน

    When are we reaching ε₀, ζ₀, η₀, φ(ω, 0) and Γ₀ for our set? And what does each of the set mentioned above contains?

  • @Wielorybkek
    @Wielorybkek 3 หลายเดือนก่อน

    it really feels like cheating xD

  • @hehehepaitachato9184
    @hehehepaitachato9184 3 หลายเดือนก่อน

    adoro seus videos, mas esse foi FODA...

  • @rafelfarre3185
    @rafelfarre3185 3 หลายเดือนก่อน +1

    no is wrong. no should be {0,1,2...,n-1}

  • @alexchan4226
    @alexchan4226 3 หลายเดือนก่อน

    wrong proof

    • @bjornfeuerbacher5514
      @bjornfeuerbacher5514 3 หลายเดือนก่อน +2

      Could you please be more specific?

    • @alexchan4226
      @alexchan4226 3 หลายเดือนก่อน

      @@bjornfeuerbacher5514 with equal not unequal

    • @bjornfeuerbacher5514
      @bjornfeuerbacher5514 3 หลายเดือนก่อน

      @@alexchan4226 He shows in the video that 1 + omega and omega + 1 are not equal. Didn't you watch the video?!

  • @oliverherskovits7927
    @oliverherskovits7927 3 หลายเดือนก่อน +2

    The property mentioned at 5:42 is called transitivity

  • @badmanjones179
    @badmanjones179 3 หลายเดือนก่อน

    I don't follow how n+⍵ always equals ⍵; i get that for the U(all n+1) you can just U in 0 since its elements are empty, but how about 2+⍵? wouldnt that be U(all n+2)? then youd get 2U3U4... which wouldnt be the same as ⍵, right?

    • @aperson6821
      @aperson6821 3 หลายเดือนก่อน

      Notice 1 ⊂ 2. Thus 1U2 = 2, so ”skipping 1” in the big union doesn’t matter

    • @jpdiegidio
      @jpdiegidio 2 หลายเดือนก่อน

      @23:12 : U_{n in |N} (2n) = 0 U 2 U 4 U 8 U ..., but 2n = {0,1,2,...,2n-1}, so all natural numbers are in the union.