A New Way to Measure Sets! (How to build a strictly monotone measure)

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  • Опубликовано: 5 авг 2022
  • List of corrections/clarifications (including "clopen", "topologist", and "measure") below
    This is a summary of about a year or two of my own personal math research. It is not complete overview, nor does it constitute peer-reviewed research. I am a teacher by trade, not a professional mathematician. This is just a fun intro to my research to see if people are interested. I also do claim to have invented some of these ideas myself, but definitely don't claim that it hasn't been done before! There's a lot of research out there and I am not an "insider" to the fields I studied.
    This video is based on my findings in this preprint paper: arxiv.org/abs/2008.09969. This is not published or peer-reviewed work.
    You can support me by sending a Venmo tip to the handle @MathTrain. Please don't donate unless you are able.
    Check the pinned comment for an ongoing list of minor mistakes people have caught!
    By popular demand, I've made a discord server so we can discuss the results of this video. If you're interested in discussion about the results and trying to further this research, join in! / discord
    Thanks to the Summer of Math Exposition 2 #SoME2 for being the impetus for me making this passion project!
    Correction list:
    1. Whenever I say "clopen", I'm not referring to a set that is both closed and open (which is the correct use). I'm referring to half-open intervals, and I'm misusing terminology here. My mistake!
    2. It's debatable what a topologist would actually say here, and it's not as cut-and-dry as I make it out to be in the video. Many have pointed out that topologists might say B is larger because it is open, while A is compact. Valid point, but dividing math up into subcategories isn't really the point of the video, so I chose to hand-wave this one and say "topology" since it's a broad topic in which every point matters.
    3. The width of the interval at 5:30 should be 12, not 8. But this doesn't change the end result.
    4. Technically, when I say "measure" I mean "polynomial-valued measure". An actual measure has codomain R, while I use codomain R[omega]. Also an actual measure is defined on a sigma-algebra, while this is only on a non-sigma algebra of sets. But that would've been a bit esoteric to include in this introduction. The actual algebra on which mu works is the algebra of definable sets in an o-minimal structure.

Комментарии • 538

  • @MathTrain1
    @MathTrain1  2 года назад +109

    Some important corrections:
    1. Whenever I say "clopen", I'm not referring to a set that is both closed and open (which is the correct use). I'm referring to half-open intervals, and I'm misusing terminology here. My mistake!
    2. It's debatable what a topologist would actually say here, and it's not as cut-and-dry as I make it out to be in the video. Many have pointed out that topologists might say B is larger because it is open, while A is compact. Valid point, but dividing math up into subcategories isn't really the point of the video, so I chose to hand-wave this one and say "topology" since it's a broad topic in which every point matters.
    3. The width of the interval at 5:30 should be 12, not 8. But this doesn't change the end result.
    4. Technically, when I say "measure" I mean "polynomial-valued measure". An actual measure has codomain R, while I use codomain R[omega]. Also an actual measure is defined on a sigma-algebra, while this is only on a non-sigma algebra of sets. But that would've been a bit esoteric to include in this introduction.

  • @crazyspider17
    @crazyspider17 2 года назад +99

    you say in the video a "topologist view" but you really just meant a "set theorist view". containment is just a notion of set theory, and a lot of different subjects use set theory, including measure theory, so containment doesn't really have anything to do with topology in particular.

    • @MathTrain1
      @MathTrain1  2 года назад +34

      I did consider saying "set theorist" instead, but when you get to the details of the proofs behind the scenes, it's much closer to topology (plus I was trying to make a bit of clickbait lol). Good point though!

    • @jakubszczesnowicz3201
      @jakubszczesnowicz3201 2 года назад +20

      The whole point of topology is for us to be able to describe the notion of “space” using purely set theory. Set theory is at the heart of Topology so I would say he did nothing wrong by labelling it as such. Even if you disagree with what I just said, it’s tough to argue that measure theory and topology employ set theory to a comparable degree.

    • @nicholasandrzejkiewicz
      @nicholasandrzejkiewicz 2 года назад +4

      @@jakubszczesnowicz3201 you're saying that measure theory and topology are different "distances" away from using set theory?
      (This doesn't make sense)

    • @Grassmpl
      @Grassmpl 2 года назад +1

      In this case he should have mentioned a homeomorphism from B to a subset of A.

    • @matron9936
      @matron9936 2 года назад

      @@jakubszczesnowicz3201 that‘s not the point of topology I‘d argue. Consider locales for example. Set theory is just the tool used for means of formalisation I‘d say.

  • @andriesvancauwenberge5182
    @andriesvancauwenberge5182 Год назад +5

    This is the best #some2 video I have seen really love it. Keep up the great work. I would love to see more of this.

  • @RuleAndLine
    @RuleAndLine 2 года назад +4

    Holy cow, I wasn't expecting original math research when I started watching this. Really interesting result!

    • @MathTrain1
      @MathTrain1  2 года назад +1

      Tricked you into learning something new lol

  • @oceannuclear
    @oceannuclear Год назад +6

    Damn, you're out here making a video about your primary research at a level that is understandable by the (not so general) public. This is insanely cool!

  • @harsinsinquin4032
    @harsinsinquin4032 2 года назад +9

    9:12 my first thought was “why not just use omega” and the second I said it you said it. It made me feel very delighted.

  • @ellismaddox2776
    @ellismaddox2776 2 года назад +31

    Why would the large square not simply eat the small one?

    • @MathTrain1
      @MathTrain1  2 года назад +9

      They're fighting to see who's larger - have you learned nothing??

    • @ellismaddox2776
      @ellismaddox2776 2 года назад +4

      @@MathTrain1 I’m just saying they’d settle it the fastest that way

    • @clairecostelloe5741
      @clairecostelloe5741 2 года назад +5

      I would like to see your next video exploring this further. How does this work relate to the Cinnamon Toast Crunch squares that eat each other in mid-2000s commercials?

    • @MathTrain1
      @MathTrain1  2 года назад +6

      @@clairecostelloe5741 This is an interesting point to explore. I suggest you read the Toast section of the original paper - it also mentions homeomorphisms onto Frosted Flakes.

  • @rcteg
    @rcteg 2 года назад +4

    This was fantastically well made! I'm an undergrad in CS and have a hobbyists interest in math but no actual experience in analysis, measure theory, algebra, or topology, and you did a fantastic job of explaining this to a layman. Seriously well done and thanks for sharing, seeing this kind of math research done at an actually understandable level is super cool!

    • @MathTrain1
      @MathTrain1  2 года назад +3

      Thanks so much! I'm really glad to see it's having a positive impact and people are enjoying it. The level of explanation is probably because I only have an undergrad degree myself, lol

  • @ilonachan
    @ilonachan 2 года назад +12

    Great video, what you failed to consider is that as a topologist the open square is just ℝ² and hence larger than the closed square/disc. Checkmate scaleheads!

    • @MathTrain1
      @MathTrain1  2 года назад +2

      Folks, this commenter has ascended - we are not worthy

  • @monsieuralexandergulbu3678
    @monsieuralexandergulbu3678 2 года назад +7

    Dudes really be like inventing new maths and doing papers for SoME ;)
    Really nice video and theory tho!

  • @DynestiGTI
    @DynestiGTI 2 года назад +5

    The use of ω, ω^2, etc. to label the progressively larger infinities reminds me of John Conway's *surreal numbers.* Also is this seriously the first time someone has created a strictly monotone measure that is independent of how the set is measured? I'm honestly surprised no one has thought about this before... good job!

    • @MathTrain1
      @MathTrain1  2 года назад +4

      I doubt it's actually the first time, but it is the first time I know of, and I've done a lot of research. So I feel confident that if someone else has done this, it's not well known.

  • @justinofelipeimbert4694
    @justinofelipeimbert4694 Год назад +2

    outstanding video, you introduced the question in a very simple way, so simple that people without any knowledge of measure theory or topology could be able to understand, your solution is very intuitive and you did a very good job maintaining the interest throughout the video.
    again, outstanding video.

  • @NixoticaTTV
    @NixoticaTTV 2 года назад +7

    Hidden gem of a video, its really nice that we live in an age where people can communicate existing theorems or even laymanize their own theorems in a single digestible video. Really hope to see more :)

    • @MathTrain1
      @MathTrain1  2 года назад +2

      Hoping I can make more! Thanks so much

  • @laszloliptak611
    @laszloliptak611 2 года назад +5

    The word "larger" makes sense only if we specify the partial order that we want to use for the comparison. Without specifying it, the question will always be ambiguous.

    • @MathTrain1
      @MathTrain1  2 года назад +1

      Absolutely - that's why the video is about building a different measure that obeys the subset partial order

  • @eli0damon
    @eli0damon Год назад +6

    This is fascinating, It makes perfect sense to me to think of a shape owning a half share of its boundary, with its complement owning the other half share. But as others have already expressed, it seems like this would only work for finite collections of finite cells. I'm imagining problems with even some fairly mundane example sets. Like, how would you combine rational points with irrational points on a unit interval?

  • @captain-carre
    @captain-carre Год назад +8

    Very interesting. How would you measure infinite discrete sets with this measure ? Like the rationals on [0,1]. If you start from the empty set and add all the points you get infinity. If you start from [0,1] and remove the irrationals you get omega - infinity ?

    • @giuseppepapari8870
      @giuseppepapari8870 Год назад

      I was going to ask the same question

    • @kennethhedden5970
      @kennethhedden5970 Год назад +2

      Even simpler: what would be the measure of the set of integers on the real number line, or of the whole number line itself? In standard measure theory (e.g., Lebesgue), a measure is defined on a sigma-algebra, which consists of exactly those sets which are "measurable". What is sigma-algebra for this measure?

    • @toniokettner4821
      @toniokettner4821 Год назад

      it's just not a measurable set

    • @ribozyme2899
      @ribozyme2899 Год назад

      @@kennethhedden5970 Quote from the pinned comment: "Also an actual measure is defined on a sigma-algebra, while this is only on a non-sigma algebra of sets. But that would've been a bit esoteric to include in this introduction."
      So yes, they took years to get the details right, but did not manage to create an actual measure.

  • @zaringers
    @zaringers Год назад

    This is super super interesting! What a great job you did here, for the math but also for the video, the editing etc!
    Congratz!!

  • @ikhu6042
    @ikhu6042 2 года назад +2

    Hey this is actually the coolest thing I've seen in a while. I *always* wanted to be able to consolidate how we could compare volumes of different dimensions while still preserving the information abt lower dimensions. Like adding a line to an are always gives (inf+1) lines so it just disappears but using these half-open intervals is an amazing way to do it, and how you can the get the span of the polynomial measure space from squaring/cubing the open interval.

  • @MichaelFJ1969
    @MichaelFJ1969 Год назад

    Great video! Thanks for taking the time to explain your findings in such a patient manner.

  • @code_explorations
    @code_explorations Год назад +1

    This is a really great video, and I will eagerly watch any follow-up you make.

  •  2 года назад +4

    Very interesting. For me, both squares were equal. After the video, I still think they're equal. But I really love this perfectly monotonic function you build. So clever and interesting.

    • @MathTrain1
      @MathTrain1  2 года назад +1

      Thank you! It's awesome how everyone can have their own perspectives

  • @jaafars.mahdawi6911
    @jaafars.mahdawi6911 Год назад +1

    Math is all about creativity and this video does just that! Keep it up, my friend!

  • @spiffinn_music_lists
    @spiffinn_music_lists Год назад +1

    I'm excited to get more information/content on this measure! Especially explanations for the points at the end of the video, and how to deal with and distinguish sets countably and uncountably infinite sets.

  • @rishikaushik8307
    @rishikaushik8307 2 года назад +3

    That's a very interesting system, it is more intuitive for me to visualize it slightly differently though. We can imagine each line segment as the complete length with 2 half points at each end, so ω^2 can be thought of as (the area + half parameter + 4 one-fourth points at each vertex), this also makes visualizing tiling areas easier. Any 2d shape we can deconstruct into - > the area + 0.5 parameter + (internal angle/360)th of a point at each sharp vertex.
    We can even extend this to 3D with any space being broken into -> the volume + 0.5 surface area + (angle between faces/360)th of a segment at each sharp edge + (internal solid angle / complete sphere)th of a point at each vertex. In 3D the "density" of an edge can change continuously along its length if the angle between faces changes.
    imagining like this every shape fits nicely

  • @neoztar
    @neoztar 2 года назад +3

    The -1/2 coefficient that you end up with for the boundaries is very reminiscent of the concept of "half-edges" from computational geometry where an edge between two vertices A and B can be broken up into two directed half-edges, one from A to B, and one from B to A.
    So if we consider an arbitrary region within some larger tesselation to be intrinsically surrounded by half-edges, then it makes sense that when you cut it out to form its own shape, you have to either add in the missing half-boundary to get a closed set, or remove it to get an open set.

    • @MathTrain1
      @MathTrain1  2 года назад

      This is really working towards the practical applications that I'd love to find more of. I have done a bit of work by using half-edges, but didn't know they were a pre-existing idea! That's something I'll try to look into - thanks!

  • @tonaxysam
    @tonaxysam 2 года назад +1

    This is absolutely bizzare! I love it
    Just the idea of such a measure to exists is crazy! I love how the video kept simple but the details covered dive in very deeply without losing the easy approach. That was very nice!

    • @MathTrain1
      @MathTrain1  2 года назад +1

      Thanks so much, I'm glad you enjoyed it! Share it if you'd like to support :)

  • @stanleydodds9
    @stanleydodds9 Год назад +5

    I feel like you'll get a lot of problems if you try to make this work for all sets, considering that most sets are just some uncountable, fuzzy, indescribable mess. I mean, the obvious thing that comes to mind is the Banach Tarski paradox; if you really can measure all sets, and the measure is preserved under rotations, translations, and partitions, then you have a problem: measure a sphere, then you can split it into just a few (I think 5) parts, which can be rotated and translated to construct 2 spheres, each identical to the first, which should measure twice as much, contradicting the measure being preserved. I'm fairly sure there need to be quite a lot of non-measurable sets, same as the usual Lebesque measure which can't measure most sets (e.g. it can't measure something like a subset of R that is a basis for the reals over the rationals).

  • @calyodelphi124
    @calyodelphi124 Год назад +2

    This is a very neat and intuitive idea! :)

  • @azai.mp4
    @azai.mp4 2 года назад +3

    This is so cool! I did a project on nonstandard analysis in uni, and since then I've been wondering if measure theory would benefit from having more numbers to use as sizes. It's nice to see someone had the same idea and worked it out successfully

  • @hughobyrne2588
    @hughobyrne2588 2 года назад +3

    Fantastic debut. As to the aspects you can talk more about, I say go for it. All of it - well, all of it you're interested in sharing. The passion for the subject matter makes the video solid.

    • @MathTrain1
      @MathTrain1  2 года назад

      Thank you, that's kind of you to say!

    • @hughobyrne2588
      @hughobyrne2588 2 года назад +1

      @@MathTrain1 There's a RUclips video called "Kill the Mathematical Hydra". Your polynomials 'in infinity', so to speak, reminded me of it.

    • @MathTrain1
      @MathTrain1  2 года назад

      @@hughobyrne2588 thanks, I'll check it out!

  • @columbus8myhw
    @columbus8myhw 2 года назад +6

    I appreciate how for the first poll, if we say B is larger, the two options are: we're an idiot, or we have ascended

    • @MathTrain1
      @MathTrain1  2 года назад +6

      Many in the comments have ascended

  • @cristinaaugegiribet1236
    @cristinaaugegiribet1236 2 года назад +5

    How does this not conflict with the Banach-Tarsky paradox? In the sense that one would ovously assume that two balls would have double the measure of a single one yet you can break one into 5 pieces and rearrange them into 2 balls. What did I miss that solves this problem?

    • @MathTrain1
      @MathTrain1  2 года назад +6

      Great question! It doesn't solve the problem - it's just that not every set is measurable in this paradigm. The same is true with non-measurable sets in usual measure theory. If you're referring to when I say at the end it can be extended to any set, the answer is that when you do that extension using nonstandard analysis, rotations and translations no longer preserve measure. Since you have to rotate and translate in Banach-Tarski, they don't conflict.
      However, on the sets I laid out here (the technical limitation is that the sets be "definable in an o-minimal structure") it is translation and rotation invariant.

  • @lexinwonderland5741
    @lexinwonderland5741 2 года назад +2

    WOW! this was a brilliant video, very accessibly explained (the nod to omega as a limit ordinal is clever as well), and especially i'm amazed that you came up with this yourself!! i look forward to your future research (as someone who wishes they could be a professional mathematician but unfortunately with no current chance to do so), and your future educational videos!! keep up the great work, friend!

    • @MathTrain1
      @MathTrain1  2 года назад +2

      Let it be known that I, too, am not a professional mathematician! Just a HS teacher. Thanks so much for your support! I'll try to update this with additional research but we'll see how it goes (I did most of this when I was not working full time)

    • @lexinwonderland5741
      @lexinwonderland5741 2 года назад +1

      @@MathTrain1 I get you. Math is my side-love, most of my free time is spent doing neuropsych research and that's only when I'm not swamped with work (as an engineer... yeah, my career path has been weird). Taking a summer off to make a thorough lit review on the pathology and treatment of fibromyalgia was amazing but it was the only time I got anywhere beyond just taking notes. Hopefully in a few years I can move forward to academia in my preferred field... until then, let's keep dreaming!

  • @MatthewWroten
    @MatthewWroten Год назад +5

    Hi, great work, but I think I see problems with extending this to any set. What about the set of rational numbers on the number line? What if you add a point?

  • @johnchessant3012
    @johnchessant3012 2 года назад +2

    Great video! Where I thought you were going with it was from a set theory perspective, A and B are the same size because there's a (not continuous, of course) one-to-one correspondence between them; perhaps a bit of a chore to describe but the idea one dimension lower is that (0,1) and [0,1] are the same size because you can map 1/2 to 0; 1/4 to 1; 1/2^(n+2) to 1/2^n for n > 0; and x to x for all other x. Glad to learn something new; I hadn't seen anything like that before and it seems really interesting to think about!

    • @MathTrain1
      @MathTrain1  2 года назад +1

      Yes! Perhaps I could make a follow up that points out cardinality as well, since of course most geometric sets have cardinality of the continuum, so in that sense every non-finite definable* set has the same size!
      And thanks for the feedback - glad you think it's cool too!
      *Definable in the sense of an o-minimal structure - basically means it has integer dimension

  • @juansamuelarbelaezrocha4113
    @juansamuelarbelaezrocha4113 2 года назад +2

    This is amazing!!! Well done!!!

  • @mattschoolfield4776
    @mattschoolfield4776 2 года назад

    Very well done. Looking forward to see what you do next

    • @MathTrain1
      @MathTrain1  2 года назад +1

      Thank you :). Here's hoping I have the time!

  • @zuzaaa1998
    @zuzaaa1998 2 года назад +4

    Nice video and interesting concept. I have spent some time on similar idea once, but I haven't gone so far with it as you did. I am looking forward for next videos about this

    • @MathTrain1
      @MathTrain1  2 года назад

      Just curious - what was your idea you spent time with?

    • @zuzaaa1998
      @zuzaaa1998 2 года назад +2

      @@MathTrain1 The more I think about it, the more I see that I was thinking about exactly the same thing as you. During my investigations I didn't come up with your very insightful idea of using half-open half-closed intervals as building blocks.
      I am curious if you have investigated what are measurable sets of your measure. How would you measure for example the Cantor Set? I mean one can assign value of infinity to it but then strict monoticity would break

    • @MathTrain1
      @MathTrain1  2 года назад +1

      Yes, I have! It's described in the arxiv paper in the description. There's a few ways to extend it. The one I've shown here works for sets that are "definable in an o-minimal structure". You can use nonstandard analysis to extend to literally all subsets of R^n, but if you do that there are some weird philosophical implications and it becomes much harder to actually do calculations.
      I'd love to figure out a way to calculate the measure of the cantor set and I've thought about it extensively but just haven't come up with how to do it

  • @mileszeweniuk2999
    @mileszeweniuk2999 2 года назад +5

    My knowledge of your system and of surreal numbers is mostly surface level so this could be (probably is) completely wrong, but I couldn't help but notice some parallels between your ω and what ω means in surreal numbers:
    in both systems ω is analogous to infinity - ω is larger than any finite number (or any countable infinity for that matter), ω^2 is larger than any finite amount of ω, etc.
    in both systems ω is meaningfully distinct from ω + 1, despite ω being analogous to infinity
    in both systems, despite the distinctness of ω and ω + 1, ω is not an ordinal number (to my knowledge)
    I wonder if surreal numbers could in any way help with the idea of extending to fractional or even irrational exponents

    • @MathTrain1
      @MathTrain1  2 года назад +4

      You're correct in seeing a connection - I was using omega as a hyperreal number in my research. The hyperreals are contained in the surreals, so it's not just a parallel, they are the same thing :)

    • @mileszeweniuk2999
      @mileszeweniuk2999 2 года назад +1

      @@MathTrain1 that's really neat! And shows how extensive the surreal numbers are. TBH I haven't really dived deep into any branch of mathematics yet so I'm surprised and fascinated by these types of connections quite often :)

    • @user-hh5bx8xe5o
      @user-hh5bx8xe5o 2 года назад

      @@MathTrain1 Indeed, your system can be thought as a geometric ordinal number for certain shapes.

  • @orangeguy5463
    @orangeguy5463 2 года назад +5

    Definitely thought B was larger because I thought this would be a topology video. I thought B is homeomorphic to a 2 manifold of infinite measure while A is not, as it's compact like you said.

    • @MathTrain1
      @MathTrain1  2 года назад +1

      Pulled a bit of a switcheroo on ya

  • @perle.1715
    @perle.1715 Год назад +1

    just noticed that the constant term of your polynomial is exactly baez's so-called "euler-schanuel characteristic"! fascinating stuff!!

  • @sayantakrkuila2602
    @sayantakrkuila2602 2 года назад +2

    Very nice. Motivated me. Thank you 😊.

  • @arbeladm
    @arbeladm Год назад +7

    Does this work for sets that ate not lines?
    Consider The rational numbers, or cantor set, they are both measurable

    • @deinauge7894
      @deinauge7894 Год назад +1

      maybe there are solutions including log(omega), sqrt(omega) etc.
      i suspect it wouldn't be easy to do this, and that there will be sets whose measure cannot be expressed by standard math language... even it's possible to have a strictly monotone measure for all sets. but don't trust my instinct, try it ;-)

  • @RSLT
    @RSLT 2 года назад +3

    Interesting! Great job

  • @gabriellasso8808
    @gabriellasso8808 2 года назад +1

    Awesome math content!!!

  • @78Mathius
    @78Mathius Год назад +3

    What happens is you use a closed square and space filling curve to define the same set of points?
    Would this give a translation from an infinite length line to an area?

  • @scentoni
    @scentoni 2 года назад +1

    We kind of count fractions of a point in crystallography. An atom on the face of a cube counts 1/2; on an edge 1/4; and at a corner counts 1/8, but in the end the multiples of the fractions always add up to integers. In a lattice it’s equivalent to think of a closed finite ball cut in half and paired with another on the other side of the cell, or an infinitesimal point that’s slightly displaced to the side.

  • @landy4497
    @landy4497 Год назад +2

    this is amazing dude

  • @columbus8myhw
    @columbus8myhw 2 года назад +5

    This is like the Ehrhart polynomial (en.wikipedia.org/wiki/Ehrhart_polynomial ) of a polytope! And the constant term is the geometric Euler characteristic (also called the combinatorial Euler characteristic), which - for compact shapes - has a nice property called "homotopy invariance".
    I will say, though - for "wild" sets you can have weird things happen. For example, consider the infinite earing (what Wikipedia calls the "Hawaiian earring", though there are reasons not to use that name en.wikipedia.org/wiki/Hawaiian_earring ). Or the Cantor set. Or a double spiral union a line segment through its center. What are their mu measures?
    EDIT: Another test case is Osgood curves.

    • @MathTrain1
      @MathTrain1  2 года назад +2

      They're very closely related! The answer is that mu is not well-defined on such wild sets. But it is defined on a larger set than ehrhart polynomials (which only exist for polytopes). Mu is defined on the collection of sets that are definable in an o-minimal structure. Check the paper in the description if you would like details :).
      Or you can extend it further with nonstandard analysis and get the values for wild sets, but then you will definitely get some weird properties. I just proved such a measure exists, but have no idea how to do that many calculations with it!

    • @columbus8myhw
      @columbus8myhw 2 года назад +1

      @@MathTrain1 Ah, neat! O-minimal structures are fun. (PS see some edits to my comment)
      Wait, if you have that background, then I can give you a fantastic puzzle.
      Define an _arc_ to be a subset of the plane homeomorphic to a compact interval. The _multiset sum_ of two arcs is basically their union, except points in their intersection are double-counted (it's basically union with multiplicity).
      Can the multiset sum of two arcs equal the multiset sum of three arcs?
      The surprising answer is, yes! You can't do it if they come from an o-minimal structure (mu it, look at the constant term), but you can do it with wild sets (where we like to party). I know at least three different-ish solutions (though only one can be reasonably described through text).

    • @MathTrain1
      @MathTrain1  2 года назад +2

      @@columbus8myhw @columbus8myhw I'm now thoroughly engrossed in this puzzle - thanks for telling me it's not possible in an o-minimal structure otherwise I'd have just assumed it's not possible! Honestly I'm not sure if I'm good enough of a topologist to figure it out but we'll see

    • @MathTrain1
      @MathTrain1  2 года назад +1

      @@columbus8myhw okay I've been trying for an hour, can I have a solution? So far I've been trying some stuff with the function xsin(1/x) to no avail

    • @columbus8myhw
      @columbus8myhw 2 года назад

      @@MathTrain1 Alright. Here's the simplest solution of the three that I know of.
      Take a double spiral (or a finite segment of it, rather) plus a line segment through its center. A double spiral is, funnily enough, an arc (with a "bad point" in the middle), so this is the sum of two arcs. Some fiddling around will give you a way to decompose it into three arcs.
      See these images for the decomposition into three arcs (sorry that the color scheme isn't consistent between them):
      i.stack.imgur.com/palPy.jpg
      i.stack.imgur.com/bhpiH.jpg

  • @lock_ray
    @lock_ray Год назад +2

    I had a thought about what the zeros of the polynomials mean... They tell you how big a marker you would need to draw your shape clearly, in other words they give a measure of how detailed the shape is (kind of sort of, not really, you need negative marker sizes, it's a bad analogy, stay with me).
    Think about what happens when you evaluate some measure m_A(x) at a particular value, say x=2. You're essentially lumping together all the different measures (count, length, area etc.), with x=2 as a kind of "exchange rate" between higher measures and the counting measure. You're cashing in each half-open interval for twice its length's worth of counts, and a half-open square for 4 times its area etc. One way to visualize this is to imagine having a marker with a very wide tip, in this case width 1/2. Drawing a 1/2x1/2 dot is the best you can do to represent a single point, for a unit line (half-open, as it must tessellate) the best you can do to represent it is by drawing in a straight line for 1 unit, but because of the width you will actually get a 1/2x1 rectangle, or 2 dots worth of ink. Finally for a unit square (half-open so it can tessellate) you can actually draw it perfectly as a 1x1 square using this marker, or 4 dots worth of ink (this analogy kind of breaks down if you try to draw a square with side-lengths less than 1/2 though :( ). Similarly, you can think of negative terms in the polynomial as using a wide eraser, but this only works if there's already something to erase, so instead we'll think of it as negative ink, which deletes and is deleted by positive ink.
    Think about a AxB open rectangle. This has measure (Ax-1)(Bx-1) = ABx^2-(A+B)x+1, with roots at x=1/A and x=1/B, why is this? Well, think of trying to draw this using a wide marker of width W. First you have to draw a half-open AxB rectangle, then you must erase from it two half-open edges of length A and B respectively and finally add back a point. The best you can do is to draw an AxB box, then with some negative ink erase from the left side of it a AxW strip , leaving a Ax(B-W) rectangle. Next you are supposed to remove a WxB strip from the bottom, which erases the Wx(B-W) strip that remains there, but also leaves behind a leftover WxW spot of negative ink. Finally you have to draw a point, so you use this to perfectly annihilate that unsightly negative ink. What you are left with is a perfectly good (A-W)x(B-W) rectangle, which is a pretty good representation of the 1x1/2 open rectangle... as long as W isn't too big. In fact you want W to be strictly less than min(A,B), or the reciprocal of the largest root.
    In general, if the polynomial has a very large root, that means that you need a very fine marker tip to draw the shape - in some sense it's very intricate. This makes sense because there is a very high value of x for which the linear and/or constant terms still manage to undo the quadratic term, i.e. there is a lot of boundary per unit area, or the shape is "squiggly". You need to value area very highly (high "exchange rate") in order to get something out of it.
    Just some food for thought, though this breaks down in all sorts of cases I'm sure. Also it only addresses the largest root, and I haven't thought about higher dimensions/degrees at all.

  • @tyzonemusic
    @tyzonemusic 2 года назад +7

    Would a countably infinite set of points be measurable here? What would its size be? Really interesting video btw, one of my favorite from SoME2 so far

    • @MathTrain1
      @MathTrain1  2 года назад +2

      Thanks! And no, countable sets are not measurable with the tools I show here. But it can be extended to include them using nonstandard analysis.

  • @omrireved2
    @omrireved2 2 года назад +5

    I'm not a mathematician at all, and I'm not sure whether it's meaningful or not, but your formula for the measure reminds me of Pick's theorem - maybe there is some connection?

    • @MathTrain1
      @MathTrain1  2 года назад +5

      There is certainly a connection. In fact, Pick's theorem is equivalent to the statement that the number of interior points i of an integer-vertex polygon is mu(omega=1). I believe (like 80% sure) you can also then calculate the number of interior points when you add in other rational coordinates (like the number of half-integer coordinates in the interior would be the same polynomial evaluated at omega=2). Not sure how deep the connection goes but it's certainly there!

  • @Tletna
    @Tletna 2 года назад +4

    I will say though, either your math or my math doesn't add up, which means either you're confused, or I'm confused. If I'm confused, you might need to re-explain your concept here.
    The main point of confusion I have here is it seems like you're forcing your math to add up. Why is it when assigning a measure to the perimeter-less triangles you set it to (w^2)/2 - (2w + sqrt(2)w)/2 + 1 when you should measure one half the total area with perimeter included minus one half the original perimeter (4w/2 = 2w) then minus one half the diagonal ((sqrt(2)w + 1)/2)? Essentially (w^2 + 2w + 1)/2 - (2w) - (1/2)sqrt(2)w) -(1/2) = (w^2)/2 + w + (1/2) - 2w - sqrt(2)w/2 - 1/2 = (w^2)/2 - w - sqrt(2)w/2. That's not the same as what you had. It's a difference of (w + 1), which is pretty significant. Even if I did something wrong, the way you're doing it still confuses me.
    Would you clarify, how exactly are you 'measuring' the triangles with no perimeter? Can you show it step by step rather than using a something like a polynomial table forcing the numbers to align as expected? Show me using a 'ruler' and measurement of sorts using your rules how the triangles without perimeter are (w^2)/2 - (2w - sqrt(2)w)/2 + 1. I'm not seeing it. Until I do, I cannot understand or agree with the rest of your suppositions in your video past 17:18. Maybe, that doesn't matter so much, but I'm sure some others were confused as well.
    It's still an interesting topic either way. Thanks for the video!

  • @cmilkau
    @cmilkau 2 года назад +1

    It is absolutely possible to give the border of the square a nonzero measure. You can even give the border if every square a nonzero measure, however if you want real numbers as measures, this will force the areas of squares to have infinite measures. If you can live with more exotic "numbers" as measure, you can have finite nonzero measures for borders and areas at the same time (the measures of the borders will be smaller than every positive real number but still positive).

  • @guywholovesmath
    @guywholovesmath 2 года назад +5

    Inventing... Ooh so are you not a platonist (such as myself)? Still this was a fantastic video!

    • @MathTrain1
      @MathTrain1  2 года назад +3

      I could go either way on discovered/invented, but it's fun to think about. And thanks I'm glad you enjoyed it!

  • @Namgnuoy123
    @Namgnuoy123 Год назад +3

    A very intriguing idea, Im not sure if this does really fully distinguish orders of sets. For example if you look at a line like the one generated by sin(1/x) as x ->0, or any 'infinitely long' line segment, in this system this would have measure of order w as a line segment it occupies no area, but has infinite length. You can think of a subset of this segment which would also have infinite length, but with a monotone measure should have lower measure. I dont think these objects are measurable in this system, but I wonder if this is reconcilable. Thought provoking!

  • @MayankKumar-so6sr
    @MayankKumar-so6sr Год назад +2

    I was a little confused with the example of circle at 17:48 (might have to re-watch this a few times and experiment myself a little bit) with 2 points not included... because I understand that is we are including a circle with it's circumference, the circumference when opened up into a line segment, it would have a open hole on one end and closed one on other but when closed it would have no hole since the closed end would overlap with closed one hence completing the circumference, but I still cannot wrap my head around how would I do it if I was not including the circumference like you did at 17:48. But you did intrigue my brain and at the end I do agree with the way this formula works... I might come back to this video again and spend a couple of hours sketching and scribbling things around on a pad....
    Though the most important thing I might have taken from this is my brain at this point at least works like a topologist, cos it feels like when we are trying to add area and circumference and the points together what we are doing is we are removing any additional points or like that we might have counted twice... like we do when we are counting items in a Venn diagram, we add the number of elements in both sets and subtract the elements common in both to get the count of a union of both sets.... And this approach of sets the my brain prefers to take is what makes me think that I might be thinking like a topologist...
    Though I am pretty sure I am no way as good an intellectual as an actual topologist....
    But thanks a tonne again for this video.

  • @quadrannilator
    @quadrannilator 2 года назад +2

    Thank you very much for this...
    This video both closed some of my open rabbit holes of thought and also opened new ones....
    now my head is utterly clopen with rabbit holes, in the wrong sense of the word.
    if I could ever contact you and engage you in a conversation, I'd be delighted
    one of my long standing inner unresolved paradox/dichotomy always had to do with the concept of the infinitesimal (the epsilon we use in calculus to shrink things down) and the attribute of equivalence we assign to things in mathematical logic - in my inner FELT experience, I was much more comfortable with Ω because I felt it was much more additive in nature - than €, which seems divisive in nature only trailing the line between arbitrarily small and non-existence,
    and as in Cantor's paradox, of how we measure and assign size equivalence to different sets, their cardinality correspondence, etc
    I have never found myself unable to use topological ways of working but there is always a measure-theoric itch I need to scratch in my head.

    • @MathTrain1
      @MathTrain1  2 года назад +1

      I'd love to correspond with you about this - I know few/no people who are actually willing to discuss it in detail, haha. We can chat on the discord server for this video: discord.gg/kgqcR9sq

    • @MathTrain1
      @MathTrain1  2 года назад +1

      Updated my reply with the discord server

  • @tinymints3134
    @tinymints3134 Год назад

    Just gotta say that as someone who is mostly illerate in math speak, this video is helping me a lot to comprehend what's happening in the paper.

  • @johneonas6628
    @johneonas6628 2 года назад +1

    Thank you for the video.

  • @PhrontDoor
    @PhrontDoor Год назад

    I was thinking of two boxes.. inside one are all the integers. inside the other we have the same thing, but we add 4 more.
    Which one is bigger now?

  • @gilbertbernstein7373
    @gilbertbernstein7373 Год назад +5

    Hi Joseph, have you seen/read Gian-Carlo Rota's "Introduction to Geometric Probability"? I think you would appreciate it quite a bit.
    Some colleagues (I do Computer Graphics research) pointed me to this video after we had a discussion about how the valuation Rota discusses in that book (and which ought to more or less be the polynomial measure you're discussing) can't quite be extended to a measure in traditional measure theory. It seems very interesting (and compelling) if that limitation can be removed simply by building up everything constructively instead (i.e. via non-standard analysis).
    You may find Rota's book very interesting as well from the standpoint of applications. For example, Let X subset of Y both be compact convex sets in R^n, and let S be a random k-dimensional affine subspace of R^n. Then the probability of S intersecting X given that S intersects Y is V_{n-k}(X) / V_{n-k}(Y), where V_{n-k} is the (n-k)-th dimensional invariant measure. For example, the odds of a random line striking a box X given that the line strikes an enclosing box Y is given by the ratio of their surface areas. This fact is known in Computer Graphics as the "Surface Area Heuristic" and is used to accelerate ray tracing algorithms.

    • @MathTrain1
      @MathTrain1  9 месяцев назад

      Hi Gilbert, I know you commented this a while ago but let me know if you'd have any interest in communicating outside RUclips about this. I would be interested to hear your take on a lot of these topics.

  • @Udidda
    @Udidda 2 года назад +2

    Really good video, discusses an original and organic subject and makes it very easy to understand.
    I still think it would've been neat to include a short tangent section on how B can be called larger since it's isomorphic to R^2 but it's not that big of a deal.

    • @MathTrain1
      @MathTrain1  2 года назад

      Given all the comments about that, I bet I should have done that too! Thought it would be too hard to keep people's attention that long though before getting to what I really wanted the video to be about (mu)

  • @lachlanperrier2851
    @lachlanperrier2851 2 года назад

    I would really vibe a live stream where people throw questions up and we look at them - really interesting topic!

  • @SodaliteSabre
    @SodaliteSabre Год назад +2

    What's the mu value of a countably infinite set of zero-dimensional points? Maybe something like log_2(omega)?

  • @angeldude101
    @angeldude101 2 года назад +3

    When thinking about how to include lengths and counts without areas completely dwarfing them, I thought of hyperreal numbers, which uses ω as an infinite number whose powers can be distinguished from each other. In a way, you could call this "Hyperreal Geometry."
    Regarding the initial question, I came to the conclusion that they were the same size, but by comparing the cardinality of the sets. While one has "more" points than the other, those extra points when added are completely dwarfed by the existing points making the sets have the same cardinality even if one is a strict subset of the other.
    Working through the size of a closed cube gave ω³ + 3ω² + 3ω + 1, which might look familiar as (ω + 1)³, or just row 3 of Pascal's Triangle. In hindsight, this makes perfect sense, ω + 1 is the length of one closed edge, so squaring that gives the cube. Similarly, an open cube (which I calculated first before adding back the sides) gave ω³ - 3ω² + 3ω - 1, which is (ω - 1)³. This means that calculating the size of a cube, and likely any parallelotope, is just the same as the normal volume formula. Circles look like they'll take a bit more work to apply that rule to.

    • @MathTrain1
      @MathTrain1  2 года назад

      You got it! All your calculations are correct, and yes, this video is all just disguised hyperreal analysis :)
      Though actually the calculation does not work for parallelotopes. They have different surface areas from cubes which changes the game
      Also hyperreal geometry seems like a dope name

    • @angeldude101
      @angeldude101 2 года назад +1

      @@MathTrain1 I actually considered trying to use this with a system with directed measures, and while it worked for rectangles in a way, I wasn't able to get it to work with a parallelogram. I thought it might have a chance of working since it can take into the two sides being non-perpendicular, but I guess not. Maybe with some more work.

    • @MathTrain1
      @MathTrain1  2 года назад

      @@angeldude101 what is a directed measure?

    • @angeldude101
      @angeldude101 2 года назад

      @@MathTrain1 In this case, it's in the form of a vector, though I did have a way to combine it with a scalar (akin to a point) and multiply them together into a directed area. Using unit vectors in place of ω, I can do (x̂ + 1)^(ŷ + 1) = x̂ŷ + x̂ + ŷ + 1, which if you replace x̂ and ŷ with ω, you get the right result for a closed square. When I tried with a parallelogram (x̂ + 1)^(x̂ + ŷ + 1) = x̂ŷ + 2x̂ + ŷ + 1. The expected value with ω would be ω² + (1+√2)ω + 1. This specific case would work by square rooting the coefficients first, but then it wouldn't work for rectangles.
      On the other hand, 2x̂ + ŷ is the sum of the two vectors, which could be interpreted with its own meaning. It is related to the desired value by the triangle inequality. Half the perimeter would be |v₁| + |v₂|, but I got |v₁ + v₂|.
      Edit: Rather amusingly, it works perfectly fine for two in-line vectors. (x̂ + 1)^(x̂ + 1) = 2x̂ + 1. Since the triangle inequality doesn't cause problems in the case, trying to form a square with two identical vectors just extends the line like you'd expect. It's a rhombus with 0 height.

  • @isaacdeutsch2538
    @isaacdeutsch2538 Год назад +2

    Very nice video. I'm not sure a topologist would say that A is larger though. The problem is that "larger" or "smaller" automatically implies a measure, something that point-set topology doesn't touch at all--a topologist doesn't generally endow their space with a metric, and so size is a nonsensical concept. They'd definitely say that A contains B, but they'd also say the points of A are in bijection with the points of B, and thus have the same cardinality. It would thus be up to the individual topologist to define what she means by "larger". Personally, as someone pursuing topology and currently studying it and adjacent fields, I'd say A and B are the same size, but A is compact while B is not (presuming of course I give the squares the standard product topology of the real numbers). And perhaps this is the measure theorist in me seeping in. But I can totally see how set containment could be an alternate way of defining "larger". Great video!

  •  2 года назад +2

    I want to give here an approach from computer science, in case anyone likes it. It will have lots of flaws, but it seems interesting to me.
    We could define that if something exists (a point) it must occupy space. For example, if an electron exists, it must have a volume greater than zero. In the same manner, if a point exists, it must have a volume greater than zero.
    Then, for a 2D case, define the space as a grid of very squares that can be either used (lit) or unused (unlit).
    A point is then just a square lit. A line is a set of continuous squares lit, and so on.
    In this space, you no longer have real numbers, because any length or area needs to be multiple of the space "plank size".
    Now, you draw a square. Then put a perimeter on it. First problem becomes: where you do place the perimeter? inset or outset of the square?
    If you place the perimeter inset, you need to do double-counting. The edges to be accounted 2 times when computing the perimeter.
    If this way seems "okay", then A and B are the same size, and are exactly the same set. Or, if you remove "the perimeter", the area must shrink, and you get a new perimeter, area and points.
    You could place the perimeter "outset", and this seems to work for the square case - each point in this space will be accounting for a single thing, the outset edges for the points, the outset lines for perimeter and the inset for the area.
    With this idea, you could imagine taking a limit where the size of the plank length approaches zero from positive and everything should still work out.
    But it has lots of problems for sure that I don't see. One that is clear is that I'm not sure how to deal with euclidean distances in diagonals, or curves where pi is involved. Technically if all that exists is a tiny square, PI would equal 4; which is obviously incorrect.
    But this line of thought can give a lot of insight: First, the fact that you have to take some arbitrary choices that will change the result. Second, that either the A set and B set are still the same set and you removed zero, or B is smaller (because it lost it's perimeter or maybe even lost area in the process, depending on counting).
    And third, more interestingly, it seems this also builds a sort of perfectly monotonic function as your system and my intuition tells me that it would give similar results if the problem with diagonals and curves are solved.

    • @MathTrain1
      @MathTrain1  2 года назад +1

      Ideas similar to what you describe were huge in my intuition when developing my system. In my head, omega is actually the "number of points on a unit line" - basically dividing it into pixels and taking a limit. You're right that it falls short with respect to curved shapes or anything that doesn't follow the gridlines. But literally the exact same formulas are given in your idea and mine for any set where every boundary is parallel to the axes.

  • @Joffrerap
    @Joffrerap Год назад +4

    What is the mesure of the cantor set, the cantor set minus a point. What about the set of even numbers, and the set of natural numbers ?

    • @MathTrain1
      @MathTrain1  Год назад +3

      I haven't been able to calculate the measure of the cantor set - it exists in an extension of this theory, but I haven't figured out what it is. I suspect it's of the order omega^(log2/log3). Natural numbers and other countable sets pose a similar issue of existing but not being calculable (by me, not saying it can't be done)

  • @mathlitmusic3687
    @mathlitmusic3687 2 года назад +1

    You can eliminate the "B is larger" more naturally by examining what you mean by "larger". The fact that you can put B on top of A and see that B sits (is strictly contained) in A, means B should be smaller than (or at most, equal to) A. This is the translation invariance axiom of Lebesgue measure.

  • @lock_ray
    @lock_ray Год назад

    My first though was, either this is infinite for most sets, or it's not real-valued, and you went ahead and covered both counting measures (mostly infinite) and this polynomial-valued solution, so... Yes and yes I guess?
    This was really nicely explained, good job

  • @andriesvancauwenberge5182
    @andriesvancauwenberge5182 Год назад +2

    Can anyone explain where the - Cw/2 comes from in the circle. 19:00

  • @bogiokas7426
    @bogiokas7426 Год назад

    Hi, interesting video! The measure of the different squares you compute seems to be the erhart polynomials of the unit squares with the boundary added or removed. May there be a connection there?

  • @tomkerruish2982
    @tomkerruish2982 2 года назад +1

    Any chance this could help crack the puzzle of surreal integration? I'm working my way through your arxiv paper, and I definitely noticed some familiar names in the bibliography.

  • @kj4242
    @kj4242 Год назад

    The best expalnation of this topic to date so for on you tube

  • @kenchan2377
    @kenchan2377 2 года назад +1

    Good video. You deserve more subs.

  • @davidnoll9581
    @davidnoll9581 Год назад +1

    Love this. Trying to figure out what it means for this polynomial to have complex vs real roots.

  • @ddystopia8091
    @ddystopia8091 Год назад +1

    What about bijection? If there are continuously many points in A and B, then we could find a bijection from A to B...

  • @TheBasikShow
    @TheBasikShow 2 года назад +2

    Slight correction at 12:27: the line pictured is not actually a clopen set, instead it’s a set that’s neither closed nor open. The term “clopen set” is used for sets that are _both_ closed _and_ open. In a connected space (like the plane), the only clopen sets are the empty set and the set of every point in the space.

    • @MathTrain1
      @MathTrain1  2 года назад

      Very much right! I've corrected this in the pinned comment.

    • @columbus8myhw
      @columbus8myhw 2 года назад

      @@MathTrain1 I've heard the name "half-open" for this.
      I propose the name _clopen't._ ... Never mind. I take it back. I don't propose this.

  • @rubberduck2078
    @rubberduck2078 2 года назад

    I would like some clarifications on the question 3 at the end. Is the measure only defined on an algebra generated by the sets bounded by a jordan curve? There was a phrase about "any definable set" somewhere in the middle of the video, but that is probably very very impossible.

    • @MathTrain1
      @MathTrain1  2 года назад

      The term I used is short for "definable in an o-minimal structure" which is a very particular collection of well-behaved geometrical sets!

  • @djsmeguk
    @djsmeguk 2 года назад +1

    Does this have a relationship to the flux integral through the shape at its boundary? I believe that's another way to get at the area/perimeter relationship. So I wonder if there's something related here...

    • @MathTrain1
      @MathTrain1  2 года назад +1

      I wonder too. No clue, but it'd be cool if there was

  • @Stdvwr
    @Stdvwr 2 года назад +1

    at 18:23 how did the length of the seam on the top transform into the perimeter of the shape on the bottom?

    • @MathTrain1
      @MathTrain1  2 года назад

      More of an intuition thing. You just imagine that every open set is a cut-out of some larger set, and every boundary that gets created will act like the seam, which gives half the perimeter in total. But this is just an intuitive guess - not a formal argument! It just happens to work, and it mirrors how I actually came up with the idea.

  • @SirTravelMuffin
    @SirTravelMuffin 2 года назад +1

    Very thought provoking! Wonder if this could be extended to fractals/fractional dimension objects. The fundamental "clopen" set also reminds me of a vector, so I get clifford algebra vibes when you add scalars (w^0) to vectors (w^1) to bivectors (w^2). Maybe these thoughts are a stretch, but either way I found this really interesting

  • @xnor5644
    @xnor5644 2 года назад +2

    Very interesting work! I have a question on what sets your polynomial-valued measure applies to. From the definition of definable (tame) sets in your paper, I don't immediately see how these would necessarily include subsets like the unit disk or even a line segment of slope 2:1, which appear in the video. How do we deal with these formally? Similarly, in the definition for Euclidean invariance, can the domain itself not be closed under rigid motions (rotations)?

    • @MathTrain1
      @MathTrain1  2 года назад +1

      Great question. I suggest you read "an introduction to o minimal geometry" by Coste. The first page defines the semi algebraic sets, which include both of the sets you mention. This is the smallest o-minimal structure containing the basic arithmetic operations. Also I'm not sure I understand your question re: the domain being closed under rigid motions. But if you mean whether a tame set can be made un-tame by a rigid motion, I believe that is not possible. Linear maps are polynomials, so as long as you include the semi algebraic sets in your structure, any rigid motion (a kind of linear map) of a tame set will be tame as well.
      Edit: corrected my answer to the second question

    • @xnor5644
      @xnor5644 2 года назад +1

      ​@@MathTrain1 Thanks for the response. Looking more closely, it seems my issue was missing your line "We additionally stipulate that S includes all singletons of R as well as the addition and multiplication operations", which as you note implies it contains all semi-algebraic sets and is closed under rotation. It looks like I'm blind to text not in italics when looking for a definition.

  • @evanev7
    @evanev7 2 года назад +3

    I'm sure someone has pointed this out but that's not what clopen means!! A clopen set is closed AND open simultaneously, a half-open interval is neither closed nor open (in R with the standard topology, anyway)

    • @MathTrain1
      @MathTrain1  2 года назад +2

      Pinned comment has that correction :)

    • @evanev7
      @evanev7 2 года назад

      @@MathTrain1 Hm, I couldnt actually see the pinned comment anywhere

    • @MathTrain1
      @MathTrain1  2 года назад +1

      Thanks for letting me know! RUclips is not letting me keep it pinned for whatever reason. I'll put it in the description :)

  • @oida10000
    @oida10000 2 года назад +5

    But wouldn't a set theorist also say "eventhough B is a proper subset of A, as the points are dense both sets have the same cardinailty as you can demonstrate the following bijection: ??." Well I am a little bit blank on that but basically both squares have 2^N0 many points right?

    • @MathTrain1
      @MathTrain1  2 года назад +2

      Yes - the squares do have the same cardinality. Just another paradigm for measuring size I wish I'd had time to get into

  • @JeffACornell
    @JeffACornell 2 года назад

    I've thought about this exact question off and on for about a decade. I got as far as solving it generically up to 3 dimensions, and solving it for polytopes in n dimensions, and that the generic solution in n dimensions would involve some complicated sum of curvature integrals, but that's where I ran out of steam. I could never quite wrap my head around higher dimensional curvature well enough to figure out the right integrals.
    For what it's worth, the polytope solution I found was to think in terms of real-valued functions rather than sets, with sets embedded as functions with values of only 0 and 1, and to find the 'natural' function for a given n-dimensional polytope that would give it a pure ω^n measure, with no lower-order terms. Analogous to the observation that the 'natural' 1-dimensional interval is the half-open one. An arbitrary polytope set could then be built as a weighted sum of these natural functions. And the natural function for an n-dimensional polytope simply assigns to each point a value equal to the limiting proportion of an n-ball centered at the point that intersects the set, as the radius of the ball tends to 0 (that is, the angle subtended by the polytope at that point, generalized to n dimensions, and normalized so the maximum possible angle is 1).
    Generic unions of polytopes could then be measured with an appropriate summation involving the various angles on its boundary. With an appropriately defined limit, this could be generalized to sets with smooth curvature using a sequence of approximating polytopes, and the summations of boundary angles would become integrals of curvature. But as I said, that's where I ran out of steam. I worked out the curvature integrals in R^3, but higher dimensions broke my brain trying to figure out the right notions of curvature reached in the limit.
    Assuming I was on the right track, it sounds like these "Intrinsic Volumes" might capture the final consequences of all those curvature integrals. I would definitely appreciate more information on that aspect.

    • @MathTrain1
      @MathTrain1  9 месяцев назад

      I know it's now a decade and an extra year, but the intrinsic volumes are often calculated via curvature integrals and sometimes called curvature measures.

  • @3141minecraft
    @3141minecraft Год назад +4

    11:00 there is a much simpler way to solve the problem:
    The problem is size of the set depends the way you construct the set. The way we solve it is we think about every way to construct this set and pick the one with smallest mesure.
    (In this case, we will assume its constructed with a single quadragonal instead of 2 triangles because that has smallest mesure, so the linear term will be 10.72)
    Are there any problems with that?

    • @geheimemartha
      @geheimemartha Год назад +4

      My guess is that, while it works, it's not always clear what the smallest mesure is. Here it's obvious but what about more complicated sets, or in n dimensions. I think for the simpler cases, his solution ends up being more complicated, but in general you would want it because it is just a couple of formulae.

    • @MathTrain1
      @MathTrain1  9 месяцев назад

      I believe you absolutely could do that. As the other commenter said, that becomes super unclear as soon as your set is more complicated. For instance, what about a circle? Can that not be divided? Because the only way to divide it would be to cut along curves. But the smallest way we construct the set can get smaller if we allow ourselves to cut in places that aren't vertices. But why can't we? But if we only use vertices, the circle cannot be simplified. But then as soon as you cut out corners it drastically changes the measure? That seems discontinuous. What about vertices that are half curved and half straight? Can we split at those? That about higher dimensions? My point is it gets really complicated lol

  • @kasuha
    @kasuha 2 года назад +1

    Very interesting idea. I wonder if the measure would hold if we try to cut the object using some fractal line such as Koch curve. Might be interesting to explore.
    When I saw the beginning, first thing I was thinking about was some kind of Banach-Tarski action on the square. All of this actually assumes all we're allowed is translation or rotation in euclidean space, otherwise there's no dispute there exists bijection between the two.

    • @MathTrain1
      @MathTrain1  2 года назад +1

      I really hope it would, but have no clue. If you'd like to work on it, there's people who are interested in that in the discord linked in the description.

  • @alexismiller2349
    @alexismiller2349 2 года назад +3

    I totally expected you to say that B was larger, in that it's homeomorphic to the entire plain, whereas A can't be homeomorphic to an unbounded subset or R^2

    • @MathTrain1
      @MathTrain1  2 года назад

      That would've been pretty similar to the idea I brushed off at 0:56. It's a great one, but we'd be here all day if I went over every paradigm :) you were just operating on a higher level

  • @Wielorybkek
    @Wielorybkek 2 года назад +4

    I had this thought how would that work with integrating functions? I think usual approach uses the "measure theory" way of thinking, as we don't take boundries into account, right? Would be cool to have some new kind of integral which yields this omega-based polynomial as a result. :D
    Disclaimer: I'm not a mathematician

    • @MathTrain1
      @MathTrain1  2 года назад +1

      Oh boy I wish I had time to get the details right on that! It's definitely doable it's just a lot of trial and error to see what a reasonable definition would be

  • @keremkelleboz6959
    @keremkelleboz6959 2 года назад +3

    Well, a topologist also can say that square B is larger since it is homeomorphic to the whole plane.

    • @MathTrain1
      @MathTrain1  2 года назад +3

      1:00 - you are on a higher math playing field. Since A is compact it's smaller 🤷‍♂️

    • @YitzharVered
      @YitzharVered 2 года назад

      That's what I was thinking lol. What topology does to a mfer...

  • @focoma
    @focoma 2 года назад +4

    Adding objects with different dimensionalities together reminds me a lot of Clifford algebra. Were you influenced by that in any way?

    • @MathTrain1
      @MathTrain1  2 года назад +1

      Short answer - nope! Never heard of it except in passing. Would be cool if they relate though!

    • @focoma
      @focoma 2 года назад +1

      @@MathTrain1 It's also called geometric algebra, and it has something called a geometric product where you can, for example, multiply two different oriented line segments (vectors) and get an oriented area (bivector), or multiply a vector and a bivector to get an oriented volume (trivector). I'm not a mathematician but it's a topic that's fascinated me ever since I saw videos about it by RUclipsrs like Bivector and sudgylacmoe.

    • @angeldude101
      @angeldude101 2 года назад +2

      I actually tried applying this to Geometric Algebra. The main issue I ran into is honestly just the triangle inequality. |v| + |u| ≥ |v+u| Otherwise, it seems to work pretty well.

  • @jaireetchahal7557
    @jaireetchahal7557 2 года назад

    What animation software do you use?

    • @MathTrain1
      @MathTrain1  2 года назад +1

      Manim community edition :)
      I created a couple of custom modules to make the boundaries do weird stuff though

  • @jimktrains0
    @jimktrains0 2 года назад +3

    To nit pick (since it doesn't change the outcome of the section), at 5:31, should the line length be 12 since it's supposed to be the perimeter of the square?

    • @MathTrain1
      @MathTrain1  2 года назад +1

      Huh, you're right! That actually happened because the coordinate system I used in the code of the video actually had the side length as 2, but I changed it so the area and semiperimeter would be different. Good catch!

  • @PhilipSmolen
    @PhilipSmolen 2 года назад +3

    Do the coefficients have to be finite? For example, could this measure make sense of the hotel with an infinite number of rooms?

    • @MathTrain1
      @MathTrain1  2 года назад +2

      In the case of the measure I've shown in the video, the coefficients do have to be finite. However, if you extend using nonstandard analysis they don't have to be. But it's much more difficult to calculate that way.

    • @xXJ4FARGAMERXx
      @xXJ4FARGAMERXx 2 года назад +2

      I'm thinking of the pipe with finite volume but infinite area, which now can finally be compared to a bigger cube, we can say that 10ω³+∞ω² < 15ω³

    • @MathTrain1
      @MathTrain1  2 года назад

      This is discussed precisely (not that exact shape, but the same concept) in my arxiv paper linked in the description! That is indeed how it's done (until the nonstandard analysis part)

  • @johnsalkeld1088
    @johnsalkeld1088 Год назад

    Have you thought about fibrations - surface * circle but where the surface has a twist along the circle? The surfaces i am thinking of are Seifert surfaces of knots

  • @eeee69
    @eeee69 Год назад +3

    is it possible for the degree-zero term to not be an integer?

    • @MathTrain1
      @MathTrain1  9 месяцев назад

      Late reply but no. One of my proofs that it extends to more types of sets requires the fact that the constant term is an integer

  • @timurpryadilin8830
    @timurpryadilin8830 2 года назад +2

    this is a very impressive exploration ! from a technical point i am interested in how would you rigourously define all sets in R2 for which this measure can be calculated (as you mentioned, it doesn't quite work for fractals, but where is the line here ?)

    • @MathTrain1
      @MathTrain1  2 года назад +2

      Great question - that's the hardest part! You can read the paper linked in the description, but the short answer is this works for sets that are "definable in an o-minimal structure"

  • @pepkin88
    @pepkin88 2 года назад +5

    I wonder how it behaves for fractals, like Koch snowflake. The perimeter is infinite, but I guess that infinity would be still smaller than omega, for it not to mix different dimensions with each other. Or maybe the exponent would be different, taking into account the Hausdorff dimension.
    Also I wonder how it behaves in the context of the Banach-Tarski paradox.

    • @MathTrain1
      @MathTrain1  2 года назад +5

      Answer 1: I would be willing to bet that it's omega^hausdorff dimension, but have no way to prove that yet.
      Answer 2: If you extend it to all sets, it behaves very weirdly with the Banach Tarski sets. What happens is that the measure of the subsets changes when you rotate/translate them.

    • @vladyslavkorenyak872
      @vladyslavkorenyak872 2 года назад

      Well, the measure for the Area + half the perimeter would just be A*w^2, where A is the finite area of the snowflake, no?

    • @MathTrain1
      @MathTrain1  2 года назад +2

      @@vladyslavkorenyak872 yes, but then what about the Koch snowflake boundary only? Or the snowflake with its boundary included vs excluded?