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

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/.... 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.

[@MathTrain1]

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.

[@ellismaddox2776]

Redo the audio fraud.

[@MathTrain1]

No u

[@tonyennis1787]

For the slow kids, what's the answer to the original question?

[@andriesvancauwenberge5182]

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]

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

[@MathTrain1]

Tricked you into learning something new lol

[@crazyspider17]

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]

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]

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]

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

[@Grassmpl]

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

[@matron9936]

@@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.

[@oceannuclear]

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!

[@rcteg]

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]

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

[@justinofelipeimbert4694]

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]

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]

Hoping I can make more! Thanks so much

[@harsinsinquin4032]

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.

[@MathTrain1]

Glad to delight :)

[@jaafars.mahdawi6911]

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

[@spiffinn_music_lists]

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.

[@hughobyrne2588]

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]

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

[@hughobyrne2588]

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

[@MathTrain1]

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

[@monsieuralexandergulbu3678]

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

[@MathTrain1]

Thanks :)

[@ikhu6042]

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.

[@ellismaddox2776]

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

[@MathTrain1]

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

[@ellismaddox2776]

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

[@clairecostelloe5741]

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]

@@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.

[@azai.mp4]

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

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]

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

[@ilonachan]

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]

Folks, this commenter has ascended - we are not worthy

[@tonaxysam]

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]

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

[@code_explorations]

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

[@rishikaushik8307]

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

[@MathTrain1]

Love this intuition

[@lexinwonderland5741]

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]

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]

@@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!

[@DynestiGTI]

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]

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.

[@neoztar]

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]

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!

[@eli0damon]

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]

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]

I was going to ask the same question

[@kennethhedden5970]

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]

it's just not a measurable set

[@ribozyme2899]

@@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.

[@zuzaaa1998]

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]

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

[@zuzaaa1998]

@@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]

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

[@laszloliptak611]

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]

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

[@zaringers]

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

[@johnchessant3012]

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]

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

[@MayankKumar-so6sr]

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.

[@calyodelphi124]

This is a very neat and intuitive idea! :)

[@juansamuelarbelaezrocha4113]

This is amazing!!! Well done!!!

[@MathTrain1]

[@perle.1715]

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

[@MichaelFJ1969]

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

[@scentoni]

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.

[@MatthewWroten]

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?

[@cristinaaugegiribet1236]

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]

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.

[@columbus8myhw]

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]

Many in the comments have ascended

[@stanleydodds9]

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).

[@Namgnuoy123]

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!

[@angeldude101]

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]

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]

@@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]

@@angeldude101 what is a directed measure?

[@angeldude101]

@@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.

[@RSLT]

Interesting! Great job

[@MathTrain1]

Thank you!

[@sayantakrkuila2602]

Very nice. Motivated me. Thank you 😊.

[@MathTrain1]

:)

[@cmilkau]

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).

[@Udidda]

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]

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)

[@quadrannilator]

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]

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]

Updated my reply with the discord server

[@tinymints3134]

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.

[@arbeladm]

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

[@deinauge7894]

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 ;-)

[@tyzonemusic]

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]

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.

[@mattschoolfield4776]

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

[@MathTrain1]

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

[@gilbertbernstein7373]

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]

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

[@Tletna]

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!

[@lock_ray]

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.

[@orangeguy5463]

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]

Pulled a bit of a switcheroo on ya

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]

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.

[@lachlanperrier2851]

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

[@lock_ray]

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

[@johneonas6628]

Thank you for the video.

[@MathTrain1]

Thanks for watching!

[@isaacdeutsch2538]

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!

[@mathlitmusic3687]

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.

[@3141minecraft]

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]

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]

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

[@timurpryadilin8830]

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]

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"

[@gabriellasso8808]

Awesome math content!!!

[@MathTrain1]

Muchas gracias!

[@landy4497]

this is amazing dude

[@mileszeweniuk2999]

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]

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]

@@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]

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

[@kasuha]

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]

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.

[@omrireved2]

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]

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!

[@guywholovesmath]

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

[@MathTrain1]

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

[@Joffrerap]

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]

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)

[@davidnoll9581]

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

[@78Mathius]

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?

[@TheKivifreak]

This is super amazing! So the open equilateral triangle would be sqrt(3)/4 w^2 - 3/2 w + 1 !

[@N00BB123]

lmao of course I find you here kevin

[@TheKivifreak]

and the koch snowflake 8/5 w^2 +/- oo w + 1!

[@naringrass]

this is as good as any 3blue1brown. which is to say, this is excelent.

[@MathTrain1]

Now that's the highest compliment I've gotten! Thank you!

[@TheBasikShow]

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]

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

[@columbus8myhw]

@@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.

[@oida10000]

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]

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

[@evanev7]

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]

Pinned comment has that correction :)

[@evanev7]

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

[@MathTrain1]

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

[@writerightmathnation9481]

You misused the terminology. “Clopen” doesn’t mean what you indicated.

[@kj4242]

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

[@alexismiller2349]

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]

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

[@JeffACornell]

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]

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.

[@xnor5644]

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]

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]

​@@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.

[@keremkelleboz6959]

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

[@MathTrain1]

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

[@YitzharVered]

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

[@Tumbolisu]

Right before you showed your first take on μ, I immediately started thinking about the same idea. However, instead of using the letter ω to differentiate between different dimensional parts, I simply thought of an infinitely long vector where the nth element is for the nth dimension. (With the count starting at zero, of course.) So what you call 20ω²+5ω+1, I would call (1, 5, 20, 0, 0, 0, ...). To compare two vectors, you just use a lexicographic ordering. The multiplication of two polynomials is the only non-obvious thing here. Two vectors can be multiplied by calculating the outer product and then summing up elements along the minor diagonals. I wonder if there is an established operation for that? Other than that, this vector representation does not open up any further questions about derivates, integrals, fractal dimensions, etc. I can't really see how those would work or what they would mean anyway, tbh.

[@angeldude101]

Well, polynomials are vectors! Really the only difference beyond how they're usually visualised is that polynomials have specific multiplication rules that don't apply to other types of vectors, though they don't conflict with them being vectors either.

[@columbus8myhw]

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]

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]

@@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]

@@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]

@@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]

@@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

[@MrRyanroberson1]

here's what i thought: mapping an open 1x1 square onto a closed 1x1 square: let f be a solution to the problem of mapping an open line segment onto a closed line segment (aka, producing two new points from 'nothing'). sweep f continuously along the square in one dimension, and then sweep it in the other dimension. problem of the square solved. what is f? let g be a solution to the problem of mapping an open line segment onto a half-open line segment (aka, producing one new point from 'nothing'). f is simply g followed by a flipped g. what is g? for every point with a position of 10^-n, for integers n > 0, move it to the position 10^(1-n). thus an open square has the same number of points as a closed square with no loss.

[@alomirk2812]

what about weird and undefined behavior like the se of all rational point between (0,0) and (1,1) or lines from infinityor you have the line from 0 to 1 and remove the points which are 1/2^n

[@lake5044]

Very interesting questions... I was just thinking, how does this affect the sizes of infinities. The finite number of points is clearly enumerable, while the omega's are not. How do we write w*w as a line with w length, vs w*w as a square with length 1. Same for squares, w*w*w as an infinite plane vs a cube of sides 1. And can we even compare the half plane x>0 with the whole plane? I'd argue that it's impossible since translations shouldn't affect area but translation will inevitably affect any half plane "area".

[@Azyo64]

Good question about the size and proprieties of infinities. The set of all rationals Q in countably infinite, while ω is uncountably infinite (it is equivalent to the real numbers R), so ω > Q. In that sense, the set of all rationals are all points and no 1D line ω. Say C is a countable infinity, then the polynomial for Q would be p(ω) = C. (countably infinite points) Also, ω represent a line segment of size 1, so an infinite line could be measured with a countably infinite number of these: p(ω) = Cω. It is also important to distinguish between a measure of length (here ω) and an amount of something (C or U maybe for a uncountable infinity). There are case where they could be considered equal (or equivalent) but that would not always be the case. You could for example take an infinite line defined by p(ω) = Uω, then warp the line into a square of side length 1 so that every point is covered by the line. But you could just as well warp it into a rectangle of 1 x 2, or even a square. As long as U is not defined in term of the measure ω, then Uω has arguably no other measure the Uω and its dimension is undefined. (ω was created specifically to define the size and dimensionality.)

[@thegozer100]

Very nice video! The fact that points contributed zero to the area was a fact that I slowly accepted but which still irked in my head for years. You made an elegant way to write the point, length, area etc. contributions in one line together. I have one question though: wouldn't it make more sense to define open intervals as having measure w, clopen intervals as having measure w+1 and closed intervals as having measure w+2. This way it is a little more explicit what you are displaying: a closed interval is a line (measure w) + 2 points (measure 2). The area formula would become A*w^2+P*w+? for closed shapes and A*w^2+? for open shapes. Have you thought about this? Would this work?

[@Milan_Openfeint]

Review the part where he's joining several lines together. You'd have to add 1 for each segment. But just looking at a line, you can't really say how many segments there were.

[@thegozer100]

@@Milan_Openfeint I hadn't thought about it in that way, good point!

[@simonflavioibanez7715]

Really good job my friend! One of the fundamental aspects of math research is to implement new mathematical or real applications in science’s toolbox, wich you just did with your mu function (i’m more of a topologist…so measure theory has always been a pain in the ass for me…). Maybe it will be helpful in someone’s research project (or yours). I look forward to see the extension of the measure up to homomorphisms… definitely not an easy task.

[@MathTrain1]

Thank you! I hope there's someone who gets use out of it. I wonder what you mean by extension up to homomorphism?

[@simonflavioibanez7715]

@@MathTrain1 , it’s actually a good place to start a research (i think… i repeat, i’m more the metrics-not-measure type of mathematician). On your video you talked about mu’s uniqueness, and that is (sometimes) generalizable to any set… So, let f:A -> B be a homomorphism from a mu-measurable set A to another measurable (possibly not mu-measurable) set B, and let mu* be a measure on B. Is it always possible to find some mu* that satisfies the same properties that mu satisfies? If it isn’t possible, why not?

[@MathTrain1]

@@simonflavioibanez7715 Man, that would be cool to find out! I do know a bit of the basics to that answer (if you take mu up to just o-minimally definable homomorphism, I believe it preserves measurability) but that's only in a very limited case. Would love to chat about it if you're interested. Check out the discord in description

[@BongoFerno]

A d-volume with dimension D has a perimeter with dimension D-1. For example, a cube (3-volume) has a 2D surface as perimeter, and a circumference (2-volume) has a 1D circumference perimeter. So a fractal like a cantor set with dimension p/q, has to have a perimeter with dimension p/q-1. For example, a cantor set with dimension 1/2 has to have a perimeter with dimension -1/2, and a cantor set with dimension 1/3 has to have a perimeter with dimension -2/3 So, to extend it to fractals is necessary to figure what a negative dimension is.

[@SirTravelMuffin]

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

[@annaclarafenyo8185]

I believe the proof of existence is clearly only in set theory with powerset and axiom of choice. I think it is not true in the Platonic real products. Not so much a criticism, this is original math on YT, and mad respect on that.

[@Wielorybkek]

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]

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

[@PhilipSmolen]

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

[@MathTrain1]

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]

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]

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)

[@SodaliteSabre]

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

[@robertdeaton4720]

my first idea when watching was i guess a subquestion of the "can this be extended to all sets?" question. i was wondering, what if you have an infinite set of disjoint points/segments/any term? say you want to measure the set of integers on the cartesian plane. my first guess at that (to preserve the strict monotonicity even with infinite disjoint sets) is that you'd have to assign a well-ordering to the set of k-dimensional maximal connected subsets (not sure if i'm using the right vocabulary) and then use their order class as the coefficients. but then it's so far removed from the real numbers that it's worth asking if that even can make sense in measure theory XD

[@biblebot3947]

Search up “measurable set”

[@MathTrain1]

The answer to your question is complicated, but the short answer is that it's very hard to get this to work with infinite unions. There's some very elementary counterexamples that's show mu isn't even countably additive, just finitely additive. For instance, consider a countable collection of disjoint half-open intervals whose union forms an open unit interval. The union approaches omega but the open interval has size omega - 1. Though it would depend what you even mean by "approach" when the domain isn't R. That's why technically I'm grossly abusing the term "measure" here - mu is not one because it's only finitely additive and doesn't have codomain R. If you ignore this all and just use nonstandard analysis though, you can show and extension does exist - it's just really hard to do calculations with!

[@focoma]

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

[@MathTrain1]

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

[@focoma]

@@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 KZbinrs like Bivector and sudgylacmoe.

[@angeldude101]

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.

[@BleachWizz]

13:13- oh I think I see the answer; it's one of those where you imagine: "but what of the problem is just not a problem" things. When you're adding shapes you're not adding the boundaries. So depending on what 2 areas you choose to add they'll be missing the line, thus to build the entire shape by parts you're forced to add the missing line term who makes the difference. Well I hope I got it; it was the sart of the videos that lead me to this direction.

[@MathTrain1]

I think you got it. The intuition is that every piece will have a nonzero size, whether it's infinitely thin or not. So when you build the shape you have to account for that boundary. Also wondering what you mean by "but what if the problem is not a problem"? Not sure I understand that part.

[@BleachWizz]

@@MathTrain1 oh, I was wrong; I meant to edit after the video or reply myself but I forgot hahaha. Because when you were at first adding the shapes you were considering lines and points and everything. So I thought the solution would be to add just an area with 0 points and 0 lines as a constant w^2 and it would magically solve itself as if it never happened. But you need to actually subtract a few w's and add 1. So I missed the point. But that would've been my 2nd try.

[@BleachWizz]

To be more specific about the "what if the problem is not a problem". They're those moments where you're stuck at something because there's a problem then you realize you could define the same thing in a different way and the problem you were facing just isn't a thing for the new structure. Happens mostly while programming.

[@ddystopia8091]

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

[@elderpastitsios1569]

I'm waiting for the cantor set, and Sierpinsky triangle

[@MathTrain1]

Working on it!