Proving Brouwer's Fixed Point Theorem | Infinite Series

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  • Опубліковано 17 січ 2025

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  • @pbsinfiniteseries
    @pbsinfiniteseries  7 років тому +73

    Hey everyone! Today’s episode is a classic illustration of algebraic topology, a branch of math that uses tools of abstract algebra to solve problems in topology. The name of the functor (a.k.a. “portal”) that appears in the proof of Brouwer’s Fixed Point Theorem is “the fundamental group.” We’ve put some links in the description where you can learn more about algebraic topology, the fundamental group, category theory, and today’s proof. Enjoy!

    • @pbsinfiniteseries
      @pbsinfiniteseries  7 років тому +5

      Yes, exactly! Although not mentioned in the video, the same proof works in higher dimensions using the higher homotopy groups. We only showed the case for 2 dimensions.

    • @columbus8myhw
      @columbus8myhw 7 років тому +3

      While the higher homotopy groups are relatively easy to define (instead of loops, aka circles, aka 1-dimensional spheres, you use n-dimensional spheres), they're a lot harder to compute. A real surprise is the fact that π_3(S^2) is not zero; you can map a 3-sphere into a 2-sphere (aka a regular sphere) in such a way that it can't be contracted to a point! This is the Hopf map.

    • @columbus8myhw
      @columbus8myhw 7 років тому +2

      Joseph Van Name. I'm OK with that
      For the curious, there's something else called the _homology_ groups that are much harder to define (and motivate) but are much easier to compute. This is why the Brouwer Fixed Point Theorem in higher dimensions is usually proven using homology rather than homotopy.

    • @FantasticExplorers
      @FantasticExplorers 7 років тому

      PBS Infinite Series Umm... Did you guys just explain that if I walk into a room and a molecule of air moves from one side of the room to the other... it's still the same molecule!??!? lol science

  • @ChazyK
    @ChazyK 7 років тому +103

    Disc is zero and circle is integers? Why? The rest of the proof is obvious, but the most important thing is not explained.

    • @RandomBurfness
      @RandomBurfness 7 років тому +29

      It's about the composition of loops. If you have a loop in the disc, you can always continuously retract that into a point. Now, consider any two loops on the disc, say α and β. Since you can continuously retract any loop on the disc to a point, we can continuously retract α and β in two points. Call the point that we have retracted the loop α into γ, and call the point that we have retracted the loop β into ω. What we can now do is we can create a path that goes from γ to ω, and we can then continuously retract that into a single point, which we can without loss of generality assume to be the origin. Since you can do this procedure for /every loop in the disc/, they are the same up to a concept called "homotopy".
      As for the circle, if you have a loop that goes around the circle, say, 4 times clockwise, there is no way you can continuously deform that loop into a loop that goes around the circle n times clockwise if n is different from 4, or m times anti-clockwise for any m. It's also not possible to continuously deform it into a loop that doesn't go around the circle. So if we have a loop that goes around the circle 4 times, we cannot change the fundamental property that it loops around the circle 4 times without considering discontinuous deformations. As such, when we /compose two loops/ defined on the circle, say λ and μ, we first go around the circle as many times and in the same direction as the loop λ tells us to do, and we then go around the circle as many times and in the same direction as the loop μ tells us to do.
      The effective result of the multiplication property of loops λ and μ on the circle is that, if we know how many times and in which direction both loops λ and μ go around the circle, we know how many times and in which direction the loop μλ goes. Furthermore, to find this number of times we go around the circle, and if we write -3 for a loop that goes around the circle 3 times in an anti-clockwise manner, for instance, and we call the number of times we go around the circle for φ(μ) and φ(λ), then we get the remarkable equality that φ(μλ) = φ(μ) + φ(λ). And /THAT IS THE FUNDAMENTAL PROPERTY OF THE INTEGERS/! And that is why, algebraically speaking, a circle is just the integers.
      I hope this wasn't too complicated! If it was, I can clarify some things for you.

    • @ericherde1
      @ericherde1 7 років тому +10

      RandomBurfness Your explanation of topological properties makes perfect sense, but seems to have no connection whatsoever to fixed-points.

    • @user-iu1xg6jv6e
      @user-iu1xg6jv6e 7 років тому

      وفسّر الماء بعد جهد بالماء

    • @NightHawk787878
      @NightHawk787878 7 років тому +2

      its called fundamental group. there is a video on this channel u could watch, maybe that will make things clearer. If there a still questions, just ask =)
      Topology Riddles | Infinite Series watch?v=H8qwqGjOlSE&t=340s

    • @ChazyK
      @ChazyK 7 років тому

      What about the case we go around circle 0 times, it makes no sense to me. (2:43)

  • @telotawa
    @telotawa 7 років тому +159

    Gonna agree with everyone else here, you're gonna have to explain how the "portal" actually works and WHY it works, otherwise this is magical nothingness and is not useful to people who don't already understand why it works

    • @alexsaiz3741
      @alexsaiz3741 4 роки тому +14

      Don't think she can explain the fundamental group and the basis od algebraic topology in a video like this

    • @micayahritchie7158
      @micayahritchie7158 Рік тому

      ​@@alexsaiz3741Lol I don't think she could even in 10 videos like this

    • @Ddr4rtx
      @Ddr4rtx 11 місяців тому

      It's a metaphor

  • @jagoandlitefoot
    @jagoandlitefoot 7 років тому +43

    My 8th-grade math teacher liked to invite his former students back to give guest lessons. When it was my day to teach, I attempted to explain Brouwer's fixed point theorem using Sperner's 1928 proof (and, admittedly, a lot of simplification and hand-waving at the end). I have fond memories of that day, and this theorem by association, so it made me happy to see this video in my subscription feed :D

  • @tuchapoltr
    @tuchapoltr 7 років тому +75

    Personally, I feel like this should had been a small follow-up after a video about the relation between topology and algebra, or something. It may not be self-contained, but I'm sure I speak for more than myself when I claim that we are not satisfied with a huge part of a proof being "hidden away" as a "portal."

    • @rearea784
      @rearea784 7 років тому +1

      tuchapoltr It is still a valid attention paratoner for the main big subject imo. Fixed point theorem was quite unintuitive at first to me. However your approach makes sense too.

    • @semikolonne
      @semikolonne 5 років тому +1

      I agree with you that she could have been way more specific. But nevertheless, those "portals" are what I especially love about mathematics as well. (And that such "fantastic" words grip the essence of mathematical situations in so many cases - I love to speak about maths in that way!) Those portals show that in the end everything is connected even if different subcategories of mathematics often seem so far away from each other.

  • @martinkuffer5643
    @martinkuffer5643 7 років тому +132

    You should explain the analogies you are using before using them, or it gets very hard to understand you. Also, you have not explained how it works.
    I used to love this channel, but I have not seen any of her videos and left with the feeling I really understood her.
    The nice thing about Space Time and Infinite Series is that it simplifies the topic they are explaining without turning it into something else as much as they can but not more. Like the series on Hawkin's radiation that Space Time has just started. She doesn't do that, and has really disappointed me

  • @timh.6872
    @timh.6872 7 років тому +38

    I was excited to see this pop up, but dissapointed in the execution. While the proof presented is valid, the pedagogical value is missing due to one crucial omission. When explaining a proof by contradiction, I find it extremely useful to "unravel" the chain of assumptions and show why the one we're trying to prove false is the only one that can possibly be false. Ex.
    We have shown that a pair of functions from the circle to the (closed) disc and back that commute with identity cannot exist due to the functor between topology and abstract algebra. Because we always have a function from the circle to the (closed) disc by inclusion, the thing that cannot exist is the map back that we called h(x). Since we now know h cannot exist, one of its assumptions must be false. Since assigning a point in the disc to a point on the boundary by drawing a ray through some *other* point always works, the assumption that ∀x, g(x) ≠ x is the only thing that can be incorrect. Thus ∃x, g(x)=x, proving the Fixed Point Theorem.

    • @timh.6872
      @timh.6872 7 років тому +5

      Matt Olkowski It's exactly the same as cleaning up the call stack when evaluating a functional program. Proper proofs have a symmetry to them in which we start somewhere with an idea of where we're going, get there through a chain of modus ponens, then return with our nugget of truth, wrapping it in contex and meaning as we move back to the place from which we came. By skipping the "second half" a lot of people (as can be seen from the other comments) are left with a "wait, why does that prove the theorem?" feeling.

    • @Demki
      @Demki 7 років тому

      Yes, but neither you nor the presenter proved that h(x) is actually a continuous function, you've only shown how to construct if from g(x). This should be part of the proof since the continuity of h(x) is part of the contradiction

    • @letstalkaboutmath2121
      @letstalkaboutmath2121 7 років тому +2

      Actually your extension of the proof in the video, makes it a lot clearer. I understood anyway, but you're right in saying that this should have be done

    • @timh.6872
      @timh.6872 7 років тому +3

      DarkChaos110 Fair enough, but the continuity of h is rather straightforward from the continuity of g, which is assumed. For any given ε of distance from h(x) on the circle, there's very obviously a δ within which we can adjust x to get a g(x) so that the ray from x through g(x) intersects the circle within ε of h(x). The exact calculation depends on the disc we're working in, but it should be evident that h(x)'s construction cannot introduce discontinuity.

    • @TheLastScoot
      @TheLastScoot 7 років тому +1

      I didn't really get this video, but your explanation here made it click for me.
      One thing that was keeping my from understanding the video is that the functor wasn't really explained too well, so my first instinct when it was shown that the functor wouldn't work properly was to say that the functor doesn't necessarily hold true all the time as a model, not that ∀x, g(x) ≠ x could be wrong.

  • @conoroneill8067
    @conoroneill8067 7 років тому +115

    This video seems to be attracting quite a bit of negative feedback, scrolling through the comments. I... like what they're trying to do with it, but I can also see how it can fall down in practice. The videos under the new hosts are trying to squish more complicated ideas into the same time slot, and something, unfortunately, has to give. This explanation of Brouwer's Fixed Point Theorem, for example, skims over several important details, leading to the explanation feeling unsatisfying. Under the old videos, you had simpler concepts, but explained thoroughly, and, I feel, allowed you to take the ideas presented and use the ideas more generally as well - the knowledge being imparted also felt like deeper knowledge rather than the shallow knowledge that is hard to replicate elsewhere that's presented in videos like this one. (Whether or not that's actually the case, though, is likely an entirely different story, but that's what it feels like to a viewer.)
    I think this is a part of the growing pains of the new hosts, and I'm sure this is something that will get better with time - while there are problems, I think some of the comments are perhaps overly dramatic.

    • @david21686
      @david21686 7 років тому +1

      I felt the same exact way when Matthew O'Dowd replaced Gabe in PBS SpaceTime. And when Jay Leno replaced Johnny whats-his-name on the Late Night Show.

  • @michelfug
    @michelfug 7 років тому +37

    As an intuitionist mathematician, Brouwer would turn in his grave if he saw this proof using contradiction.
    Nice introduction to functors though :)

    • @Czeckie
      @Czeckie 7 років тому +4

      is it true, that the original proof was via calculus (for smooth functions) and then there was employed some sort of smooth approximation of continuous maps? Do you know about some modern account on the original proof?

    • @GreRe9
      @GreRe9 7 років тому

      +

    • @semikolonne
      @semikolonne 5 років тому +2

      Did he proove it in another way? I thought he turned away from this theorem and other results afterwards, just because it was not intuitive enough for him :D

    • @willnewman9783
      @willnewman9783 4 роки тому

      @@semikolonne I believe this is correct

    • @Achill101
      @Achill101 3 роки тому +1

      @@semikolonne - I read he turned away from topology and gave only lecture about intuitionism, because the latter interested him much more. I doubt that he had only non-constructive proof of his fixpoint theorem.

  • @jeffreybernath6627
    @jeffreybernath6627 7 років тому +55

    So of course, if a disc is represented by 0, it will be a 'black hole," i.e. any function that maps to it will only map to one thing. But you didn't convince me that the disc should be mapped to 0. If a circle maps to the integers, then I would think the disc should map to some bigger set, like the real numbers. Mapping the disc to 0 is the heart of your proof by contradiction, and I would love to understand that choice better.

    • @mswruns
      @mswruns 7 років тому +12

      The mapping from topological spaces that is referenced in the video is called The Fundamental Group Functor. As the name suggests, it always outputs a mathematical object known as a group. The basic idea is that a group is created from the collection of fundamentally different loops that can be formed in the topological space. Because all of the loops in the disk are contactable to a single point, no two loops are fundamentally distinct. Hence, the group assigned to the disk must have a single element. Since all groups contain an identity element, a one element group must be simply the identity element, namely {0}.

    • @SKyrim190
      @SKyrim190 7 років тому +14

      Jeffrey Bernath exactly! Although I am not demanding a rigorous proof of why the disc can only be mapped to zero it is necessary to have constructed a better appreciation of what we are even doing here! "All loops are points, therefore zero!" doesn't cut it!

    • @pbsinfiniteseries
      @pbsinfiniteseries  7 років тому +13

      Hi Jeffrey, I'm glad you asked. The functor ("portal") assigns a very particular group to each topological space, called the fundamental group. I didn't mention this in the video, but this group is constructed by taking "homotopy classes of continuous functions of the circle into the space." In other words, a group element is an equivalence class of loop in the space (with a fixed base point), where we consider two loops to be equivalent if one can be "wiggled" (i.e. homotoped) to another. Once you unwind all this, it turns out that the fundamental group of the disc is 0, i.e.the trivial group. In other words, there is only *one* equivalence class since (and here's the key) *every loop in a disc can be "wiggled" (i.e. homotoped) to a point.* Therefore the functor assigns the disc to 0. Interestingly enough, the proof that the fundamental group of the circle is the integers is high non-trivial, so I'm glad it made sense intuitively! If you're interested in more details, be sure to check out the links in the description!
      -Tai-Danae

  • @Theraot
    @Theraot 7 років тому +31

    This is what I hear on the video: Let us assume there is a portal and that the circle are the integers and the disk is 0 thefore Brouwer's Fixed Point Theorem is true, and we are not going into any detail to explain it.

    • @zuloo37
      @zuloo37 7 років тому +1

      That's what a lot of categorical proofs feel like, but it is actually true that any functor between these categories with these properties will be sufficient for this proof, regardless of how it's defined.

    • @johnnyrainbolt76
      @johnnyrainbolt76 7 років тому

      Math is math and making up numbers to give value to show proof of a theory isnt math. Its is imagination that isn't solving a problem its creative one.

    • @MattrsMore
      @MattrsMore 7 років тому +1

      That's not what I hear people are hung up on the contradiction and assumption it's false to then prove itself

  • @subh1
    @subh1 7 років тому +50

    Too much technicality with too much hand-waving. Either clearly explain the technical details/terminologies/concepts (which, arguably, is next to impossible in a youtube video for a subject like algebraic topology), or find a way to simplify the explanation in "intuitive" terms. If neither is possible, the best thing to do will be to choose a different topic altogether where such a feat may be possible. Not much thought was put into choosing the topic nor in creating the presentation. Brouwer's fixed point theorem has a host of different proofs, some that even high-school students can understand. Why would you choose the proof that uses homology and group homomorphism, and that too presented using all the technical jargons without clearly explaining them?!

    • @somasundaramsankaranarayan4592
      @somasundaramsankaranarayan4592 7 років тому +1

      its not homology that is being used... its homotopy

    • @MattrsMore
      @MattrsMore 7 років тому +3

      Ya know part of me feels like man this is tough for a YT video. But part of me comes to infinite series to be challenged and learn something difficult. I'd rather they err to the latter than not.

    • @Deguiko
      @Deguiko 7 років тому +1

      I'm fine with the video not beeing self-sufficient. It's like a trailer for a movie, it draws attention to a subject that may lack attention simply because it's hard to completely explain in a youtube video.
      Now I may look more into it on my onw.

  • @deepdata1
    @deepdata1 7 років тому +17

    I am very disappointed with this channel since the hosts changed. I am not familiar with algebraic topology but I have a master's degree in computer science, so I do know some maths. This video did not provide any insights into algebraic topogogy, since there was no mathematical argument to follow. The "proof" was based on the claim that there is a "bridge" between topology and algebra but instead of properly definig it or illustrating why this "bridge" would make sense with examples, they show an example where it does not make sense and state that this is a contradiction.
    While the hosts appear as they know what they are talking about, they lack some very fundamental didactic skills.

    • @AzrgExplorers
      @AzrgExplorers 7 років тому +5

      This comment expresses most clearly why I found the video unsatisfying. We need some indication of why we should believe that mapping the disk to 0 and the boundary to the integers is meaningful in the context of continuous functions, and the best way to do that (for an audience who aren't already topologists) would be to show some examples of how various intuitively obvious properties of the disk and the boundary map naturally to 0 and the integers.

  • @blackflan
    @blackflan 7 років тому +30

    Really dissapointing explanation of a beautiful concept... you're too afraid of giving details where they're really needed and leave other technicalities unclear by trying too hard with empty intuition arguments.

    • @davethesid8960
      @davethesid8960 Рік тому

      I think it wasn't meant to be a rigorous proof, only the intuition. Anyone with the required mathematical background can put it into precise statements, however this video gave away the gist for the non-mathematically inclined.

  • @arthur980807
    @arthur980807 7 років тому +11

    So... why exactly do these scenarios correspond? Why does a circle get "assigned" Z? Why does the disk get "assigned" 0? What does this have to do with loops? Why this assignment in particular, and not something else? What's the motivation behind it, if any? The video left me more confused than I was about BFPT.

    • @kamilkoczurek484
      @kamilkoczurek484 7 років тому +1

      Vsauce video about BFPT provides some really good intuition and insight, you may want to see it. This one was terrible.

    • @pbsinfiniteseries
      @pbsinfiniteseries  7 років тому +2

      Good question. I gave an bird's-eye-view explanation in the video, but you're right to ask for more! Here's the answer in full detail: This functor assigns a very particular group to each topological space, called the fundamental group. This group is constructed by taking "homotopy classes of continuous functions of the circle into the space" - a.k.a. equivalence classes of loops - and the group operation is determined by loop concatenation. The two topological spaces we see in the video - a disc and a circle - each have their own fundamental group. And once you unwind the definition, you discover that the fundamental group of the disc is isomorphic to 0, the trivial group, and the fundamental group of the circle is isomorphic to the integers. But the proof of this is highly non trivial - it involve some very heavy machinery. (If you're interested though, definitely check out the links in the description!)
      The motivation for using the fundamental group is that it allows us to *probe* topological spaces with loops. Loosely speaking, it is a "hole-detector" - it's how we can tell topological spaces apart from each other. Formally: if the fundamental groups of two spaces are different, then the two spaces must different (i.e. they are not homeomorphic). This is a key theorem in algebraic topology.
      -Tai-Danae

  • @Theraot
    @Theraot 7 років тому +10

    See also "Who (else) cares about topology? Stolen necklaces and Borsuk-Ulam" by 3Blue1Brown

  • @philmaggiacomo
    @philmaggiacomo 7 років тому +18

    I am missing how you can call a function that takes any number to a single one (0) an analogue to a function that takes any point and maps it to a discrete point. i.e., two points on the circle map to to different points on the disc, but two integers both map to a single integer.

    • @jafceeadnakjebxownrnsjsydo
      @jafceeadnakjebxownrnsjsydo 7 років тому

      Phil Maggiacomo I tried thinking about it from a set theory perspective, and ended up coming to the conclusion that it makes no sense to say that converting an infinite set to another infinite set them back to the original infinite set is the same thing as converting an infinite set to a finite set them back to an infinite set. But I’m not sure whether that proves the video, the points out a flaw.
      The way that she explains things sometimes makes them more difficult to understand than I feel they should be.

  • @SultanLaxeby
    @SultanLaxeby 7 років тому +16

    To explain it more rigorous: We assume the theorem is false. That way we construct a continous function h from the disk to the circle. Let i be the function from the circle to the disc that maps each point on the boundary to itself. Then h(i(x)) is a function from the circle to the circle that maps each point to itself, h(i(x))=x.
    Now homotopy theory comes into place: Let i* be the function that maps each class of loops [c(t)] in the circle to the class [i(c(t))] of loops in the disc. Since there is only one class of loops in the disc (represented by the number 0), i* maps every class of loops in the circle (represented by a whole number) to 0, so i*([c])=0 for every loop c in the circle. Let h* be the function that maps each class of loops [c(t)] in the disc to the class [h(c(t))] in the circle. Since there is only one class of loops in the disc (0), h* takes only on one value (which is also 0, because the trivial loop gets mapped to the trivial loop). So h*(i*([c(t)]))=0 for every loop c in circle. But that can't be if h(i(x))=x, because it should be [c(t)]=[h(i(c(t)))]=h*(i*([c(t)]))=0 for every loop in the circle, but the classes of loops in the circle are represented by the entire whole numbers, not just 0. Contradiction.

    • @Tehom1
      @Tehom1 7 років тому +1

      Excellent. Thank you, that makes it much clearer.

    • @Drachensslay
      @Drachensslay 7 років тому

      Wonderful explanation, thank you

    • @Achill101
      @Achill101 3 роки тому +1

      I still don't get it completely, but I can finally see what the proof is built on. Thank you.
      I will ponder a bit longer and look for more videos that explain it like you did. I wish they had it done in this video.

  • @AngryArmadillo
    @AngryArmadillo 7 років тому +19

    I am having trouble understanding the video after 5:10. Why are you constructing this arrow between x and g(x)?. I also don't understand the 'boring' function you reference at 5:35. It seems to me that this function leads to the loss of information, not h(x).

    • @johngalmann9579
      @johngalmann9579 7 років тому +13

      Russell Schwartz she draws an arrow between g(x) and x to make a mapping from the disk to the boundary. Note that this is only possible because g(x) never equals x.
      The boring function is simply the function that takes as input points on the circle and places them on the boundary of the disk. I'll call it i(x).
      Now we have a continuous function from the circle to the disk i(x), and we have a continuous function from the disk to the circle.
      With our portal we turn continues functions into group-maps. So i gets turned into some group-map from Z to 0 let's call it i' and h gets turned into a map from 0 to Z let's call it h'.
      We know that i(x) = x for all x on the circle and h(x) = x for all x on the boundary, that means h(i(x)) = x for all x on the circle. This means that h'(i'(a)) = a for all integers. This is impossible and hence the contradiction.
      The real magic of this proof is in the way the portal works so it's a shame she doesn't go more into that, but you can read about it in the references in the description.

    • @alexanderradoslavov1459
      @alexanderradoslavov1459 7 років тому +8

      John Galmann when this comment does better job explaining the idea of the video than the video itself

    • @NightHawk787878
      @NightHawk787878 7 років тому +1

      you do wanna construct this function h, to get an application of the "portal" the disk of its own is just zero but the boundary, the circle, is Z so u get something to work with. Furthermore the "boring" fuction provides the same as the function h, instead of a function from the Disk to the circle, which correspondence to 0 and the integers, u want something to work with, so u first consider the inclusion of the circle to the disk and then use the function h, so u get a function from the circle to itself, which correspondence to the function from Z to Z. Then we get the observation that on topological level the map, first inclusion then h, is just x goes to x, therfore on algebraic level it is also x goes to x. But this is a contradiction because on algebraic level we are going through 0 first.
      Hope this cleared things up for you

    • @huge_letters
      @huge_letters 7 років тому +2

      John Galmann
      But if we are assume the theorem is true we can still construct functions h(x) and i(x), it's just that h(x) will be undefined for a single value of x meaning h(i(x))=x for almost every x while the h'(i'(a))=a still would hold only for a single value of a meaning the "portal" still doesn't work. What did I get wrong?

    • @NightHawk787878
      @NightHawk787878 7 років тому +1

      Женя Пермин
      note that the theorem doesnt say that there is exactly one fixpoint, for example the map x goes to x from the disk to the disk has only fixpoints. For this map u also cant build the function h.

  • @michaelnovak9412
    @michaelnovak9412 7 років тому +29

    Please do more videos on Topology

  • @AtzenGaffi
    @AtzenGaffi 7 років тому +16

    First of all. This isn't a proof for the Brouwers Fixed Point Teorem, since the theorem is formulated for every ball of the R^n. This video just shows the case n=2.
    Second, the correlation between topology an the integers is for a not math student very hard and not well explained.
    Third, we don't identify the (points of the) circle with the integers, we identify the homotopy-classes of a circle with the integers. There is a big difference and I don't see why we can do things like in the video.
    And last, the idea of the proof (identifying circels with integers) cannot be applied on higher dimensions, since the fundemantal group of higher spheres is π=0.

    • @okuno54
      @okuno54 7 років тому +1

      I keep seeing homotopy come up in various places. Perhaps it's time I studied it.

    • @zechordlord
      @zechordlord 7 років тому +1

      Yes, the leap from a circle to integers was introduced way too fast and without adequate justification. Physicist talking here so I'm not mathematically novice either. Still I felt that I could not understand the proof :/

  • @Demki
    @Demki 7 років тому +9

    What I feel is missing is the proof that h(x) is continuous (and thus a morphism in the category of topological spaces, so we can indeed map it using the functor described to a morphism in the category of groups).
    This proof relies on the fact that there can't be a continuous function like h(x) (since that leads the to the contradiction described), however she failed to prove that the constructed h(x) is indeed continuous.
    I can easily think of hand-wavy ways she could have at least explained the continuity of h(x), and how it relies on the continuity of g(x) (with simple animations).

  • @rkpetry
    @rkpetry 7 років тому +3

    ...'oh boy', the cowlick theorem, the combing hair on a sphere theorem, (or in Professors' Spring Fling, the undergrad's question about collapsing a sphere while tracing Brouwer's fixed point to any interior point when there's only one such fixed point it can't return)...

  • @Aquitanius
    @Aquitanius 7 років тому +57

    DIsappointing. Brushing over "the portal" makes everything completely meaningless in this video. It proves nothing. The whole theory of homotopy groups is reduced to magic Nothing insightful here. I'm afraid for the future of this channel.

    • @kamilkoczurek484
      @kamilkoczurek484 7 років тому

      Exactly. The only thing that one can learn here is how proofs by contradiction work, that's not what I came for.

    • @GreRe9
      @GreRe9 7 років тому

      +

    • @elraviv
      @elraviv 7 років тому

      +

    • @zuloo37
      @zuloo37 7 років тому +2

      I think the point is to show the power of category theory. Even without the details of how the fundamental group of a topological space is defined, the fact that it is a functor between the category of topological spaces and the category of groups (and knowing the fundamental groups of the circle and disk) is enough to prove this theorem, and that's supposed to be motivation for using functors in other contexts. Introducing it as a "portal" just makes it easier to digest, and less intimidating, for people who don't already know category theory.

    • @elraviv
      @elraviv 7 років тому +2

      there is no point of showing "the power of category theory", without showing at least one example it working. meaning she didn't show that *there should be* a transformation from A to B, she just showed that *she didn't find* a transformation from A to B.
      basically saying " *believe me* when I failed here, it is not because I've failed, it is because I've proven what should be proven".

  • @eshneto
    @eshneto 7 років тому +108

    Infinite Series is going down the hole. It used to be precise, although simplified. Now, it seems that mathematical rigour is left aside in favour of cute "mathematical portal" nonsense explanations.

    • @ARVash
      @ARVash 7 років тому +6

      Not digressing down rabbit holes that are an entire branch of mathematics in the middle of a discussion is actually not a bad technique. She also named it for you, it's a functor. Did you watch the video?

    • @youtou252
      @youtou252 7 років тому +2

      f u

    • @slboss123
      @slboss123 7 років тому

      eshneto o

    • @ARVash
      @ARVash 7 років тому

      Actually 0 and black holes have a lot of similarities. Pray tell which number does "0" go to? Functor is a meaningless word until it has meaning, metaphors are powerful and can help people link understanding, don't let words hang around your neck like a millstone.

    • @eshneto
      @eshneto 7 років тому +2

      Matt Olkowski I have never said nothing about the "black hole" metaphor.
      The problem is with the "mathematical portal" for functors. The video says nothing about how it works. Indeed it offers links in the description, but the video should still explain the concept, which is, from the beginning, said to be the important tool for the proof.
      Not only that, but also the composition leading to a function from the circle to the circle is also not well explained. Although it is possible to guess what to do, she totally missed a precise explanation which would be simple enough to fit this kind of video.
      The quality is noticeably low with this new host. Unfortunately, not everyone is a good communicator and they should definitely pay attention to the constructive criticism that many of us are doing.

  • @Macieks300
    @Macieks300 7 років тому +5

    I really like the presentation and pacing now, but the only thing that I have a problem is that this channel is still trying to cram too much content like Brouwer's Fixed Point Theorem proof into a less than 10 minute video resulting in skimming over most of the details in the proof.

  • @morgengabe1
    @morgengabe1 Рік тому +1

    Hausdorff Groups are all you need!

  • @dylanrambow2704
    @dylanrambow2704 7 років тому +5

    Future episodes about category theory would be an excellent idea!

    • @pbsinfiniteseries
      @pbsinfiniteseries  7 років тому +1

      We think so too :)

    • @dylanrambow2704
      @dylanrambow2704 7 років тому

      Terrific. What little I've learned about Category Theory has been fascinating, and would love to learn more.

  • @zachsmith9841
    @zachsmith9841 7 років тому

    I can finally help!!! After getting told numerous times during my complex analysis course the difference between a disk and a circle @1:41 you mentioned a disk but displayed a circle. A circle has genus 0 where a disk has genus-1, due to the puncture.

  • @amaarquadri
    @amaarquadri 7 років тому +6

    Way too hand-wavy of a proof. Didn't get a deeper understanding. 3blue1brown has a good video on this topic.

  • @Trakkabase
    @Trakkabase 7 років тому +3

    It's very brave to try to explain homology in an elementary way, but I fear that on this occasion you have not been successful. Some suggestions: 1) prove BFPT in 1D by elementary means using the intermediate value theorem, this is very easy to illustrate pictorial, and will motivate the result in higher dimensions. 2) in the proof I think it's vital to explicitly show why the function h is not possible of three is a fixed point; is not clear, from what you say, how the assumption had been used when constructing h, and why the same can't be done with a map with fixed points. 3) this is the most important. Starting there is a "portal", that the circle is like the integers and the disc like 0, and that topological properties must manifest as arithmetic ones, is a claim entirely void of meaning, explanative power, or insight. It explains to exactly no-one why the proof works or the theorem is true. I think that some significant time needs to be spent justifying the claim of a corresponds between algebra and topology before an explanation of this proof to this audience will be possible. Keep on trying! You make really good videos generally

  • @canteatpi
    @canteatpi 7 років тому

    Whatever happens in the land of topology stays in the land of topology
    -no mathematician ever

  • @bunnyben5607
    @bunnyben5607 Рік тому

    Who knew someone who made such great icecream was also good at math

  • @xjuhox
    @xjuhox 6 років тому +1

    That "proof" makes no sense. One visual proof uses that same retraction method and the idea is that you can't drag points onto boundary circle in a continuous fashion unless you remove a one point from the disk - and that creates a point of discontinuity. So the non-existence of a continuous retraction from the disk onto its boundary equals the Brouwer's fixedpoint theorem. Another equivalent way to state the theorem is to say that you can't contract a circle to an one point inside the circle in a continuous way, i.e. you have to remove one point if you want to do it.

  • @Piffsnow
    @Piffsnow 7 років тому +4

    the fact that, topologically, a disc is like the number 0 and a circle is like the intergers is very unclear to me. Therefore I didn't get the proof... And I'm a math teacher (I admit : I forgot a looooot from college).

    • @satiethetutor3337
      @satiethetutor3337 7 років тому

      Piffrock it has to do with what's called the fundamental group. The fundamental group of a circle is (isomorphic to) the integers. I'm sure there is a satisfactory video on UA-cam of the Fundamental Group.

  • @viktort9490
    @viktort9490 7 років тому +6

    People are all complaining because they like to complain, but you are doing a very amazing work (although very amibitious).

    • @SKyrim190
      @SKyrim190 7 років тому

      Vik Tort People are complaining because they found something hard to understand or "didn't get" a particular step. Don't hide yourself in a bubble of illusion!
      An advice worth for our live as well, because almost invariably we will experience negative feedback in our lives! It's important to know how to deal with it and use it to improve!

    • @viktort9490
      @viktort9490 7 років тому

      People have many reasons to complain. Some constructive, some very impulsed by their passions.
      I found in this video's comments way too many "it was better before" guided with passions and impulsivity more than "I didn't get it".
      I remember a video from the girl who used to present this channel (I don't know her name", which the comments were "oh, I feel a bit lost, you should go slower, explain more.
      When I wrote my comment, there were wayyy to many "was better before", "bring her back", "damn this new presenter sucks" just because it's hard to understand, too fast and/or some lack of experience reasons. And I wanted to give a bit of love in all this hate.
      At the end, people are asking for "slower-explained", "shorter" (unconsciously) and "with more details" videos ; see the contradiction...
      Problem is that in a good video, still tons of people need more details, more explanations, and some other will already know everything. Then people have to use comments to ask questions, get some answers from some other people. Instead, they loose time on complain stuff. It can be fixed only by having these people quiting, the arrival of some other, and the adaption of the rest to the new formula (and also, following what you just said, and I completely agree with you, by some improvement of the content). Meaning : it needs some time.

  • @FernandoVinny
    @FernandoVinny 6 років тому

    3:20 Is that some kinda algebra-topology isomorphism?

  • @jdaerthe
    @jdaerthe 6 років тому

    I know you can't please everyone, but this is exactly the sort of video I am hoping to see from this channel. There is some irreducible complexity present in these ideas after all...

  • @Jackal_Blitz
    @Jackal_Blitz 7 років тому +75

    I mean this with nothing but love and respect: This new host feels like she's just rushing through reading off of cue cards or a textbook. The whole video feels very unnatural and awkward to watch, and it's disappointing.
    I'm sure it's difficult to talk to a camera the way you would talk to a person, but it's part of creating online content. The math content is here, but the delivery needs some more enthusiasm and excitement or it just ends up being boring.
    Again, I love this channel and I want it to do well, so I only offer this as a constructive criticism with the best of intentions.

    • @gooomaaal
      @gooomaaal 7 років тому +2

      set the speed to 0.75

    • @zalikster
      @zalikster 7 років тому +7

      I got a similar feeling, but I think that it's partly due to the fact that this proof is particularly un-enlightening, especially since they don't give any material on category theory (understandable, since it's a whole topic that Infinite Series hasn't touched before).

    • @brianhorne820
      @brianhorne820 7 років тому +5

      Totally disagree. For whatever it's worth, I find Tai-Danae Bradley's pedagogical approach refreshing, enthusiastic, encouraging and enlightening. And I especially love that Infinite Series has become a channel featuring alternating styles and voices!

    • @theflamingsword
      @theflamingsword 7 років тому

      I felt exactly the same way. I get they had to brush past a lot but if u can't fit it all in your time constraints rather do something else then. This felt like a very disappointing video even tho the theorem seemed so interesting.

    • @HotPepperLala
      @HotPepperLala 6 років тому

      Shes black what did you expect?

  • @jacobspear2195
    @jacobspear2195 7 років тому

    This was awesome! I like how you got across the idea that it is important that a functor has to do with the groups/shapes and with functions. Wanting to understand this proof more fully is great motivation to learn some category theory!

  • @lunafoxfire
    @lunafoxfire 7 років тому +2

    I feel like you guys should be making longer videos for topics like this... I ended up with more questions than understanding. I really like that you're tackling tough concepts that go way above other UA-cam pop-math channels, but many of your recent videos have felt like they have quite a bit of hand-waving that hurts more than helps. I mean, PBS Spacetime has been getting away with 15-ish-minute-long videos for quite a while now and they've done an excellent job of explaining some really high-level physics. I don't think you'll have a problem with viewer retention or anything if you move to a slightly longer, more in-depth format.
    That's just my two cents, and don't get me wrong--this is one of my favorite channels on UA-cam.

  • @neilmasson3609
    @neilmasson3609 7 років тому

    I think that Tai-Danae has improved her presentation and the video doesn't feel as rushed as previous ones, but she really needs to explain all the steps in her proof. The correspondence between the circle and the integers is stated, but no clue is offered as to how this works.

  • @yash1152
    @yash1152 Рік тому

    8:00 whats difference in category theory & group theeory?

  • @SpittedDusk
    @SpittedDusk 7 років тому

    Really great episode!! Perhaps the best as of yet.

  • @christophem6373
    @christophem6373 7 років тому

    Why not a video about Theorema Egregium ?

  • @EdwardBliffin
    @EdwardBliffin 7 років тому

    i could listen to her all day

  • @plasmaballin
    @plasmaballin 7 років тому +2

    The proof in this video left out too many steps to actually make sense. There is no explanation for why the groups that loops on a disc or circle correspond to should have anything to do with points on the disc or circle. All that is really proven in the video is that any function that maps a loop on the circle to a loop on the disc and then maps the loop on the disc back to a loop on the circle will have a fixed point (assuming that loops that can be continuously deformed into each other are mapped to the same result), which is obvious. I know that the proofs in these videos are meant to be concise, but they can't be made so breif that the audience doesn't get a satisfactory explanation.

  • @yash1152
    @yash1152 Рік тому

    6:51 I was expecting a bit more ramification of how the presence of that point would affect the sequence upto here.

    • @yash1152
      @yash1152 Рік тому

      presence of that point which maps back to itself

  • @sasha-2574
    @sasha-2574 Рік тому +1

    Does this mean I can't completely mix my coffee no matter how hard I try?

    • @oldcowbb
      @oldcowbb 7 місяців тому

      you can do it by alternating between two mixing methods

  • @helios7170
    @helios7170 7 років тому

    Loved the video, very well presented. Can't understand why so many people aren't happy in the comments. I'm glad there's a channel that doesn't spoon feed us like primary school students. Keep up the good work :D

  • @KurtButler-s5o
    @KurtButler-s5o 7 років тому +2

    This would have worked better if you elaborated how the circle and disk act like groups, and by extension the isomorphism from from D2 to 0 and from S1 to Z.

  • @KaiKunstmann
    @KaiKunstmann 7 років тому

    Pick any ball game with two half-times: There always exist two opposing points on the ball's surface that will have exactly the same position in space at the beginning of both halves -- no matter how much you turn the ball during the first half, as long as you put it on the same spot for the second half. This is true because any two successive rotations of a sphere above any two axes can be combined into a single rotation above a third axis. Thus, any number of successive rotations (i.e. playing with the ball) can be reduced to a single rotation, and any single rotation of a sphere by definition will have two fixed points, namely the two points where the axis of rotation pierces the sphere's surface. In German this theorem is called "Satz vom Fußball" (Football Theorem).

  • @wongwanchap
    @wongwanchap 7 років тому +4

    The biggest mystery or fail of this video is that it never explain clearly why a disc is zero and circle is integer. It make the following part of the video meaningless

    • @patrickwienhoft7987
      @patrickwienhoft7987 7 років тому +3

      Exactly. It feels weird that the full circle is part of the disc, but the analogy to the circle (the integers) is not completely included in the analogy of the disc (zero), which is basically what makes the proof work. Really just raises the question of how/why that analogy works even more.

    • @zuloo37
      @zuloo37 7 років тому +1

      The fundamental group of a disk is zero because any loop can be shrunk continuously to a point without leaving the disk. The fundamental group of a circle is the integers because there are distinct loops which wrap around the circle some integral number of times which can't be continuously transformed into each other without leaving the circle. The video did try to explain this, but algebraic topology is a pretty difficult subject.
      As for the circle being part of the disk but the fundamental group being larger, the fundamental group is essentially a tool for detecting holes in a space, and the disk has no holes, while the circle has a hole made from removing the interior part of the disk, so the circle contains more information about holes than the disk, making its fundamental group larger. At least that's a (hopefully) intuitive explanation.

    • @andrestifyable
      @andrestifyable 7 років тому

      Thanks Seán O'Neil! Your explanation gave me more intuition on what's going on. Can you explain me why those two different (topologic and algebraic) descriptons relate to one another though? I feel like this is what I'm failing to understand in order to get why this proof works

    • @zuloo37
      @zuloo37 7 років тому

      Andre Angelo, there are many relations between algebra and topology, and the whole field of algebraic topology is dedicated to them. The one used here is the fundamental group. Given some topological space, this is an operation which produces a group from that space in a natural way, as the equivalence classes of loops with a fixed base point up to homotopy (which can be thought of as a continuous deformation). The group operation here takes two loops and produces a new loop which goes through one loop followed by the other loop. Going through the loop in reverse gives the inverse element in the group, and staying at the base point gives the identity. Loops that can be continuously shrunk to a point without leaving the space are equivalent to the identity, up to homotopy. The hard part of proving that this forms a group is that it behaves well under the homotopy equivalence (if [a] gives the equivalence class of the loop a, show that [ab] = [a][b], [a]^-1 = [a^-1], etc), and that is something which involves a lot of tedious detail checking, which is probably why the video didn't want to go through it. Proving that the fundamental group of a topological space is a functor requires checking that any continuous function between any two topological spaces induces a group homomorphism between their fundamental groups. This is a pretty general idea in category theory, and the purpose of functors. A functor is a map between two different categories which preserves their structure (morphisms... Morphisms of topological spaces are continuous functions and morphisms of groups are group homomorphisms).
      If you're asking why topology and algebra are connected in this way, that's a harder question which algebraic topology attempts to answer, but it goes much deeper than just the fundamental group.

  • @luis5d6b
    @luis5d6b 7 років тому

    This is great! I love the way that you teach, very clear and deep at the same time :)

  • @kd1s
    @kd1s 7 років тому

    You know in my Catholic high school we covered Alegbra I, Algebra II, Geometry and Calculus. I so wish they'd have taught this back then. You see in all the testing they did (This is prior to the use of tests as penalty, when they were more diagnostic.) - I turn out to have abstract reasoning skills out the wazoo. One of the biggest issues I had with math was the why it works the way it works.

  • @stashdsouza4602
    @stashdsouza4602 3 роки тому

    I have a question!
    In the argument, we create a map J:S¹ --> D where x |--> x in D. Then assuming h is well defined on S¹, we create a contradiction by showing that no map composed with a constant map from Z to 0 can send each element to itself.
    But in this proof, we do not use any fact pertaining to the action of h in the interior of D. So does this not show that the fixed point not only exists, but that it should exist on the boundary of D, i.e., on S¹?

  • @tunyash
    @tunyash 7 років тому +1

    Can anyone explain how to do this proof properly?

  • @grzegorzk5242
    @grzegorzk5242 7 років тому

    I’m currently binge-watching Bartosz Milewski’s lectures on Category Theory (find it out on your YT) and it’s the best piece of maths I’ve ingested in a while.

  • @mswruns
    @mswruns 7 років тому

    Very nice video. The piece that I think I am not understanding: I thought that the integers corresponded to loops on the circle, not to points on the circle? Is there an obvious correspondence between the loops in the circle and the points on the circle that I am missing?

  • @tadessebekeshie7231
    @tadessebekeshie7231 2 роки тому

    category theory, topology and the tutor are all exciting.

  • @petrusboniatus
    @petrusboniatus 7 років тому +1

    I think the video should have a bigger duration in order to understand it without stopping the video several times. I think that explains the bad feedback on this great video.

  • @BillFlann9
    @BillFlann9 7 років тому

    First infinite series I've watched since Kelsey left, Still great stuff. Kudos!!

  • @CorbinSimpson
    @CorbinSimpson 7 років тому

    I like your \cup clasp; it's shiny and draws the eye. I can't help but imagine your "portal" as a functor from Top, the category of topological groups, to Ab, the category of abelian groups. This provides a possible context for generalizing to other categories.

  • @eruyommo
    @eruyommo 7 років тому

    I didn't understand why I could asign a zero to a disc's area and the integers to the border. I wish you only used this episode to show how to make this particular functor and a following video to using it for the theorem.

  • @ComputerNerd98234616
    @ComputerNerd98234616 6 років тому

    Wait how do we draw a projection from g(x) through x onto the boundary in a pure topological space? wouldn't you need some additional structure, such as an inner product in order to do that?

  • @JayVal90
    @JayVal90 7 років тому

    Please do more in-depth analysis of category theory, especially as it relates to programming and compiling / type systems.

  • @waylandsmith8666
    @waylandsmith8666 5 років тому

    A) The statement of the theorem needs to say 'A continuous mapping of a CLOSED disc onto itself has a fixed point'. If it's an open disc, the theorem is not true (drag the center point to the right and all other points move with it proportionally.) B) the theorem applied to any finite-dimensional Euclidean space, not just a 2-dimensional plane, and the statement should reflect that.

  • @kuijpers5870
    @kuijpers5870 4 роки тому

    h is a function of a disk D (0, r) with radius r and center 0 to a circle C (0, r). h(C(0,r))=C(0,r). Because of the continuity of h, each circle C (0, x) with radius x

  • @wg9601
    @wg9601 7 років тому

    Infinite Series is my jam! 100%

  • @shalvagang951
    @shalvagang951 3 роки тому

    OH SO THATS WHY ALGEBRAIC TOPOLOGY WAS CREATED

  • @dmr11235
    @dmr11235 4 роки тому

    Did you just refer to a bridge between geometry and algebra and not mention the nullstellensatz?

  • @Grassmpl
    @Grassmpl 6 років тому

    Why can use 0 vs integers to model circle vs disc? I don't really get it. Why is it not possible to map from x to x to h(x)=x? Sound trivially possible to me

  • @katetranscribes
    @katetranscribes 7 років тому

    How is it possible to have a mapping from the integers to the perimeter of a circle when the perimeter is continuous? Wouldn't you need the reals? (Not that the underlying logic is any different)

  • @theflaggeddragon9472
    @theflaggeddragon9472 7 років тому +33

    Oh wow I'm early, lemme make a joke....
    My understanding of this video :'(

  • @bardokobama1035
    @bardokobama1035 7 років тому

    Those who are complaining about speed they should know .....this is not hold your hand and draw and learn maths ......this is an high profile content if you don't understand this then get lost .....😤😤😈

  • @nikolaichow4663
    @nikolaichow4663 6 років тому

    You guys are doing great things!

  • @rhyswells8725
    @rhyswells8725 7 років тому

    i like the content of these recent videos , very good

  • @Frank-qd5yt
    @Frank-qd5yt 7 років тому +1

    why does 1 go to -3?? h(x) = x, so why is zero(x) not x? Doesn't h conflict with the definition of a disk?

    • @patrickwienhoft7987
      @patrickwienhoft7987 7 років тому +1

      There are two steps. Map from circle to disc, then map from disc to circle. Combining those you mapped circle to circle such that every point got mapped onto itself. Because of the "portal" thing, you should be able to do an analogous thing in algebra. However, the disc corrosponds to 0. So the first part (circle to disc) corrosponds to "all integers mapped onto zero" in algebra. But now when you try to do the second step in algebra, you have to map from zero back to integers, such that all integers end up where they started However, that's not possible because we can't distinguish 1 and -3 anymore once they were both mapped onto 0.
      I think what she meant to say was the following:
      1. Analogous to circle to disc, -3 hast to be mapped to 0 (as 0 is the only thing analogous to the disc)
      2. Because after the second step -3 needs to end up at -3 again to keep up the analogy with topology, we have to map 0 to -3. That way -3 gets mapped onto itself when combining both parts, as we want it to.
      3. Now 1 must be mapped to 0 as well in the first step. No other choice. But we already established that 1 has to map to -3 just now. So 1 must get mapped to -3 as well. But that means 1 doesn't get mapped onto itself, which means the whole analogy breaks down. That means our initial assumption must be false, because we "know" this analogy holds (by the magic portal that unfortunately wasn't explained further...)

  • @porushyadav
    @porushyadav 5 років тому

    Thanks for this wonderful video!!

  • @HouseofObiwan
    @HouseofObiwan 7 років тому

    Two travelers--polar opposite each other in time, set out to change the others history.
    There will be some event at some fixed point in time can never be changed by either of them.

  • @soumyadipsarkar5078
    @soumyadipsarkar5078 5 років тому

    a great intro towards algebraic topology

  • @JuBerryLive
    @JuBerryLive 7 років тому +1

    3blue1brown is soooo much more interesting and comprehensive

  • @ankitagarwal7896
    @ankitagarwal7896 7 років тому

    Can someone explain to me why you can just replace a disc with 0 and the corresponding circle with the integers, and why does it make sense that mapping from the integers to 0 is analogous to mapping a boundary to the disc itself??

  • @pablodemetri4660
    @pablodemetri4660 5 років тому

    Holy shit.. wondered in here on accident. Everyone in the comments is so unbelievably smart to me. I've always believed myself sufficiently intelligent (which is to say I can start fires and feed myself) but I can honestly say that I didn't understand one damn word this lady said. My favorite math trick is to turn 9 toothpicks into 10 by spelling TEN with the 9 toothpicks. That's the level of intelligence I bring to the table. Thanks for reminding me I'm a caveman. Goodluck proving whatever you were talking about. ✌

  • @aneikei
    @aneikei 7 років тому

    I think she speaks and most importantly, delivers the "idea" very well. People absorb knowledge and information at different rates so there will never be one set rate that pleases everyone and in understanding that leads to a natural and kind conclusion - have patience. Just as we all ask and expect others to have with each of us. Also time (as with all things) is relative and all that's required to move between different reference frames is to change your state of mind which isn't always easy. I myself absord new knowledge that I'm grasping for the first time slowly. Not because I'm slow, my IQ is well over 140, but because I take the time to take it in and form the new concepts visually within my mind. To understand them, but most importantly to expand upon them as knowledge is nothing without imagination.

  • @hexagon5610
    @hexagon5610 7 років тому

    I don't understand how you can get h(x). You have to draw a arrow from g(x) through x to the border. But x and g(x) are the same, aren't they? How you can draw a well defined line through just exactly one point?

    • @pbsinfiniteseries
      @pbsinfiniteseries  7 років тому

      Ah! We start the proof by *assuming* that g(x) and x are never ever the same. This is the key step to a proof by contradiction: You start by assuming the fact you want to prove (namely, that g(x) = x) is actually false. Then see where that assumption takes you.
      In this case, because we are assuming g(x) and x are not equal, we can draw an arrow from one to the other.
      -Tai-Danae

  • @wsperat
    @wsperat 7 років тому

    aren't there two possible contradictions? one would be that there are no fixed points, and the other one would be that there is more than one fixed point...

  • @jondreauxlaing
    @jondreauxlaing 7 років тому

    The channel 3blue1brown did a great video on this. They took a different approach, but I think the more important thing is that they took their time. I'm sure it's a lot of work putting these videos together, and I'm sure it's more complicated doing it for PBS as opposed to a personal labor of love, but I think a lot of your viewers would tolerate longer videos that went more in depth if it were doable. I also think your viewers would tolerate a more technical explanation. You've already gone down the path of using the term "fixed point theorem" in the title, so I think anyone afraid of vocabulary would have already jumped ship.
    Anyway, I'm only saying this because I genuinely enjoy the channel, and respect how hard it is to transpose complicated material like this into digestible videos.
    All that said, I'm really looking forward to future videos on the subject. My ears perked up when I heard Category Theory and functors. I've been getting more into functional programming as of late, and while I can use concepts like functors and monads et al in my software, I feel like I lack the formal mathematical understanding, so I'm looking forward to that.

  • @gabrielmello3293
    @gabrielmello3293 7 років тому

    Nice pace, keep it up!

  • @Grassmpl
    @Grassmpl 6 років тому

    Say couldn't we use dimension reduction and ivt?

  • @maxmusterman3371
    @maxmusterman3371 2 роки тому

    Please resurrect this awesome channel 😭

  • @huge_letters
    @huge_letters 7 років тому

    But if we are assume the theorem is true we can still construct functions h(x, g(x)) and g(x), it's just that g(x) will be undefined for a single value of x meaning h(x, g(x)) = x for almost every x while the h'(a, g'(a)) = a still would hold only for a single integer of a meaning the "portal" from topology to algebra still doesn't work.
    What did I get wrong?

  • @chillproductive
    @chillproductive 2 роки тому

    Beautiful explanation!

  • @kenhaley4
    @kenhaley4 7 років тому +2

    I'm sure the proof is valid, or PBS wouldn't be presenting it. But there is definitely something missing or unsatisfying. I think it's because I did not understand the "portal"; i.e. the correspondence between integers and zero and the circles and the disc. We don't need to understand group theory to get this, she said, but something is definitely missing.
    I sort of expected to see how the existence of a fixed point in g(x) would undo the contradiction, but I didn't. Maybe some some examples of how the portal works would've helped.

    • @pbsinfiniteseries
      @pbsinfiniteseries  7 років тому

      Hi Ken, you're right on both accounts! The proof is definitely valid, but there is definitely something unsatisfying. I left out tons of detail! Jeffrey Bernath had a similar comment, so I'll recycle my answer here.
      The functor ("portal") assigns a very particular group to each topological space, called the fundamental group. I didn't mention this in the video, but this group is constructed by taking "homotopy classes of continuous functions of the circle into the space." In other words, a group element is an equivalence class of loop in the space (with a fixed base point), where we consider two loops to be equivalent if one can be "wiggled" (i.e. homotoped) to another. Once you unwind all this, it turns out that the fundamental group of the disc is 0, i.e.the trivial group. In other words, there is only one equivalence class since (and here's the key) every loop in a disc can be "wiggled" (i.e. homotoped) to a point. Therefore the functor assigns the disc to 0. On the other hand, the fundamental group of the circle is the integers. The proof of this is high non-trivial, so I only gave an intuitive reason in the video. But if you're interested in more details, be sure to check out the links in the description! (There you'll also find other examples, such as: the fundamental group of the torus is Z×Z, the fundamental group of the Mobius band is Z/2Z, etc.)
      -Tai-Danae

  • @lucasa.8223
    @lucasa.8223 7 років тому

    I NEED a PBS Infinite Series T-Shirt!

  • @theevilmathematician
    @theevilmathematician 3 роки тому

    Dear PBS Infinite series community, do any of you have experience in mathematics competitions and olympiads?

  • @johnnyrainbolt76
    @johnnyrainbolt76 7 років тому

    I have so many questions i don't know where to begin.

  • @altrag
    @altrag 7 років тому

    Not sure we got the full story there.. the x=>x=>h(x) route would be the equivalent of 0=>0=>0 which is perfectly fine as far as I can tell, since all of those x's were on the boundary (which you assigned to be equivalent to algebraic zero, for reasons that also weren't fully made clear..)
    Need the tie in from g(x)=>x=>h(x) in order to properly demonstrate.. I think.. assuming I'm following along to any degree of accuracy at all..

  • @BlueRock704
    @BlueRock704 7 років тому

    VSauce is probably the reason I'm on this channel today

  • @nikanj
    @nikanj 7 років тому

    These analogies such as gadgets and portals make the video feel a tad condescending. The great thing about PBS Space Time and PBS Infinite Series is that they're not afraid to get a little bit technical. There are already dozens of great channels that are accessible to everyone (SciShow, Vsauce, Veritasium). I like PBS videos because they usually assume you have at least a vague understanding of what what a group or proof by contradiction is etc. or are willing to look it up.
    Still I know that the channel in going through a transitionary period with the new hosts so it'll need some time to finds its feet. I'm confident it will return to its former glory soon.