Is this one connected curve, or two? Bet you can't explain why...

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  • Опубліковано 9 тра 2024
  • Go to brilliant.org/Morphocular to get started learning STEM for free. The first 200 people get 20% off an annual premium subscription.
    One of the most fundamental properties of geometry is connectedness: when a shape is a single continuous entity. But how do you define this idea precisely so that you can apply it even to extremely bizarre shapes in very strange spaces?
    =Chapters=
    0:00 - The Topologist's Sine Curve
    1:54 - The goal of this video
    3:06 - Path-Connectedness
    6:56 - A new definition?
    7:37 - Topology basics
    10:27 - A snag
    12:04 - Connectedness 2.0 and the Topologist's Sine Curve
    15:48 - Who's right?
    17:38 - The Ordered Square
    19:32 - The connection between the definitions
    21:18 - The underappreciated art of crafting definitions
    22:16 - Brilliant ad
    ===============================
    CREDITS
    The songs used in this video are (in order of first appearance) "Dream Escape", "Checkmate", "Rubix Cube", and "Twinkle in the Night".
    The song "Rubix Cube" comes courtesy of Audionautix.com
    ===============================
    Want to support future videos? Become a patron at / morphocular
    Thank you for your support!
    ===============================
    The animations in this video were mostly made with a homemade Python library called "Morpho". If you want to play with it, you can find it here:
    github.com/morpho-matters/mor...

КОМЕНТАРІ • 811

  • @morphocular
    @morphocular  Рік тому +384

    Hi all! I wanted to quickly address a misconception that I've been seeing in a lot of comments. A number of you seem to be confused as to why a path is defined using a function, since that would seemingly exclude paths such as circles which fail the vertical line test. The thing to realize is a "path" here is not defined simply as a function that takes a real number input and then outputs another real number, but it outputs points in the space, which for a 2D space means entire (x,y) pairs. So a path in 2D space is a function that takes real number inputs in the range [0,1] and outputs (x,y) pairs, not just y-values. So for example, a full circular path can be defined as f(t) = (cos t, sin t) which describes a point moving counter-clockwise around a circle of radius 1 (if t ranges from 0 to 2pi). If you think of t as representing time, then this literally describes a point moving around a circle with speed 1.
    I hope that helps!

    • @vinny5004
      @vinny5004 Рік тому +14

      It’s called a parametric function. That’s all.

    • @viliml2763
      @viliml2763 Рік тому +19

      @@vinny5004 Every function has a parameter. You're thinking of parametric equations

    • @AnEnderNon
      @AnEnderNon Рік тому +3

      @@viliml2763 he said parametric though

    • @atlasbailly5439
      @atlasbailly5439 Рік тому +3

      @@viliml2763 actually, its not always possible to find a parameterization of multivariable functions. and indeed, when you parameterize something you define a "function" not an "equation".
      so what you said doesnt abide by the actual definitions of parameterizations, at least the ones i have seen.

    • @Zander10102
      @Zander10102 Рік тому

      Ok so the topologists sine curve doesn't have a limit as it approaches zero, or I guess it is undefined. Which makes sense that therefore the curve is "disconnected" by the first definition. But what if we add that line segment stretching from (0,-1) to (0,1)? Does it then become connected? Let's say you picked the points on the curve (-1,sin-1) and (1,sin1) and let them be t = 0 and 1 respectively. As such when t = 0.5 then f should = (0,0) and there is some range of values of t that will return points on the curve with an x value of zero. Right? What have I missed?

  • @officially_certified_nerd
    @officially_certified_nerd Рік тому +977

    my one semester of analysis has finally paid off in the form of understanding this youtube video and knowing the definition of connectedness

    • @zapazap
      @zapazap Рік тому +24

      Hee! I did my analysis course after studying topology. The assignment grader hated be for solving problems in terms of (eg) 'coverings of open sets'.
      Three cheers for soft analysis!

    • @micayahritchie7158
      @micayahritchie7158 Рік тому +5

      @@zapazap Lol I did both and to be honest I never even though of the analysis definition

    • @zapazap
      @zapazap Рік тому +11

      @@micayahritchie7158 I"m a programmer and I do a lot of code refactoring -- rewriting expressions so that they are more elegant. I was doing this with the epsilon_felta d definition one day. To my surprise, it brought me to the thteshold of the more general topological definition.
      The intuition guiding my steps was not mathematical at all, but (as far as I could tell) solely with making the case expression more elegant.
      It amazed (and amazes) be to think: can concern for (and practice with) code refactoring, over the space of a few hours, replicate a fifty(?) plus year jump From Wierstrass to topology? Might he (who is attributed with the epsilon-felts definition), simply by obsessively 'polishing his code', have been led to considerations of general topology?
      The thought is tantalizing.
      What might it entail for research today?
      Cheers! :)

    • @TheMemesofDestruction
      @TheMemesofDestruction Рік тому

      ^.^

    • @manstuckinabox3679
      @manstuckinabox3679 Рік тому +2

      ALL HAIL STEPHAN ABBOT FOR ADDING A TOPOLOGY CHAPTER!

  • @micheltenvoorde
    @micheltenvoorde Рік тому +436

    Sin(1/x) is the maths equivalent of a black hole. It's just freaking awesome.

    • @alextaunton3099
      @alextaunton3099 Рік тому +25

      Its a good visual for what happens to light near a black hole. Any frequency of light gets redshifted to a wavelength impossible to detect even against the CMB

    • @alextaunton3099
      @alextaunton3099 Рік тому +16

      With the event horizon being the point at which any escaping light becomes too redshifted to possibly detect

    • @lUnderdogl
      @lUnderdogl Рік тому +4

      @@alextaunton3099 what about sin(1/x^2)

    • @user-te3ii8ru1m
      @user-te3ii8ru1m 11 місяців тому +2

      *yes*

    • @philawsonfur
      @philawsonfur 10 місяців тому

      my favorite comment thread of all time

  • @jenbanim
    @jenbanim Рік тому +144

    The structure of this video is fantastic! Every time I thought something was confusing or needed an example, it was immediately followed by further explanation or an example. Anticipating how people are going to follow along is difficult, so I'm very impressed with how well this video manages to do that

  • @DebashishGhoshOfficial
    @DebashishGhoshOfficial Рік тому +43

    Exactly!! We need to know the motivation in arriving at a particular definition in the textbooks. This would make learning more interesting.

    • @randomdude9135
      @randomdude9135 Рік тому +5

      But I guess it's practically impossible. Like here in this beautiful video it takes roughly 22 minutes to explain the motivation behind just 2 definitions. That's cool but we've encountered hundreds, if not thousands of definitions/theorems in various branches of maths so far from school to master degree level. So I think it's obviously why our education system is the way it is. We don't have infinite time to explain every possible concept the way this guy or 3b1b explains!!!! It just isnt practical

    • @Littleprinceleon
      @Littleprinceleon Рік тому +3

      @@randomdude9135 As a biologist I also learned a lot of concepts, but two decades after university I realize that at least half of them could be omitted at the benefit of delving into more depth regarding the more basic ideas. Even at the expense of being less specialized...
      I would guess that in math to be able to connect between different branches is even more important either for practical or for theoretical creativity.
      Maybe some specializations teaching and exploring the historical development of individual sciences strictly focusing on the cognitive processes necessary for a fuller understanding of the concepts

  • @daniloelias9887
    @daniloelias9887 Рік тому +51

    I am finishing my masters in physics and I must say this is the clearest explanation on this subject I've ever seen. Keep up!

  • @MasterHigure
    @MasterHigure Рік тому +155

    The topologist's sine curve is a very cool topological space, and it is a very natural place to get into the nitty-gritty of what connectedness means in all its forms. I think you have done that very well here.
    However, one point of criticism: I think it's wrong to entirely fail to mention around 5:30 the possibility that there may be *more* than one value t that gives f(t) = (0, y). It doesn't change the result, but it does slightly complicate the argument (you need to choose your t* carefully), so I can understand glossing over it. But not mentioning it at all is far beyond just glossing over it, and I personally think that falls into the territory of "incorrect". You have a note for the pedantic when talking about neighbourhoods at 8:25, that could've been a good fit here too.

    • @morphocular
      @morphocular  Рік тому +35

      I've certainly been known to err, but from what I can understand, I'm not sure I'd glossed over all that much? In the case of multiple possible t* values, I think the basic argument I used works if we take t* to be the minimum such t* (and I think you can justify via continuity of f that such a minimum exists). Then the argument proceeds as normal (though you might have to swap out the two-sided limits for left-handed limits). Is that the subtlety you were referring to?

    • @MasterHigure
      @MasterHigure Рік тому +21

      @@morphocular Yeah, more or less. Continuity says the set of ts where f(t) has zero x-coordinate must be closed, the intermediate value theorem says it must be non-empty, and Heine-Borel (or something) tells you it is compact and therefore contains its minimum. If you pick t* to just be an arbitrary such t, rather than the minimum, then the limiting argument tells you nothing.

    • @morphocular
      @morphocular  Рік тому +37

      @@MasterHigure Gotcha. And thanks for the feedback! Figuring out where to draw the line at how much detail to skip is a never-ending battle for me, so I appreciate your input. To be honest, I hadn't considered that multiple t* values could be much of an issue until I realized (just today!) that there could be a whole interval of them!

    • @MasterHigure
      @MasterHigure Рік тому +16

      @@morphocular There is no objective place to draw the line. Heck, I don't even know that it make sense to speak about *consistency* with where you draw the line, much less correctness. I have told you one place where I disagree with your line drawing, and one where I agree, but that doesn't really mean I'm right and you're wrong.

    • @morphocular
      @morphocular  Рік тому +13

      ​@@MasterHigure True, but I can always use more data.

  • @SuperCornstock
    @SuperCornstock Рік тому +7

    eyeopening and excellently narrated. The visuals and transitions were just perfect as well. Perfect video on the subject, thank you for the hard work it took to make this!

  • @zapazap
    @zapazap Рік тому +37

    Next topic: compactness. You can present different notions of compactness like you presented different notions of connectedness.

    • @dekippiesip
      @dekippiesip Рік тому +5

      I just love the ingenuity in that definition! A set is compact if every open cover has a finite subcover. That is actually such a cool definition of a concept that is intuitively so easy.

    • @jetison333
      @jetison333 8 місяців тому +1

      Nice prediction!

  • @josvanderspek1403
    @josvanderspek1403 Рік тому +3

    Absolutely fantastic video. Even though I encountered this example 5 years ago, I can honestly say this is the first time I actually truly capture the subtle difference between the two definitions.
    Also, kudos on the definition-motivation plea at the end. Great work!

  • @dazperson8228
    @dazperson8228 Рік тому +1

    I love how in the last moments of the video he showed a graphic of the epsilon delta definition of continuity that instantly made me understand it despite eluding me for years at this point. Excellent video if anyone's reading this.

  • @brucea9871
    @brucea9871 Рік тому +16

    Interesting video. And I can certainly appreciate the point you made at the end about the importance of definitions and the difficulties mathematicians sometimes have formulating good definitions. I recall from my studies of calculus that it took a while for mathematicians to derive the definitions of limit and derivative that we take for granted today. Before our current definition of limit was proposed nobody could define limits in terms that did not verge on the mystic.

    • @Littleprinceleon
      @Littleprinceleon Рік тому +1

      There's an excellent YT video called something like: the lore of calculus

  • @cloudshaveiteasy
    @cloudshaveiteasy Рік тому +1

    Such a good watch! Very interesting topic and a well made video. Cheers!

  • @blacklistnr1
    @blacklistnr1 Рік тому +7

    Great video! I especially like the bit at the end when you compare semantics.
    I think math's greatest problem is that definitions usually throw away all context and motivation which makes them happen in the first place.
    It's like trying to describe an algorithm in assembly/binary, yeah it's useful for execution, but beyond terrible for understanding the ideas at play.

  • @raynaldilalang2003
    @raynaldilalang2003 Рік тому +3

    this was such a beautiful video i HAD to leave a comment, like, and subscribe.
    what i've found really beautiful in mathematics, is when we start coming up with definitions to seemingly answer an extreme case of an intuitive problem, the solution is just "whatever you so desire". it took me sometime to appreciate this beauty, because at first it feels like a slap in the face, like when you're in an adventure story to find the truth of the world, meets an all-knowing-being find what is the meaning of life and they ask back at you "what is your meaning of life?". and this video really encapsulates that feeling.
    also that 1st textbook definition of "connected" really blew me away. it's almost like they used the aforementioned (two open disks touching at their boundary, a.k.a. a union of two non-empty disjoint open sets) to come up with a definition. fascinating.

    • @vinny5004
      @vinny5004 Рік тому

      Yes. Much of mathematics is about the language to represent concepts. There are definitions, and not knowing them makes it impossible to solve a problem that requires knowing it. But does not make you any more or less smart or intuitive. However, once one finds out and understands a definition, there are many important and useful things that result (e.g., theorems that can be applied to large classes of problems), and then “correct” and “incorrect” answers arise from the definitions (so that our language remains consistent). This last point is sometimes lost on students, who get angry and argue against definitions (they are arbitrary, but must be accepted, like we accept the meaning of words in spoken language… but with much more rigor in mathematics!). This is futile, “by definition,” so don’t do it, and instead focus on understanding the deep meaning of the definitions and their applications. Great video.

    • @Littleprinceleon
      @Littleprinceleon Рік тому

      Even more so... I think that a sufficiently intelligent being less occupied with limitations of its individual existence would ask back:
      What do you mean by "meaning" and why is it important for you?

  • @williammanning5066
    @williammanning5066 Рік тому +17

    After watching the intro, here's my guess: to connect the two "ends" of the curve around 0, those individual ends must have well-defined limits. Even though the overall curve does not have a well-defined limit at 0, you could imagine each individual end having one. However, for this curve, each side of the curve near 0 is just as badly-behaved as 0 itself.

    • @FishSticker
      @FishSticker Місяць тому

      Pretty good summary of the path portion

  • @maxpetschack3342
    @maxpetschack3342 Рік тому +1

    Loved this video! I managed to guess the path definition but the subset definition was really cool.

  • @omerbar7518
    @omerbar7518 Рік тому +3

    Brilliant concept, wonderful video. Great job.

  • @alexbourlis
    @alexbourlis 9 місяців тому

    One of the best videos on connectedness i've seen so far, and i've seen a lot

  • @jdcdelossantos
    @jdcdelossantos Рік тому

    I subscribed immediately 'cause of this awesome vid! As a physics student, it's been difficult for me to have an intuitive grasp of topological properties like this. I hope you do a video about COMPACTNESS!!!

  • @HazemA1
    @HazemA1 Рік тому +1

    These are extremely excellent videos. Hope you do way more especially about the weirdest math topics! Thank you

  • @meccamiles7816
    @meccamiles7816 2 місяці тому

    This is a lovely video. It took my right back to my first topology course. Thanks for sharing.

  • @sambhavgupta4653
    @sambhavgupta4653 Рік тому +2

    Awesome explanation! keep going!! looking forward for more such amazing Math Content

  • @Boringpenguin
    @Boringpenguin Рік тому +10

    21:12 A follow up video on this equivalence would be awesome.

    • @viliml2763
      @viliml2763 Рік тому +1

      For #1, see that if you view the set (say, the two (almost) touching open/closed disks) not as a subset of a larger space but as a space itself, it has no boundary (there are no points outside of it as it is everything), and furthermore every connected component has no boundary and so it is an open set. If it has multiple components then it is the union of disjoint open sets.
      For #2, see that being both open and closed at the same time means having no boundary (contain all boundary while containing none of the boundary = boundary is nothing)

  • @zapazap
    @zapazap Рік тому +2

    So nice to see a video on point set topology. The _best_ topology. :)

  • @lowerbound4803
    @lowerbound4803 2 місяці тому

    Very well-made!!! Thank your for your beautiful piece of explanation!! 😻😻

  • @dyld921
    @dyld921 Рік тому +6

    I love point-set topology for how completely insane and seemingly contradictory it is despite being entirely consistent. You didn't even get into locally connected and how it's separated from connected.
    Also, I don't think the textbook definition of connected is that hard to explain, since a closed set is defined to be the complement of an open set. So if a non-trivial set in the space is both open and closed, then it's a non-empty open set, whose complement is also non-empty and open.

  • @PhysicsRaja-ul3kc
    @PhysicsRaja-ul3kc 10 місяців тому

    Fantastic !! as usual for your standard !!

  • @theultimatereductionist7592
    @theultimatereductionist7592 Рік тому +2

    15:57 My mind went immediately to the Touching Subsets definition of connectedness and concluded the Topologists' Sine Curve is connected, before you reminded me about the path-connected definition.

  • @skyking0356
    @skyking0356 Рік тому +1

    I have been studying civil engineering since last October and am struggling to grasp the concept of continuous functions. Thanks to your videos, some things have started to click and the gears are starting to move.
    That may be as well because I have been studying for the upcoming math exams for the entirety of last week, but hey, i’ll take what I can get.

  • @abekolko7143
    @abekolko7143 Рік тому +1

    My attempt at coming up with a definition for "connectedness," as you mention at 2:30. I wanted to see how I'd do before I see the answer.
    Define a Connector set as being constructed by the following process. Start with a well ordered set of m points with m>1. Call this the seed set A = {a_i| i from 1 to m inclusive}. The connector set is the union of all of the line segments connecting a_i to a_i+1 for i ranging from 1 to m-1.
    Now suppose we want to assess if a region R is connected. The set R is connected iff for all r_1 and r_2 in R, there exists a connector set A containing r_1 and r_2 st A is contained within R.
    Edit: Ah I now see this definition only works if R is of at least 2 dimensions everywhere. Tricky!

  • @peksn
    @peksn Рік тому

    This was actually a really good video wow, never thought I'd waych it for 20minutes, thank you s lot ^^

  • @happy_ice_cream
    @happy_ice_cream Рік тому +1

    I'm taking topology this semester, so seeing this video made me very happy :)

  • @shpensive
    @shpensive Рік тому

    thanks for making the end of this video, you didn't have to include that boundary, but you did ❤

  • @algorithmizer
    @algorithmizer Рік тому +6

    "Just as every theorem needs a proof, every definition needs a motivation"
    It feels like I'm the only one in my math class thinking this, including the teacher ;)

  • @gandodiallo4146
    @gandodiallo4146 10 місяців тому

    I totally agree with you, each definition should be motivated before diving into the proof

  • @cameronmartin6969
    @cameronmartin6969 Рік тому +1

    You are one of the only math creators I can understand without getting bored

  • @benjaminlee9964
    @benjaminlee9964 Рік тому

    fascinating video, when i first saw the proposed question i immediately thought about continuity and came to the conclusion that the curve is not connected, as the limit of the function at x=0 is undefined therefore the curve is not continuous at x=0. but the rest of the video really intrigued me as it gave me a new perspective on the idea of "connected"

  • @FareSkwareGamesFSG
    @FareSkwareGamesFSG Рік тому +31

    Another question: instead of the topologist's sine curve, if we modify it such that we only add the point (0, 0) instead of a whole strip from (0, -1) to (0, 1), would the resulting set still be connected?

    • @koenschouten7994
      @koenschouten7994 Рік тому +18

      Yes, this is actually also how the topologists sine curve is usually defined.

    • @ordinaryshiba
      @ordinaryshiba Рік тому +4

      It still has the same properties as the version in the video (which is the closed topologist sine curve) It's disconnected in some definition but connected in others

    • @alonamaloh
      @alonamaloh Рік тому +2

      Yes, it would be. I think the proof is essentially the same.

    • @benjaminpedersen9548
      @benjaminpedersen9548 Рік тому +5

      It does not even have to be (0, 0). You can add any non-empty subset of the line segment to make it connected, as you just need to guarantee one point of the common boundary in either parts of a split.

    • @LadislausKallig
      @LadislausKallig Рік тому +2

      No. It won't. There is still no continuous path that can connect them.
      Edit: nvm

  • @timetravellingblockhead2122

    What a surprisingly cleanly-explained video

  • @Kumurajiva
    @Kumurajiva Рік тому

    This is just fascinating and mysterious, and maddening to me, so entertaining, beats all tv programs, tbh

  • @joshuamark1601
    @joshuamark1601 Рік тому +3

    I'd say they don't agree because the topological definition doesn't include the requirement of a limit, only that the sets share a boundary which is in the whole set, which I interpret to mean the function has (at least one) y value for each x value.
    Even in the 'path definition', if you remove the requirement that the limit exists, you can define a (discontinuous) function that works (but again that is not the definition of "path")

  • @neloyghosh9010
    @neloyghosh9010 Рік тому

    Excellent explanation loved it ❤️

  • @p2beauchene
    @p2beauchene 8 місяців тому

    Very nice and clear as usual.
    Just to nitpick, the sentence _"the sets contain at least part of their common boundary"_ *implies* that two sets touch, but it's not a *definition* since two sets can touch without having a common boundary, e.g. the sets of points at a distance of at most 1 and 2 respectively from the origin.

  • @vinca43
    @vinca43 Рік тому

    The topologist' sine curve is defined as the closure of the set (x,sin(1/x)), 0

  • @takizeghida4303
    @takizeghida4303 Рік тому

    It will be amazing it you do a similar video about compactness

  • @marshallsweatherhiking1820
    @marshallsweatherhiking1820 Рік тому

    As a hint for getting to the third definition I came up with an intermediate definition that is still somewhat intuitive. Instead of defining a boundary point you simply define a limit point of a set A as a point x ( either inside or outside A) in which every neighborhood of x contains some member of A. Then the closure of A is just A plus all of A's limit points (which, by definition, will include all the boundary points of A as well as interior points). Then the equivalent definition of "touching" between two sets A and B will be that either 1.) A contains a point in the closure of B, or 2.) B contains a point in the closure of A. The proof that this definition of "touching" is equivalent to the boundary-based definition shouldn't be too hard. This is the only missing link I have yet to think through.
    Then the next second step is to refine the definition of touching for the case where we just assume nothing exists outside of A and B. This is a little cleaner because limit points outside of A U B no longer exist. They are now outside of the domain, i.e. outside the universe we are considering. They become topologically analogous to infinite limits (kind of an aside, but interesting to think about). In any case, if we throw away all the limit points external to A U B, then the definition of "touching" reduces to either A or B having ANY external limit point ( since if a point is not in A, it must be in B, and vice verse ).
    At this point the only thing left to do is show that "both open and closed" is equivalent to "contains all of its limit points" within the new restricted universe where nothing exists outside A U B. Once you do that you have the third definition. I know I'm close because "contains all of its limit points" is a pretty popular alternate definition of closed in real analysis texts.
    This would be so much easier to show using illustrations but I don't have the talent of the author of this video unfortunately. :(

  • @benjaminpedersen9548
    @benjaminpedersen9548 Рік тому +1

    Just a truly great video!

  • @UnrivaledLimit0500
    @UnrivaledLimit0500 Рік тому +3

    Very nice video. I hope you'll soon get a lot more subscribers.

  • @agranero6
    @agranero6 Рік тому +2

    This sine function is used in Physics in what is called "asymptotic methods" usually in integrals and usually disguised as a complex exponential.

  • @user-bj5xr
    @user-bj5xr 3 місяці тому

    Topologist's sine curve is one of my favourite so I watch this! If I don't misunderstand, the definition of the curve T you used is T:=the union of left part L, right part R and Y={(0,y): -1

  • @marshallsweatherhiking1820
    @marshallsweatherhiking1820 Рік тому

    It's a slightly more advanced topic that I never fully got through, but another really interesting topic involving definitional motivation is the concept of dimensionality. Hausdorff dimension vs. topological dimension would have a discourse roughly analogous to path-wise connectedness vs topological connectedness. Hausdorff dimension is more complicated, but also more intuitive IMO. In the case of Hausdorff dimension, instead of the topologists sine curve, pathological cases where intuition fails are fractals! You'd think that the boundary of a 2 dimension object must be a 1 dimensional object, but the boundary of the Mandelbrot Set has Hausdorff dimension 2. In other cases you can get a fractional Hausdorff dimension ( hence the name "fractal" ).

  • @nok9355
    @nok9355 Рік тому +1

    This was a really well made video

  • @nikolaudio
    @nikolaudio Рік тому

    This channel needs a few million subs, So good

  • @NeTfLeXr6s
    @NeTfLeXr6s 8 місяців тому

    Really nice introduction to these concepts! However, an important thing I always try to teach my students is that topological notions are always relative. An Intervall in R may be open, however the same interrval in R^2 will not be w.r.t. the standard topology on R^2. Usually the topologist's sine curve is not just considered as a subset of R^2 with its standard topology, but rather as a topological space in it's own right, by giving it the subspace topology. In this case your argument, that all points of the curve are boundary points, is incorrect. That's also why connectedness of a subset of a topological space is defined with open subsets of the subspace topology, not with open sets of the bigger space.

  • @aieousavren
    @aieousavren Рік тому

    Incredible video! ❤

  • @benjaminlum5894
    @benjaminlum5894 Рік тому

    16:12 Holy shit! That philosophy hit me deeper than it should

  • @MaxPicAxe
    @MaxPicAxe Рік тому

    Very well-made video!!

  • @EPMTUNES
    @EPMTUNES Рік тому

    Fascinating!

  • @Sokobansolver
    @Sokobansolver Рік тому

    I played around with Desmos and input other trig functions of 1/x, with similarly brain breaking results. I also input tan(x!) and part of the right side of the graph looks like a solid colored rectangle and inputting csc(x!) and sec(x!) look like a similarly solid rectangle with a square cut out of it.

  • @dorol6375
    @dorol6375 4 місяці тому

    Before seeing the definition, my definition of connected for a graph is that for any point there's another point arbitrarily close to it from both sides of it and my definition for connected in topology is that for any two points there's at least one conneted (this meaning of connected would be the definition from the graph) path that starts at the first point and ends at the second and doesn't have any points not in the set inside of it

  • @rashiro7262
    @rashiro7262 Рік тому +1

    It's simple. We have a graph defined as f(x)=sin(1/x); at x=0 it has no value because decision by 0 is not possible. Therefore the curve is disconnected at a single vertex, exactly at x=0.

  • @AJ-et3vf
    @AJ-et3vf 7 місяців тому

    Awesome video. Thank you

  • @YusufTANA
    @YusufTANA 8 місяців тому

    Hi, the graph (sin(1/x)) is connected at zero because it is a normal graph just like (sin x). They are both normal graphs, I used Desmos to graph (sin(1/x). If we graph (sin x) and (sin(50x)) then we get 50 oscillations per unit of length than with the 1 oscillation with (sin x) .And we can continue up to (sin (infinity*x) which will get us an infinite number of oscillations per unit of length which is the same thing as (sin(1/x)). Thanks for your educational videos.

  • @retrorocketman6969
    @retrorocketman6969 Рік тому

    Oh yeah that is good Stuff man. thanks, i hope there is more. Keep up the good work!!thanks

  • @Rocherz
    @Rocherz 7 місяців тому

    21:18 “But in any case, one of the things I hope you take away from this video is an appreciation of the ingenuity that goes into crafting a good mathematical definition. So often in Math, the focus is on the big theorems, proofs, and calculations, but I feel we don’t often pause to appreciate all the creativity and cleverness that go into some of the definitions. Maybe it’s partly because textbooks and lectures often just state definitions unmotivated before quickly moving on to the theorems and proofs. *But just as every theorem needs a proof, I think every definition needs a motivation.* And it’s an important mathematical skill to cultivate too! One that I worry too often goed unexercised. So the next time you encounter a strange mathematical definition out in the wild, take a minute to flex that mathematical muscle and try to understand what motivated it, and also appreciate the work that went into crafting it.”

  • @geraldsnodd
    @geraldsnodd Рік тому

    Your animations look good, helpful too.🙂

  • @asaferguson1141
    @asaferguson1141 Рік тому +1

    Wonderful video!

  • @c4ashley
    @c4ashley Рік тому +1

    Before I watch the rest of the video, you asked for my thoughts. I think the graph... is both connected and disconnected. On one hand, in the same way that there are infinitely many values between 0 and 1, now we find that no matter how far we zoom in on the y-axis, we can always zoom in further and insert a new point on either side that will have a different y-value. There is no meaningful "final" value of y for the "final" position x immediately below and above 0, so there is no well defined "attachment point" to which the y-axis can be "connected" with the curves on either side, hence disconnected. But, this oscillation continues infinitely, so there is no break or discontinuity between the y-axis and the curves, hence connected. Can limits help with this, maybe? As they say on "Hollywoo Stars and Celebrities: What do they know, do they know stuff? Let's find out," let's find out!

  • @schweinmachtbree1013
    @schweinmachtbree1013 Рік тому

    Excellent video!

  • @slowpnir
    @slowpnir Рік тому +2

    There's always the "room" for rationals (b/c compact rows), and more that infinitely much space for real numbers.
    The boundaries got duplicated at 10:04. Before that, they we correctly shown belonging to only one set on cutting.

    • @Littleprinceleon
      @Littleprinceleon Рік тому

      To which subset will the boundary belong to?

    • @slowpnir
      @slowpnir Рік тому

      @@Littleprinceleon if you cut existing "continuous" space, then only to one of them, I guess.

  • @KhoaNguyen96
    @KhoaNguyen96 Рік тому

    Thanks for the video.
    I just want to know that what is the name of the song in 3:48. Thank you so much

  • @kotted
    @kotted Рік тому +1

    I like to think of functions in terms of sets, so for me "connectivness" is the ability of sets to have common members (aka intersect)

  • @KyzenEX
    @KyzenEX Рік тому +3

    The definition I came up for "Connectedness" was:
    Any two shapes of bodies that share, at any point in their area/volume, the same spatial coordinate/s, as infinitesimally small said coordinate is, as long as both shapes/bodies share this coordinate, they're connected. The moment any two shapes/bodies stop sharing their coordinates, they're disconnected.
    I know it's kinda weird talking about connectedness with plural examples, but it is that way so it can be as general as it can be whilst also being as precise as I could make it.

    • @javierflores09
      @javierflores09 Рік тому +2

      This assumes the shape has been split in the first place, hence why it sounds weird

    • @Littleprinceleon
      @Littleprinceleon Рік тому

      @@javierflores09 connectedness also assumes that a state of disconnectedness exists...
      For me, as a layperson it's more weird to talk about two or more evidently separate objects as if they were one. Just as weird as quantum entanglement... Nonlocal connection

  • @gideonk123
    @gideonk123 Рік тому +4

    Very interesting video. But please kindly clarify regarding the Ordered Square (17:41)
    1. Why a space-filling curve, such as the (infinite iteration of the) Hilbert curve cannot be considered an ordered square?
    2. If it is indeed possible to use the Hilbert curve as a 1D path through the ordered square, then would this also enable us to use the path idea to define connectedness?
    Thanks in advance to anyone explaining these questions.

    • @decare696
      @decare696 Рік тому

      I'm not sure, but I think a space filling curve _could_ be used, but in topology, there's always a definition hiding somewhere of what topological space you're dealing with, and that includes a so-called topology. The ordered square is using the order topology for the given ordering, while using a space filling curve would give you a different topology (if even possible, idk), so yok'd be dealing with a different topological space, just like how the ordered square is different from the regular square topologically, even though they are the same when only looking at what points there are

    • @stevenfallinge7149
      @stevenfallinge7149 Рік тому

      It has to be continuous in the order topology, not the usual Euclidean topology. The order topology is the topology where open intervals or any union of open intervals are considered open sets.

    • @gideonk123
      @gideonk123 Рік тому

      @@decare696 Thanks

    • @gideonk123
      @gideonk123 Рік тому

      @@stevenfallinge7149 Thanks

    • @zapazap
      @zapazap Рік тому

      The Hilbert space filling curve preserves 'closeness' of points with respect to the standard square, but not to the ordered square.

  • @ilikedogs6425
    @ilikedogs6425 9 місяців тому

    I would say according to physics they are connected in a state that is similar to a particle in superposition, it's specific values is not 1 or zero like normal particles, but in fact both (and everything in between at the same time). However this property doesn't make the particle different, and it's not defined as one or another, just both. Assuming math can be stated to in some way describe the physical world, then therefore this wave could represent a particle in said superposition.

  • @tubebrocoli
    @tubebrocoli Рік тому

    iirr the issue with path continuity for the topologist's sine function isn't quite that "no value could work", it's that you don't have enough domain space in a regular function to reach the origin while going through all the oscilations. If you had a long function, i.e., one defined over the domain [0,+inf[ + [-inf,0] you can find a "long path" that connects the two halves for any number you choose as the connection.

    • @LadislausKallig
      @LadislausKallig Рік тому

      [0;1] is isomorphic to the (-inf; +inf). (Their cardinality is the same aleph-1). We have "enough space". For example the set defined by function y(x) = x^2 * sin(1/x), y(0) = 0 is connected, even though the path between some points has an infinite length. The problem with the function from the video is that the path is "undefined" in x=0. Its limit doesn't exist, so it can't be continuous. Which contradicts with the definition of the path. Therefore path doesn't exist.

  • @latenter1428571
    @latenter1428571 10 місяців тому

    2:31 The first definition that comes to mind is:
    A shape S is connected iff all of its points are connected. Points x1 and x2 that belong to S are connected iff there is a point x3 that also belongs to S and distance(x1,x2) is greater than both distance(x1,x3) and distance(x3,x2).
    Not sure if I can define it without a distance function. And not sure if it's correct definition yet.

  • @lowellrindler9454
    @lowellrindler9454 10 місяців тому

    I think it is connected. As you move toward the origin, y is always between 1 and -1 so no matter where it actually is, it will always touch the line at x=0. Of course the problem is that lim x->0 of sin(1/x) from the left or right doesn’t converge so it kind of doesn’t make sense to claim that it’s between -1 and 1.

  • @andehhhhhhh
    @andehhhhhhh Рік тому +1

    Masterpiece of a video

  • @yelmoralardclaw
    @yelmoralardclaw Рік тому

    11:46 but according to our defenition of open sets, they do not include boundaries, therefore even if their visible boundaries connect in a single tabgent point, this point is not included in the set and therefore this set can be cut in two. Thus, this tw open tangent circles set is disconnected.

  • @inseptus712
    @inseptus712 Рік тому +1

    I would argue that the path definition actually leads to Connected. The problem with the argument in the video was that it assumed we could only include 1 point in the center line on the path, while we can actually include multiple. A parametric function, as used here, is able to have different outputs with the same x-coordinate.

    • @davidellsworth4203
      @davidellsworth4203 Рік тому +3

      This was my first thought... but in order for this to work, a parametric function (x(t),y(t)) must be defined leading from some point (a,sin(1/a)) to (b,sin(1/b)), where a0, such that its speed sqrt(x'(t)^2 + y'(t)^2) is finite for all values of t∈[0,1]. I don't think that's possible, because the parametric function must traverse an infinite distance when it crosses at x=0.

  • @sebbes333
    @sebbes333 Рік тому +1

    1:37 Connected, but NOT in the classical sort of math way.
    This connection is like what "imaginary-numbers" are to regular numbers, in this case that would be an "imaginary-connection". (or similar to what quantum entanglement is.)
    In what I call "classical math" it mean that a line is singular & continuous & infinitely thin. But in this connection you can think of the line as becoming WIDER, to the point that EVERY point within the oscillation is the same point on the same line, regardless of what Y-value you choose.

  • @mev680
    @mev680 9 місяців тому

    It’s two unconnected curves. The sine function is odd,
    -sin(x)=sin(-x)
    LH and RH values have opposite signs. True even as
    x-> infinity.
    Conversely, the cosine is even,
    cos(x)=cos(-x)
    LH and RH values are always equal, and “connect” as
    x-> infinity.

  • @silversoul21000
    @silversoul21000 Рік тому +1

    awesome video as always

  • @columbus8myhw
    @columbus8myhw Рік тому +5

    There's a fun puzzle related to this.
    Consider a (closed) square ABCD, with the corners labeled so that A is opposite C and B is opposite D.
    Find two disjoint subsets of the square, one containing A and C and one containing B and D, such that each subset is _connected_ (in the touching subsets sense).
    Hint: both subsets must be neither open nor closed.

    • @mmmmmmmmmmmmm
      @mmmmmmmmmmmmm Рік тому

      What's the solution?

    • @columbus8myhw
      @columbus8myhw Рік тому +1

      @@mmmmmmmmmmmmmSearch "Counterintuitive topological result concerning connectedness" (including quotation marks). It should lead you to a post on Math Stack Exchange discussing this. There are some images in the comments

    • @mmmmmmmmmmmmm
      @mmmmmmmmmmmmm Рік тому

      @@columbus8myhw Thank you!

    • @cubing7276
      @cubing7276 10 місяців тому

      ​@@columbus8myhwi cant find it anymore 😭

    • @BaalThondral
      @BaalThondral 9 місяців тому

      You could take the set of all points with rational coordinates excludind A and C, and the set of all points with both coordinates irrational, merged with A and C

  • @Triszious
    @Triszious Рік тому +9

    The function is clearly undefined at x = 0, but its behaviour infinitely close to zero is interesting. f(-𝛿) = -1 and f(𝛿) = 1 for 𝛿 -> 0, both sides approaching zero infinitely fast towards the origin. So in a sense, f(±x -> 0) maps to every value between -1 and 1 except 0.

    • @andrewkarsten5268
      @andrewkarsten5268 Рік тому

      You can change it to be f(x)=sin(1/x) for x≠0 and f(0)=0, then it would be defined and continuous everywhere

    • @andrewkarsten5268
      @andrewkarsten5268 Рік тому +2

      Whoops nevermind, that for xsin(1/x), not this function. I got confused. Nevermind

  • @Obbsnja
    @Obbsnja Рік тому

    0:21 is that a good way to think about the reciprocal in all cases and if not, which ones?

  • @KaliFissure
    @KaliFissure Рік тому

    I do think that by inserting a segment along y axis you join the sides. in fact one could see the two sides like matter and anti, joined by the plane of gravity itself, the plane of the vacuum energy. 90° to regular space. the tightening of the sine is like the increase in velocity in free fall from gravity. the way everything falls towards singularity.

  • @SpadeWolfo
    @SpadeWolfo Рік тому

    In my head, I define connected shapes as shapes that, at all points, have a continuous link. For line and curves, each point must connect with exactly two points to count. More and that's seen as passing through itself at at least one point.
    For geometric shapes you define an inside and outside. Outside is defined as having access to infinity without touching inside. In the case of "hollow" shapes like a ring, inner "outsides" are considered their own shape and these follow their own rules for defining in/out, and their inside is considered outside for the larger shape.
    If both insides can continuously connect without breaching a border or the outside, they are connected. So if two points are seperated by the border of a shape, they cannot connect.
    By my definitions, this sine function is connected as long as no points from either half touch x=0 aside from the final "infinity" points of each wave.
    I should mention that I am, in fact, a gay dog on the internet with no qualifications in this field, so my definitions may be incomplete or flawed. This is just an intuition opinion born of boredom

  • @notjerrett
    @notjerrett Рік тому +1

    Great video! Doesn't the second definition require a metric? Given that a ball is defined as points within a certain distance of another point. How could we define "connectedness" without relying on metrics?

    • @tommimuller8602
      @tommimuller8602 Рік тому +2

      It is possible to define "connectedness" without a metric, and it is based on the idea of an "open set" of a topological space (which doesn't have to be a metric space). A topological space is a set together with a collection of subsets satisfying some axioms which generalise how the open balls of R^n behave. We call these subsets "open sets", and in this general setting, we can define a topological space to be connected if it is not the disjoint union of two open sets.

  • @pendragon7600
    @pendragon7600 Рік тому +1

    The "textbook definition 1" is pretty clearly equivalent to the one you explained imo.

  • @nolanwright3949
    @nolanwright3949 9 місяців тому

    Just wanted to provide an inital guess(im at the 1:43 mark). My guess is it is conected because it is continous at all points except x=0 so we only have to worry about the funtion being discontinous at 0. Its connected because for any distance from 0 N their will always be a point P on the function within N of 0. If we choose P to be (N, f(N)) we will have a point N away. And N can be arbitrarily small

    • @nolanwright3949
      @nolanwright3949 9 місяців тому

      Another observation i have made is that for any point on the funtion, f(x), the distance from the function to the line segment at 0 is x because the line segment will always have a point with the same y value as the function. This should hold as x goes to 0

  • @raduetsya
    @raduetsya Рік тому

    18:12 "this picture is a lie". Love it!

  • @zeotex2851
    @zeotex2851 Рік тому

    Such a good mini-lesson! Thats exactly how you would ideally motivate a definition: What is your intuitive understanding of this concept? What issues arise with it when we try to apply it to certain situations? How do we tweak our definition to solve these issues? And repeat. Chefs Kiss!

  • @RavenLuni
    @RavenLuni Рік тому +6

    Because sine is a cyclical function you can substitute 0 for 1/0 at the origin. Or another way of looking at it is if you take the mean value of any 'path' of length 0, the minimum and maximum (and therefore the mean) will be the same. The 0 length path at the origin has a different minimum and maximum but as it covers the full range of the functions values, they are by definition the positive and negative extremes for that function and the mean value of 0 is correct.

  • @erikpannocchia1426
    @erikpannocchia1426 Рік тому

    A function similar to this was in my first exam in college, mathematical analysis.
    It was f(x) = { 0 if x=0; max(0, x^2 sin (1/x)) if x!=0}

  • @HoSza1
    @HoSza1 Рік тому +1

    @Morphocular regarding the addition of the points of the segment [-1,1] on the y axis to either the right side or the left side: I understand that when we add it, we end up having a connected set, but that set is not the set of points that are on the curve y=sin(1/x) (i.e. the graph of the curve). Again, the graph of sin(1/x) does _not_ contain the segment, so why are we adding points to it that don't belong to the set in the first place? As I understand you showed that an augmented set is connected but that doesn't mean that the original set was also connected. Could you elaborate on this please?

    • @alonamaloh
      @alonamaloh Рік тому +1

      He is building a topological space that has the property that it is connected, but not arcwise-connected, to demonstrate that these notions are not equivalent. If you don't add any points with x coordinate 0, the set is disconnected, and therefore not interesting for the purpose of the video.

  • @aeugh4200
    @aeugh4200 4 місяці тому +1

    short answer:
    lim x->0 (x) => infinity => infinite oscillations
    remediable gap in definition at x = 0

  • @hjalmarhellstrom9658
    @hjalmarhellstrom9658 4 місяці тому

    A shape were any point in the shape can be conected to any other point with a line that does not leave the shape is conected, if there are at least 2 points in the shape that can't be conected without the line going outside of the shape then the shape is disconected.