@CogitoErgoCogitoSum It's been a while since I've seen this video and so, I don't really understand all your objections. Anyway, his video was not too rigorous, but when you present a proof in a short amount of time, you usually need to pass the key ideas and a satisfying description of a generalisable proof. As far as I remember this is what he did. No?
@CogitoErgoCogitoSum - I think his proof is rigorous, at least for two dimensions, for which he shows the existence of a fixpoint. The extension of Sperner's Lemma to higher dimensions is NOT obvious to me now.) . . . You write: "When he gets to sub-sub-triangulations at 16:11, I wonder why he picks a small triangle outside of the medium one, rather than inside the medium one" - There is NO guarantee that there is a small triangle with three differently colored corners in the medium one. Yes, the medium one has three differently colored corners, too, but its sides may have the color from the opposing corner, too. . . . "if you're just picking red points randomly throughout the domain in an infinitely sub-triangulated space then I don't see why convergence is guaranteed." - The convergence is NOT guaranteed. But the Bolzano-Weierstrass theorem guarantees us a convergent SUBsequence. Because the blue and green points belonging to this red convergent subsequence have smaller and smaller distance to their red points and will converge against the same point that is then the fixpoint. . . . "I'm not sure how it applies to the identity function" - He includes the identity function when he describes how the function, f, is defined in the point (1;0;0): If f decreases in this point in x, then we color the point red. But if it doesn't decrease, it's constant, and we have a fixpoint, and we're done.
Thanks a lot. U save me. I study a master degree in mathrmatics. My teacher request me expose this theorem, but im no familiarized with this particular theme beacuse is not my regular study area.
thanks a bunch! a little note: I wished you would have explained a little more about the convergens at the end, but overall as it is, the video is fantastic!
Here is my proof for the triangle problem: Consider the region of green vertices (nodes) that are connected to the green topmost vertex, either directly or by other green vertices that also are in the region. In the example triangle at 7:00, this region contains the five green vertices at the top, but it could be anywhere from 1 to 15 of them. Then, make a path from the left edge of the large triangle to the right one where all the vertices are connected to the green region i.e. its outline. In the previous example, it has two red vertices and three blue ones. This path starts with a red vertex and ends with a blue one (and it always does), therefore, there has to be a place on that path that where blue and red meet. And since all vertices on that path are connected to the green region, the place where blue and red meet creates a triangle with one of each color on its vertices.
Beautiful explanation! One quip: when you explain how to color the triangle using the function, couldn't you end up with a point having two colors? It seems that you could have f(p) decreasing two coordinates.
@14:22 he says one thing but the image shows another: If f strictly decreases in x at p, then p is colored red. ELSE, if f strictly decreases in y at p, then p is colored blue. ELSE, if f strictly decreases in z at p, then p is colored green. The ELSE makes the function unambiguous: If f decreases in x and in z at p, then p is colored red He could have exchanged the order of the colors, of course, and the conclusions would still be valid.
Beautiful proof of a beautiful theorem. I find it hard to grasp why the 3 sub-series must converge to the same point. Also, if you put milk foam in the tea, on the opposite side of the globe and Italian and a Japanese slap their face in the exact same point.
Excellent video. Though, I wonder if at around 15:50 you could have instead applied Sperner's Lemma to the yellow triangle with a finer triangulation (instead of staying on the big triangle all the time), then repeating this with the new little yellow triangle... At the end, you would have a limit point with the desired properties. If I'm not mistaken ! Other than that, perfect. I needed only one viewing and it's crystal clear !
Thanks Dr. Trefor Bazett, really nice. I've got one question: Why if points are all around the triangle 17:14, and because there are infintely many, they converge to a point? I can picture a triangle with infinitely many points sampled uniformly at random.
The idea is they have to start getting clustered together somewhere. If you put them uniformly, then they actually are getting clustered everywhere. But whatever epsilon area you choose, eventually some point needs to get in there.
but this theorem reminded me of waves in physics like when we create a disturbance in a string the string moves up na ddown but the particle that was there will be there it was just a disturbace that was created can we prove it like this
What a beautiful proof! Thank you. I understand the proof for two dimensions now, but it isn't clear to me yet how the proof could be transferred to higher dimensions. IF we can show Sperner's Lemma for higher dimensions, THEN it seems to be straightforward to come to Brouwer's fixpoint theorem also. BUT how can we show Sperner's Lemma in higher dimensions, like three dimensions? We would have the tetrahedron with four differently colored corners: are then the edges colored in the two colors of the corners at the end and the faces in the three colors of the corners? If we enter on a face through one of the triangles with three differently colored corners (of which there must be at least one), we can continue through the volume until we leave through a face again or hit a four colored mini-tetrahedron. But can there not be FOUR of the those starting triangles on the FOUR faces and we get in and out without hitting a four colored tetrahedron?
The higher-dimensional proof of sperner's lemma in n dimensions with n+1 colours is more or less the same, but you just have to be more careful with your parity arguments.
@@DrTrefor - Sorry, I don't know which parity I have to consider here. I understood the 2-d case, without explicitly thinking of a parity - do you mean with parity that the number of "doors" in and out of the triangle had to be odd and not even? And what would correspond to that in the 3-d case? (I saw that we must have at least one "door" on each of the four faces of the tetrahedron, but I haven't seen yet that the numbers of these "doors" would have to be odd.) . . . If you have not time to explain, do you have a link to a publicly available source, like paper or video?
I don't like the example either. The physical world of atoms and molecules is different. But it doesn't make this beautiful proof any less valid, does it?
My problem with the theorem is that the notion of "point" is pointless, 'point' is not a defined concept. Euclid's Elementa does give perfectly sensible and intuitively valid definition of 'point'. But this has nothing to do with Euclid's definition (end of a line). When you work in the limitations of a coordinate system, what many people refer to as "point", really means a meet of lines aka vertex. Here from nested subtriangulation somehow forllows "sequenses of coloured points", which I'm afraid is just mumbo jumbo to me. As is implying a "finished infinite process", when assuming that LNC holds. I've been trying to really understand Brouwer's claim. Was not convinced by this attempt.
Nice video, although I had a little confusion at around 6:47. I didn't clearly understand what you said about filling the middle vertices. Did you say you don't care what they are as long as they are filled with using all three red, blue, and yellow, or did you mean that any order is allowed including all filled with same colour?
Hello I have a question. So my profesor also used this lemma to prove BFT, but we used it to prove BFT's equivalent statement ((n-1)-sphere is not a retract of a n-ball) , while you used it to prove BFT directly. And seems much easier this way or is it basicaly the same?
Thank you professor Trefor Bazett, the explanation was really clear and use of diagram illustrations was helpful. Came across a reference to this theorem while reading graph theory material for the IMO.
Couldn't you prove Brouwer's fixed point theorem simply by applying the intermediate value theorem n times? I've never looked at the proof, but it always seemed intuitive to me (and no need of Sperner or Bolzano-Weierstrass). Less complicated than this while perhaps less entertaining.
@@DrTrefor I just watched it, great vid. I think I had something similar in mind. Here is the sketch for the 2d case. Consider a continuous mapping [x',y']=f(x,y). For any y, we know by the intermediate value that there exists at least one x* such that [x*,y']=f(x*,y). Define x*(y) as the set of fixed points for every value of y. Trivially, if x*(y) is single valued for every y, it is continuous too since f is continuous too, hence by the i.v.t., it has a fixed point. The tricky bit: x*(y) may not be continuous. If x*(y) contains 3, 5, 7... fixed points for some y, two of them could essentially "merge and disappear". But to do so, they must also have the same y' (either y'>y, y'
Would a one-dimensional example help? Consider the function f with f(x)=x^2 and the sequence 1/(n+2) that converges against zero. At all points of the sequences: f(x)
I like the video but it definitely feels like a few key details are getting glossed over at the end of the proof. For example, why should it be that the limit of the convergent subsequence of x-decreasing points should be (non-strictly) x-decreasing itself? Bolzano Weierstrass only guarantees that such a limit point would be contained in the standard 2-simplex. I also noticed that you never actually appealed to the continuity of your function in the proof. I suspect that continuity in the epsilon-delta sense would come in handy to prove the former point
You're right that the continuity of the function, f, is essential to prove that the function at the point of convergence cannot be increasing. To prove it, I would assume that f increases in one direction, then define a function f* that picks the value of f in that direction and lead it to a contradiction, using epsilon-delta mechanics. If you have difficulty with doing that, I could try to show it, but it's nothing great.
Excellent presentation; easy to follow and informative, even for the uninitiated. 👏🏻 P.S. I have always been curious if there is a deeper (albeit not immediately obvious) connection with the modular forms, as well. Any thoughts on that? I have a feeling that the holographic description of a black hole in terms of a gauge theory living on the horizon might benefit from such an insight. We now know, for instance, that black holes behave like ordinary quantum mechanical objects, where information about them is not lost, as previously imagined, but retained on their horizons, despite the apparent chaos.
The value of the function at (1;0;0) should be denoted as (x0;y0;z0). (x0;y0;z0) could be (1;0;0), too, and we would be done. Or it is (x0;y0;z0) with y0>0 or z0>0. Because of x+y+z=1 we get x=1-y-z
@martinkunev9911 - nothing. If your question was which color the point should have, I would point to an older reply of mine below, to another comment. Does that answer your question, or did you have a different one? My old reply: @14:22 he says one thing but the image shows another: If f strictly decreases in x at p, then p is colored red. ELSE, if f strictly decreases in y at p, then p is colored blue. ELSE, if f strictly decreases in z at p, then p is colored green. The ELSE makes the function unambiguous: If f decreases in x and in z at p, then p is colored red He could have exchanged the order of the colors, of course, and the conclusions would still be valid.
@martinkunev9911 - the video is a bit older, and I had to get into it first again, before understanding enough. . . . If f decreases strictly in x at point p, I mean by it: With p=(x0;y0;z0) and f(p)=(fx(p);fy(p);fz(p)) we get fx(p)
@@Achill101 Thanks, what I needed was the fact that when we have two possibilities for coloring a node, we can choose either color and the proof goes fine.
I'm trying to see where the proof brings in the assumption that f is continuous. Is it where you go from the big x+y+z=1 triangle to the little one at about 19:29?
It's necessary to assume that f is continious for making a statement about f at the point of convergence for the sequence of triangles with three differently colored points. If f was discontinuous, then f at the point of convergence could be ANYTHING. . . . But if we assume the CONTINUOUS f would INCREASE at the point of convergence in one direction, then we get a contradiction due to the existence of a sequence converging against this point, but with f DECREASING for all points of that sequence.
Except that all mathematicians use that analogy and as a physicist I find that frustrating simply because a liquid is a group of rather independant molecules and deformation in it is precisely NOT continuous. Therefore the probability that one molecule is at the exact same place as it was before mixing is just null.
the proof which you gave of sperner's lemma is incomplete as you proved the converse of what was needed. I mean you proved that if there's a special triangle, then your path get stuck and the path intersects only one of the outer edges, and then you claimed that since there are always odd number of special(different colours) outer edges, there must be a path intersecting only one edge. But you didn't prove that if there's a path which intersects only one such outer edge and get stuck, then there must be a special triangle, you proved the converse of it. Although it's easy to show that it's and if and only if condition, I think you should explain that clearly.
I'm sorry if i haven't been able to communicate properly, but I wanted to say that you haven't shown that that hitting a special triangle is the one and only way to get stuck. I mean it's almost obvious but I thought you forgot to mention it. You mentioned that if there's a special triangle then the path would get stuck. Thanks.
I didn't understand why you said all the time "strictly less" and after suddently said "less or equal". Was that because the LIMIT points of the sequences are "ideal" points and don't have necessarily to respect the definition criteria for the associated sequence (nor even belong to the sequence!)? And you speak so fast, this is hard to follow for a non native english speaker. Anyway, great video. Guess I have to watch it many times in the future.
Yes, the point of convergence doesn't have to have the STRICTLY decreasing character of the points on the series, for which f is strictly decreasing. . . . As an example, consider the function f with f(x)=x^2 and the sequence 1/(n+2) that converges against zero. At all points of the sequences: f(x)
That was a fascinating proof! An elegant chain of arguments that match up miraculously!
@CogitoErgoCogitoSum It's been a while since I've seen this video and so, I don't really understand all your objections. Anyway, his video was not too rigorous, but when you present a proof in a short amount of time, you usually need to pass the key ideas and a satisfying description of a generalisable proof. As far as I remember this is what he did. No?
@CogitoErgoCogitoSum - I think his proof is rigorous, at least for two dimensions, for which he shows the existence of a fixpoint. The extension of Sperner's Lemma to higher dimensions is NOT obvious to me now.)
. . . You write: "When he gets to sub-sub-triangulations at 16:11, I wonder why he picks a small triangle outside of the medium one, rather than inside the medium one" - There is NO guarantee that there is a small triangle with three differently colored corners in the medium one. Yes, the medium one has three differently colored corners, too, but its sides may have the color from the opposing corner, too.
. . . "if you're just picking red points randomly throughout the domain in an infinitely sub-triangulated space then I don't see why convergence is guaranteed." - The convergence is NOT guaranteed. But the Bolzano-Weierstrass theorem guarantees us a convergent SUBsequence. Because the blue and green points belonging to this red convergent subsequence have smaller and smaller distance to their red points and will converge against the same point that is then the fixpoint.
. . . "I'm not sure how it applies to the identity function" - He includes the identity function when he describes how the function, f, is defined in the point (1;0;0): If f decreases in this point in x, then we color the point red. But if it doesn't decrease, it's constant, and we have a fixpoint, and we're done.
Thanks a lot. U save me. I study a master degree in mathrmatics. My teacher request me expose this theorem, but im no familiarized with this particular theme beacuse is not my regular study area.
lol, I have to expose this in my 5th bachelor semester XD
amazing! my first time to understand a proof of Brauwer fixed point theorem
Thank you! Your video really helped me prepare for presentation of this lemma in the class. You explain everything very intelligibly.
Wow, I didn't think that I'd find another interesting math channel like 3Blue1Brown's or Mathologer's. Great work, keep on going!
When he started orienting things and not orientating things, I knew he would complete the proof.
My mind is actually blown right now
haha nice!
Oh I've never seen it done like this; what a nifty proof!
The visuals are really awesome, thank you for the great video!
thanks a bunch! a little note: I wished you would have explained a little more about the convergens at the end, but overall as it is, the video is fantastic!
Here is my proof for the triangle problem:
Consider the region of green vertices (nodes) that are connected to the green topmost vertex, either directly or by other green vertices that also are in the region. In the example triangle at 7:00, this region contains the five green vertices at the top, but it could be anywhere from 1 to 15 of them.
Then, make a path from the left edge of the large triangle to the right one where all the vertices are connected to the green region i.e. its outline. In the previous example, it has two red vertices and three blue ones. This path starts with a red vertex and ends with a blue one (and it always does), therefore, there has to be a place on that path that where blue and red meet. And since all vertices on that path are connected to the green region, the place where blue and red meet creates a triangle with one of each color on its vertices.
17:00 "Each BOUNDED sequence"
Beautiful explanation!
One quip: when you explain how to color the triangle using the function, couldn't you end up with a point having two colors? It seems that you could have f(p) decreasing two coordinates.
@14:22 he says one thing but the image shows another:
If f strictly decreases in x at p, then p is colored red.
ELSE, if f strictly decreases in y at p, then p is colored blue.
ELSE, if f strictly decreases in z at p, then p is colored green.
The ELSE makes the function unambiguous: If f decreases in x and in z at p, then p is colored red
He could have exchanged the order of the colors, of course, and the conclusions would still be valid.
Beautiful proof of a beautiful theorem.
I find it hard to grasp why the 3 sub-series must converge to the same point.
Also, if you put milk foam in the tea, on the opposite side of the globe and Italian and a Japanese slap their face in the exact same point.
All three keep existing in smaller and smaller subtriangles!
@@DrTrefor Thanks!
Thank you for the great explanation and the clear illustrations! This video helped me a lot :)
I loved your video!!!! Is there any biography that you used for the proof of this theorem? which one could you recommend?
14:40 I don't get how you color-code a point if two of the conditions hold at the same time (e.g. decreases in both x and y directions).
A lot of arguments like this one just pick one of the two colors
Excellent video. Though, I wonder if at around 15:50 you could have instead applied Sperner's Lemma to the yellow triangle with a finer triangulation (instead of staying on the big triangle all the time), then repeating this with the new little yellow triangle... At the end, you would have a limit point with the desired properties. If I'm not mistaken !
Other than that, perfect. I needed only one viewing and it's crystal clear !
Sub-triangles can have points on their SIDES colored in the same color as the OPPOSING corner. You can NOT apply Sperner's Lemma to them.
Thanks Dr. Trefor Bazett, really nice. I've got one question: Why if points are all around the triangle 17:14, and because there are infintely many, they converge to a point? I can picture a triangle with infinitely many points sampled uniformly at random.
The idea is they have to start getting clustered together somewhere. If you put them uniformly, then they actually are getting clustered everywhere. But whatever epsilon area you choose, eventually some point needs to get in there.
Congratulations! It's generalization in Kakutani's Fixed Point Theorem would be amazing too! It uses point-to-set mapping.
Absolutely amazing video!
but this theorem reminded me of waves in physics like when we create a disturbance in a string the string moves up na ddown but the particle that was there will be there it was just a disturbace that was created can we prove it like this
To the point explanation , loved it.
4:50
It's unclear why you choose such organization of colors dots, you can select any configuration
Fantastic proof!
Brilliant explanation!
Glad you liked it!
What a beautiful proof! Thank you.
I understand the proof for two dimensions now, but it isn't clear to me yet how the proof could be transferred to higher dimensions. IF we can show Sperner's Lemma for higher dimensions, THEN it seems to be straightforward to come to Brouwer's fixpoint theorem also. BUT how can we show Sperner's Lemma in higher dimensions, like three dimensions? We would have the tetrahedron with four differently colored corners: are then the edges colored in the two colors of the corners at the end and the faces in the three colors of the corners? If we enter on a face through one of the triangles with three differently colored corners (of which there must be at least one), we can continue through the volume until we leave through a face again or hit a four colored mini-tetrahedron. But can there not be FOUR of the those starting triangles on the FOUR faces and we get in and out without hitting a four colored tetrahedron?
The higher-dimensional proof of sperner's lemma in n dimensions with n+1 colours is more or less the same, but you just have to be more careful with your parity arguments.
@@DrTrefor - Sorry, I don't know which parity I have to consider here. I understood the 2-d case, without explicitly thinking of a parity - do you mean with parity that the number of "doors" in and out of the triangle had to be odd and not even? And what would correspond to that in the 3-d case? (I saw that we must have at least one "door" on each of the four faces of the tetrahedron, but I haven't seen yet that the numbers of these "doors" would have to be odd.)
. . . If you have not time to explain, do you have a link to a publicly available source, like paper or video?
Thank you professor!
I don't like the example with mixing a cup of tea (or the foam ontop). Who says that mixing is continuous?
I don't like the example either. The physical world of atoms and molecules is different.
But it doesn't make this beautiful proof any less valid, does it?
Great video, really interesting and clear!
Great proof really nifty
Nice.... greetings from His grandson Martin Sperner 😅
My problem with the theorem is that the notion of "point" is pointless, 'point' is not a defined concept. Euclid's Elementa does give perfectly sensible and intuitively valid definition of 'point'. But this has nothing to do with Euclid's definition (end of a line). When you work in the limitations of a coordinate system, what many people refer to as "point", really means a meet of lines aka vertex.
Here from nested subtriangulation somehow forllows "sequenses of coloured points", which I'm afraid is just mumbo jumbo to me. As is implying a "finished infinite process", when assuming that LNC holds.
I've been trying to really understand Brouwer's claim. Was not convinced by this attempt.
Totally inspiring totally awesome with no decreasing in any direction of mindblowing.Thanks so much for posting this!!!
How beautiful. Thank you so much.
Nice video, although I had a little confusion at around 6:47. I didn't clearly understand what you said about filling the middle vertices. Did you say you don't care what they are as long as they are filled with using all three red, blue, and yellow, or did you mean that any order is allowed including all filled with same colour?
Thanks, i got it later on the video.
U explained so beautifully..
Hello I have a question. So my profesor also used this lemma to prove BFT, but we used it to prove BFT's equivalent statement ((n-1)-sphere is not a retract of a n-ball) , while you used it to prove BFT directly. And seems much easier this way or is it basicaly the same?
@@DrTrefor Aha, that would make sense, since we this in our undergraduate course in topology which doesnt include alg top. Thanks!
Thank you professor Trefor Bazett, the explanation was really clear and use of diagram illustrations was helpful. Came across a reference to this theorem while reading graph theory material for the IMO.
Couldn't you prove Brouwer's fixed point theorem simply by applying the intermediate value theorem n times? I've never looked at the proof, but it always seemed intuitive to me (and no need of Sperner or Bolzano-Weierstrass). Less complicated than this while perhaps less entertaining.
@@DrTrefor I just watched it, great vid. I think I had something similar in mind. Here is the sketch for the 2d case. Consider a continuous mapping [x',y']=f(x,y). For any y, we know by the intermediate value that there exists at least one x* such that [x*,y']=f(x*,y). Define x*(y) as the set of fixed points for every value of y. Trivially, if x*(y) is single valued for every y, it is continuous too since f is continuous too, hence by the i.v.t., it has a fixed point.
The tricky bit: x*(y) may not be continuous. If x*(y) contains 3, 5, 7... fixed points for some y, two of them could essentially "merge and disappear". But to do so, they must also have the same y' (either y'>y, y'
Why do the corners need to be 3 seperate colours? In terms of game theory, I'm trying to think of how strategies relate to this.
Why does these videos sounds so good and your newer one sounds like you were recording them in a cave ?
Could you please elaborate on how the convergent sequence would give "less than or equal to" instead of strictly "less than"?
Would a one-dimensional example help? Consider the function f with f(x)=x^2 and the sequence 1/(n+2) that converges against zero. At all points of the sequences: f(x)
I like the video but it definitely feels like a few key details are getting glossed over at the end of the proof. For example, why should it be that the limit of the convergent subsequence of x-decreasing points should be (non-strictly) x-decreasing itself? Bolzano Weierstrass only guarantees that such a limit point would be contained in the standard 2-simplex. I also noticed that you never actually appealed to the continuity of your function in the proof. I suspect that continuity in the epsilon-delta sense would come in handy to prove the former point
You're right that the continuity of the function, f, is essential to prove that the function at the point of convergence cannot be increasing. To prove it, I would assume that f increases in one direction, then define a function f* that picks the value of f in that direction and lead it to a contradiction, using epsilon-delta mechanics. If you have difficulty with doing that, I could try to show it, but it's nothing great.
Excellent presentation; easy to follow and informative, even for the uninitiated. 👏🏻
P.S. I have always been curious if there is a deeper (albeit not immediately obvious) connection with the modular forms, as well. Any thoughts on that? I have a feeling that the holographic description of a black hole in terms of a gauge theory living on the horizon might benefit from such an insight. We now know, for instance, that black holes behave like ordinary quantum mechanical objects, where information about them is not lost, as previously imagined, but retained on their horizons, despite the apparent chaos.
thanks for your ideoks
TREFOR'S BACK BABY!!!!!!!
14:47: one question : what is the mathematical statement corresponding to f(p) strictly decreases x?
The value of the function at (1;0;0) should be denoted as (x0;y0;z0).
(x0;y0;z0) could be (1;0;0), too, and we would be done.
Or it is (x0;y0;z0) with y0>0 or z0>0. Because of x+y+z=1 we get x=1-y-z
@@Achill101 What stops p from decreasing in both the x and y directions?
@martinkunev9911 - nothing.
If your question was which color the point should have, I would point to an older reply of mine below, to another comment. Does that answer your question, or did you have a different one?
My old reply:
@14:22 he says one thing but the image shows another:
If f strictly decreases in x at p, then p is colored red.
ELSE, if f strictly decreases in y at p, then p is colored blue.
ELSE, if f strictly decreases in z at p, then p is colored green.
The ELSE makes the function unambiguous: If f decreases in x and in z at p, then p is colored red
He could have exchanged the order of the colors, of course, and the conclusions would still be valid.
@martinkunev9911 - the video is a bit older, and I had to get into it first again, before understanding enough.
. . . If f decreases strictly in x at point p, I mean by it:
With p=(x0;y0;z0) and f(p)=(fx(p);fy(p);fz(p))
we get fx(p)
@@Achill101 Thanks, what I needed was the fact that when we have two possibilities for coloring a node, we can choose either color and the proof goes fine.
I'm trying to see where the proof brings in the assumption that f is continuous. Is it where you go from the big x+y+z=1 triangle to the little one at about 19:29?
Perhaps continuity is used to infer that the limit point is a fixed point.
It's necessary to assume that f is continious for making a statement about f at the point of convergence for the sequence of triangles with three differently colored points. If f was discontinuous, then f at the point of convergence could be ANYTHING.
. . . But if we assume the CONTINUOUS f would INCREASE at the point of convergence in one direction, then we get a contradiction due to the existence of a sequence converging against this point, but with f DECREASING for all points of that sequence.
Love that coffee analogy. Hands down.
Except that all mathematicians use that analogy and as a physicist I find that frustrating simply because a liquid is a group of rather independant molecules and deformation in it is precisely NOT continuous. Therefore the probability that one molecule is at the exact same place as it was before mixing is just null.
the proof which you gave of sperner's lemma is incomplete as you proved the converse of what was needed. I mean you proved that if there's a special triangle, then your path get stuck and the path intersects only one of the outer edges, and then you claimed that since there are always odd number of special(different colours) outer edges, there must be a path intersecting only one edge. But you didn't prove that if there's a path which intersects only one such outer edge and get stuck, then there must be a special triangle, you proved the converse of it. Although it's easy to show that it's and if and only if condition, I think you should explain that clearly.
I'm sorry if i haven't been able to communicate properly, but I wanted to say that you haven't shown that that hitting a special triangle is the one and only way to get stuck. I mean it's almost obvious but I thought you forgot to mention it. You mentioned that if there's a special triangle then the path would get stuck. Thanks.
Can't you just skip Bolzano-Weierstrass by simply applying Sperner's Lemma to one of the sub-triangles?
Sub-triangles can have points on their SIDES colored in the same color as the OPPOSING corner. You can NOT apply Sperner's Lemma to them.
I didn't understand why you said all the time "strictly less" and after suddently said "less or equal".
Was that because the LIMIT points of the sequences are "ideal" points and don't have necessarily to respect the definition criteria for the associated sequence (nor even belong to the sequence!)?
And you speak so fast, this is hard to follow for a non native english speaker.
Anyway, great video. Guess I have to watch it many times in the future.
Yes, the point of convergence doesn't have to have the STRICTLY decreasing character of the points on the series, for which f is strictly decreasing.
. . . As an example, consider the function f with f(x)=x^2 and the sequence 1/(n+2) that converges against zero. At all points of the sequences: f(x)
Gee. The Lagrange points look easier.
What kind of math is this ?
COME BACK TO UC, THE MATH DEPARTMENT NEEDS YOU.
guilty... I know you are already a dad but if you are open to adopting an 18 y/o college student please let me know...
Bridge of Konigsberg Proof ..
Wow
It’s jarring to watch this without even a short pitch or intro to your channel, and self proclaimed goal or planned depth of discussion
The sweaty armpits 😫😅👀🤷♂️
god I love this channel