A beautiful combinatorical proof of the Brouwer Fixed Point Theorem - Via Sperner's Lemma

Using a simple combinatorical argument, we can prove an important theorem in topology without any sophisticated machinery.

Brouwer's Fixed Point Theorem: Every continuous mapping f(p) from between closed balls of the same dimension have a fixed point where f(p)=p.

Sperner's Lemma: Every Sperner covering of a triangulation - that is, unique colors on exterior vertices, edges inherent the colors between the vertices, and insides inherent any of the colors - contains at least one triangle with vertices of all three colors.

We require exactly no technical machinery to prove Sperner's Lemma. To prove Brouwer from Sperner we do need one major theorem, Bolzano-Weierstrauss, which shows the existence of convergent subsequences on closed and bounded regions.

Leave the proof of the general case for n greater than 2 in the comments!

BECOME A MEMBER:

MATH BOOKS & MERCH I LOVE:

[Stelios]

That was a fascinating proof! An elegant chain of arguments that match up miraculously!

[andrewtychinin]

Wow, I didn't think that I'd find another interesting math channel like 3Blue1Brown's or Mathologer's. Great work, keep on going!

[nikaporozov]

Thank you! Your video really helped me prepare for presentation of this lemma in the class. You explain everything very intelligibly.

[amyz]

The visuals are really awesome, thank you for the great video!

[henriquedeoliveira]

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.

[אביבאברהם-די]

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!

[carlamaldonadoivankovic]

I loved your video!!!! Is there any biography that you used for the proof of this theorem? which one could you recommend?

[jonesbetty-zp]

amazing! my first time to understand a proof of Brauwer fixed point theorem

[MrThemalena]

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.

[TheTessatje]

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.

[josealvim]

Oh I've never seen it done like this; what a nifty proof!

[sanjursan]

When he started orienting things and not orientating things, I knew he would complete the proof.

[Kolinnor]

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 !

[Bia]

Congratulations! It's generalization in Kakutani's Fixed Point Theorem would be amazing too! It uses point-to-set mapping.

[inbalcohen]

Thank you for the great explanation and the clear illustrations! This video helped me a lot :)

[martinkunev]

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).

[Achill]

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?

[ivarangquist]

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.

[FranFerioli]

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.

[alex_everget]

17:00 "Each BOUNDED sequence"