Talk:Sperner's lemma

Latest comment: 8 months ago by MVDVPRSD in topic Question's about the proof

The other lemma edit

Note to Axel - the other Lemma is an unrelated result proved earlier by Sperner, rather than the result of Borsuk et al.

Charles Matthews 09:07, 28 Feb 2004 (UTC)

Off-hand, I'm not so sure about the above comment. Sperner proved two lemmas on his way to Lebesgue's theorem. The second one is in Lyusternik Convex Figures and Polyhedra, along with Sperner's proof of Lebesgue's theorem. I'd have to find the reference.

As it is, I removed this line:

For the 'other' Sperner lemma dealing with set systems, see antichain.

In fact, it's known as Sperner's theorem, not lemma, see Sperner family and my comments in Talk:antichain. Perhaps a disambiguation remark at the top would be appropriate.--192.35.35.36 00:15, 18 Feb 2005 (UTC)

Dimension of a subface edit

In the article it says at this time: "The vertices of T located on any given k-dimensional subface

 

are colored[...]"

However, it should be (k-1) dimensional subface, or we should speak of k+1 points and colors. Am I wrong? Joerg Bader (talk) 15:20, 3 October 2012 (UTC)Reply

Please note that the dimension of the whole simplex is n, not k. k is a free variable; it can be anything less than n. So offsetting its value by one makes no difference. —David Eppstein (talk) 16:25, 3 October 2012 (UTC)Reply
Yes, I understand. Maybe I'm missing something, but for any fixed k, 1<=k<=n, the k points here form a k-1 dimensional simplex/subface, I think. Joerg Bader (talk) 16:43, 3 October 2012 (UTC)Reply
Ah, it's a mismatch between the number of points and the dimension of the subspace. Another way to fix it would be to start indexing the points at 0, I suppose... or, we don't really need the extra variable k because the condition only needs to be true for the (n-1)-dimensional faces (and the rest follow automatically). —David Eppstein (talk) 16:58, 3 October 2012 (UTC)Reply


Question's about the proof edit

What is the "degree of a triangle"? I was able to find a definition of the degree of a vertex, but no luck with the degree of a triangle. Without understanding the term, I do not find it to be "easily seen . . ." Thanks. Maurice Fox (talk) 00:56, 11 May 2014 (UTC)Reply

As the text at the start of the section describes, each triagle corresponds to a vertex in G. It is the degree of that vertex. —David Eppstein (talk) 01:08, 11 May 2014 (UTC)Reply

Can someone help clear the ambiguity in this for me? what is capital A and B the proof is referring to? is it the regions "small" a and b ? also doesn't this proof need some citations? — Preceding unsigned comment added by MVDVPRSD (talkcontribs) 14:35, 29 June 2023 (UTC)Reply

Ok.. it seem to be referring to the vertices of the triangle; if someone could label the image and find a citation for this proof that'd be grt MVDVPRSD (talk) 12:07, 30 June 2023 (UTC)Reply

"Random colouring" left unexplained edit

One illustration's caption reads as follows:

"A random Sperner-colouring of a regular triangularisation. Each cul-de-sac is an RGB-triangle".

That *might* be interesting or worth inclusion in Wikipedia — or not, it's hard to be sure.

Especially since a "random colouring" suggests that each vertex is coloured R, G, or B independently with 1/3 probability each.

But this says nothing about how the edges are determined.

Without a better reason for inclusion forthcoming, I hope that this illustration can be deleted from Wikipedia. 2601:200:C000:1A0:8936:A4D9:37CE:3D84 (talk) 02:13, 21 November 2022 (UTC)Reply