so I am just trying to illustrate to an informal audiance how to prove the Brower Fixed Point Theorem using Sperner's lemma. I seem to have trouble with the iterative application of Sperner's lemma. The reason is that I don't think that the baby triangle (tetrahedon) obtained from Sperner can be Sperner labelled after further triangulation. (Jump to the bold face font for those familiar with the proof)
For simplicity, let's take the $2d$ case. The theorem goes along these lines: Let $\Delta$ be the triangle (i.e. a compact and convex subset of $\mathbb R^2$) and $f:\Delta\rightarrow\Delta$ a continuous function from $\Delta$ to itself. Then there exists a fixed point, i.e. $x\in \Delta$ with $f(x)=x$.
The proof constructs a sequence of triangulations whose convergent subsequence (due to compactness) converges to the fixed point. For this to work, the first triangulation receives a labelling of each vertex with baryocentric cooridinates $(a,b,c)\mapsto (a',b',c')$:
1 if $a'< a$
2 if $a'\geq a \;\;\wedge\;\;b'<b$
3 if $a'\geq a \;\;\wedge\;\;b'\geq b\;\;\wedge\;\;c'<c$
and I would also add 0 if $x$ is a fixed point
Now this labelling can be shown to be a Sperner labelling of the first triangulation (1,2,3 in the corner vertices and vertices lying on the outer edges take values in the labels of the respective corner vertices). By Sperner's lemma we know of the existence of an odd number of baby triangles that are also labelled with $1,2,3$.
The proof now goes on by constructing yet another triangulation of that baby triangle. But who can explain to me, why this can also be Sperner labelled? In other words, who can tell me, why the labels of the outer vertices take values of one of the corner vertex-labels. If they dont, then we cannot iterate the triangulation and the argument doesn't work.
I was thinking of fixing it by a continuity argument and sort of choosing a tiny triangulation as possible ($\epsilon$-$\delta$-type), but that doesn't work. Also, once we managed to successfully construct the sequence of triangulations, why do we need Bolzano Weierstrasse? The sequence is strongly convergent, we don't need to construct subsequences, or do we?
Thanks for enlightening me!
You don't need to Sperner label this triangle. Rather, you consider a sequence of triangulations $T_n$ where the diameter of the baby triangles tends to zero. For each $n$, pick a baby triangle $B_n$ such that $B_n$ is labelled with 1,2,3.
Bolzano-Weierstrass then allows you to pick a convergent subsequence $B_{n_k}$ that converges to a point $x$, since the diameter of the triangles converges to zero. The point $x$ is your fixpoint.