2.1
Given \(f\colon \Delta \to \Delta \) (where \(\Delta \) is triangulated), we construct a coloring of the triangulation.
Let \(\Delta \) be a simplex. Given an ordered list of its vertices, and a function \(f:\Delta \to \Delta \), the induced color of a point \(x\in \Delta \) is \(i=\min \{ j ~ |~ f(x)_j {\lt} x_j\} \).
The coloring above is a Sperner coloring.
Apply the above to a triangulation of \(\Delta \) of diameter \({\lt}\epsilon \), to get a panchromatic triangle of diameter \(\epsilon \).
For all \(\varepsilon {\gt}0\) exists \(\delta {\gt}0\), such that given a panchromatic simplex \(X\subseteq \Delta \) of a diameter \(\leq \delta \) then \(|f(x) - x| {\lt} \varepsilon ,\ \forall x \in X\).
By compactness, \(f\) is uniformly continuous in \(\Delta \). Then for all \(\frac{\varepsilon }{2n}{\gt}0\) there exists \(\delta _0{\gt}0\) such that \(diam(f(X)) {\lt} \frac{\varepsilon }{2n}\). We define \(\delta :=\min (\delta _0,\frac{\varepsilon }{4n})\), that preserves the properties of \(\delta \).
Let \(p_0, \ldots , p_n\) be the vertices of \(X\). Then, since \(X\) is panchromatic, for each \(i = 0,\ldots ,n-1\), \(f(p_i)_i {\lt} (p_i)_i\) and \(f(p_{i+1})_i \geq (p_{i+1})_i\). Then, for all \(x \in X\),
and
Putting them together,
Therefore, for all \(x \in X\),
Finally, for all \(x \in X\),
\( \)
\(f\colon X \to X\) continuous and \(X\) compact metric space. Suppose that \(\forall \epsilon {\gt}0\), there exists \(x\in X\) such that \(d(f(x),x) {\lt} \epsilon \). Then \(f\) has a fixed point.
Using the epsilons property, we can construct a sequance verifying the following inequality
Because of the Bolzano-Weierstrass theorem, we know there exists a convergent subsequance \( x_{k_n} \) of \( x_n \) satisfying,
Applying the limits in the inequality, we obtain the equality
Using the continuity property of limits we can conclude,
\( \)
Now we apply Sperner’s lemma to deduce that there is a panchromatic face in the subdivision.
The baricenter of points \(p_1, \dots , p_k\) is \(B(p_1, \dots , p_k)= \frac{1}{k}\sum _{i=1}^k p_i\)
Let \(S\subseteq \mathbb {R}^n\) be a \(n\)-dimensional simplex with barycentric coordinates \(x_0,\ldots ,x_n\). The barycentric subdivison is a map form permutations of the \(n+1\) vertices of \(S\) to \(n\)-dimensional simplexes defined as follows
See end of https://github.com/leanprover-community/mathlib/blob/sperner-again/src/combinatorics/simplicial_complex/subdivision.lean
Let \(S\) be a \(k\)-dimensional simplex with vertices \(v_0, \dots , v_k\). The diameter of \(S\) \( Diam(S)= \max _{x,y\in S} |x-y|\) equals \(\max |v_i - v_j|\).
Since \(v_i, v_j \in S\), we have \(Diam(S) \geq \max |v_i - v_j|\).
Let \(l\max |v_i - v_j|\), and for each vertex consider the closed ball centered at \(v_i\) and radius \(l\).
Since \(B_l(v_i)\) is convex, and it contains every vertex of \(S\), we have \(|x-v_i|\leq l\) for all \(x\in S\).
Finally we show \(|x-y|\leq l\) for all \(x, y\in S\). Given \(x\in S\), \(B_l(x)\) contains all vertices of \(S\), and since \(B_l(x)\) is convex, it contains any \(y\in S\). Therefore \(|x-y|\leq l\) for all \(x, y\in S\). This is \(diam(S)\leq l\).
Let \(\hat S = B(v_0,v_1,\dots , v_k)\) be the barycenter of \(S\). We bound the distance of any vertex \(v_i\) of \(S\) to \(\hat S\). Given a simplex \(S\) of diameter \(D\), its barycentric subdivision has diameter at most \(\frac{n}{n+1}D\)
The dinstance of any vertex of \(S\) to its barycenter is bounded by
Therefore, the closed ball centered at \(\hat S\), of radius \(\frac{p}{p+1}D\), contains the vertices of \(S\), by convexity it contains \(S\).
Finally we proof the result by strong induction, for \(k=0\) the result is trivial. If \(s\) and \(s'\) are faces of \(S\) such that \(s\) is a proper face of \(s'\),
For all \(\epsilon {\gt}0\), there exists a subdivision of \(\Delta \) of diameter \({\lt}\epsilon \).