S ≡

where the t_{i} are uniquely determined and the map x → t is continuous from S to the compact set
To see this, suppose x^{k} → x in S. Let x^{k} ≡∑ _{i=0}^{n}t_{i}^{k}x_{i} with x defined similarly with t_{i}^{k} replaced with t_{i}, x ≡∑ _{i=0}^{n}t_{i}x_{i}. Then

Thus

Say t_{i}^{k} fails to converge to t_{i} for all i ≥ 1. Then there exists a subsequence, still denoted with superscript k such that for each i = 1,

which contradicts independence of the x_{i} −x_{0}. It follows that for all i ≥ 1,t_{i}^{k} → t_{i}. Since they all sum to 1, this implies that also t_{0}^{k} → t_{0}. Thus the claim about continuity is verified.
Let f : S → S be continuous. When doing f to a point x, one obtains another point of S denoted as ∑ _{i=0}^{n}s_{i}x_{i}. Thus in this argument the scalars s_{i} will be the components after doing f to a point of S denoted as ∑ _{i=0}^{n}t_{i}x_{i}.
Consider a triangulation of S such that all simplices in the triangulation have diameter less than ε. The vertices of the simplices in this triangulation will be labeled from p_{0},
Note that for the vertices which are on

because t_{k} = 0 for each of these and so r_{k} = ∞ unless k ∈
By the Sperner’s lemma procedure described above, there are an odd number of simplices having value ∏ _{i≠k}p_{i} on the k^{th} face and an odd number of simplices in the triangulation of S for which the product of the labels on their vertices, referred to here as its value, equals p_{0}p_{1}

What is r_{k}, the smallest of those ratios in determining a label? Could it be larger than 1? r_{k} is certainly finite because at least some t_{j}≠0 since they sum to 1. Thus, if r_{k} > 1, you would have s_{k} > t_{k}. The s_{j} sum to 1 and so some s_{j} < t_{j} since otherwise, the sum of the t_{j} equalling 1 would require the sum of the s_{j} to be larger than 1. Hence r_{k} was not really the smallest after all and so r_{k} ≤ 1. Hence s_{k} ≤ t_{k}.
Let S≡

r_{ks} ≤ r_{j} for all j≠k_{s} and

Do the same system of labeling for each n + 1 simplex in a sequence of triangulations where the diameters of the simplices in the k^{th} triangulation is no more than 2^{−k}. Thus each of these triangulations has a n + 1 simplex having diameter no more than 2^{−k} which has value P_{n}. Let b_{k} be the barycenter of one of these n + 1 simplices having value P_{n}. By compactness, there is a subsequence, still denoted with the index k such that b_{k} → x. This x is a fixed point.
Consider this last claim. x = ∑ _{i=0}^{n}t_{i}x_{i} and after applying f, the result is ∑ _{i=0}^{n}s_{i}x_{i}. Then b_{k} is the barycenter of some σ_{k} having diameter no more than 2^{−k} which has value P_{n}. Say σ_{k} is a simplex having vertices

Re ordering these if necessary, we can assume that the label for y_{i}^{k} is p_{i} which implies that, as noted above, for each i = 0,

the i^{th} coordinate of f

and so the i^{th} coordinate of y_{i}^{k},t_{i}^{k} must converge to t_{i}. Hence if the i^{th} coordinate of f

By continuity of f, it follows that s_{i}^{k} → s_{i}. Thus the above inequality is preserved on taking k →∞ and so

this for each i. But these s_{i} add to 1 as do the t_{i} and so in fact, s_{i} = t_{i} for each i and so f
Theorem 2.10.3 Let S be a simplex
Corollary 2.10.4 Let K be a closed convex bounded subset of ℝ^{n}. Let f : K → K be continuous. Then there exists x ∈ K such that f
Proof: Let S be a large simplex containing K and let P be the projection map onto K. See Problem 30 on Page 266 for the necessary properties of this projection map. Consider g
The proof of this corollary is pretty significant. By a homework problem, a closed convex set is a retract of ℝ^{n}. This is what it means when you say there is this continuous projection map which maps onto the closed convex set but does not change any point in the closed convex set. When you have a set A which is a subset of a set B which has the property that continuous functions f : B → B have fixed points, and there is a continuous map P from B to A which leaves points of A unchanged, then it follows that A will have the same “fixed point property”. You can probably imagine all sorts of sets which are retracts of closed convex bounded sets.