Let D denote the triangle. For each n∈N, construct a 2-d simplex Sn with (n+1)2 nodes in D, where the color of a point corresponds to the place in the disk that f carries that point to, then choose xn to be a point within a trichromatic triangle in the graph. Then (xn)n∈N is a bounded sequence having a limit point x∗. Let t be the center of the disc; suppose that f(x∗)≠t. Then there is at least one region of the disc that f(x∗) doesn’t touch. Let ϵ be the distance to the furthest side, that is, let ϵ=max{d(f(x∗),G),d(f(x∗),R),d(f(x∗),B)}. Since the sides are closed regions, we have ϵ>0. Using continuity of f, choose δ∈R+ small enough such that Bd(x∗,δ)⊂Bd(f(x∗),ϵ). Then choose n∈N large enough so that (1) all triangles in Sn have diameter less than δ2 and (2) d(xn,x∗)<δ2. Then, given any other point y in the triangle around xn in Sn, we have that d(y,x∗)≤d(y,xn)+d(xn,x∗)<δ2+δ2=δ, so that d(f(y),f(x∗))<ϵ. This proves that the triangle in Sn does not map points to all three sides of the disc, contradicting the fact that it is trichromatic.
Ex 6
(This is way easier to demonstrate in a picture in a way that leaves no doubt that it works than it is to write down, but I tried to do it anyway considering that to be part of the difficulty.)
(Assume the triangle T=¯¯¯¯¯¯¯¯¯¯¯¯ABO is equilateral.) Imbed T into R2 such that O=(0,0), A=(−12,12√3), B=(12,12√3). Let f:T↦T be continuous. Then h:T↦R2 given by h:x↦f(x)−x is also continuous. If h(x)=O=(0,0) then (0,0)=h(x)=f(x)−x⟹f(x)=x. Let R be the circle with radius 2 around O; then h(T)⊂R because ||h(x)||=||f(x)−x||≤||f(x)||+||x||≤2 (it is in fact contained in the circle with radius 1, but the size of the circle is inconsequential). We will use exercise 5 to show that h maps a point to the center, which is O, from which the desired result follows. For this, we shall show that it has the needed properties, with the modification that points on any side may map precisely into the center. It’s obvious that weakening the requirement in this way preserves the result.
Rotate the disk so that the red shape is on top. In polar coordinates, the green area now contains all points with angles between 60∘ and 180∘ , the blue area contains those between 180∘ and 300∘, and the red area those between 0∘ and 60∘ and those between 300∘ and 360∘. We will show that h(¯¯¯¯¯¯¯¯AB) does not intersect the red area, except at the origin. First, note that we have
Since both T and ¯¯¯¯¯¯¯¯AB are convex combinations of finitely many points, it suffices to check all combinations that result by taking a corner from each. This means we need to check the points
A−A and B−A and C−A and A−B and B−B and C−B.
Which are easily computed to be
(0,0) and (−1,0) and (12,−12√3) and (1,0) and (0,0) and (−12,−12√3)
Two of those are precisely the origin, the other four have angles 270∘ and 150∘ and 90∘ and 210∘. Indeed, they are all between 60∘ and 300∘.
Now one needs to do the same for the sets T−¯¯¯¯¯¯¯¯OA and T−¯¯¯¯¯¯¯¯BO, but it goes through analogously.
Ex 5 (fixed version)
Let D denote the triangle. For each n∈N, construct a 2-d simplex Sn with (n+1)2 nodes in D, where the color of a point corresponds to the place in the disk that f carries that point to, then choose xn to be a point within a trichromatic triangle in the graph. Then (xn)n∈N is a bounded sequence having a limit point x∗. Let t be the center of the disc; suppose that f(x∗)≠t. Then there is at least one region of the disc that f(x∗) doesn’t touch. Let ϵ be the distance to the furthest side, that is, let ϵ=max{d(f(x∗),G),d(f(x∗),R),d(f(x∗),B)}. Since the sides are closed regions, we have ϵ>0. Using continuity of f, choose δ∈R+ small enough such that Bd(x∗,δ)⊂Bd(f(x∗),ϵ). Then choose n∈N large enough so that (1) all triangles in Sn have diameter less than δ2 and (2) d(xn,x∗)<δ2. Then, given any other point y in the triangle around xn in Sn, we have that d(y,x∗)≤d(y,xn)+d(xn,x∗)<δ2+δ2=δ, so that d(f(y),f(x∗))<ϵ. This proves that the triangle in Sn does not map points to all three sides of the disc, contradicting the fact that it is trichromatic.
Ex 6
(This is way easier to demonstrate in a picture in a way that leaves no doubt that it works than it is to write down, but I tried to do it anyway considering that to be part of the difficulty.)
(Assume the triangle T=¯¯¯¯¯¯¯¯¯¯¯¯ABO is equilateral.) Imbed T into R2 such that O=(0,0), A=(−12,12√3), B=(12,12√3). Let f:T↦T be continuous. Then h:T↦R2 given by h:x↦f(x)−x is also continuous. If h(x)=O=(0,0) then (0,0)=h(x)=f(x)−x⟹f(x)=x. Let R be the circle with radius 2 around O; then h(T)⊂R because ||h(x)||=||f(x)−x||≤||f(x)||+||x||≤2 (it is in fact contained in the circle with radius 1, but the size of the circle is inconsequential). We will use exercise 5 to show that h maps a point to the center, which is O, from which the desired result follows. For this, we shall show that it has the needed properties, with the modification that points on any side may map precisely into the center. It’s obvious that weakening the requirement in this way preserves the result.
Rotate the disk so that the red shape is on top. In polar coordinates, the green area now contains all points with angles between 60∘ and 180∘ , the blue area contains those between 180∘ and 300∘, and the red area those between 0∘ and 60∘ and those between 300∘ and 360∘. We will show that h(¯¯¯¯¯¯¯¯AB) does not intersect the red area, except at the origin. First, note that we have
h(¯¯¯¯¯¯¯¯AB)={f(x)−x|x∈¯¯¯¯¯¯¯¯AB}⊂{y−x|x∈¯¯¯¯¯¯¯¯AB,y∈T}={T−x|x∈¯¯¯¯¯¯¯¯AB}=T−¯¯¯¯¯¯¯¯AB
Since both T and ¯¯¯¯¯¯¯¯AB are convex combinations of finitely many points, it suffices to check all combinations that result by taking a corner from each. This means we need to check the points
A−A and B−A and C−A and A−B and B−B and C−B.
Which are easily computed to be
(0,0) and (−1,0) and (12,−12√3) and (1,0) and (0,0) and (−12,−12√3)
Two of those are precisely the origin, the other four have angles 270∘ and 150∘ and 90∘ and 210∘. Indeed, they are all between 60∘ and 300∘.
Now one needs to do the same for the sets T−¯¯¯¯¯¯¯¯OA and T−¯¯¯¯¯¯¯¯BO, but it goes through analogously.