Which leaf shape predominates in rainy regions? Explain?

Prove: 5 regions of any shape and size...?

• Suppose you have five two dimensional regions of any shape and size on a plane. The only requirement for the existence of these shapes is that each and every one borders all of the others. Note that a region can not cross over another region, because that would be considered dividing the shape that is overbridged in two parts. Prove (or disprove, even though you won't be able to) that it is impossible to draw such 5 shapes.

• Answer:

  Just going to sketch a proof, and not be too detailed. By one-point compactifying the plane to the standard sphere, we can assume the five regions, A, B, C, D, and E are each open and relatively compact. (We have to assume the regions are open. else we can "squeeze" a region through the pinch between two other regions and get a counterexample.) That two regions A and B are said to border if there exist a point p on the boundary of A that is also on the boundary of B such that p is an interior point of the closure of (A union B). (Roughly speaking, the border has to be one dimensional, and not just a point). We will let p(A,B) denote one such border point between A and B. Assume each of C, D and E borders both A and B. We will show that there exists a pair among C,D,E that cannot border each other. By the connectedness of A, B, and C, there exists a closed curve g such that g passes through p(A,B), p(B, C), p(A,C); that g's restriction to A, B, and C are each connected; and that g lives in the interior of cl(A union B union C). g divides the plane into the interior and exterior regions. Since a neighborhood of g is contained in cl(A union B union C), and D cannot intersect A,B or C, we see that the sets D and E must either lie entire in the interior or entirely in the exterior. If D and E are such that one is in the interior, the other in the exterior, we are done. Since an ε tubular neighborhood of g doesn't intersect D and E, the minimum distance between D and E > 0 and they cannot border each other. Suppose D and E are both on one side of g. Let h(A) be a curve connected p(D,A) and p(E,A) such that except the the endpoints, h(A) lives in A. We can even choose h(A) such that it does not intersect g. Similarly choose h(B). choose a interior point on h(A) and connect it to g in A. Similarly connect h(B) to g in B. Connecting p(D,A) and p(D,B) we get a curve g(D) that passes through A,B and D in the interior. similarly we get a curve g(E). Call the original g g(C). By construction the points of g(E) that are not shared with g(C) must lie entirely to the interior of exterior of g(C). This holds for any pair. Therefore we can speak of the inner-most, outer-most, and middle g. with possibly a relabeling of D and E, we can assume the middle one is g(D). Then it is clear that C and E lie one on the interior and one on the exterior of g(D). By the previous argument C and E cannot touch. . Edit: Oh yeah, about the one point compactification: you don't really have to do that. I need it to guarantee that g is a bounded curve. You can circumvent that by being more careful with the definitions and not allowing a region that loops to infinity and then back.