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The Strange World of the Hausdorff Metric Geometry


XIX. When Is A Hausdorff Line Not A Line - 1?


We have discussed many properties of betweeness in H(Rn), now we move on to lines. Theorem 5 tells us that, given distinct A and B in H(Rn), there is always at least one point C in H(Rn) at each location satisfying ACB. Can we say the same about elements satisfying ABC or CAB? The fascinating answer turns out to be no. The reasons are contained in Theorems 3 and 4 of [2], which can be summarized as follows.


Theorem 7: Let A and B be elements of H(Rn) and let r = h(A, B). If d(A, B) > d(B, A) and (A)s is a subset of Nr+s(B)

for some s > 0, then there is no element C in H(Rn) satisfying BCA and h(A, C) = s1 for any s1 > s.


So if A and B are in H(Rn) with


d(A, B) > d(B, A) and (A)s a subset of Nr+s(B) for some s > 0


for some s, then there can be no Hausdorff line containing elements farther than s units to the right of A. In other words, some Hausdorff lines have holes in them. Does this ever really happen?


Consider the example illustrated in the next applet. The set B is a two-point set {b1, b2} where b1 = (0, z) and b2 = (0, -z) for some z > 0. The set A is also a two-point set {a1, a2}, where a1 is the point at the origin and a2 = (x, y) is a point in the first quadrant with y < z and dE(a2, b2) > z. Then r = h(A, B) = d(A, B) = dE(a2, b2) > d(B, A). The set (A)s is colored light red and (B)r+s is light blue. Note that Nr+s(B) is the interior of (B)r+s. For some values of s, the single point c is in (A)s but not Nr+s(B). You can move every point except a1 to find configurations in which (A)s is a proper subset of Nr+s(B). In these examples there is a transition point where (A)s is not contained in Nr+s(B) for values of s less than some s0 but (A)s is a proper subset of Nr+s(B) for s greater than or equal to s0. It turns out that this is always the case and that, in fact, our Hausdorff lines actually become rays in these situations. In fact, we can construct configurations in which our rays stop as far to the right of A as we like.



How can we be sure that our Hausdorff lines really are rays? The answer is next.


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