FractionalGeometry-Chap2

See the left side of fig 221 c the next largest holes

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Unformatted text preview: are not equal, in fact, they are disjoint, yet the Hausdorff distance between them is 0. On the other hand, the proof of positivedefiniteness is convincing. Have we found a logical inconsistency that will bring down all of mathematics, and so all of physics? Will airplanes fall out of the sky because of this? No: recall we defined the Hausdorff metric on compact sets, and Q and I are not compact, because neither is closed. FiThe part of the proof of Prop. 2.2.1 that relies on compactness of the sets will reveal why this argument cannot be applied to Q and I . PrxsSol 2.2.2 (a) The Sierpinski carpet consists of 8 copies of itself, each scaled by r = s = 1/3, without rotation or reflection. Following the image of the lower left corner (0, 0) of the carpet in each small copy of the carpet, we find the translations and consequently the IFS rules in the table below. (b) For an ǫ-nbhd Oǫ to be simply-connected, we must fill all the holes in O. The largest is the 1/3 × 1/3 hole in the middle, and the point in that hole farthest from O is (1/2, 1/2). The distance between th...
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