FractionalGeometry-Chap2

# B find the hausdor distance between the unit circle

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Unformatted text preview: is point and the carpet 42 CHAPTER 2. ITERATED FUNCTION SYSTEMS is 1/(2 · 3), so O1/6 is simply-connected, and for every ǫ < 1/6, Oǫ has a hole in the middle. See the left side of Fig. 2.21. (c) The next-largest holes in the carpet have side length 1/9. Arguing as in (b), these holes are ﬁlled in O1/18 , and ǫ = 1/18 leaves the middle hole open. See the middle of Fig. 2.21. (d) The next-largest holes in the carpet have side length 1/27, so O1/54 ﬁlls these holes, and leaves open the 1 largest and 8 second-largest. See the right side of Fig. 2.21. (e) By now the general pattern is clear, once we recall the carpet has 1 hole of side length 1/3, 8 of side length 1/32 , 82 of side length 1/33 , and so on. The minimum ǫ for which Oǫ has 1 + 8 + 82 + · · · 8k holes is the minimum ǫ ﬁlling the 8k+1 holes of side length 1/3k+2 . That is, ǫ = 1/(2 · 3k+2 ). r 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 s θ ϕ e 1/3 0 0 0 1/3 0 0 1/3 1/3 0 0 2/3 1/3 0 0 0 1/3 0 0 2/3 1/3 0 0 0 1/3 0 0 1/3 1/3 0 0 2/3 The IF...
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## This document was uploaded on 02/14/2014 for the course MATH 290B at Yale.

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