FractionalGeometry-Chap2

# 232 iterating this argument we obtain ht k d a rk

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Unformatted text preview: , T1 (B ) ∪ (T2 (B ) ∪ · · · ∪ Tn (B ))) ≤ max{h(T1 (A), T1 (B )), h(T2 (A) ∪ · · · ∪ Tn (A), T2 (B ) ∪ · · · ∪ Tn (B ))} ≤ ··· ≤ ≤ max{h(T1 (A), T1 (B )), h(T2 (A), T2 (B )), . . . , h(Tn (A), Tn (B ))} ≤ max{r1 · h(A, B ), r2 · h(A, B ), . . . , rn · h(A, B )} = max{r1 , r2 , . . . , rn } · h(A, B ) = r · h(A, B ) where the ﬁrst inequality follows from Lemma 2.3.1, the next string of inequalities by continuing to apply Lemma 2.3.1, and the last inequality by Lemma 2.3.2. 48 CHAPTER 2. ITERATED FUNCTION SYSTEMS Proof of Theorem 2.3.1. Take B a compact set large enough that Ti (B ) ⊆ B for i = 1, ..., n. The eﬀect of the translation should be negligible compared with the eﬀect of the contraction. For the gasket with vertices (0, 0), (1, 0), and (0, 1), the square with vertices (0, 0), (1, 0), (1, 1), and (0, 1) can be used for B ; the square with vertices (0, 0), (3/4, 0), (3/4, 3/4), and (0, 3/4) cannot. Then B ⊇ T1 (B ) ∪ ... ∪ Tn...
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## This document was uploaded on 02/14/2014 for the course MATH 290B at Yale.

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