FractionalGeometry-Chap2

Xy b wk z k xy c figure 229 showing x y

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Unformatted text preview: of A guarantees there is a minimum distance between (x0 , y0 ) and A. That is, min{d((x0 , y0 ), (x, y )) : (x, y ) ∈ A} = δ0 > 0. Recall the contraction ratio r satisfies r < 1, so we can take k large enough that rk · h((x0 , y0 ), A) < δ0 . In this setting, eq (2.9) of the proof of Theorem 2.3.1 becomes h(T k (x0 , y0 ), A) ≤ rk · h((x0 , y0 ), A). Recalling (x0 , y0 ) ∈ 59 2.4. CONVERGENCE OF RANDOM IFS T k (x0 , y0 ), we see that (x0 , y0 ) ∈ Ark ·h((x0 ,y0 ),A) . That is, there is some point (x, y ) ∈ A for which d((x0 , y0 ), (x, y )) ≤ rk · h((x0 , y0 ), A) < δ0 . This contradicts min{d((x0 , y0 ), (x, y )) : (x, y ) ∈ A} = δ0 , and so we conclude (x0 , y0 ) ∈ A. Proof of Lemma 2.4.2. From Theorem 2.3.1 recall T (A) = A, and by Lemma 2.4.1, the fixed point (x0 , y0 ) ∈ A. Now (x1 , y1 ) = Ti1 (x0 , y0 ) ∈ T (x0 , y0 ) ⊆ T (A) = A (x2 , y2 ) = Ti2 (Ti1 (x0 , y0 )) ∈ T (T (x0 , y0 )) = T 2 (x0 , y0 ) ⊆ T 2 (A) = A Continuing in this fashion proves the lemma. Proof of Theorem 2.4.1. Because Ti is a d-contraction with contraction factor ri , it is easy to see for an...
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