FractionalGeometry-Chap2

Because r a this shows p a1 2 this argument holds for

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Unformatted text preview: hyperlinked to its proof, allowing the general structure of ideas to be seen quickly, with details available whenever desired. Not being able to figure out a physical realization of a hyperlink, we settled for this. First, we establish a result necessary for proving the Hausdorff distance h satisfies the triangle inequality. Lemma 2.2.1 For any compact set A and for any positive numbers δ1 and δ2 (Aδ1 )δ2 ⊆ Aδ1 +δ2 40 CHAPTER 2. ITERATED FUNCTION SYSTEMS With this, we prove the Hausdorff distance is a metric. Proposition 2.2.1 The Hausdorff distance is a metric. Proofs Proof of Lemma 2.2.1. For every point p ∈ (Aδ1 )δ2 , we show p ∈ Aδ1 +δ2 . First, from the definition of thickening we see p ∈ (Aδ1 )δ2 implies there is a point q ∈ Aδ1 with d(p, q ) ≤ δ2 . Next, q ∈ Aδ1 implies there is a point r ∈ A with d(q, r) ≤ δ1 . By the triangle inequality for d, d(p, r) ≤ d(p, q ) + d(q, r) and so d(p, r) ≤ δ1 + δ2 . Because r ∈ A, this shows p ∈ Aδ1 +δ2 . This argu...
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