FractionalGeometry-Chap2

Then b t1 b tn b t b applying t to this inclusion

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Unformatted text preview: V )δ . Combining these, we see h(U ∪ V, W ∪ X ) ≤ δ = max{h(U, W ), h(V, X )}. Proof of Lemma 2.3.2. Let δ = h(A, B ). We want to show T (A) ⊆ (T (B ))r·δ and T (B ) ⊆ (T (A))r·δ . That is, for every point a′ ∈ T (A), there is a point b′ ∈ T (B ) with d(a′ , b′ ) ≤ r · δ (and similarly for the reverse inclusion). For every a′ ∈ T (A), there is some point a ∈ A with T (a) = a′ . Because A ⊆ Bδ , there is some b ∈ B with d(a, b) ≤ δ . Let b′ = T (b). Certainly, b′ ∈ T (B ) and d(a′ , b′ ) = d(T (a), T (b)) ≤ r · d(a, b) ≤ r · δ = r · h(A, B ) That is, for all a′ ∈ T (A), there is b′ ∈ T (B ) with d(a′ , b′ ) ≤ r · δ , and so T (A) ⊆ (T (B ))r·δ . Similarly, T (B ) ⊆ (T (A))r·δ , and so h(T (A), T (B )) ≤ r · δ = r · h(A, B ). Proof of Prop. 2.3.2. Apply the definition of T and observe h(T (A), T (B )) = h(T1 (A) ∪ T2 (A) ∪ · · · ∪ Tn (A), T1 (B ) ∪ T2 (B ) ∪ · · · ∪ Tn (B )) = h(T1 (A) ∪ (T2 (A) ∪ · · · ∪ Tn (A))...
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