B first t 2 0 1 t 0 1 t 0 10 to nd

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Unformatted text preview: T (A)) ≤ r · h(D, A), by Prop. 2.3.2. Iterating this argument, we obtain h(T k (D), A) ≤ rk · h(D, A) (2.9) Because r < 1, we see limk→∞ h(T k (D), A) = 0. Alternative proof of Theorem 2.3.1 Another approach, uses the Contraction Mapping Theorem. Suppose d is a metric on a space X . A sequence of points {x1 , x2 , x3 , . . . } in X is Cauchy if for every ǫ > 0 there is an N for which d(xi , xj ) < ǫ whenever i, j > N . The space X is called complete in the metric d if every Cauchy sequence of points in X converges to a point in X . For example, X = R is complete in the metric d(x, y ) = |x − y |. This is a familiar example from calculus. On the other hand, Y = (0, 1) is not complete in d because the sequence {1/2, 1/3, 1/4, . . . } is Cauchy, but converges to 0 ∈ Y . / The Contraction Mapping Theorem states that if X is complete in the metric d and f : X → X is a contraction, then f has a unique fixed point in X . With some work we can show K(R2 ) is complete in t...
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This document was uploaded on 02/14/2014 for the course MATH 290B at Yale.

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