104_Fall09_hw_5_sol

# 104_Fall09_hw_5_sol - Solutions for Homework 10 &amp; 11,...

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Homework # 10 p. 193, # 33 Consider the characteristic function f of the dyadic rational numbers Z [1 / 2]. (a) What is the set of discontinuities of f ? The discontinuities of f will be R . Let x R - Z [1 / 2], then for any ± < 1, and for any δ > 0, there exists y Z [1 / 2], such that | x - y | < δ . Let n be such that 1 / 2 n < δ . By the Archimedean principle, for some k , k/ 2 n < x < ( k + 1) / 2 n , and therefore | x - k/ 2 n | < 1 / 2 n < δ . Then | f ( x ) - f ( y ) | = 1 > ± , so f is not continuous at x . If x Z [1 / 2], then again there exists y R - Z [1 / 2] such that | x - y | < δ for any δ , and therefore | f ( x ) - f ( y ) | = 1 > ± . Applying the previous case to x - 2, we see there is y 0 Z [1 / 2] such that | x - 2 - y 0 | < δ . Then let y = 2 + y 0 R - Z [1 / 2], so | x - y | < δ . (b) At which points is its oscillation κ ? Since f takes on only values of 0 , 1, osc x f 1, for all x . The above argument shows that osc x f = 1, for all x R , since for any δ , diam ( f ( x - δ,x + δ )) = 1. Therefore, if κ > 1, then osc x f < κ for all x , and if κ 1, then osc x f κ for all x . (c) Is

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## 104_Fall09_hw_5_sol - Solutions for Homework 10 &amp; 11,...

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