hom3sol - Solutions to Homework 2 February 5, 2009 19.2(b)...

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Unformatted text preview: Solutions to Homework 2 February 5, 2009 19.2(b) Prove that f ( x ) = x 2 is uniformly continuous on [0 , 3] using the - definition: Proof: Note that | x + y | 6 for all x, y [0 , 3] by the triangle inequality. For epsilon > 0 let = 6 so that for all | x- y | < and x, y [0 , 3] we have | f ( x )- f ( y ) | = | x 2- y 2 | = | x- y | ( x + y ) 6 | x- y | < 6 = . Therefore f ( x ) is uniformly continuous on [0 , 3]. 19.4(a) Prove that if f is uniformly continuous on a bounded set S, then f is a bounded function on S. Proof by contradiction: Assume that f is uniformly continuous on a bounded set S, but f is not bounded on S. Let f (S) = { y | y = f ( x ) for any x S } , then f (S( is not bounded since f is unbounded on S. Then ( y n ) f (S) such that y n < y n +1 n and lim n y n = . For each y n x n dom f such that y n = f ( x n ) (this follows immediately for the fact that y n f (S).) So ( x n ) is a sequence in S, and since S is bounded so is ( x n ). Thus ( x n ) has a convergent subsequence ( x n k ) by Bolzano-...
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hom3sol - Solutions to Homework 2 February 5, 2009 19.2(b)...

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