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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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 Spring '09

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