This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 Ç« Î´ definition: Proof: Note that  x + y  â‰¤ 6 for all x, y âˆˆ [0 , 3] by the triangle inequality. For epsilon > 0 let Î´ = Ç« 6 so that for all  x y  < Î´ and x, y âˆˆ [0 , 3] we have  f ( x ) f ( y )  =  x 2 y 2  =  x y  ( x + y ) â‰¤ 6  x y  < 6 Î´ = Ç« . 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 âˆˆ S } , then f (S( is not bounded since f is unbounded on S. Then âˆƒ ( y n ) âŠ‚ f (S) such that y n < y n +1 âˆ€ n and lim n â†’âˆž y n = âˆž . For each y n âˆƒ x n âˆˆ dom f such that y n = f ( x n ) (this follows immediately for the fact that y n âˆˆ 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...
View
Full Document
 Spring '09
 Sin, Metric space, Xn, BolzanoWeierstrass theorem

Click to edit the document details