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Unformatted text preview: Math 220A Complex Analysis Solutions to Homework #3 Prof: Lei Ni TA: Kevin McGown Conway, Page 24, Problem 5. Let X be the set of all bounded sequences of complex numbers together with the metric d induced by the sup norm k x k = sup { x k  : k Z + } , i.e. d ( x,y ) = k x y k . Show that for each x X and > 0 the set B ( x ) is not totally bounded, although it is complete. Proof. Fix x X and > 0. First we show that B ( x ) is complete. If we can show that X is complete, then B ( x ) is a closed subspace of a complete space and therefore complete. Let x ( j ) be an arbitrary Cauchy sequence in X . Consider the following array of complex numbers: x (1) 1 x (1) 2 ... x (1) k ... x (2) 1 x (2) 2 ... x (2) k ... . . . . . . . . . x ( j ) 1 x ( j ) 2 ... x ( j ) k ... . . . . . . . . . Define f j : Z + C by f j ( k ) = x ( j ) k . Since x ( j ) is Cauchy in X , it follows that f j is uniformly Cauchy and hence converges uniformly to a function f : Z + C .....
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 Three '09
 Cong
 Math, Complex Numbers

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