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metric space topology

Name: Anonymous 2007-11-09 20:17

halp, how do I show that the set of bounded functions under the supremum norm is a complete metric space?

Name: Anonymous 2007-11-10 8:16

>>1

If a sequence is cauchy in the sup norm that implies the sequence of functions converges uniformly.

Then use the definition of uniform convergence to proof that a sequence of bounded functions cannot converge to an unbounded one uniformly, which is pretty obvious.

I'll sketch a proof.

Ok, assume there exists a sequence -> f such that f is not bounded.

Now for any fn in the sequence there is a max |fn| , call this M.
However as f is unbounded there exists a point x s.t f(x) > M + 1
Therefore for all n sup|fn - f| > 1 at least. but then fn is not a cauchy sequence, contradiction.



Hmm, didn't use uniform convergence, oh well.

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