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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 2007-11-11 20:45

>>10
That's what I meant by saying that a proof by contradiction might put us on the path to understanding why a theorem holds. But often, especially in more advanced mathematics, proof by contradiction may lead you round a merry path that, while indeed giving a contradiction, offers your insight little more than a string of abstract routes.
I'm trying to think of an example to illustrate my point but it's getting on 2am over here and I'm tired :(

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