Re: Banach/Analysis Proof

In article <1159463222.262722.269100@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Jason Pawloski <jpawloski@xxxxxxxxx> wrote:
I need to prove the following:

For D a subset of R^n+1, let f:D -> R^n be continuous, and let K be a
compact subset of R^n+1 such that K is a subset of D. Show that there
exists an open subset U of R^n+1 such that K is a subset of U and f is
bounded on U intersect D.

I'm further told to try to construct a proof that would work on an
arbitrary Banach space. I've never had a class that involved Banach
spaces, so I wiki'd it to find its a complete normed vector space.
Since the norm satisfies the triangle inequality, I assume you use the
norm as your distance function in the metric space. Is this correct?

Yes.

Also, I'm not sure what I should take to be an "arbitrary Banach
space." Is there some convention that its the image or the preimage of
a function? In other words, is f:D-> R^n mean that D should be the
arbitrary Banach space?

D is an arbitrary subset of an arbitrary Banach space X. You could
also replace R^n by another arbitrary Banach space Y.

Either way, I don't see how to proceed with this proof. Obviously if f
is continuous on a compact set, then it achieves its maximumand minimum
and hence must be bounded. But I don't see how to extend this to an
open set larger than K...

Hint: for each x in K, there is an open ball B_r(x)
= {v in X: ||v - x|| < r} such that ||f(v) - f(x)|| < 1
for all v in B_r(x) intersect D.

Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.