Banach/Analysis Proof

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?
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?

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...

Thanks, Jason

.

Relevant Pages

  • Metric Spaces Question
    ... Hi I've been struggling with this problem and I would really appreciate any help or hints anybody could give me. ... Let and be metric spaces ... Suppose C is a compact subset of X such that: ... prove that there exists a real number R>0 such that on the open subset G of X ...
    (sci.math)
  • Re: Metric Spaces Question
    ... f:X->Y is a continuous function ... Suppose C is a compact subset of X such that: ... prove that there exists a real number R>0 such that on the open subset G of X ... R is indeed positive since points where injection of f fails cannot be limit points of C. ...
    (sci.math)
Quantcast