Re: Banach Spaces / equivalent norms



On Tue, 03 Apr 2007 23:22:03 EDT, craig <ctcowan@xxxxxxxxxxx> wrote:

Does anyone know of an easy example of a vector space with two non-equivalent norms and for which both spaces are complete?

I doubt that an easy example exists, in the sense in which I suspect
you mean.

One can give an easy proof from the Axiom of Choice that such an
example exists - if that's not clear I can show you that. But
I doubt that the easy AC proof of existence counts as an easy
_example_; it doesn't really "give" the example...

thanks


craig


************************

David C. Ullrich
.



Relevant Pages

  • Re: Equivalent to Axiom of Choice?
    ... The easy proof I've found of this uses the axiom of choice (well, ... No, it's called the order extension principle, and it's weaker than ... look up the relevant literature. ...
    (sci.math.research)
  • Re: Equivalent to Axiom of Choice?
    ... In article, Thomas Andrews ... The easy proof I've found of this uses the axiom of choice (well, ... Rubin and Jean E. Rubin, ...
    (sci.math.research)
  • Re: Equivalent to Axiom of Choice?
    ... Thomas Andrews wrote: ... The easy proof I've found of this uses the axiom of choice (well, ... Stefan Wehmeier ...
    (sci.math.research)
  • Re: Matrix Algebra question
    ... Let U_2 be its transpose. ... I am looking for an easy proof of the following fact: ... The group (with matrix multiplication) generated by ... David C. Ullrich ...
    (sci.math)
  • Equivalent to Axiom of Choice?
    ... Every partially ordered set can be extended to a total order. ... The easy proof I've found of this uses the axiom of choice (well, ...
    (sci.math.research)