Date: Feb 2, 2013 12:32 AM
Author: Butch Malahide
Subject: Re: looking for example of closed set that is *not* complete in a<br> metric space
On Feb 1, 11:10 pm, "Daniel J. Greenhoe" <dgreen...@yahoo.com> wrote:
> On Saturday, February 2, 2013 12:52:55 AM UTC+8, peps...@gmail.com wrote:
> > ...To say that a space is "closed"
> > (as in your statement "closed -> complete") doesn't really mean anything.
> > To make progress replace "closed -> complete" by something more
> > formal and rigorous and precise.
> This is certainly good advice and many apologies for my sloppy original posting. Is the following any better?...
> Let (X,d) be a metric space.
> Let T be the topology induced by d and
> (X,T) be the resulting topological space.
> Let Y be a subset of X.
> (Y,d) is complete ==> Y is closed in (X,d).
> (Y,d) is complete ==> Y is closed in (X,T).
> But what about the converse? That is, is this true?
> Y is closed in (X,d) ?==>? (Y,d) is complete
Is (X,d) complete? If (X,d) is a complete metric space, then every
closed subspace of (X,d) is complete. If (X,d) is not complete, then
it has at least one closed subspace which is not complete, namely,
(X,d) is a closed subspace of itself.