```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.> Then>   (Y,d) is complete ==> Y is closed in (X,d).> Alternatively,>   (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 completeIs (X,d) complete? If (X,d) is a complete metric space, then everyclosed subspace of (X,d) is complete. If (X,d) is not complete, thenit has at least one closed subspace which is not complete, namely,(X,d) is a closed subspace of itself.
```