On Feb 2, 2:36 am, William Elliot <ma...@panix.com> wrote: > On Fri, 1 Feb 2013, Butch Malahide wrote: > > On Feb 1, 9:14 pm, William Elliot <ma...@panix.com> wrote: > > > On Fri, 1 Feb 2013, Daniel J. Greenhoe wrote: > > > > > Let (Y,d) be a subspace of a metric space (X,d). > > > > > If (Y,d) is complete, then Y is closed with respect to d. That is, > > > > complete==>closed. > > > > > Alternatively, if (Y,d) is complete, then Y contains all its limit > > > > points. > > > > Would anyone happen to know of a counterexample for the converse? > > > > That is, does someone know of any example that demonstrates that > > > > closed --> complete is *not* true? > > > > No. Assume K is a closed subset of the complete space (S,d). > > > But the original poster did not say that his metric space (X,d) was > > complete. > > Oh, so any closed subset of Q is an example.