On Feb 1, 10:37 am, "Daniel J. Greenhoe" <dgreen...@yahoo.com> 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? I don't know for sure that it is not true, but I might > guess that it is not true.
Do you happen to know an example of a metric space which is not complete? If so, let (X,d) be that metric space, and let Y = X.
If not, do you know an example of a metric space in which some subset is not closed? In that case, let (W,d) be that metric space, let X be a non-closed subset of W, and let Y = X. Then (X,d) is an incomplete metric space, (Y,d) is a closed subspace of (X,d), and (Y,d) is not complete.