In <Pine.NEB.firstname.lastname@example.org>, on 02/01/2013 at 07:14 PM, William Elliot <email@example.com> said:
>No. Assume K is a closed subset of the complete space (S,d).
You're answering a different question. The answer to the question he asked is yes.
>Conclusion. K subset complete S implies (K closed iff K complete).
The question "does someone know of any example that demonstrates that closed --> complete is *not* true?" does not assume completeness. If (Y,d) is a closed subspace of the metric space (X,d), it need not be complete. In fact, if (X,d) is not complete then (X,d) is a closed subspace of (X,d) that is not complete.
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