Date: Feb 2, 2013 9:52 PM
Author: Shmuel (Seymour J.) Metz
Subject: Re: closed but not complete
In <Pine.NEB.4.64.1302011852290.7448@panix3.panix.com>, on 02/01/2013

at 07:14 PM, William Elliot <marsh@panix.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.

--

Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel>

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