```Date: Feb 3, 2013 4:32 AM
Author: Achimota
Subject: Re: looking for example of closed set that is *not* complete in a<br> metric space

On Sunday, February 3, 2013 1:24:54 AM UTC+8, quasi wrote:> Suppose (X,d) is not complete.  Then there must exist a > Cauchy sequence in X which does not converge. Let Y be the > set of distinct elements of that Cauchy sequence. Then any > infinite subset of Y is closed in X but not complete. Sorry to bother you again. I still don't follow.Why is Y closed in (X,d)?On Sunday, February 3, 2013 1:24:54 AM UTC+8, quasi wrote:> Daniel J. Greenhoe wrote:> > >Butch Malahide wrote:> > >>quasi wrote> > >>>Butch Malahide wrote> > >>>>> > >>>>If (X,d) is not complete, then it has at least one closed> > >>>>subspace which is not complete, namely, (X,d) is a closed> > >>>>subspace of itself.> > >> > >Understood.> > >> > >>>Moreover, if (X,d) is not complete, it has uncountably many> > >>>subsets which are closed but not complete.> > >>> > >> Oh, right. At least 2^{aleph_0} of them.> > >> > >Not understood. Can someone help me understand this one?> > > > Suppose (X,d) is not complete.  Then there must exist a > > Cauchy sequence in X which does not converge. Let Y be the > > set of distinct elements of that Cauchy sequence. Then any > > infinite subset of Y is closed in X but not complete. Since > > Y is countably infinite, the cardinality of the set of > > infinite subsets of Y is 2^(aleph_0).> > > > quasi
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