
Re: looking for example of closed set that is *not* complete in a metric space
Posted:
Feb 3, 2013 4:32 AM


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

