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