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