Date: Dec 13, 2012 3:11 PM
Author: Kaba
Subject: Re: Precompactness

13.12.2012 9:52, William Elliot wrote:
> On Wed, 12 Dec 2012, Kaba wrote:
>> 12.12.2012 5:24, William Elliot wrote:
>>> On Tue, 11 Dec 2012, Kaba wrote:
>> Related, let X be a Hausdorff space. Royden (Real analysis) defines E subset X
>> to be _bounded_ if it is contained in a compact set. It seems to me that
>> precompact and bounded are equivalent properties.

> It's not for R with the include 0 topology, { U | 0 in U } \/ {empty set}.
> {0} is bounded but not precompact.

Therefore R is not Hausdorff:) Interesting example though of what can
wrong without the Hausdorff requirement.

>> Assume E is precompact. Then cl(E) is a compact set which contains E.
>> Therefore E is bounded. Assume E is bounded. Then there is a compact set K
>> such that E subset K. Since X is Hausdorff, K is closed.

> Why is X Hausdorff?

By definition.

>> subset K. Since cl(E) is a closed subset of a compact set K, cl(E) is compact.
>> Therefore E is precompact.
>> Unless I am missing something?

> Assuming X is Hausdorff.
> For Hausdorff spaces, or more general, kc spaces (compact sets are
> closed), precompact and bounded are equivalent. Otherwise, they're not.

That's a nice term to know, thanks.