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Re: Precompactness
Posted:
Dec 13, 2012 4:19 PM
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On Dec 13, 1:52 am, William Elliot <ma...@panix.com> wrote:
> On Wed, 12 Dec 2012, Kaba wrote:
> > 12.12.2012 5:24, William Elliot wrote:
> > > On Tue, 11 Dec 2012, Kaba wrote:
> > > > Let X be a locally compact Hausdorff space. Is every open set of X
> > > > precompact (compact closure)?
> > 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.
>
> > 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?
>
> > 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.
More generally, for any space X in which every compact set has a
compact closure (this includes compact spaces and kc spaces), a subset
of X is precompact if and only if is bounded.
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