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Re: Precompactness
Posted:
Dec 13, 2012 4:19 PM


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.



