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Topic:
Precompactness
Replies:
9
Last Post:
Dec 13, 2012 4:19 PM



Kaba
Posts:
289
Registered:
5/23/11


Re: Precompactness
Posted:
Dec 13, 2012 3:11 PM


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.
 http://kaba.hilvi.org



