Timeline for Stone-Čech compactification of $\mathbb R$

11 events

when toggle format	what		by	license	comment
Jun 15, 2020 at 7:27	history	edited	CommunityBot		Commonmark migration
Mar 4, 2012 at 15:35	comment	added	George Lowther		So $\beta X\setminus U=\beta(X\setminus U)$ for all locally compact $X$ and relatively compact open $U$. Looks good to me.
Mar 4, 2012 at 14:56	comment	added	Matthew Daws		@George, @BS: I've added the hypothesis that U is relatively compact. Thanks! George's 2nd comment is what's works for general $U$; and so I guess if $U$ isn't relatively compact, then $\beta X\setminus U$ is not $\beta(X\setminus U)$.
Mar 4, 2012 at 14:55	history	edited	Matthew Daws	CC BY-SA 3.0	added 70 characters in body
Mar 4, 2012 at 14:34	history	edited	Matthew Daws	CC BY-SA 3.0	Added hypothesis to (hopefully) correct argument
Mar 4, 2012 at 10:32	comment	added	George Lowther		I think $C(\beta X\setminus U)$ should be identified with $C_b(X)/C_0(U)$.
Mar 4, 2012 at 10:29	comment	added	BS.		I think $U$ is implicitly assumed relatively compact: "... then $F\in C_0(X)$"
Mar 4, 2012 at 9:44	comment	added	George Lowther		So, why doesn't this argument also work for $U=X$?
Mar 4, 2012 at 9:05	vote	accept	Mariarty
Mar 4, 2012 at 8:35	comment	added	Matthew Daws		In some sense this is nothing but a careful elaboration of George Lowther's hint!
Mar 4, 2012 at 8:33	history	answered	Matthew Daws	CC BY-SA 3.0