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From: dushya on 20 Jun 2010 08:34 suppose a space X be such that closure of every open set U (=! X) in X is compact. does it imply that the space X is compact ?
From: José Carlos Santos on 20 Jun 2010 09:56 On 20-06-2010 13:34, dushya wrote: > suppose a space X be such that closure of every open set U (=! X) in X > is compact. does it imply that the space X is compact ? If X contains at least one point _p_ such that the set {p} is closed then, yes, it is true. Simply write: X = {p} U (X\{p}). Then X = (closure of {p}) U (closure of (X\{p})) = {p} U (closure of (X\{p})). It is the union of two compact sets, and therefore it is compact. Best regards, Jose Carlos Santos
From: dushya on 20 Jun 2010 10:38 On Jun 20, 6:56 am, José Carlos Santos <jcsan...(a)fc.up.pt> wrote: > On 20-06-2010 13:34, dushya wrote: > > > suppose a space X be such that closure of every open set U (=! X) in X > > is compact. does it imply that the space X is compact ? > > If X contains at least one point _p_ such that the set {p} is closed > then, yes, it is true. Simply write: > > X = {p} U (X\{p}). > > Then > > X = (closure of {p}) U (closure of (X\{p})) > > = {p} U (closure of (X\{p})). > > It is the union of two compact sets, and therefore it is compact. > > Best regards, > > Jose Carlos Santos and we can replace {p} with any nonempty open set not equal to X. thank you Jose :-)
From: José Carlos Santos on 20 Jun 2010 10:58 On 20-06-2010 15:38, dushya wrote: >>> suppose a space X be such that closure of every open set U (=! X) in X >>> is compact. does it imply that the space X is compact ? >> >> If X contains at least one point _p_ such that the set {p} is closed >> then, yes, it is true. Simply write: >> >> X = {p} U (X\{p}). >> >> Then >> >> X = (closure of {p}) U (closure of (X\{p})) >> >> = {p} U (closure of (X\{p})). >> >> It is the union of two compact sets, and therefore it is compact. >> >> Best regards, >> >> Jose Carlos Santos > > and we can replace {p} with any nonempty open set not equal to X. Why? Let A be such a set. Then X is the union of its closure and its complement and the closure of A is compact. But X\A doesn't have to be (or, at least, I don't see why it should be). Best regards, Jose Carlos Santos
From: dushya on 20 Jun 2010 11:18 On Jun 20, 7:58 am, José Carlos Santos <jcsan...(a)fc.up.pt> wrote: > On 20-06-2010 15:38, dushya wrote: > > > > > > >>> suppose a space X be such that closure of every open set U (=! X) in X > >>> is compact. does it imply that the space X is compact ? > > >> If X contains at least one point _p_ such that the set {p} is closed > >> then, yes, it is true. Simply write: > > >> X = {p} U (X\{p}). > > >> Then > > >> X = (closure of {p}) U (closure of (X\{p})) > > >> = {p} U (closure of (X\{p})). > > >> It is the union of two compact sets, and therefore it is compact. > > >> Best regards, > > >> Jose Carlos Santos > > > and we can replace {p} with any nonempty open set not equal to X. > > Why? Let A be such a set. Then X is the union of its closure and its > complement and the closure of A is compact. But X\A doesn't have to be > (or, at least, I don't see why it should be). > > Best regards, > > Jose Carlos Santos- Hide quoted text - > > - Show quoted text - i don't know but please let me know if anything is wrong with following argument -- B=clr(A) is compact.(and we can assume that it is not equal to X fot if it is then X is compact) => X- clr(A) is open and is niether empty nor equal to X. => C=clr(X-clr(A))is compact. And X is union of B and C. so X is compact.
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