
G_delta
Posted:
Jan 24, 2013 3:33 AM


On Mon, 21 Jan 2013, Butch Malahide wrote: > On Jan 21, 2:59 am, William Elliot <ma...@panix.com> wrote:
You've proven all of this and yet claim you don't know (point set) topology?
> > Ok, the other difference is > > a) regular & every point a G_delta > > and > > b) every point the countable intersection of closed nhoods of the > > point. > > > > That is for all p, there's some closed Kj, j in N with > > for all j in N, p in int Kj & {p} = /\_j Kj. > > > > Yes, a) > b) but I doubt the converse. > > In other words, is a) strictly stonger than b). > > > > Now if we assume a Lindelof space, is
> > a) still strictly stronger than b? > > > > I think it is. Are you of the same opinion? > > Let Y be the real line with the topology generated by the usual open > sets and the set Q of all rational > numbers. Offhand, it seems to me that Y is a Lindelof space which has > property b) but is not regular. Let's see. > > Is Y Lindelof? Q is Lindelof because it's countable. Y\Q is Lindelof > because the subspace topology on Y\Q is just the usual topology of the > irrational numbers. Y is Lindelof because it's the union of two > Lindelof subspaces, Q and Y\Q. > > Does Y have property b)? The real line with the usual topology has > property b). If K is a closed neighborhood of p in the real topology, > then it is a closed neighborhood of p in the stronger topology of Y. > Hence Y has property b). > > Is Y regular? No, because Q is a neighborhood of zero which contains > no closed neighborhood of zero. >

