
Re: G_delta
Posted:
Jan 21, 2013 5:00 AM


On Jan 21, 2:59 am, William Elliot <ma...@panix.com> wrote: > 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.

