
Re: compact
Posted:
May 15, 2013 3:06 AM


On Tue, 14 May 2013, Butch Malahide wrote: > On May 14, 11:15 pm, William Elliot <ma...@panix.com> wrote:
> > A point x is an accumulation point of A > > when for all open U nhood x, U /\ A is infinite. > > > > If S is compact, then every infinite set A has an accumulation point. > > Proof. > > If not, then for all x, there's some open U_x nhood x with finite U /\ A. > > Since C = { U_x  x in S } covers S, > > there's a finite subcover { U_x1,.. U_xj } > > Thus A = A /\ (U_x1 \/..\/ U_xj } = (U_x1 /\ A) \/..\/ (U_xj /\ A), > > is finite, which of course it isn't. > > > > If every infinite A subset S has an accumulation point, is S compact? > > omega_1 A point x is a saturation point of A when for all open U nhood x, U /\ A = A
If S is compact, then every infinite set A has saturation point. Proof. If not, then for all x, some open U_x nhood x with U_x /\ A < A. Since C = { U_x  x in S } covers S, . . there's a finite subcover { U_x1,.. U_xj } Thus A = A /\ (U_x1 \/..\/ U_xj } = (U_x1 /\ A) \/..\/ (U_xj /\ A), is finite which of course it isn't.
If every infinite A subset S has a saturation point, is S compact?

