
Re: compact
Posted:
May 15, 2013 10:34 PM


On Wed, 15 May 2013, dullrich@sprynet.com wrote: > On Wed, 15 May 2013 00:06:25 0700, William Elliot <marsh@panix.com> > > > >> > 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, > > "has cardinality less than A", not "is finite". > Wrong place for that comment. See ** below.
> > 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),
** which since A is infinite, is < A, which of course it isn't.

