
Re: compact
Posted:
May 15, 2013 11:24 AM


On Wed, 15 May 2013 00:06:25 0700, William Elliot <marsh@panix.com> wrote:
>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,
"has cardinality less than A", not "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?

