
Re: compact
Posted:
May 15, 2013 12:27 AM


On May 14, 11:15 pm, William Elliot <ma...@panix.com> wrote: > A point x is an axxumulation 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

