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Topic:
compact
Replies:
9
Last Post:
May 18, 2013 9:56 PM




Re: compact
Posted:
May 15, 2013 7:30 AM


On May 15, 2:06 am, William Elliot <ma...@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, 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?
Yes. Suppose S is not compact. If kappa is the minimum cardinality of an open cover of S which has no finite subcover, then there exists A subset S such that A = kappa and A has no saturation point.



