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




compact
Posted:
May 18, 2013 9:56 PM


On Wed, 15 May 2013, Butch Malahide wrote: > On May 15, 2:06 am, William Elliot <ma...@panix.com> wrote:
> > 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. > Let C be a cover of S with minimal cardinality. Let eta = C, well order C = { U_xi  xi < eta }. and for all nu < eta, let C_nu + { U_xi  xi <= nu }.
By minimality, for all nu < eta, there's some a_nu in S  \/C_nu. A = { a_nu  nu < eta } is infinite. Otherwise, there'd be some nu < eta with A subset \/C_nu, so a_nu not in A.
Let x be a saturation point of A. Some nu < eta with x in U_nu.
x in V = \/{ U_xi  xi <= nu } V /\ A = { a_xi  xi <= nu }
Thus, provided A = C can be proved, the contradiction, V /\ A < A.



