
compact
Posted:
May 15, 2013 12:15 AM


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?

