Claim: A family of sets A of S can be extended to a non-rpincipal ultrafilter iff every finite family of A has as an intersection an infinite set.
(1) I have been trying to prove this by showing that every set of A is infinite.
Then I argue that every finite collection of set belonging to A have the finite intersection property. So $A_1, \cap …, \cap A_n = X \neq \emptyset$. Since $X \in U$ we have immediatley that X is infinite by (1). |Does this make sense ?
Best Answer
Use your family of sets $\mathscr{A}$ as a base of a filter $\mathscr{F}$ on $S$, that is, define $$\mathscr{F} = \{B\subseteq S\colon B\supseteq A\text{ for some }A\in \mathscr{A}\}.$$ (Check that this is a filter indeed.) Now use the Kuratowski-Zorn lemma to argue that $\mathscr{F}$ is contained in some maximal filter (ultrafilter).