If $\mathcal{F}$ is a filter on $X$, will the below conditions be equivalent?

(1) $\mathcal{F}$ is an ultrafilter.

(2) For every $ \emptyset \neq M \subset X$, either $M \in \mathcal{F}$ or $X - M \in \mathcal{F}$.

(3) If $F_1 \cup \ldots \cup F_n \in \mathcal{F}$, then there is $j$ s.t $F_j \in \mathcal{F}.$

I know the proofs of (2‎)$\implies$(3) and (3)$\implies$(1).

Could you help me to prove (1)$\implies$(2)?

  • $\begingroup$ How do you define ultrafilter without using (2)? $\endgroup$ – Martin Argerami Sep 14 '13 at 1:15

Suppose that $\mathscr{F}$ is an ultrafilter on $X$, and let $\varnothing\ne M\subsetneqq X$. Suppose that $M\notin\mathscr{F}$. Because $\mathscr{F}$ is an ultrafilter, it is a maximal filter, and therefore $\mathscr{F}\cup\{M\}\subseteq\mathscr{G}$ cannot be extended to a filter on $X$; this implies that $F\cap M=\varnothing$ for some $F\in\mathscr{F}$. Clearly $F\subseteq X\setminus M$; since $\mathscr{F}$ is a filter, this implies that $X\setminus M\in\mathscr{F}$. Thus, either $M\in\mathscr{F}$, or $M\notin\mathscr{F}$, in which case (as we’ve just proved) $X\setminus M\in\mathscr{F}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.