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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)?

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  • $\begingroup$ How do you define ultrafilter without using (2)? $\endgroup$ – Martin Argerami Sep 14 '13 at 1:15
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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) $X\setminus M\in\mathscr{F}$.

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