If a set X has the finite meet property, then there is an ultrafilter such that X is a subset of it.

Question

I need to prove that if $X \subseteq B$ is a set with the finite meet property, then there exists an unltrafilter $U$ of $B$ such that $X \subseteq U$. I know that the finite meet property means that the meet of all the elements in the subset is not 0. I think I have to use the ultrafilter theorem, that says that if F is a proper filter, then there an ultrafilter, so I need to prove that $X$ is a prime filter, am I right? There I start to have trouble, I think I need to prove that for every $a \in B$ either $a \in X$ or the complement of $a$ is. I can prove that just one is because if it wasn't then the finite meet condition wouldn't hold. I would really appreciate any input on both, how to continue, and some ideas to write it formally. Thank you.

Answer 1

HINT: Let

$$A=\left\{\bigwedge F:F\subseteq X\text{ is finite}\right\}\;,$$

and let $F=\{x\in B:\exists a\in A(a\le x)\}$. Show that $F$ is a proper filter, and apply the ultrafilter theorem to $F$.