 Solution

A filter on a set $I$ means a proper family $\mathcal{F}\subseteq \mathcal{P}(I)$ with $I\in \mathcal{F}$, $\varnothing \notin \mathcal{F}$, closure under finite intersections, and upward closure under inclusion. An ultrafilter is a maximal proper filter; equivalently, for every $X\subseteq I$, exactly one of $X$ and $I\setminus X$ belongs to it. A principal ultrafilter is $\{X\subseteq I:i\in X\}$ for some $i\in I$. A nonprincipal ultrafilter is one not of that form, equivalently one containing no finite set.

To prove the ultrafilter lemma using Zorn's lemma, order all proper filters containing $\mathcal{F}$ by inclusion. The union of any chain is a proper filter: finitely many members lie together in one filter of the chain, and no chain member contains the empty set. Thus every chain has an upper bound, and Zorn's lemma gives a maximal member $\mathcal{U}$.

For completeness, its decision property follows from maximality. If $X\notin \mathcal{U}$, adjoining $X$ cannot generate a proper filter. Hence some $F\in \mathcal{U}$ has $F\cap X=\varnothing$. Upward closure gives $I\setminus X\in \mathcal{U}$. Both complements cannot be members, since their intersection is empty. Therefore $\boxed{\mathcal{F}\subseteq \mathcal{U}\text{ for an ultrafilter }\mathcal{U}}$. Properness is essential; a family containing the empty set cannot extend to an ultrafilter.