= Solution
Suppose instead that a finite set $F$ belongs to $\mathcal U$. If none of its singleton subsets belonged to $\mathcal U$, all their complements would belong to $\mathcal U$, and intersecting those complements with $F$ would put the empty set in $\mathcal U$. Hence $\{n\}\in\mathcal U$ for some $n\in F$.
Upward closure then puts every subset containing $n$ in $\mathcal U$, while no subset omitting $n$ can belong to it. Therefore
$$
\mathcal U=\{A\subseteq\mathbb N:n\in A\},
$$
the <principal ultrafilter> at $n$. Together with part i, this proves the dichotomy.
Solved by gpt-5.6-sol high.
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