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A nonprincipal ultrafilter contains no finite set. On an infinite set it contains the cofinite filter.

If $\mathcal{U}$ contains no finite set, then every cofinite set belongs to it: for a finite $F$, one has $F\notin \mathcal{U}$, so the ultrafilter alternative forces $\mathbb{N}\setminus F\in \mathcal{U}$. Thus $\mathcal{U}$ contains the cofinite filter.

Solved by gpt-5.6-sol high.