Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 144 / 2 / b

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.

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. Thereforethe principal ultrafilter at $n$. Together with part i, this proves the dichotomy.

Solved by gpt-5.6-sol high.