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

A filter on a set $\mathcal{U}$ on $\mathbb{N}$ is a nonempty family of subsets of $\mathbb{N}$ such that:

These conditions imply $\mathbb{N}\in \mathcal{U}$.

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An ultrafilter is a proper filter maximal under inclusion. Equivalently, for every $A\subseteq \mathbb{N}$, exactly one of $A$ and $\mathbb{N}\setminus A$ belongs to $\mathcal{U}$.

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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.

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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.

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Yes. Choose the principal ultrafilter at any $j\ge 2$. Evaluation at the $j$th coordinate giveswhich is a finite cyclic group of order $j$.

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No. Suppose a formula with parameters defined a function $f:\mathcal{C}\to \mathcal{C}$ that was surjective and not injective. The assertions that the formula defines a function, that the function is surjective, and that it is not injective are all first-order statements about that formula and those parameters. By Łoś theorem, they would hold simultaneously in ${\mathcal{C}}_i$ for $\mathcal{U}$-almost every $i$.

Every surjective self-map of a finite set is injective, so no finite factor can satisfy those statements. This contradiction shows that the ultraproduct has no such definable function.

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Yes. Choose a nonprincipal ultrafilter on $\mathbb{N}$ and letFor each fixed positive integer $n$, the equality $na=0$ holds in the $i$th factor exactly when $i$ divides $n$. Only finitely many positive integers divide $n$, so the cofinite set of indices for which $na\ne 0$ belongs to $\mathcal{U}$. Łoś theorem gives $na\ne 0$ in $\mathcal{C}$ for every $n>0$. Thus $a$ has infinite order of a group element.

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