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

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

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.