Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 116 / 1

A strongly inaccessible cardinal is an uncountable regular cardinal $\kappa$ that is also a strong limit cardinal:

A theory is $\kappa$-satisfiable when every subtheory of size below $\kappa$ has a model. An uncountable cardinal $\kappa$ is weakly compact when every $\kappa$-satisfiable theory in an infinitary language $L_{\kappa ,\kappa }$ with at most $\kappa$ nonlogical symbols is satisfiable.

Equivalently, $\kappa \to (\kappa {)}_2^2$: every coloring $c:[\kappa {]}^2\to 2$ has a homogeneous subset of cardinality $\kappa$.

A filter $U$ is $\kappa$-complete when intersections of fewer than $\kappa$ members remain in $U$. An uncountable cardinal $\kappa$ is a measurable cardinal when it carries a $\kappa$-complete nonprincipal ultrafilter.

An uncountable cardinal $\kappa$ is a strongly compact cardinal when every $\kappa$-satisfiable theory in any $L_{\kappa ,\kappa }$ language is satisfiable. Unlike weak compactness, there is no cardinality bound on the language.

Let $U$ witness that $\kappa$ is measurable. Regularity is given, so it remains to prove the strong limit cardinal property. First, every $A\in U$ has cardinality $\kappa$: if $|A|<\kappa$, thenby nonprincipality and $\kappa$-completeness, contradicting $A\in U$.

Suppose $\lambda <\kappa$ and $2^{\lambda }\ge \kappa$. Choose an injection $f:\kappa \to \mathcal{P}(\lambda )$. For each $\xi <\lambda$, exactly one oflies in $U$. Their chosen intersection lies in $U$ by $\kappa$-completeness. On that intersection every $f(\alpha )$ is the same subset of $\lambda$, contradicting injectivity because every member of $U$ has size $\kappa$. Thus $2^{\lambda }<\kappa$, and $\kappa$ is strongly inaccessible.

Let $F$ be a $\kappa$-complete filter on $\kappa$. Use an $L_{\kappa ,\kappa }$ propositional language with a sentence $P_A$ for every $A\subseteq \kappa$. Form a theory containing $P_A$ for $A\in F$, the Boolean identitiesand, for every $\delta <\kappa$,

Every subtheory of size below $\kappa$ mentions fewer than $\kappa$ required members of $F$. Their intersection is nonempty by $\kappa$-completeness; choosing a point in it and interpreting $P_A$ as membership of that point satisfies the subtheory. The theory is therefore $\kappa$-satisfiable. Strong compactness supplies a model. Thenis an ultrafilter, contains $F$, and is $\kappa$-complete by the infinitary intersection axioms.

For regular $\kappa$, the cobounded filter on a regular cardinal is $\kappa$-complete. Part (c) extends it to a $\kappa$-complete ultrafilter $U$. Since $\kappa \setminus \{\alpha \}$ is cobounded for every $\alpha <\kappa$, no singleton belongs to $U$; hence $U$ is nonprincipal. Thus every strongly compact cardinal is measurable.