Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 116 (Large Cardinals)

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.

Let $\lambda$ be inaccessible and let $j:V_{\lambda }\to M$ be an elementary embedding into a transitive set with critical point $\kappa$. It is $\beta$-strong whenA formula $\Phi (x,\kappa )$ is a beta-stable cardinal property when it is absolute between the universe and every transitive set containing $V_{\kappa +\beta }$.

To say that the embedding reflects such a property means that whenever $\Phi (\kappa )$ holds,is unbounded in $\kappa$: for every $\gamma <\kappa$ there is such a $\mu$ with $\gamma <\mu <\kappa$. Indeed, stability gives $M\models \Phi (\kappa )$, so $M$ sees the witness $\kappa$ between $\gamma$ and $j(\kappa )$. Elementarity reflects a witness between $\gamma$ and $\kappa$. This is reflection by a beta-strong embedding.

The standard closure lemma for the ultrapower embedding says that every $\kappa$-sequence of members of $M$ which belongs to $V_{\lambda }$ is itself in $M$. If $\kappa <\alpha <{\kappa }^{+}$, choose in $V_{\lambda }$ a surjectionAll ordinal values of $s$ belong to the transitive model $M$, so closure gives $s\in M$. Thereforeand $M$ does not regard $\alpha$ as a cardinal.

Every ordinal belowhas an ultrapower representative $f:\kappa \to {\kappa }^{+}$: by Łoś theorem, a representative below the successor of $j(\kappa )$ may be chosen below ${\kappa }^{+}$ on a set in the ultrafilter. Hence, in $V_{\lambda }$,The ultrapower is closed under $\kappa$-sequences, so it contains every subset of $\kappa$ and computes $2^{\kappa }$ correctly. By elementarity $M$ regards $j(\kappa )$ as measurable, hence strongly inaccessible by Question 1(b). ConsequentlyThus $V_{\lambda }$ has a set of cardinality at most $2^{\kappa }$ whose order type is $\eta >2^{\kappa }$, so

Letbe the supremum of the Kunen critical sequence, and putThe required choices areIndeed $X$ has rank $\hat{\kappa }$, so $X\in V_{\hat{\kappa }+1}$, while the Kunen lemma gives $X\notin M$ once the domain contains the omega-Jonsson function required by the proof, which is ensured by $\alpha \ge \hat{\kappa }+2$.

Let $\hat{\kappa }={\sup }_{n<\omega }{j}^n(\kappa )$. Every term of the critical sequence is below $\delta$. If $\mathrm{cf}(\delta )>{\aleph }_0$, then its countable supremum also satisfies $\hat{\kappa }<\delta$, and because $\delta$ is a limit ordinal,Apply the Kunen lemma to the restriction available inside $V_{\delta }$. It giveswhich is impossible because $V_{\hat{\kappa }+1}\subseteq V_{\delta }$. Hence $\mathrm{cf}(\delta )\le {\aleph }_0$. A nonzero limit ordinal has infinite cofinality, so