Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 116 (Large Cardinals)

An uncountable cardinal number $\kappa$ is weakly compact when every $\kappa$-satisfiable theory in an infinitary language $L_{\kappa ,\kappa }$ with at most $\kappa$ nonlogical symbols is satisfiable. A cardinal is inaccessible when it is uncountable, regular, and a strong limit cardinal.

Two standard results supply the proof. First, every weakly compact cardinal is inaccessible. Second, every weakly compact $\kappa$ has the Keisler extension property: there is a transitive set $X⊋V_{\kappa }$ such thatand $\kappa \in X$. Since $\kappa$ is inaccessible, the relevant downward absoluteness makes $X\models$ “$\kappa$ is inaccessible”. Hence $X$ satisfies “there is an inaccessible cardinal”. By elementarity, $V_{\kappa }$ satisfies the same sentence, so it contains some inaccessible $\lambda$. Every ordinal in $V_{\kappa }$ is below $\kappa$, and inaccessibility is absolute here, giving

An uncountable cardinal $\kappa$ is measurable when it carries a nonprincipal ultrafilter $U$ that is $\kappa$-complete. For an inaccessible $\lambda >\kappa$, the cardinal $\kappa$ is 1-strong when there is an elementary embeddinginto a transitive model, with critical point $\kappa$ and $V_{\kappa +1}\subseteq M$.

The fundamental theorem on measurable cardinals constructs from $U$ the well-founded ultrapower and its ultrapower embedding $j:V_{\lambda }\to M$, whose critical point is $\kappa$. The embedding fixes $V_{\kappa }$. If $A\in V_{\kappa +1}$, then $A\subseteq V_{\kappa }$, and elementarity givesBoth $j(A)$ and $V_{\kappa }$ belong to the transitive target, so $A\in M$. Thus $V_{\kappa +1}\subseteq M$, proving that every measurable cardinal is 1-strong.

Let $U$ be a nonprincipal $\kappa$-complete ultrafilter on the measurable cardinal $\kappa$, and fix $\lambda <\kappa$. Suppose for a contradiction that $2^{\lambda }\ge \kappa$. Choose distinct subsets $A_{\alpha }\subseteq \lambda$ for $\alpha <\kappa$. For each $\xi <\lambda$, the ultrafilter property chooses exactly one ofas a member $H_{\xi }$ of $U$. Since $\lambda <\kappa$, $\kappa$-completeness gives $H={\bigcap }_{\xi <\lambda }{H}_{\xi }\in U$.

Any two indices in $H$ give subsets having the same membership decision at every $\xi <\lambda$, so they give the same $A_{\alpha }$. The chosen subsets were distinct, hence $|H|\le 1$. This contradicts the fact that a small set is absent from a complete nonprincipal ultrafilter. Therefore $2^{\lambda }<\kappa$ for every $\lambda <\kappa$, which is precisely the strong limit cardinal condition. This is the measurable cardinal is a strong limit cardinal argument.

Suppose that a first-order formula $\phi$ described the inaccessible cardinal $\kappa$, so that $\kappa$ were the least ordinal with $V_{\kappa }\models \phi$. Since $\kappa$ is inaccessible, $V_{\kappa }$ is a model of ZFC. Apply the Lévy reflection theorem inside this model to the single formula $\phi$. There is some $\alpha <\kappa$ for whichThe right side holds, so the left side contradicts the asserted minimality of $\kappa$. Hence no first-order formula describes an inaccessible cardinal, as recorded by ordinal described by a first-order formula.

Let $I(\kappa )$, $W(\kappa )$, and $M(\kappa )$ mean respectively that $\kappa$ is inaccessible, weakly compact, and measurable. Define a fourth cardinal propertyTake ${\Phi }_0=I$, ${\Phi }_1=W$, and ${\Phi }_2=\Theta$.

If inaccessible and weakly compact cardinals exist, Question 1a shows that an inaccessible lies below every weakly compact cardinal. Thus $I{<}_{1}W$. If $W\mathbf{C}$ and $\Theta \mathbf{C}$ both hold, then $\Theta$ is exactly measurability. Every measurable cardinal is weakly compact, and the usual ultrapower reflection theorem gives weakly compact cardinals below every measurable cardinal. Therefore $W{<}_1\Theta$.

Now assume the consistency of ZFC with an inaccessible cardinal but no weakly compact cardinal. In such a model $\Theta (\kappa )$ is exactly $I(\kappa )$, soConsequently $I/{<}_1\Theta$. This is an explicit nontransitivity of the least-occurrence order on cardinal properties, even though $I{<}_{1}W$ and $W{<}_1\Theta$.

WriteIf $\kappa$ is a worldly cardinal, then $V_{\kappa }\models \mathrm{ZFC}$. The existence of this set model proves $\mathrm{Con}(\mathrm{ZFC})$ in the universe. By arithmetic absoluteness for a rank-initial model, the same formal consistency statement holds in $V_{\kappa }$. Hence $V_{\kappa }\models T^{\ast }$, so $T$ proves $\mathrm{Con}(T^{\ast })$.

Conversely, suppose $T^{\ast }$ proved $\mathrm{Con}(T)$. The theory $T$ proves every axiom of $T^{\ast }$, since a worldly cardinal proves $\mathrm{Con}(\mathrm{ZFC})$. It would therefore also prove $\mathrm{Con}(T)$, contrary to the Gödel second incompleteness theorem when $T$ is consistent. Thus $T^{\ast }$ cannot prove $\mathrm{Con}(T)$, andThis is the consistency strength of a worldly cardinal comparison.

Because ${\kappa }_0<{\kappa }_1=\mathrm{crit}(j_1)$, the definition of the critical point of an elementary embedding immediately gives

To evaluate $j_0({\kappa }_1)$, first use the regularity of the measurable cardinal ${\kappa }_1$. Every function $f:{\kappa }_0\to {\kappa }_1$ has bounded range, so every ordinal below $j_0({\kappa }_1)$ lies below $j_0(\beta )$ for some $\beta <{\kappa }_1$. HenceThe strong-limit property of ${\kappa }_1$ gives ${\beta }^{{\kappa }_0}<{\kappa }_1$ for every $\beta <{\kappa }_1$. There are therefore fewer than ${\kappa }_1$ functions ${\kappa }_0\to \beta$, which implies $j_0(\beta )<{\kappa }_1$. On the other hand $j_0(\beta )\ge \beta$. Taking suprema yieldsthe two measurable cardinals under an ultrapower embedding formula.

Fix $i\in \{0,1\}$ and write $\kappa ={\kappa }_i$, $j=j_i$, and $M=M_i$. Every ordinal below $j(\kappa )$ is represented in the ultrapower by a function $\kappa \to \kappa$. Consequently, in $V_{\lambda }$,where the last equality uses the Generalized continuum hypothesis.

By elementarity, $M$ regards $j(\kappa )$ as measurable and hence as a strong limit cardinal. Moreover $V_{\kappa +1}\subseteq M$, so $M$ and $V_{\lambda }$ have the same subsets of $\kappa$ and the same ${\kappa }^{+}$. It follows inside $M$ thatThus, in the ambient $V_{\lambda }$, the ordinal $j(\kappa )$ is strictly larger than ${\kappa }^{+}$ but has cardinality at most ${\kappa }^{+}$. It cannot be a cardinal number. Applying this argument to both $i=0$ and $i=1$ proves that neither $j_0({\kappa }_0)$ nor $j_1({\kappa }_1)$ is a cardinal in $V_{\lambda }$, exactly as in moved critical point is not an ambient cardinal under GCH.