An uncountable <cardinal number> $\kappa$ is <weakly compact cardinal>[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 <strongly inaccessible cardinal>[inaccessible] when it is uncountable, <regular cardinal>[regular], and a .

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 $X\supsetneq V_\kappa$ such that
$$
(V_\kappa,\in)\prec(X,\in)
$$
and $\kappa\in X$. Since $\kappa$ is inaccessible, the relevant <set-theoretic absoluteness>[downward absoluteness] makes $X\models$ “$\kappa$ is inaccessible”. Hence $X$ satisfies “there is an inaccessible cardinal”. By <elementary substructure>[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
$$
\boxed{\lambda<\kappa\text{ and }\lambda\text{ is inaccessible}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 116 / 1 / a / Solution

{title2=$L_{\kappa,\lambda}$}
{wiki=Infinitary_logic}

The infinitary language $L_{\kappa,\lambda}$ permits conjunctions and disjunctions of fewer than $\kappa$ formulas and simultaneous quantification over fewer than $\lambda$ variables. Ordinary first-order logic is $L_{\omega,\omega}$.


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 Incoming links: Infinitary language