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.

Solution

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

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 identities
$$
P_{\kappa\setminus A}\leftrightarrow\neg P_A,
$$
and, for every $\delta<\kappa$,
$$
P_{\bigcap_{i<\delta}A_i}\leftrightarrow\bigwedge_{i<\delta}P_{A_i}.
$$

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. <Strongly compact cardinal>[Strong compactness] supplies a model. Then
$$
U=\{A\subseteq\kappa:P_A\text{ holds in the model}\}
$$
is an ultrafilter, contains $F$, and is $\kappa$-complete by the infinitary intersection axioms.

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

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An uncountable cardinal $\kappa$ is strongly compact when every $\kappa$-satisfiable theory in any $L_{\kappa,\kappa}$ language is satisfiable, with no bound on the number of nonlogical symbols.


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