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An inaccessible cardinal $\kappa$ has the Keisler extension property when there is a proper transitive set $X⊋V_{\kappa }$ for whichSuch a $\kappa$ is not the least inaccessible cardinal: $X$ regards $\kappa$ as inaccessible, elementarity reflects the existence of an inaccessible into $V_{\kappa }$, and strong-inaccessibility absoluteness from rank agreement makes the resulting witness genuinely inaccessible below $\kappa$.

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

Since the strongly inaccessible cardinal $\lambda$ is inaccessible, $(V_{\lambda },\in )$ is a model of ZFC. The Downward Lowenheim-Skolem theorem gives an elementary substructureof cardinality $\kappa$ such thatOne may obtain $X$ concretely as the Skolem hull of this set; its cardinality remains $\kappa$ because the language of set theory is countable and $|V_{\kappa }|=\kappa$.

Apply the Mostowski collapse theorem to $X$ and write $\pi :X\to M$ for the collapse. Then $M$ is a transitive set, $|M|=\kappa$, and $\pi$ fixes $V_{\kappa }$ pointwise. It also fixes $\kappa$, because it fixes every ordinal below $\kappa$. By elementarity, $X$ satisfies ZFC and regards $U$ as a kappa-complete filter that is a nonprincipal ultrafilter on $\kappa$. Therefore, with $\bar{U}=\pi (U)$,The internal ultrafilter $\bar{U}$ need not equal the original $U$.

After choosing a Skolem function for each existential formula, the Skolem hull of $A\subseteq N$ is the smallest subset containing $A$ and closed under those functions. The Tarski-Vaught test makes it an elementary substructure of $N$, and in a countable language an infinite hull has cardinality at most $|A|+{\aleph }_0$.