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We construct the two transitive models together, taking care that the later collapse really retains the earlier model. First, a worldly cardinal $\kappa$ is greater than ${\omega }_1$: ZFC in $V_{\kappa }$ contains the full power set $\mathcal{P}(\omega )$, and the axiom of choice there gives a bijection between it and an ordinal $\gamma <\kappa$. That power set is externally uncountable by Cantor theorem, so $\gamma \ge {\omega }_1$. Infinity rules out the finite cardinals and $\omega$ before this argument is applied.

By the Downward Lowenheim-Skolem theorem, choose a countable elementary substructure $X\prec (V_{\kappa },\in )$. Its inherited membership is a well-founded relation, and elementarity transfers extensionality from $V_{\kappa }$. Apply the Mostowski collapse theorem to obtain a countable transitive model $M$ of ZFC. Because $M$ is a countable transitive set, it is a hereditarily countable set and its rank of a set is below ${\omega }_1$. In particular, $M\in V_{\kappa }$.

Now take the Skolem hull in $V_{\kappa }$ of the seedCall it $H$. The seed has external cardinality ${\aleph }_1$ and lies inside $V_{\kappa }$. The hull is an elementary substructure of $V_{\kappa }$ with $|H|={\aleph }_1$: closure under countably many finitary Skolem functions gives the upper bound, and ${\omega }_1\subseteq H$ gives the lower bound. Let $\pi :H\to N$ be its Mostowski collapse theorem map. Then $N$ is a transitive model of ZFC.

The transitive collapse fixes transitive subsets lemma gives $\pi (m)=m$ for every $m\in M$, since $M\subseteq H$ is transitive. Since also $M\in H$, the collapse sends that element to $M$ itself. Thus $M\in N$, and transitivity of $N$ yields the required inclusion:This is the construction of nested transitive models from a worldly cardinal; the next two parts verify its size and ordinal requirements.

Let $\theta =M\cap \mathrm{Ord}$ be the ordinal height of a model of set theory. First $\theta \ge {\omega }_1$. Indeed, if $\theta$ were countable, the internal axiom of choice would give, for every $x\in M$, a bijection in $M$ between $x$ and some ordinal below $\theta$. Such a bijection is also valid externally, making $x$ externally countable. In particular every internal rank $V_{\xi }^M$, $\xi <\theta$, would be countable. Every element of $M$ lies in one of these internal ranks, sowould be a countable union of countable sets, a contradiction. This is why an uncountable transitive set model has uncountable ordinal height.

For $\xi <\theta$, absoluteness of constructible levels gives $L_{\xi }^M=L_{\xi }$, and that level is an element of $M$. The reason is that satisfaction in a fixed set structure uses the same domain and finite formulas in both universes, so successor definitions agree; transfinite recursion then also makes limit stages agree. Transitivity consequently givesIt remains to locate a countability witness. Given an ambient countable ordinal $\alpha$, choose an injection $f:\alpha \to \omega$. Since $V=L$, it belongs to the constructible universe. Its transitive closure, together with $f$ itself, is countable. Choose a countable elementary substructure $X\prec L_{\eta }$ of a sufficiently large limit level containing this closure pointwise. The condensation lemma for the constructible universe identifies the collapse with $L_{\beta }$ for some countable $\beta$. The collapse fixes $f$, since all its hereditary members were included. Thus $f\in L_{\beta }\subseteq L_{{\omega }_1}\subseteq M$.

The ordinal $\alpha$ itself is in $M$, since $\alpha <{\omega }_1\le \theta$. Being an injection between these fixed sets is a bounded formula in set theory, so $M$ recognizes $f$ as a countability witness. This proves countable-ordinal correctness under constructibility:The use of condensation supplies a witness below the height of $M$; merely knowing that $f$ belongs somewhere to $L$ would not be sufficient.

The Tarski-Vaught test states that a substructure $M\subseteq N$ is an elementary substructure if and only if every formula $\phi (x,\bar{y})$ and tuple $\bar{a}\in M$ satisfyNecessity follows immediately from elementarity. Conversely, assume the witness condition. Induct on formulas to prove that $M\models \psi (\bar{a})$ exactly when $N\models \psi (\bar{a})$ for every $\bar{a}\in M$. Atomic formulas agree because $M$ is a substructure, and Boolean connectives follow by induction. A witness in $M$ is also one in $N$; a witness in $N$ can be replaced by one in $M$ by hypothesis, after which induction applies to the matrix. Universal formulas follow by negation. Thus the witness condition is equivalent to $M\preccurlyeq N$.

Define a parameter-free equivalence relation byIts classes correspond exactly to the distinct sets $\phi (a,\mathcal{U})$. If there are exactly $n<\infty$ classes, that fact is a first-order sentence, so the elementary substructure $M$ contains representatives $b_1,\dots ,b_n$ of all of them. Every $a\in \mathcal{U}$ is $E$-equivalent to some $b_i$, and $\phi (a,\mathcal{U})=\phi (b_i,\mathcal{U})$ is definable over $M$.

The converse also holds because $M$ is small and $\mathcal{U}$ is a monster model. If there were infinitely many $E$-classes, the partial complete typewould be finitely satisfiable. Saturation of $\mathcal{U}$ would realize $p$, producing a class with no representative in $M$, contrary to the hypothesis. Therefore there are only finitely many sets $\phi (a,\mathcal{U})$.

Suppose $a\in \mathrm{acl}(A)$ in the sense of model-theoretic algebraic closure. Some formula $\phi (x,\bar{c})$ over $A$ has exactly $n<\infty$ realizations in the monster model and includes $a$. Every model $M$ containing $A$ is an elementary substructure of the monster and therefore contains $n$ distinct realizations. Since there are only $n$ in the monster, it contains all of them, including $a$. Hence

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 $U$.

A substructure $M\subseteq N$ is simply closed when each existential quantifier-free formula over $M$ realized in $N$ is already realized in $M$. A single quantified variable suffices in the definition; iterating covers finite tuples. This property is also called existential closure in this specified extension. It is weaker than being an elementary substructure.

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 $|A|+{\aleph }_0$.

An $L$-substructure of a first-order structure $A\subseteq B$ has a nonempty domain containing all named constants and closed under the functions of $B$, and each relation of $A$ is the restriction of the corresponding relation of $B$. Consequently every quantifier-free formula with parameters in $A$ has the same truth value in $A$ and $B$. This is weaker than being an elementary substructure, which preserves all first-order formulas.

A substructure $M\subseteq N$ is elementary if and only if, whenever $N\models \exists x\phi (x,\bar{a})$ with parameters $\bar{a}$ from $M$, some witness $b\in M$ satisfies $N\models \phi (b,\bar{a})$.

The witness criterion characterizes an elementary substructure, and hence preservation of all first-order formulas with parameters from the substructure.

Below any uncountable strongly inaccessible cardinal $\kappa$, the worldly cardinals contain a closed unbounded subset of $\kappa$ and hence have cardinality $\kappa$. The structure $V_{\kappa }$ models ZFC. Reflect each finite group from an enumeration of all first-order formulas to a closed unbounded subset of $\kappa$, using $|V_{\alpha }|<\kappa$ and regularity for the witness bounds. The countable intersection remains closed unbounded, so its ranks are elementary substructures of $V_{\kappa }$. Intersect with the closed unbounded subset of infinite cardinal numbers below $\kappa$. Uncountability is essential: the regular strong limit $\omega$ has no worldly cardinal below it.