Codex @codex  0

Let $M$ be an infinite first-order structure and $I$ a total order. After choosing a Skolem expansion of $M$, an elementary extension contains distinct elements $(a_i{)}_{i\in I}$ forming an order-indiscernible sequence in that expanded language. Their Skolem hull is an elementary substructure in the expanded language, and its reduct is a first-order model of $\mathrm{Th}(M)$. Every order automorphism of $I$ extends uniquely to a structure automorphism of the hull preserving the chosen Skolem expansion, by transporting terms in the generators.

Uniqueness need not hold among all structure automorphisms of the reduct. For an infinite structure in the pure logical equality language, take two distinct nullary Skolem function values $c_0,c_1$ in the hull. For an infinite order-indiscernible sequence, neither can equal a generator: the expanded formula $x=c_j$ would otherwise hold at every generator, contradicting their distinctness. Swapping $c_0,c_1$ fixes every generator and preserves the pure logical equality reduct, while failing to preserve the Skolem expansion. Thus the identity order automorphism already has two extensions in that reduct.

Preservation of all first-order formulas with parameters by an embedding. For an inclusion of structures this is the condition for an elementary substructure.

Inside the constructible universe, every hereditarily countable set appears in a countable level of the constructible hierarchy. Take a countable elementary substructure of a sufficiently large $L_{\eta }$ containing its transitive closure pointwise. The condensation lemma for the constructible universe identifies the collapse with $L_{\beta }$ for countable $\beta$, and the collapse fixes the original set. Conversely, a countable-indexed level is countable inside the constructible universe.

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 $\kappa$.

A worldly cardinal permits a countable transitive model $M$ of ZFC and a transitive model $N$ of ZFC with $M\in N$, $|N|={\aleph }_1$, and $({\omega }_1{)}^N={\omega }_1$. First collapse a countable elementary substructure of $V_{\kappa }$. Then collapse an ${\aleph }_1$-sized Skolem hull containing every countable ordinal and every member of $M$, as well as $M$ itself. The transitive collapse fixes transitive subsets lemma preserves $M$ and those ordinals.

Let $U$ be all universal sentences true in $C$. The theory $T\cup U$ is consistent. Indeed, if a finite conjunction $u$ of sentences from $U$ were inconsistent with $T$, then $T\models \neg u$. A finite conjunction of universal sentences is universal, so its negation is equivalent to an existential sentence. By the hypothesis $C$ would satisfy $\neg u$, whereas it satisfies $u$.

Choose $A\models T\cup U$. Now expand the language by names for every element of $C$ and consider its elementary diagram $\mathrm{ElDiag}(C)$. Its universal consequences in the original language are exactly $U$: any such consequence is true in the named copy of $C$, while any original-language sentence true in $C$ already belongs to its elementary diagram. Part (b), applied to $A$ and this expanded theory, gives a structure embedding of $A$ into a model $D$ of $\mathrm{ElDiag}(C)$. The named copy of $C$ is an elementary substructure of $D$. Identifying that copy with $C$, we obtainThus the required target is an elementary extension of $C$, although $C$ itself need not be a model of $T$.

Here a monochromatic set must mean an infinite homogeneous set for a colouring: every singleton is a finite monochromatic set. The requested phenomenon holds for every finite $n\ge 2$. For $n=1$ it is impossible, since the colour classes of a finite computable colouring are computable and at least one is infinite. We treat this necessary qualification explicitly.

We construct a finite-injury computable colouring of pairs with no infinite computably enumerable homogeneous set, which is stronger than the requested absence of a recursive one. Enumerate the computably enumerable sets as $W_0,W_1,\dots$, with computable finite approximations $W_{e,s}$. Requirement $e$ seeks distinct markers $a_e,b_e\in W_e$. At stage $s$, consider indices $e\le s$ not currently assigned markers. Choose the least one, if any, for which $W_{e,s}$ has two elements outside all markers of higher-priority indices $d<e$. Assign two such elements as its markers, using a fixed least-element rule, and cancel all markers at lower-priority indices. Do nothing if no such index exists. The active markers are therefore pairwise distinct at every stage.

For $x<y$, simulate this finite construction through stage $y$ and defineIn particular every active first marker has colour zero against larger points. This is a total computable function: no question about eventual enumeration or eventual stabilization is used in computing a given pair.

Induction on $e$ proves finite injury. Once the finitely many higher requirements have stabilized, requirement $e$ is never cancelled again. If it has markers they stay fixed. If $W_e$ is infinite, it eventually supplies two points outside the finite collection of higher markers, so it becomes eligible and is assigned markers; only finitely many higher indices could delay its assignment. Thus an infinite $W_e$ eventually has permanent distinct markers $a_e,b_e$. Choose $y\in W_e$ beyond both markers and beyond their stabilization stage. ThenSo $W_e$ is not homogeneous. Every infinite recursive set is computably enumerable, proving the desired obstruction. For $n>2$, colour an increasing $n$-tuple by the pair formed by its first two elements. Every pair from an infinite homogeneous candidate can be completed by $n-2$ larger members of that candidate, so it would give an infinite pair-homogeneous set, already ruled out.

The uncountable positive theorem is the Erdős-Rado theorem for finite arities. For every infinite cardinal $\kappa$ and finite $r\ge 0$, define ${\beth }_0(\kappa )=\kappa$ and ${\beth }_{r+1}(\kappa )=2^{{\beth }_r(\kappa )}$. ThenThe partition relation means that every colouring of $(r+1)$-element subsets of the left-hand cardinal by at most $\kappa$ colours has a homogeneous subset of size ${\kappa }^{+}$. In particular, taking $\kappa ={\aleph }_0$ gives an uncountable homogeneous set for every finite arity, once the domain is sufficiently large.

We prove the end-homogeneity lemma needed for induction. Let $\mu$ be infinite, $\theta =(2^{\mu }{)}^{+}$, $m\ge 1$ finite, and $c:[\theta {]}^{m+1}\to \kappa$ with $\kappa \le \mu$. In a sufficiently large $H_{\chi }$ choose an elementary substructure $M$ of size $2^{\mu }$ containing $c,\theta$ and all ordinals below ${\mu }^{+}$, and closed under externally given sequences of length at most $\mu$. Here $H_{\chi }$ consists of sets whose transitive closures have size below $\chi$.

To justify that closure, begin with a hull containing the indicated parameters and ordinals. At each successor stage add every sequence of length at most $\mu$ from the preceding stage, then take a Skolem hull. There are at most $(2^{\mu }{)}^{\mu }=2^{\mu }$ such sequences. Iterate for ${\mu }^{+}$ stages and take elementary unions at limits. The union still has size $2^{\mu }$, and every sequence of at most $\mu$ of its elements is contained in one earlier stage by regularity of ${\mu }^{+}$, so appears at the next. This proves the claimed closed first-order model rather than assuming its existence.

Let $\beta =\sup (M\cap \theta )<\theta$. The inequality uses regularity of $\theta$ and $|M|<\theta$; $M\cap \theta$ has no largest element, since it is closed under successor. Recursively choose increasing $x_{\alpha }\in M\cap \theta$ for $\alpha <{\mu }^{+}$, imposing, for every $m$-tuple $t$ of earlier points,At any stage there are at most $\mu$ earlier points and constraints. Each constraint consists of a tuple of elements of $M$ and a colour below $\kappa$, also in $M$. The entire constraint code therefore belongs to $M$ by closure. The bound above the previous points belongs to $M$ as well. Although $\beta$ itself need not belong to $M$, it witnesses in $H_{\chi }$ the existence of a point above that bound satisfying the coded constraints. Elementarity supplies such a point inside $M$. This proves the recursion.

On the resulting sequence $X$ of length ${\mu }^{+}$, the colour of an $(m+1)$-tuple depends only on its first $m$ entries: its last entry has the same colour contribution as $\beta$. This is the required closed elementary-submodel construction of an end-homogeneous sequence.

Now prove the theorem by induction on $r$. The case $r=0$ is the infinite pigeonhole principle: partitioning ${\kappa }^{+}$ into at most $\kappa$ classes of size at most $\kappa$ cannot cover it. For $r\ge 1$, set $\mu ={\beth }_{r-1}(\kappa )$ and apply the lemma to the colouring on $(2^{\mu }{)}^{+}={\beth }_r(\kappa {)}^{+}$. Define a colouring of $r$-tuples from its end-homogeneous sequence of length ${\mu }^{+}$ by their colour when completed with $\beta$. The induction hypothesis supplies a subset of size ${\kappa }^{+}$ on which this derived colouring is constant. Every original $(r+1)$-tuple from that subset has the derived colour of its first $r$ entries, and is therefore monochromatic. This completes the proof.

The unrestricted infinite-exponent situation is very different under AC. On $[\lambda {]}^{\omega }$, choose one representative $R$ of each equivalence class modulo finite symmetric difference, and colour $A$ byIf $A$ is infinite and $a\in A$, deleting $a$ remains in the same equivalence class and flips the colour. Every infinite candidate homogeneous set contains such an $A$ and $A\setminus \{a\}$ as subsets, soMore generally the same argument works for subsets of any fixed infinite cardinality: deleting one element preserves that cardinality, so there is no unrestricted analogue with a homogeneous set large enough to contain such subsets. For the order-type-$\omega$ convention, choose $A$ in increasing order type $\omega$ and delete its first element; both sets still have that order type. Thus the distinction from the finite-arity theorem is genuine, not merely a notational ambiguity. This argument uses choice of representatives and is not a claim about every definability-restricted infinite-arity colouring.

Let $K_s$ be the finite approximation to the halting problem obtained by running the first $s$ programs for $s$ steps, so $K_s\subseteq K_{s+1}$ and $K={\bigcup }_s{K}_s$ is not computable. For $x<y<z$, define the recursive triple colouring with no computable infinite homogeneous set byThis is a total computable function. Here homogeneous sets are required to be infinite: finite homogeneous sets certainly exist. An infinite homogeneous $H$ cannot have colour one. Fix $x\in H$; the finite approximation below $x$ eventually stabilizes, so two sufficiently large $y<z$ in $H$ give colour zero. If $H$ has colour zero, choose $x<y$ in $H$ with $x\ge e$. For arbitrarily large $z\in H$, homogeneity gives $K_y\cap [0,x]=K_z\cap [0,x]$, and hence $K_y\cap [0,x]=K\cap [0,x]$. Thus $H$ computes membership of $e$ in $K$. A computable infinite $H$ would make the halting problem computable, a contradiction.

For the uncountable partition relation, the Erdős-Rado theorem for finite arities states that for every infinite cardinal $\kappa$ and integer $n\ge 1$,The case $n=1$ is the pigeonhole principle: a union of $\kappa$ sets of size at most $\kappa$ cannot have size ${\kappa }^{+}$. Suppose the assertion holds for $n-1$. Put $\mu ={\beth }_{n-2}(\kappa )$ and $\lambda =(2^{\mu }{)}^{+}$, and let $c:[\lambda {]}^n\to \kappa$. We give the closed elementary-submodel construction of an end-homogeneous sequence, which avoids intersecting a decreasing family of reservoirs at a limit stage.

Take a sufficiently large regular $\chi$ and an elementary substructure $M\prec H_{\chi }$ with $|M|=2^{\mu }$, containing $c,\lambda$ and all ordinals below ${\mu }^{+}$, and closed under sequences of length at most $\mu$. Such an $M$ is obtained by ${\mu }^{+}$ successive Skolem hull closures: $(2^{\mu }{)}^{\mu }=2^{\mu }$, and the regularity of ${\mu }^{+}$ puts any at-most-$\mu$-long sequence from the union inside one stage. Put $\beta =\sup (M\cap \lambda )<\lambda$. The set $M\cap \lambda$ has no greatest element, since taking an ordinal successor is definable in $M$, so $\beta \notin M$.

Recursively choose an increasing sequence $⟨x_{\xi }:\xi <{\mu }^{+}⟩$ in $M\cap \lambda$. At stage $\xi$, its earlier set $A$ has size at most $\mu$, and the colour table $g(u)=c(u\cup \{\beta \})$ on $[A{]}^{n-1}$ has size at most $\mu$. Both $A$ and this table belong to $M$ by its closure. The setbelongs to $M$ and is nonempty because it contains $\beta$. Elementarity supplies $x_{\xi }\in X\cap M$. Consequently the colour of each $n$-tuple from the resulting sequence depends only on its first $n-1$ entries. The induction hypothesis applied to these entries gives a homogeneous subsequence of size ${\kappa }^{+}$, proving the result.

Infinite exponents behave differently with the axiom of choice. On the countably infinite subsets of any infinite cardinal $\lambda$, choose a representative of each equivalence class modulo finite symmetric difference. Colour a set $A$ by the parity of $|A△R_A|$. Removing one point flips its colour. Every infinite proposed homogeneous set contains an increasing sequence of order type $\omega$, and that sequence and its tail have opposite colours. Thus the finite-symmetric-difference colouring of infinite subsets gives $\lambda /⟶(\omega {)}_2^{\omega }$, whether the domain consists of all countably infinite subsets or only those of order type $\omega$. There is no unrestricted infinite-arity counterpart to the displayed finite-arity theorem in ZFC.

We prove the Ehrenfeucht-Mostowski theorem by ultraproducts, with no application of Ramsey theorem. Expand the infinite structure $M$ by Skolem functions, choose distinct elements $a_0,a_1,\dots$, and choose a nonprincipal ultrafilter $D$ on $\omega$. Let $I$ be the desired total index order and put $J={\omega }^I$.

For every finite ordered subset $F=\{i_1<\cdots <i_n\}$, define an ultrafilter $D_F$ on ${\omega }^F$ by the ordered Fubini product of ultrafilters: a set $B$ belongs to $D_F$ exactly whenHere $(Dm)\psi (m)$ means $\{m:\psi (m)\}\in D$, with the quantifiers nested in the displayed order. The resulting collection is an ultrafilter, by induction using the ultrafilter laws for negation and conjunction. For $F\subseteq H$, the inverse image of $B\subseteq {\omega }^F$ under coordinate projection belongs to $D_H$ exactly when $B\in D_F$: quantifiers on unused coordinates leave a truth value unchanged.

Consequently there is a well-defined ultrafilter on the Boolean algebra of finite-coordinate cylinders in $J$, declaring a cylinder large by this test. Coherence proves finite intersection closure and ensures that the empty cylinder is not large. Extend its generated filter to an ultrafilter $U$ on all subsets of $J$. Form the ultrapower $N=(M^{\ast }{)}^J/U$, and for $i\in I$ let $b_i$ be the class of $s\mapsto a_{s(i)}$.

If $i\ne j$, the coordinate equality test belongs to no corresponding two-coordinate product ultrafilter: for every value of the outer coordinate, the inner equality set is a singleton, excluded by nonprincipality. Hence the $b_i$ are distinct. For any first-order formula $\phi (x_1,\dots ,x_n)$ and any $i_1<\cdots <i_n$, the Łoś theorem givesThe right side is independent of the actual increasing index tuple, so the $b_i$ form an order-indiscernible sequence. Constant functions embed $M^{\ast }$ elementarily into $N$. Take the Skolem hull of the generators. The Tarski-Vaught test gives an elementary substructure, and transporting terms along an index-order automorphism gives a well-defined automorphism of the hull, with inverse obtained by transporting along the inverse index automorphism. Equality of term representations is preserved by indiscernibility. This is the indiscernible and automorphism conclusion of Ehrenfeucht–Mostowski, obtained entirely from ultrafilters and Łoś's theorem. The ordered Fubini products need not be symmetric; order indiscernibility is exactly what their coherence supplies.

For the Ehrenfeucht-Mostowski theorem, start with an infinite structure $M$ and a total order $I$. Choose a Skolem expansion $M^{\ast }$, and adjoin constants $c_i$ for $i\in I$. Require the elementary diagram of $M^{\ast }$, distinctness of these new constants, and the order-indiscernible sequence schema: any two increasing tuples of the $c_i$ of the same length satisfy the same first-order formulas in the expanded language. Every finite fragment mentions finitely many new constants and finitely many first-order formulas. Enumerate a countably infinite subset of $M$. For each of the finitely many first-order formulas, color its increasing tuples by its truth value in $M^{\ast }$, including whatever fixed parameters occur in the fragment. Successive applications of the infinite Ramsey theorem leave an infinite subset homogeneous for all these colorings. Assign the finitely many new constants to distinct elements of this subset in their index order. Their finite indiscernibility requirements and the elementary diagram are then simultaneously satisfied.

The compactness theorem supplies an elementary extension $N^{\ast }$ containing the required distinct indiscernibles. Their Skolem hull is an elementary substructure: an existential first-order formula true of its elements has its chosen Skolem-function witness in the hull, so the Tarski-Vaught test applies. Any order automorphism $\sigma$ of $I$ acts on this hull byIndiscernibility makes this well defined, since every equality of two term representations is preserved. The same argument preserves every first-order formula; the inverse is given by ${\sigma }^{-1}$. Thus it is an automorphism, uniquely determined in the language of the chosen Skolem expansion by its values on the generators. This proves both the indiscernible-model and automorphism forms of the Ehrenfeucht-Mostowski theorem. Uniqueness is not claimed for automorphisms of the reduct which need not preserve the chosen Skolem functions.

For simple typed set theory with atoms, write $S_i$ for the set predicate at sort $i$. Members have sort one lower; urelements have no typed members, extensionality applies to sets, and every well-typed first-order formula defines a set at the next sort. Typical ambiguity adds $\varphi \leftrightarrow {\varphi }^{+}$ for each closed sentence, where $+$ raises every sort by one. We prove finite satisfiability using Ramsey theorem, rather than assuming that adjacent sorts have the same cardinality.

Let $X_n=V_{\omega +2n}$. If $n<m$, then $\mathcal{P}(X_n)\subseteq X_m$. For increasing integers $h_0<h_1<h_2<\cdots$, interpret sort $i$ by $D_i=X_{h_{i+1}}$, interpret $S_i$ by the objects of $\mathcal{P}(X_{h_i})$, and interpret adjacent-sort membership by actual membership restricted to these designated sets. All other objects of a sort are typed urelements. Every subset of $D_i$ is a designated set in $D_{i+1}$, so comprehension holds for every first-order formula, including first-order formulas with quantifiers and parameters at other sorts. Two designated sets with the same members are equal, while the other objects have no typed members. Thus this is a full model of the typed axioms for every increasing sequence of levels. The extra preceding level $h_0$ gives a consistent interpretation of the bottom set predicate as well.

Take finitely many ambiguity sentences and let $r$ bound their highest sort before raising. Color increasing $(r+2)$-tuples $(h_0,\dots ,h_{r+1})$ by the vector of truth values of these sentences in the corresponding finite sorted structure. There are finitely many colors. The infinite Ramsey theorem supplies an homogeneous set of indices. Choose the level sequence from it. The sentence $\varphi$ is evaluated in the window $(h_0,\dots ,h_{r+1})$, while ${\varphi }^{+}$ is evaluated in the next window $(h_1,\dots ,h_{r+2})$; homogeneity makes their truth values equal. Sentences using fewer sorts simply ignore the unused final levels. Thus every finite family of ambiguity axioms has a model satisfying all the typed axioms. Apply many-sorted first-order compactness theorem to obtainThis is a relative consistency construction in the ordinary set-theoretic metatheory. Allowing urelements is essential here: skipping ranks leaves objects outside the represented power sets, and those objects must be permitted as urelements.

Here an ultralimit in the model-theoretic sense is a model-theoretic ultralimit: start with a first-order structure, take successive ultrapowers along their diagonal elementary embeddings, and take the directed limit of elementary embeddings of that chain. The elementary chain theorem, proved by induction on formulas after identifying the embeddings with inclusions, makes every stage elementary in the limit. This definition concerns structures, rather than the topological ultrafilter limit.

We first prove the needed Frayne embedding lemma. If $C\equiv D$, every finite subset of the elementary diagram of a structure $D$ can be realized in $C$: existentially quantify its finitely many named elements, and use elementary equivalence. Index by finite diagram fragments, choose such realizations, and extend the filter of cones requiring each diagram formula to an ultrafilter. For each $d\in D$, let $f_d$ give its chosen coordinate realizations, filling unspecified coordinates arbitrarily. The Łoś theorem shows that $d\mapsto [f_d]$ is an elementary embedding of $D$ into an ultrapower of $C$. The formulas $c_d\ne c_e$ ensure injectivity.

There is a relative version: if $C⪯D$, keep all named elements from $C$ constant in every coordinate. Each finite diagram fragment is realizable in $C$ over those parameters by elementarity, so the resulting embedding extends the diagonal embedding of $C$ into its ultrapower.

Now use alternating ultrapowers give isomorphic ultralimits. First embed $B$ into an ultrapower $A_1$ of $A$. Apply the relative lemma to $B⪯A_1$, embedding $A_1$ into an ultrapower $B_1$ of $B$ over its diagonal copy. Next embed $B_1$ into an ultrapower $A_2$ of $A_1$ over $A_1$, and continue. After coherent identifications this givesEach $A_{n+1}$ is an ultrapower of $A_n$ along its diagonal embedding, and each $B_{n+1}$ is an ultrapower of $B_n$ likewise. The two subsequences are cofinal in the same chain of elementary substructures, so their limits are the same union. Hence $\boxed{A\equiv B⟹A_{\infty }\cong B_{\infty }}$. No isomorphism theorem for single ultrapowers has been assumed.

A model-theoretic ultralimit is the elementary direct limit of a chain of successive ultrapowers along their diagonal embeddings. We will build two such chains with isomorphic limits. This is not a topological ultrafilter limit and does not require the stronger claim that one ultrapower of each structure is already isomorphic.

First prove the Frayne embedding lemma, including the relative form needed to make the diagrams commute. If $C\equiv D$, expand the language by constants naming every element of $D$, and let $\mathrm{ED}(D)$ be its complete elementary diagram. Index by finite subsets $s$ of this diagram. The finitely many constants in $s$ can be interpreted in $C$ so that $s$ holds: existentially quantifying them gives a sentence true in $D$ and hence in $C$. Choose such interpretations, and give unused constants an arbitrary fixed value in the nonempty domain of $C$.

For each diagram sentence $\delta$, its cone $\{s:\delta \in s\}$ has the finite intersection property with the other cones. Extend the generated filter to an ultrafilter $U$. For $d\in D$, let $f_d(s)$ be its chosen interpretation at coordinate $s$. The mapis elementary by Łoś: every formula true of a tuple from $D$, or its negation, is a diagram sentence and holds on its corresponding cone. In particular distinct named elements stay distinct.

If $C⪯D$, the construction can keep every constant naming $c\in C$ equal to that same $c$ at every coordinate. For a finite diagram fragment, existentially quantify just the names outside $C$; the resulting formula with parameters in $C$ is true in $D$, hence true in $C$. Thus the finite realizations still exist with all those old names fixed. ConsequentlyThis relative extension of the diagonal map is the essential strengthening.

Now let $A_0\equiv B_0$. The nonrelative lemma embeds $B_0$ into an ultrapower $A_1$ of $A_0$. Identify the image with a copy of $B_0$ inside $A_1$. Since $B_0⪯A_1$, the relative lemma embeds $A_1$ into an ultrapower $B_1$ of $B_0$ extending its diagonal map. Rename this larger structure so that the displayed copy of $A_1$ is a literal elementary substructure. Next apply the relative lemma over $A_1⪯B_1$ to embed $B_1$ into an ultrapower $A_2$ of $A_1$ fixing its diagonal copy. Continue alternately. We obtain a coherent elementary chainwith $A_0⪯A_1$ as well. Each inclusion $A_n\to A_{n+1}$ is, up to these coherent identifications, the diagonal embedding into its ultrapower; the same holds for $B_n\to B_{n+1}$. It is the relative clause, not merely the existence of arbitrary elementary embeddings, which guarantees this.

By the elementary chain theorem, the unions of the two subsequences are elementary limits. They are the same structure: each $A_n$ lies in a subsequent $B_n$, and each $B_n$ lies in $A_{n+1}$. Hence the alternating ultrapowers give isomorphic ultralimits construction provesConversely each ultrapower and each elementary direct limit has the same first-order theory as its starting structure. Isomorphic ultralimits therefore imply $A_0\equiv B_0$ as well. Thus the conclusion exactly characterizes elementary equivalence at the level of ultralimits.

The printed inclusion-and-elementarity assertion is false if transitivity is required of the same submodel. Take the theorem of ZFC that combines the axiom of infinity and the Axiom of power set. Whenever $V_{\delta }$ satisfies this sentence, it contains $\omega$, every subset of $\omega$, and their actual power set $\mathcal{P}(\omega )$. A submodel $M\prec V_{\delta }$ contains $\omega$ and $\mathcal{P}(\omega )$, since these are uniquely definable in $V_{\delta }$. If $M$ were transitive, it would contain every element of $\mathcal{P}(\omega )$, contradicting countability by Cantor theorem.

The corrected conclusion uses an elementary embedding rather than elementary inclusion. By Lévy reflection theorem, choose $\delta >\omega +2$ with $V_{\delta }\models \phi$ and with Extensionality true there. The Downward Lowenheim-Skolem theorem says that an infinite structure in a countable language has a countable elementary substructure. Apply it to obtain a countable $N\prec (V_{\delta },\in )$. The membership relation on $N$ is externally well-founded, and elementarity makes it extensional. The Mostowski collapse theorem gives an isomorphism $\pi :N\to M$ onto a countable transitive set. HenceIndeed $M\subseteq V_{\delta }$: rank induction gives $\mathrm{rank}(\pi (x))\le \mathrm{rank}(x)$ for every $x\in N$. The crucial correction is that $j$, rather than the inclusion of $M$, is elementary. For a formula with free variables, apply this argument to its universal closure.

Put $\mathcal{V}=⟨V_{\kappa },\in ,R⟩$. Because $\kappa$ is inaccessible, $|V_{\xi }|<\kappa$ for every $\xi <\kappa$, and $\kappa$ is regular. Choose one witness in $V_{\kappa }$ for every existential formula of this expanded language and every finite parameter tuple for which a witness exists. These are Skolem functions; they can be chosen in the ambient universe even when $R$ is not definable inside $V_{\kappa }$.

For each $\xi <\kappa$, the ranks of all chosen witnesses with parameters from $V_{\xi }$ have supremum below $\kappa$: there are fewer than $\kappa$ such parameters and only countably many formulas. Choose $F(\xi )<\kappa$ strictly above those ranks. The limit closure pointsform a club set. At $\alpha \in C$, every existential assertion in $\mathcal{V}$ with parameters in $V_{\alpha }$ has a witness in $V_{\alpha }$. The Tarski-Vaught test yieldsThus the set $E$ of all such elementary levels is unbounded. It is also closed: the union at a limit of increasing elementary levels is an elementary substructure by the elementary chain theorem, and the union of their ranks is $V_{\alpha }$ with predicate $R\cap V_{\alpha }$. ThereforeThis is club reflection below an inaccessible cardinal.

If $\kappa$ is a Mahlo cardinal, its inaccessible ordinals form a stationary set. For any $R\subseteq V_{\kappa }$, intersect that set with the club set of elementary levels from club reflection below an inaccessible cardinal. This supplies an inaccessible $\alpha <\kappa$ with the required elementary substructure.

Conversely, take an arbitrary club set $C\subseteq \kappa$ and use it as the predicate $R$. In $⟨V_{\kappa },\in ,C⟩$, the sentence asserting that predicate-marked ordinals occur above every ordinal is true. Any inaccessible elementary level therefore satisfies that $C\cap \alpha$ is unbounded in its ordinals, which are precisely the ordinals below $\alpha$. Since $C$ is closed and $\alpha$ is a limit ordinal, $\alpha \in C$. The assumed reflection property consequently gives an inaccessible ordinal in every club set. Hence the inaccessible ordinals are stationary and

A suitable relative condensation lemma fixes the base pointwise. Let $\alpha$ be a nonzero limit ordinal, and let $Y\prec (L_{\alpha }(X),\in )$ be an elementary substructure withThen the Mostowski collapse theorem gives an isomorphism $\pi :Y\to T$ onto a transitive set, andFor the collapse to apply, the restricted membership relation is well-founded because it is actual membership. It is extensional: if two members of $Y$ differ, an element distinguishing them in $L_{\alpha }(X)$ can be chosen in $Y$ by elementary substructure structure. Since $X$ is transitive and all its members are in $Y$, transfinite induction on their rank of a set shows that $\pi$ fixes every member of $X$, and then $\pi (X)=X$.

By the preceding relative constructible level recognition, $L_{\alpha }(X)\models \Phi (X)$. Elementary substructure agreement transfers this to $Y$, and the collapse isomorphism transfers it to $T$. Thus $T\models \Phi (X)$ and $T=L_{\beta }(X)$ for a nonzero limit ordinal $\beta$.

For the bound, let $\rho =X\cap \mathrm{Ord}$. The relative constructible hierarchy has ordinal height $L_{\gamma }(X)\cap \mathrm{Ord}=\rho +\gamma$: at a successor the definable power set adds exactly the previous ordinal height as a new ordinal, and at limits take unions. The order type of $Y\cap \mathrm{Ord}$ is at most $\rho +\alpha$, so $\rho +\beta \le \rho +\alpha$; strict increase of ordinal addition in the right argument gives $\beta \le \alpha$. The requirement $X\subseteq Y$ is essential to this version: merely having $X\in Y$ need not make the collapse fix the base.

The Tarski-Vaught test says that $(M,\in )$ is an elementary substructure of $(N,\in )$ if and only if every existential statement true in $N$ with parameters from $M$ has a witness in $M$ that satisfies its matrix in $N$. Explicitly, for every first-order formula $\psi (x,\bar{y})$ and every parameter tuple $\bar{a}\in M^{|\bar{y}|}$, requireThe truth of the matrix in this test is evaluated in the larger structure $N$. Under this criterion, $(M,\in )\prec (N,\in )$ means that all first-order formulas with parameters from $M$ have identical truth values in the two structures.

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.