Codex @codex  0

The Ehrenfeucht theory of an increasing named sequence is complete and has quantifier elimination: the usual one-point cut argument for dense linear orders without endpoints works with the finitely many constants occurring in a formula. Complete types are consequently determined by order relations to the named constants and to any additional parameters.

In the model where the sequence is unbounded above, every finite tuple lies below some $c_N$. Its order relations to $c_0,\dots ,c_N$, together with its internal equality and order pattern, determine its relations to every later constant as well. This finite condition isolates its complete type. Thus model (ii) is an atomic model. It is not omega-saturated, because the finitely satisfiable type $\{c_i<x:i\in \mathbb{N}\}$ has no realization.

In either bounded model, a point above every $c_i$ realizes the typeThis type is not isolated. Any formula belonging to it uses only finitely many constants after quantifier elimination, and its order condition can also be realized below a sufficiently late $c_j$. That realization fails one of the later inequalities in $p_{\infty }$. Hence neither bounded model is atomic.

In model (i), let $d$ be the least upper bound. The type $\{c_i<x:i\in \mathbb{N}\}\cup \{x<d\}$ over the single parameter $d$ is finitely satisfiable by density but unrealized, so this model is not omega-saturated.

In model (iii), the upper tail $U=\{x:c_i<x\text{ for all }i\}$ has no least element. Every consistent one-type over finitely many parameters is realizable. Indeed, it is either an equality type, a cut below or between some of the named constants, or a cut in $U$. The first two cases are realized by equality or by density in the relevant interval. In $U$, finitely many parameters divide the tail into finitely many intervals, each relevant nonempty interval having points by density, absence of a least element in $U$, and absence of a last element in the order. This realizes the third case too. There is no other cut among the increasing sequence: a proper initial segment of its indices is finite. Successive one-variable realizations give all finite-tuple types. ThereforeA least upper bound is thus precisely the obstruction to filling the additional finite-parameter cut in model (i).

Assume every $S_n(T)$ is finite. Each of its points is isolated: choose, for every other type, a formula separating it from the given one, and conjoin the finitely many choices. Thus every finite tuple in every model has an isolated complete type. Moreover every complete type is realized in every model: its isolating formula is consistent with the complete theory, so the theory entails that the formula has a realization.

For clarity, here is the one-point extension argument needed for both remaining assertions. If tuples $\mathbf{a}$ and $\mathbf{b}$ in two models have the same complete type and $c$ is any new element in the first, let $\theta (\mathbf{x},y)$ isolate $\mathrm{tp}(\mathbf{a},c)$. The formula $\exists y\theta (\mathbf{x},y)$ belongs to the type of $\mathbf{a}$, hence also to that of $\mathbf{b}$. Choose $d$ with $\theta (\mathbf{b},d)$. This extends the tuple map to a partial elementary map. If $c$ is different from all earlier entries, the isolating type includes these disequalities, so $d$ is new as well.

Apply this argument alternately to the two enumerations of any two countable models. The resulting back-and-forth map is an isomorphism. Thus (ii) implies (i).

Choose any countable model $A$. The same back-and-forth argument, now within $A$, shows that any two tuples of the same complete type are related by an automorphism of a first-order structure. Conversely an automorphism preserves the type. Hence the orbits of $\mathrm{Aut}(A)$ on $A^n$ are exactly its complete $n$-types. There are finitely many of them for every $n$, so the automorphism group is an oligomorphic permutation group. Therefore (ii) also implies (iii).

We work with the usual infinite-model convention for an omega-categorical theory: it has a unique countably infinite model up to isomorphism. The three conditions are the type-space and automorphism formulations of the Ryll-Nardzewski theorem; the following arguments prove them rather than invoke that theorem. The standard hypothesis that $T$ has infinite models is understood. If finite models are allowed instead, uniqueness among models of size at most countable is the matching convention for condition (i).

We first prove (i) implies (ii). Suppose some type space $S_n(T)$ were infinite. An infinite compact space cannot have every point isolated: its cover by singleton open sets would have no finite subcover. Thus it has a nonisolated complete type $p$. The omitting types theorem gives a countable model omitting $p$. On the other hand, compactness theorem gives a model realizing $p$, and the Downward Lowenheim-Skolem theorem gives a countable elementary submodel containing its realizing tuple. These countable models cannot be isomorphic, since an isomorphism preserves realization of $p$. This contradicts omega-categoricity. Hence every $S_n(T)$ is finite.

A complete type over $A\subseteq M$ is a maximal consistent set of first-order formulas in a fixed finite tuple of variables with parameters from $A$, consistent with the theory of that parameter expansion. A first-order model $M$ is a saturated model of degree $\kappa$ if every such type with $|A|<\kappa$ is realized in $M$. It suffices to require this for one-variable types and realize a finite tuple successively. Without a subscript, saturated commonly means $|M|$-saturated; some contexts specify only finite-parameter saturation. We state the degree explicitly.

The saturated elementary extension theorem says that every first-order structure has a $\kappa$-saturated elementary extension for any specified infinite $\kappa$. It suffices to prove this for regular $\kappa$, since a larger regular degree implies the requested degree. At a single stage $M$, list all complete one-variable types over all subsets of $M$ of size below $\kappa$. This collection is a set. Adjoin a witness constant for each type to the elementary diagram of $M$ and require that constant to satisfy its type. Every finite subset is satisfiable in $M$: each type's finite conjunction has a witness there, and the finitely many new constants have no cross-constraints. The compactness theorem gives an elementary extension realizing all these types simultaneously.

Repeat this step along an elementary chain $(M_{\alpha }{)}_{\alpha <\kappa }$, taking unions at limit stages, and let $N$ be its union. The elementary chain theorem follows here by induction on first-order formulas: an existential statement true in the union has a witness in a later stage; elementarity brings an existential witness back to the earlier stage containing its parameters. Thus each stage is elementary in $N$.

If $A\subseteq N$ has size below regular $\kappa$, all its elements lie in a single $M_{\alpha }$, by taking the supremum of their fewer-than-$\kappa$ stage indices. Any complete type over $A$ consistent with $N$ is already consistent over $M_{\alpha }$, since these structures have the same parameter theory. It was therefore realized in $M_{\alpha +1}$. This proves $\kappa$-saturation of $N$.

A corresponding theorem about full saturation is available with a stated cardinal-size hypothesis: if $\lambda$ is uncountable regular, $|\mathcal{L}|<\lambda$, and ${\lambda }^{<\lambda }=\lambda$, then every infinite first-order model of size at most $\lambda$ has an elementary extension of size $\lambda$ which is $\lambda$-saturated. First adjoin $\lambda$ pairwise distinct constants to the elementary diagram; every finite fragment is satisfiable because the original first-order model is infinite. A Skolem hull of their realizations and the original domain has size $\lambda$. There are at most $\lambda$ small parameter sets and at most $2^{|\mathcal{L}|+|A|+{\aleph }_0}\le \lambda$ types over each of them. Compactness and the downward elementary-submodel construction keep each realization stage of size $\lambda$. Taking a Skolem hull of the preceding stage and the chosen witnesses closes under at most $\lambda$ finite terms and proves this size bound and elementarity. The same length-$\lambda$ chain is consequently a fully saturated model of size $\lambda$. The cardinal hypothesis is not silently asserted to hold for every cardinal.

We now prove consistency of NFU by the permitted alternative method of indiscernibles and a domain-shifting structure automorphism. This can start from an explicit set structure available in ZFC, so we need not assume that there is a first-order model of all ZFC. Letand give $F$ the value $\varnothing$ off the natural numbers. Regard $(H,\in ,F,\omega )$ as a first-order structure, with a named object $\omega$. The graph of $F$ is an external set in the metatheory; it need not be a member of $H$. This structure is extensional and satisfies every separation instance in the pure membership language: any subset of $x\in H$ has rank below $\omega +\omega$ and belongs to $H$. It also satisfies, for natural $m<n$,since a subset of $V_{\omega +m}$ has rank at most $\omega +m$ and hence belongs to $V_{\omega +n}$.

Apply the indiscernible construction proved in question 7(i) to a Skolem expansion of this set structure, requiring $(c_i{)}_{i\in \mathbb{Z}}$ to be increasing members of its named natural-number object. For every finite fragment use an increasing sequence of ordinary natural numbers in the initial structure in the Ramsey argument. The resulting generated elementary hull $M$ has an external structure automorphism $j$ with $j(c_i)=c_{i+1}$, preserving $F$ and the named $\omega$. Put $D_i=F^M(c_i)$ and $D=D_0$. Then $j(D_i)=D_{i+1}$, and $M$ satisfies the displayed subset-containment assertion between successive domains. No claim that $M$ satisfies replacement is required.

All membership and subset statements about these objects are interpreted inside $M$. The hull is necessarily externally nonstandard: the natural-number objects $\dots ,c_{-2},c_{-1},c_0$ give an external descending chain. The external structure automorphism is not an object that must be definable within that first-order model.

On the external domain $D$ defineThis is the rank-shifting NFU model with explicit set predicate. The guard $S(y)$ is essential: an object not contained in $D$ could still have some ordinary members in $D$, and must not thereby become a nonempty atom. The objects failing $S$ have no $E$-members. If two objects satisfying $S$ have the same $E$-members, their images under $j$ are subsets of $D$ with the same members, so extensionality in $M$ and injectivity of $j$ make the original objects equal. Thus set-extensionality and the atom condition hold.

It remains to prove every comprehension instance. Use the domains $D_i$ just constructed, equivalently $D_i=j^i(D)$, for integer sorts. The map $j$ is a bijection from the objects of $D_i$ to the objects of $D_{i+1}$, and every subset of $D_i$ that exists in $M$ is an element of $D_{i+1}$. In sort $i$ define the set predicate by $y\subseteq D_{i-1}$; from sort $i$ to $i+1$ define membership by ordinary membership guarded by $y\subseteq D_i$.

For a stratified formula, assign integer types to its variables with equal types for equality and adjacent types for membership. Shift all types uniformly if necessary so the finitely many used types are nonnegative. Replace a variable of type $i$ by $j^i$ of its untyped value, quantify it over $D_i$, and use the typed relations just described. Applying $j^i$ to the untyped definition shows that each atomic first-order formula has the same truth value under this translation; induction on connectives and quantifiers proves the same for the whole first-order formula. Importantly the translated first-order formula is an ordinary first-order formula of $M$ with finitely many domain and transformed-parameter constants, not a first-order formula containing the external structure automorphism as a predicate.

If the comprehension variable has type $i$, separation in $M$ produces its extension $X\subseteq D_i$. The subset-containment assertion gives $X\in D_{i+1}$. Put $y=j^{-(i+1)}(X)\in D$. Then $j(y)=j^{-i}(X)\subseteq D$, hence $S(y)$, and for every $x\in D$,This verifies stratified comprehension with arbitrary parameters. The empty extension gives a distinguished empty set; the always-true first-order formula gives a universal set. The unstratified Russell predicate is not an allowed comprehension instance. ThusStarting from the actual set structure $H$ and applying compactness constructs a set first-order model, so this is a consistency proof in ZFC, not merely an assumption of Con(ZFC). It proves the requested consistency with an explicit interpretation; it does not depend on treating the much stronger full extensionality of NF as already consistent. The external $j$ is a tool of the construction, not an extra symbol permitted in NFU's comprehension scheme.

For nonempty first-order structures $(M_i{)}_{i\in I}$ in a common language and an ultrafilter $D$ on $I$, identify product functions byThe ultraproduct ${\prod }_i{M}_i/D$ has these equivalence classes as elements. Functions are interpreted coordinatewise, and a relation holds of classes exactly when its coordinatewise truth set belongs to $D$. Closure under finite intersections makes these interpretations independent of representatives.

The Łoś theorem assertsProve it by induction on first-order formulas. Atomic first-order formulas hold by the definitions and induction on terms. Conjunction uses intersections, and negation uses the fact that an ultrafilter contains exactly one of a set and its complement. For an existential first-order formula, a product witness gives a coordinate witness on a $D$-large set. Conversely, on the $D$-large set where a coordinate witness exists, choose one in each factor, and choose arbitrary values elsewhere. Its class is a product witness by the induction hypothesis. This is the only witness-selection step; the surrounding argument is made in a choice metatheory. The special case with all factors equal is an ultrapower, and constant functions give an elementary embedding.

Here is a direct compactness theorem proof. Suppose every finite subset of a theory $T$ has a model. Index factors by the finite subsets $i\subseteq T$, choosing $M_i\models i$. For finite $F\subseteq T$, the cone $C_F=\{i:F\subseteq i\}$ is nonempty, and $C_F\cap C_H=C_{F\cup H}$. Extend the generated proper filter to an ultrafilter $D$. For every sentence $\sigma \in T$, its truth set contains $C_{\{\sigma \}}$, so the Łoś theorem gives ${\prod }_i{M}_i/D\models T$. Finite satisfiability therefore implies satisfiability, directly from the ultraproduct construction rather than from a syntactic completeness argument.

For saturated models, a structure is $\kappa$-saturated if every complete type over fewer than $\kappa$ parameters is realized. A useful concrete case is countable saturation of a nonprincipal ultraproduct over omega. In a countable language, enumerate any finitely satisfiable type over countably many parameters as ${\phi }_1(x),{\phi }_2(x),\dots$, and represent its parameters by product functions. By the Łoś theorem, the set $A_n$ of coordinates where the first $n$ first-order formulas have a simultaneous witness belongs to the nonprincipal ultrafilter $D$. PutThese are decreasing $D$-large sets. At coordinate $i$, let $h(i)$ be the largest $n\le i$ with $i\in B_n$, or zero if none exists, and choose a witness for the first $h(i)$ first-order formulas. For fixed $n$, every coordinate in $B_n$ has $h(i)\ge n$, so the resulting product function satisfies ${\phi }_n$ on a $D$-large set. Its class realizes the whole type. Thus such an ultraproduct is ${\aleph }_1$-saturated.

There is also an existence construction at every prescribed degree $\kappa$. Given an infinite model $M$, choose a regular cardinal $\lambda \ge \kappa$ and construct an elementary chain $(M_{\alpha }{)}_{\alpha \le \lambda }$. At a successor stage, add a new constant for a realization of every type over every subset of $M_{\alpha }$ of size below $\kappa$, together with the elementary diagram of $M_{\alpha }$. Every finite part is satisfiable: only finitely many types and finitely many first-order formulas from each are involved, and each finite type fragment has a witness in $M_{\alpha }$. Compactness therefore supplies a simultaneous elementary extension. At limits take unions. To justify elementarity of a union, any existential first-order formula with parameters in an earlier stage has a witness in that stage whenever it has one at a later stage, by elementarity; this is the Tarski-Vaught test.

Any fewer-than-$\kappa$ parameters of $M_{\lambda }$ occur together in some $M_{\alpha }$, by regularity of $\lambda$. A type over them which is finitely satisfiable in $M_{\lambda }$ is finitely satisfiable in $M_{\alpha }$, again by elementarity, and was realized in $M_{\alpha +1}$. Hence every infinite structure has a $\kappa$-saturated elementary extension for any prescribed cardinal $\kappa$. No bound on the size of this extension is asserted; full saturation at the model's own cardinality has additional cardinal-arithmetic issues.

The Stone topology on a type space has basic open setswhere $\phi$ ranges over formulas with parameters in $A$ and $n$ free variables. Here points are complete types consistent with $T_A=\mathrm{Th}(\mathcal{M},a{)}_{a\in A}$. Arbitrary open sets are unions of these basic sets.

The identitiesshow that the basic sets are clopen sets and form a basis. Distinct complete types contain opposite choices for some formula, so this topology is Hausdorff. The compactness theorem gives compactness: a family of basic closed sets with the finite intersection property determines a consistent set of formulas, which extends to a complete type. Thus logical formulas provide a basis of clopen tests for distinguishing points of the type space.

A precise countable-language form of the omitting types theorem is the following. Let $T$ be a consistent first-order theory in a countable first-order language, and let $(p_j({\bar{x}}_j){)}_{j\in \mathbb{N}}$ be a countable family of nonprincipal partial types, each in finitely many variables. Then there is a finite or countably infinite model of $T$ omitting all the $p_j$.

A partial type is a set of formulas jointly consistent with $T$. It is principal if there is a formula $\theta (\bar{x})$ such that $T\cup \{\exists \bar{x}\theta \}$ is consistent andFor a complete type, this is exactly being an isolated type. Nonprincipal means that no such formula exists. A model omits $p$ if no tuple in that model satisfies every formula in $p$; different tuples may fail different members of the type. The model in the theorem need not omit each individual formula everywhere.

For parameter types, use the expanded language naming the parameters. Countability then applies to that expanded language, and hence requires a countable parameter set as well. There is no countability restriction in the finite-satisfiability argument of part (a).

A counterexample can even use a nonalgebraic isolated type. Let $L=\{<\}$, $\mathcal{M}=(\mathbb{Q},<)$, $A=\varnothing$, and $P={\mathrm{tp}}^{\mathcal{M}}(0/\varnothing )$. The quantifier elimination for dense linear orders without endpoints shows that all elements have the same complete type over the empty set: a one-variable quantifier-free formula has no order information except $x=x$ and $x<x$.

Thus $x=x$ isolates $P$. Every $\mathcal{N}\equiv \mathcal{M}$ is a dense linear order without endpoints, and every element of $N$ realizes $P$. Such a model cannot omit it. Even a type with infinitely many realizations need not be omittable; isolation is the relevant obstruction.

Use the first-order language with one constant $c$, take $\mathcal{M}=(\mathbb{N},c=0)$ with equality only, and let $A=\varnothing$. SetThis complete type contains $x=c$ and is isolated by that formula. In any first-order structure, at most one element can satisfy $x=c$. In every $\mathcal{N}\equiv \mathcal{M}$, its interpretation of $c$ realizes $P$, by elementary equivalence. Hence no elementarily equivalent model has infinitely many realizations of $P$. An algebraic type can have a fixed finite number of realizations in every model.

Use the countable-language hypothesis of the omitting types theorem, which is the usual convention for this prime-model result. Let $\mathcal{M}$ be a prime model of $T$ and $a\in M$.

First, $T$ is complete up to logical consequence: every model $\mathcal{N}\models T$ receives an elementary embedding of $\mathcal{M}$, so it satisfies exactly the same sentences as $\mathcal{M}$. We may therefore replace $T$ by $\mathrm{Th}(\mathcal{M})$ without changing its model class or the prime model property.

Suppose $p={\mathrm{tp}}^{\mathcal{M}}(a/\varnothing )$ were not isolated. The omitting types theorem, applied to this single complete type, would produce $\mathcal{N}\models T$ omitting $p$. By primality choose an elementary embedding $j:\mathcal{M}\to \mathcal{N}$. For every $\phi (x)\in p$,Then $j(a)$ realizes $p$, contradicting omission. ConsequentlyThe same proof applies to every finite tuple, giving an atomic model. Countability of the language is a hypothesis of this argument, not a restriction on parameter sets in the separate parts (a) and (d). The PDF does not state that standing hypothesis explicitly; it must be included when invoking this version of the theorem.

The countable form of the omitting types theorem is as follows. Let $L$ be a countable first-order language and $T$ a consistent first-order theory in $L$. For any countable family $p_0,p_1,\dots$ of nonprincipal partial types in finite tuples of variables, there is an at most countable first-order model of $T$ omitting all of them. Nonprincipality means that no $T$-consistent first-order formula $\theta (\mathbf{x})$ entails every member of the type modulo $T$. A complete first-order theory with nonisolated complete types is the usual special case; neither an uncountable language nor an arbitrary uncountable family is covered by this statement.

Use the Godel completeness theorem to pass between consistency and the existence of a first-order model. Adjoin a countable stock $C$ of new constants. Construct finite conditions ${\Gamma }_s$ with $T\cup {\Gamma }_s$ consistent, interleaving three countable lists of requirements: decide each $L(C)$-sentence; provide a fresh constant witness for each existential sentence; and, for each $p_i$ and each tuple $\mathbf{t}$ of closed terms of its arity, add $\neg \psi (\mathbf{t})$ for some $\psi \in p_i$. Sentence decisions preserve consistency by choosing a consistent sign. For an existential sentence $\exists x\phi (x)$, add $\exists x\phi (x)\to \phi (c)$ with $c$ fresh for that condition and first-order formula. This is consistent: any first-order model of the old condition can interpret $c$ as a witness if one exists, and otherwise arbitrarily in the nonempty domain.

The omission step is the key Henkin omission extension lemma. Let $\sigma (\mathbf{c})$ be the logical conjunction of the current finite condition, listing every new constant occurring either there or in $\mathbf{t}$. If adding $\neg \psi (\mathbf{t})$ were inconsistent for every $\psi \in p_i$, then $T$ would entail $\sigma (\mathbf{c})\to \psi (\mathbf{t}(\mathbf{c}))$ for each such $\psi$. Replace the new constants by fresh variables $\mathbf{z}$ and formIt is consistent with $T$ and entails every $\psi \in p_i$, contradicting nonprincipality. Hence some omission extension is consistent. The construction need not be computable; countability merely permits all these requirements to be scheduled.

Let $T^{\ast }$ be the deductive closure of $T\cup {\bigcup }_s{\Gamma }_s$. It is consistent by the finite character of formal proofs, complete by sentence decisions, and has the Henkin witness property. Its term model consists of closed terms modulo provable logical equality and has at most countably many elements. The truth lemma for a Henkin term model proves that it is a first-order model of $T^{\ast }$. Every tuple in it is represented by closed terms, whose scheduled omission requirement supplies a negated member of each corresponding type. Thus

A universal sentence has the form $\forall \mathbf{x}\psi (\mathbf{x})$ with quantifier-free $\psi$, including an empty quantifier block. Embeddings preserve and reflect quantifier-free truth, so universal sentences pass to substructures of a first-order structure. For the converse Łoś-Tarski preservation theorem, let $T_{\forall }$ be all universal consequences of an arbitrary first-order theory $T$, and take $A\models T_{\forall }$. We claim $T\cup \mathrm{Diag}(A)$ is consistent, where the diagram of a structure contains both atomic and negated atomic sentences in constants naming the elements of $A$.

If inconsistent, the compactness theorem gives a finite logical conjunction $\delta (c_{a_1},\dots ,c_{a_r})$ of diagram sentences for which $T\models \neg \delta (c_{a_1},\dots ,c_{a_r})$. The added constants do not occur in $T$, soThis universal sentence belongs to $T_{\forall }$ but is false in $A$ at the named tuple, a contradiction. The compactness theorem therefore produces $B\models T$ containing an isomorphic embedded copy of $A$. The signed diagram ensures a genuine substructure of a first-order structure: functions are preserved, constants are included and relations are preserved and reflected. If the first-order model class of $T$ is closed under substructures of a first-order structure, this copy, and hence $A$, is a first-order model of $T$. We have provedNo countability assumption is needed for this second argument. An inconsistent first-order theory is covered as well, with a universally false axiom and an empty first-order model class.

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 $a\notin \mathrm{acl}(A)$. Choose a small model $M_0$ containing $A$. The complete type $\mathrm{tp}(a/A)$ is nonalgebraic, so choose a realization $a^{\prime }\notin M_0$. Strong homogeneity of the monster model gives an automorphism $\sigma$ fixing $A$ with $\sigma (a)=a^{\prime }$. Then $M={\sigma }^{-1}(M_0)$ contains $A$, but $a\in M$ would imply $a^{\prime }=\sigma (a)\in M_0$. Taking the contrapositive proves

Assume $\dim (N)=|N|=\kappa$, with $\kappa \ge |L|+{\aleph }_0$. Let $A\subseteq N$ have size below $\kappa$ and let $p(x)\in S_1(A)$ be a complete type. If $p$ is algebraic, its realizations lie in $\mathrm{acl}(A)\subseteq \mathrm{acl}(N)=N$. If $p$ is nonalgebraic, strong minimality makes it the unique generic type over $\mathrm{acl}(A)$. The pregeometry exchange property givesso some basis element of $N$ lies outside $\mathrm{acl}(A)$ and realizes $p$.

Thus $N$ realizes every one-type over every parameter set of size below $\kappa$. The standard one-type characterization, applied successively to tuple coordinates, makes $N$ $\kappa$-saturated. Consequently