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An asymptotic power series uses powers as its asymptotic sequence. Near a finite point $a$, the condition in the title applies along the specified domain for every fixed $N$; at infinity, inverse powers are convenient. It need not be a convergent power series. Its coefficients record the successive limiting responses, as the integral example below demonstrates.

The finite-order Taylor theorem supplies an asymptotic power series for a smooth function near a point, but smoothness does not imply that its Taylor series converges to the function. Define $g(x)=e^{-1/x}$ for $x>0$ and $g(x)=0$ for $x\le 0$. Every right-hand derivative is an exponential function times a polynomial in $1/x$, tending to zero as $x\downarrow 0$. Consequently $g$ is a smooth function with every derivative zero at the origin. Its all-zero Taylor series misses $g(x)>0$ to the right. Zero and $g$ also have the same all-zero asymptotic power series on that side, despite being different functions. An asymptotic expansion can determine all its own coefficients without determining the represented function uniquely.

Start with a well-founded universe of sets over an infinite set $A$ of urelements. The full symmetric group of $A$ acts recursively on sets. Retain the objects all of whose descendants have a finite support in a permutation action. This class satisfies set-extensionality, foundation, infinity, pairing, union, separation, replacement and its internal power set axiom. A definable functional image has the finite union of the supports of its domain and parameters as a support; uniqueness makes the image invariant. The family of two-element subsets of $A$ has no choice function: a transposition of two atoms outside the function's finite support fixes their pair and moves either possible choice. For untyped membership with extensionality merely restricted to nonempty objects, atoms are also vacuous subsets; add all atoms to each internal power-set object.

Start with distinct formal atoms $q\in A$ with membership extensions $\{q\}$. Recursively adjoin a unique representative $S(X)$ of every subset $X$ of a previous stage, except that $S(\{q\})=q$ uses the atom already present. Define membership by these extensions. Each set-sized collection of objects has bounded construction rank, so it has a representative at a later stage. This gives full extensionality, unlike adjoining an ordinary singleton distinct from a Quine atom. Definable subsets and ranges, unions and power sets are set-sized collections in the ambient model and hence have representatives. The pure finite hierarchy supplies infinity. Every permutation of the atoms extends recursively; hereditary finite support restriction gives the hereditarily finite-supported Quine-atom model, whose two-element atom subsets have no choice function.

Build well-founded coded sets over a single urelement $a$ with no members, retaining a distinct empty set. Use the unmodified membership language, so both $a$ and the empty set have the same empty extension and extensionality fails. Pairs, unions, separation and functional ranges have set codes, and pure natural numbers give infinity. For the unrestricted Axiom of power set, the power-set code must contain $a$ as well as all ordinary subset codes, since the atom is vacuously a subset of every object. Rank gives foundation. Thus the other ZF axioms hold.

The independence statements are relative consistency statements. We construct models from an ambient model of ZFC; the construction is an interpretation and does not assume that the ambient model is transitive. Foundation is already true in the ordinary ambient membership structure, so its positive consistency direction is immediate.

For the negative direction, construct an extensional cumulative universe over Quine atoms. Let $A$ be a countably infinite set of formal symbols $q_a$. Give each $q_a$ the extension $\{q_a\}$. For a set $X$ of already constructed objects, let $S(X)$ be its unique representative: set $S(\{q_a\})=q_a$, and otherwise use a fresh tagged code $(1,X)$. Atom codes have a different tag. Begin with $U_0=A$ and putSet $W={\bigcup }_{\alpha }{U}_{\alpha }$. The interpreted membership relation $E$ gives $S(X)$ exactly the members $X$, and gives $q_a$ exactly the member $q_a$. This convention matters: creating a second representative for $\{q_a\}$ would violate extensionality. Every object is uniquely determined by its extension, so $E$ satisfies full extensionality. Every set-sized collection $X$ of objects has bounded construction rank, by ambient replacement, and hence has its representative $S(X)$ in $W$.

Here are the remaining axiom verifications. Empty set and pairing use $S(\varnothing )$ and $S(\{x,y\})$. If $E(y)=X$, its interpreted union is $S({\bigcup }_{x\in X}E(x))$. Its interpreted power set issince every subset extension has a unique representative. The pure, atom-free finite hierarchy supplies a copy of $\omega$ and an inductive set. For separation, translate an $E$-formula into the ambient language, with quantifiers restricted to the definable class $W$; ambient separation gives the appropriate subset of $X$, and $S$ represents it. For replacement, uniqueness in the interpreted functional relation gives an ambient definable functional image of the set $X$. Ambient replacement makes that image set-sized, and $S$ again represents it. These arguments also establish that all the operations used have bounded construction rank.

Ambient choice gives choice in $W$: choose a member from each nonempty extension in a set-sized family and represent the graph, using interpreted ordered pairs. But $q_{a}E{q}_a$, and the nonempty set with extension $\{q_a\}$ has no $E$-minimal member. ThusTogether with the ordinary model, this proves independence of foundation from the other ZF axioms, even when choice is retained.

To extend the construction to failure of choice, let the full permutation group of $A$ act on $W$ byThe singleton convention is equivariant, so this is an automorphism of $(W,E)$. An object has finite support if some finite $F\subseteq A$ has the property that every permutation fixing $F$ pointwise fixes the object. Retain objects all of whose membership descendants have finite support, stopping the hereditary recursion at the atomic self-loops. This gives the hereditarily finite-supported Quine-atom model $\mathcal{H}$; each $q_a$ belongs to it, with support $\{q_a\}$.

Extensionality persists because every retained object's members are retained. Empty set, pairing, infinity and union preserve hereditary finite support. For separation on a retained set $x$, the union of finite supports for $x$ and the formula's parameters supports the separated extension: permutations fixing those supports preserve satisfaction in $\mathcal{H}$. All its members are retained, so its representative is retained. For replacement, the same support fixes the functional range, because an automorphism must take the unique output at an input to the unique output at its transported input. Ambient replacement bounds that range; all its members are in $\mathcal{H}$, so its representative belongs to $\mathcal{H}$. Finally, the collection of all retained subset representatives of $x$ is an ambient set, contained in the full interpreted power set of $x$. A support of $x$ permutes this collection onto itself, and its members are retained. Its representative is therefore the internal power set in $\mathcal{H}$. This verifies every ZF axiom except foundation; foundation still fails at each $q_a$.

The atom set $A$ and the family $[A{]}^2$ of its two-element subsets have empty support and are retained. Suppose a choice function $f$ on $[A{]}^2$ were retained, with finite support $F$. Choose distinct $q_a,q_b$ outside $F$ and transpose them. The transposition fixes $f$ and fixes the unordered pair $\{q_a,q_b\}$, but moves whichever of its two members $f$ selects. Equivariance would require that selected member to be fixed, a contradiction. HenceThe full universe $W$ gives the corresponding positive-choice model. Choice is therefore independent of ZF without foundation. Unlike a permutation model with ordinary urelements, this model retains full pure-set extensionality: the atomic objects are self-membered sets, not empty objects exempted from extensionality.

The constructible-universe interpretation gives $\mathrm{Con}(\mathrm{ZF})\Rightarrow \mathrm{Con}(\mathrm{ZFC})$. Thus the consistency assumptions in both displayed conclusions can equivalently be taken to be $\mathrm{Con}(\mathrm{ZF})$; no additional consistency strength is being used.

An inner model of ZF is a transitive class $M$ containing every ordinal, with inherited membership, such that each ZF axiom holds with all quantifiers restricted to $M$. Here transitivity means $x\in y\in M\Rightarrow x\in M$. For a proper class this is an axiom-by-axiom assertion, not an internal truth predicate for the whole class. An inner model can have fewer subsets of a given set than the surrounding universe; it need not have the same cardinals or the same continuum function.

Inner models prove relative consistency. If a theory $T$ proves that a definable class $M$ satisfies every axiom of $S$, an inconsistency proof from $S$ could be relativized to $M$ and become an inconsistency proof from $T$. Thus $\mathrm{Con}(T)\Rightarrow \mathrm{Con}(S)$. This reasoning concerns formal interpretations and needs no assumption that a given model of $T$ is externally well-founded.

The central example is the constructible universe. Definewhere $\mathrm{Def}(X)$ contains exactly the subsets definable over $(X,\in )$ using parameters from $X$. Satisfaction for a set structure is definable, so this recursion is available in ZF. Induction shows that the levels are transitive and increasing: if $a\in L_{\alpha }$, the subset $\{x\in L_{\alpha }:x\in a\}$ is $a$, so $a\in L_{\alpha +1}$. The ordinal part of $L_{\alpha }$ is $\alpha$, giving all ordinals in $L$.

For completeness, the main axiom mechanisms explain why this example works. Finite set operations on constructible parameters are definable at finitely many further stages. For a fixed finite family of formulas and parameters, close a sequence of levels under their least constructible witness stages and take its union; induction on formulas gives a reflecting level. Separation of a constructible set then occurs at the next definability stage. For replacement, ambient replacement bounds the construction stages of the unique constructible outputs; a reflecting level above that bound makes the range definable. For the internal power set of $a\in L$, use ambient separation to form $\mathcal{P}(a)\cap L$, bound the construction stages of its members, and at a sufficiently high level define the family of all subsets of $a$ in that level. Thus the internal power set is constructible even though some ambient subsets may not be. Foundation is inherited by transitivity, and infinity comes from $L_{\omega }$ and its successor stages.

There is a canonical well-order of $L$: order by first construction stage, then by the least definition code and finite parameter tuple, using the previously constructed well-orders for parameters. Each set in $L$ can be well-ordered by the restriction of this set-like class order; hence $L$ satisfies AC. Moreover the hierarchy computed inside $L$ is the same hierarchy, because definability over a fixed set structure and limit unions are absolute. Thus $L$ satisfies $V=L$. The further cardinal analysis of this hierarchy, encapsulated in the constructible universe theorem, gives GCH. ConsequentlyThis proves the relative consistency of choice, CH and GCH; in particular none can be refuted by ZF if ZF is consistent. The assertions here are internal to $L$: they do not assert CH in the ambient universe.

Inner models also isolate definable information. The relative constructible universe built with a set parameter retains that parameter and all ordinals while discarding unrelated sets. Such constructions help compare relative consistency strengths and identify which hypotheses survive passage to smaller universes; preservation of a large-cardinal property must be checked separately, not inferred merely from transitivity.

There is an important limitation, the constructible-universe obstruction to a uniform inner-model proof of failure of choice. Every inner model $M$ of ZF contains the ambient $L$. Indeed $L_{\alpha }^M=L_{\alpha }$ by induction: the same set structure has the same definable subsets at a successor stage, and both hierarchies take the same union at a limit. If the ambient universe satisfies $V=L$, this forces $M=V$. Thus inner models cannot uniformly produce a model of $\neg \mathrm{AC}$ or $V\ne L$ from every ZF universe. To prove the opposite independence directions one needs an outer-model or other interpretation technique, such as forcing or the non-well-founded symmetry construction in Question 2.

The saturated model condition in degree $\kappa$ means that a first-order model $M$ realizes every one-variable complete type over a parameter set $A\subseteq M$ of size less than $\kappa$. Equivalently every finitely satisfiable such complete type is realized, since it can be extended to a complete type. Without a specified degree, saturated usually means $|M|$-saturated. Countable saturation means ${\aleph }_1$-saturation, including complete types over countably many parameters.

Here is a general saturated elementary extension theorem, with proof: for every infinite regular cardinal $\kappa$, every first-order structure $M$ has a $\kappa$-saturated elementary extension. At a stage $M_{\alpha }$, list all complete types over its subsets of size less than $\kappa$. There is a set of them, since their formulas lie in a set-sized language. Add a witness constant for each complete type to the elementary diagram of $M_{\alpha }$, together with every formula in that complete type evaluated at its witness constant. A finite part mentions finitely many complete types and finitely many formulas from each; each finite conjunction is realizable in $M_{\alpha }$, and the witnesses for different complete types are unconstrained relative to each other. Thus the finite part is satisfiable. The compactness theorem produces an elementary extension $M_{\alpha +1}$ realizing all those complete types. At limits take elementary unions, justified by the elementary chain theorem.

Iterate for $\kappa$ stages and put $N={\bigcup }_{\alpha <\kappa }{M}_{\alpha }$. If $A\subseteq N$ has size below $\kappa$, regularity bounds the stages at which its elements enter, so $A\subseteq M_{\alpha }$ for some $\alpha <\kappa$. A complete type over $A$ consistent with $N$ is finitely satisfiable in $M_{\alpha }$: each finite conjunction is an existential formula with parameters from $M_{\alpha }$, and elementarity transfers its truth from $N$. The complete type is therefore realized in $M_{\alpha +1}$. This provesFor a singular desired degree, apply the theorem at a larger regular cardinal. If $M$ is infinite and an uncountable regular $\kappa$ satisfies ${\kappa }^{<\kappa }=\kappa$, with $|L|<\kappa$ and $|M|\le \kappa$, the construction can keep every stage of size $\kappa$. There are then at most $\kappa$ parameter sets and complete types, and downward Löwenheim-Skolem gives that size after each extension; adding $\kappa$ distinct elements initially ensures final size exactly $\kappa$. The resulting model is saturated in the unqualified sense. The cardinal-arithmetic hypothesis belongs to this size refinement, not to the general existence theorem.

An explicit ultraproduct version is useful too. In a countable language let $U$ be a nonprincipal ultrafilter on $\omega$ and $N={\prod }_{i<\omega }{M}_i/U$, where each factor is nonempty. Enumerate a countable finitely satisfiable complete type, including its chosen parameter representatives, as ${\phi }_1(x),{\phi }_2(x),\dots$. By the Łoś theorem, the set $A_n$ of coordinates where the first $n$ formulas have a simultaneous witness belongs to $U$. PutThese are decreasing members of $U$. At coordinate $i$, let $k(i)$ be the greatest $n\le i$ with $i\in B_n$, if there is one. Choose a witness to the first $k(i)$ formulas there, and an arbitrary element when there is none. For fixed $n$, every coordinate in $B_n$ has $k(i)\ge n$, so the chosen function satisfies ${\phi }_n$ on a $U$-large set. Its ultraproduct class realizes the whole complete type. A countable language over a countable parameter set has only countably many formulas, so this proves countable saturation of a nonprincipal ultraproduct over omega.

For NFU, we give the rank-indiscernible construction of an NFU model explicitly. NFU has a unary predicate $S$ for sets, atoms have no members, extensionality applies to sets, and comprehension applies to every stratified formula. A stratification assigns integer types to variables so that equality uses equal types and $x\in y$ requires $\mathrm{type}(y)=\mathrm{type}(x)+1$. A unary set predicate adds no type difference.

Start with the actual set structure $B=(V_{\omega +\omega },\in )$. This structure satisfies extensionality and every pure-membership separation instance: the subset defined in a set $a$ of rank below $\omega +\omega$ still has rank below $\omega +\omega$. We do not assert that $B$ satisfies replacement or all of ZFC. Expand it by Skolem functions. In it use the sequence of rank objects $V_{\omega +n+6}$, $n<\omega$. For two such objects $d<e$ in sequence order, we havewith quantifiers in $B$. The second statement holds because every subset of $V_{\alpha }$ belongs to $V_{\alpha +1}$, and the later rank is at least $\alpha +1$.

Introduce constants $d_i$, $i\in \mathbb{Z}$, and require them to be order indiscernibles in the Skolem language, satisfying these two assertions whenever $i<j$. Also require $V_{\omega +5}\subseteq d_i$ for every $i$; this fixed rank object is first-order definable in $B$. Every finite part of these requirements together with $\mathrm{Th}(B^{\ast })$ is satisfiable: color finite increasing tuples of the original rank sequence by the truth values of the finitely many formulas mentioned, and use Ramsey's theorem to get a homogeneous subsequence. The rank assertions hold for every increasing tuple there. The compactness theorem gives a model of the full theory. In its Skolem hull $N$ generated by the $d_i$, the integer shift extends to an automorphism $j$ with $j(d_i)=d_{i+1}$, by the term-transport proof of the Ehrenfeucht-Mostowski theorem. Thus $N$ still satisfies extensionality and all pure-membership separation instances, andThis is a set-sized first-order model; no satisfaction predicate for a proper-class universe is being assumed.

Externally let $D_i=\{x\in N:N\models x\in d_i\}$. The automorphism bijects $D_i$ onto $D_{i+1}$. Interpret a typed structure with sort $i$ equal to $D_i$. At sort $i+1$, declare $y$ a set exactly when $N\models y\subseteq d_i$; otherwise it is an atom. Define adjacent-sort membership byThe guard is necessary: a typed atom might have some ordinary $N$-members in $D_i$, and those must not become its typed members. Set-extensionality follows from extensionality in $N$, because the ordinary members of a typed set all lie in $D_i$. For any typed formula on sort $i$, its quantifiers can be restricted to the corresponding $d_l$, its set predicates replaced by subset assertions, and its memberships replaced by the guarded relation. This is an ordinary pure-membership formula of $N$ with finitely many parameters. Separation gives its extension $X\subseteq d_i$, and the displayed closure assertion gives $X\in d_{i+1}$. It is therefore a typed set witnessing comprehension. The automorphism shifts these sorted domains and preserves the set flags and guarded memberships.

Collapse the types onto $D_0$. DefineObjects not satisfying $S$ have no $E$-members. If two sets have the same $E$-members, their images under $j$ are subsets of $d_0$ with the same ordinary members, so are equal by extensionality of $N$; injectivity of $j$ gives equality of the original objects.

To verify every stratified comprehension instance, assign a variable $v$ its stratification type $t(v)$ and translate its value $a\in D_0$ to $j^{t(v)}(a)\in D_{t(v)}$. Equality is preserved. An atomic $E(x,y)$ translates to the guarded adjacent-sort membership because $t(y)=t(x)+1$; the unary predicate $S(y)$ translates to $j^{t(y)}(y)\subseteq d_{t(y)-1}$. Induction on formulas, using the domain bijections for quantifiers, preserves truth. If the free variable $x$ has type $t$, typed comprehension produces its extension $X\subseteq d_t$, with $X\in d_{t+1}$. Put $b=j^{-(t+1)}(X)\in D_0$. Then $S(b)$ holds, and for every $a\in D_0$,This is exactly the required NFU set. No formula containing the external automorphism $j$ has been used in a separation or comprehension scheme; $j$ only transports values after the pure formula is formed.

The construction also permits the usual infinity requirement. The ordinary $\omega$ and its successor graph $f$ are definable in $N$, hence fixed by $j$, and lie below $V_{\omega +5}$, so belong to all the required domains. Under $E$, the members of $\omega$ are still its ordinary $N$-members. The new Kuratowski pair satisfiesSince $j(f)=f$, membership of this pair in $f$ under $E$ is equivalent to $⟨x,y{⟩}_N\in f$. Thus $f$ is an internal injection of $\omega$ into itself omitting zero, giving a Dedekind-infinite set.

We have built a model of NFU, with infinity if it is included in the formulation, in ordinary ZFC metatheory:In particular $\mathrm{Con}(\mathrm{ZFC})\Rightarrow \mathrm{Con}(\mathrm{NFU})$. This proof does not assume $\mathrm{Con}(\mathrm{ZFC})$ as an additional internal axiom: its starting structure is the set $V_{\omega +\omega }$, not a model of ZFC.

Independence here is a relative consistency assertion: from a ground first-order model satisfying ZFC, construct a first-order model of each remaining theory in which the omitted axiom fails. The ground axiom of choice is a tool for constructing these countermodels, not an extra axiom imposed on the theories being tested. The constructible universe supplies the usual relative-consistency implication from ZF to ZFC, so this ground choice assumption requires no additional consistency strength. These constructions can be formalized inside an arbitrary ground model; they do not require the stronger assumption that a transitive model exists.

Union. Put $\lambda ={\beth }_{\omega }$ and use the hereditarily locally small membership modelThe ground cardinal $\lambda$ is a strong limit cardinal of countable cofinality. This transitive class contains the empty set and $\omega$. Pairing preserves the bound, as does taking a subset, so separation holds for every formula relativized to this class. If an internally definable function has domain $a\in B_{\lambda }$, ground replacement collects its range $r$. Since $|r|\le |a|<\lambda$ and every descendant of an element of $r$ already has fewer than $\lambda$ members, $r\in B_{\lambda }$. Thus the full functional Axiom schema of replacement holds. For $x\in B_{\lambda }$, every actual subset of $x$ belongs to the class andConsequently its entire power set belongs to the class. Extensionality and foundation are inherited from the ground membership relation; $\omega$ witnesses infinity. However $a=\{{\beth }_n:n<\omega \}$ belongs to $B_{\lambda }$, while its only possible union is $\lambda$, which does not. All the other ZF axioms hold, and union fails. Bounding the entire transitive closure by $<\lambda$ instead would give the wrong model: functional ranges at this singular bound need not be hereditarily small.

Power set. Use the transitive class $H_{{\omega }_1}$ of hereditarily countable sets. Pairing, union and separation preserve hereditary countability. A functional image of a countable domain is countable, and its members have countable transitive closures; their countable union is countable in the ground ZFC model. This proves replacement. The remaining structural axioms and infinity are inherited. Every actual subset of $\omega$ belongs to $H_{{\omega }_1}$, but a putative internal power set of $\omega$ would contain all these subsets and hence be uncountable by Cantor's theorem. Power set fails in $H_{{\omega }_1}$, with every other ZF axiom intact.

Replacement. Use $V_{\omega +\omega }$ in the cumulative hierarchy. At a limit ordinal cutoff, pairing, union, power set and every instance of separation remain below the cutoff; foundation, extensionality and infinity hold as well. With parameter $V_{\omega }$, define $f(n)=V_{\omega +n}$ by $n$ finite iterations of power set. The finite iteration witnesses and each value are inside $V_{\omega +\omega }$, so this is an internally definable total function on $\omega$. Its range has rank of a set $\omega +\omega$ and is absent from the model. This instance of replacement fails.

Extensionality. Build well-founded coded sets over one urelement $a$, retaining a distinct ordinary empty set $e$. Membership of an atom is always false, and membership of a set code is its coded extension. Thus $a\ne e$ but they have identical empty extensions. Pairing, union, separation and replacement are performed on set codes; pure $\omega$ witnesses infinity, and the well-founded construction gives foundation. There is a small point concerning the unmodified, untyped Axiom of power set: the atom is vacuously a subset of every object. Therefore the internal power-set object for $x$ contains all ordinary set-subsets of $x$ and the atom. It has exactly the required extension. All remaining ZF axioms hold, but extensionality fails. In the usual typed atom theory, the power-set axiom instead quantifies only over set-subsets.

Foundation. In the ground universe let $\pi$ exchange $\varnothing$ and $\{\varnothing \}$ and fix all other objects. On the same domain define the Rieger-Bernays permutation modelSince $\pi$ is a bijection, two objects with the same $E$-members are equal, so extensionality holds. Any ground collection $B$ of desired members is represented by the object ${\pi }^{-1}(B)$. In particular, an $E$-pair is ${\pi }^{-1}(\{a,b\})$ and the $E$-union of $a$ isAn $E$-subset of $a$ is an object $z$ with $\pi (z)\subseteq \pi (a)$, so its $E$-power-set object isTranslate every formula by replacing membership with $E$; ground separation and replacement collect the required extensions and the same representation supplies their internal witnesses. The $E$-empty object is $e={\pi }^{-1}(\varnothing )$, and its $E$-successor operation is $s(x)={\pi }^{-1}(\pi (x)\cup \{x\})$. Ground well-founded recursion and replacement collect the iterates of $s$ from $e$; representing this collection gives an $E$-inductive object. Thus infinity holds. But for $q=\varnothing$, $\pi (q)=\{q\}$, whence $qEq$ and $q$ has precisely itself as its $E$-member. Foundation fails.

Infinity. The hereditarily finite sets $V_{\omega }$ satisfy extensionality, foundation, pairing, union, power set and separation. Each internal domain is externally finite, so every internally definable functional image is finite and belongs to $V_{\omega }$, proving replacement. No finite object can contain the empty set and all its successive Finite von Neumann ordinals. Infinity fails, and all other ZF axioms hold.

Choice with atoms. For the negative model use the Basic Fraenkel permutation model. Let $A$ be an infinite set of urelements and let every permutation of $A$ act recursively on sets. Keep precisely the hereditarily finitely supported objects: an object has a finite support $F\subseteq A$ if every permutation fixing $F$ pointwise fixes the object, and every object occurring in its membership ancestry must also have such a support. The class of these objects contains the pure universe and $A$ itself.

Supports for a pair are united; union preserves a support; a formula defining a subset uses the finite union of the supports of its domain and parameters. An internally definable functional range has the same support, since its values are uniquely specified, and its elements are already hereditarily supported. This proves the full separation and replacement schemas. All hereditarily supported set-subsets of a fixed set form its symmetric internal power set; this collection is invariant under the pointwise stabilizer of a support of the original set. For the untyped version add all the atoms, as above. Pure $\omega$, ground well-foundedness and set-extensionality give the remaining axioms.

The family $[A{]}^2$ of unordered pairs of atoms is hereditarily supported. If it had a choice function $f$, let $F$ be a finite support of $f$. Choose distinct $a,b\in A\setminus F$. The transposition exchanging $a,b$ fixes $F$ and $\{a,b\}$, but exchanges its two possible chosen elements. Invariance of $f$ would imply $f(\{a,b\})$ is fixed by that transposition, a contradiction. Conversely, the full well-founded universe over the atoms, before restricting to finite supports, satisfies choice when the ground universe does. Both choice and its negation are relatively consistent with the stated weakening of extensionality.

Forcing constructs new first-order models while controlling which ground-model properties survive. Fix a ground model $M$ and a partially ordered set $P\in M$, writing $q\le p$ when $q$ is the stronger condition. A generic filter is upward closed, directed towards stronger conditions, and meets every ground-model dense subset of a forcing order. If $M$ is externally a countable transitive model, enumerate its dense subsets and successively choose stronger conditions in them. The resulting filter is $M$-generic. External countability is crucial: this is not a claim that the ground model regards its own universe as countable.

A forcing name is built recursively from pairs $(\sigma ,p)$ of earlier names and conditions. For a generic filter $G$, its evaluation of a forcing name isThe canonical forcing name $\check{x}=\{(\check{y},1):y\in x\}$ evaluates to $x$, so $M\subseteq M[G]$. Recursive evaluation also proves transitivity. Name ranks bound the ranks of their evaluations, so an ordinal in the extension is below some ground ordinal and is already a ground ordinal. Thus forcing preserves ordinals.

The central link between syntax and generic interpretation is the forcing theorem. The relation $p⊩\phi ({\tau }_1,\dots ,{\tau }_n)$ is definable in the ground model, and the forcing truth lemma saysFor an atomic membership assertion, densely many refinements must witness equality with a member-name whose associated condition is compatible with the refinement. Equality is obtained by recursive mutual inclusion of names. For negation, $p⊩\neg \phi$ means no stronger condition forces $\phi$; for an existential quantification, witnesses must be obtainable densely below $p$. Induction on name rank establishes the atomic clauses, and induction on formulas establishes definability and the truth lemma. The generic filter meets the relevant dense sets at the existential step. An arbitrary semantic assertion about one chosen generic filter is insufficient: the definable relation must work uniformly for all generic interpretations.

This explains preservation of ZFC. Extensionality and foundation follow from transitivity; canonical names supply infinity. Pair and union names give pairing and union. For separation, restrict a name's members to conditions forcing the defining formula. For replacement, the definable forcing relation lets ground axiom schema of collection bound a set of names for all possible witnesses to the unique outputs; evaluating their collected name gives the range. For power set, let $\tau$ name the given set and let $D$ be the set of its member-names. Every subset of ${\tau }_G$ has an equivalent name drawn from $\mathcal{P}(D\times P)$, by the truth lemma, so a ground set of candidate names collects all its subsets. Finally choose a ground well-order of the relevant names; its evaluated image surjects onto the desired set, and choosing the first name for each value well-orders that set. This is why ordinary forcing over a choice model preserves choice.

For a concrete example, Cohen forcing consists of finite partial functions $\omega \to 2$, ordered by extension. The generic union $c=\bigcup G$ is total because deciding each coordinate is dense. For every ground real $x$, the set of conditions disagreeing with $x$ somewhere is dense, so $c\ne x$. This construction changes the universe by adding a genuinely new real without changing its ordinals.

The size and compatibility of conditions determine what else changes. With the countable chain condition for forcing, an ordinal-valued name has only countably many possible values at each coordinate: take a maximal antichain deciding that value. Therefore the possible range of a name for a map with infinite ground domain of cardinality $\mu$ lies in a ground set of size at most $\mu \cdot {\aleph }_0=\mu$. This prevents cardinal collapse and changes of uncountable regular cofinalities. Forcing by finite partial functions $\lambda \times \omega \to 2$ is ccc: the delta-system lemma thins any uncountable family of finite domains to a delta-system, and a further thinning makes their values agree on its finite root. Two remaining conditions are compatible. Distinct coordinates add distinct reals.

For example, start with GCH and take $\lambda ={\aleph }_2$. This forcing has size $\lambda$ and adds $\lambda$ distinct reals. A nice name for a real uses countably many antichains, and each antichain is countable. The ground arithmetic ${\lambda }^{{\aleph }_0}=\lambda$ bounds the number of such names by $\lambda$. Hencegiving a relative-consistency example of the failure of CH. The ground choice of GCH is available through the constructible universe; no claim that every starting model has this arithmetic is needed.

By contrast, countably closed forcing adds no new countable sequences of ground ordinals. Given a name, recursively strengthen a condition to decide each coordinate, then take a lower bound of the descending sequence. That condition decides the entire sequence. Closure and chain conditions thus control different features; neither should be silently substituted for the other.

Further applications include changing cofinalities, adding branches through trees, and constructing prescribed continuum patterns. If the aim is to violate choice, an ordinary generic extension of a choice model will not do: one restricts to a suitable symmetric collection of names, checking the axioms again. These examples also mark a limitation: forcing proves relative consistency, not absolute consistency of the background theory. The countable-transitive-model picture gives a concrete semantic construction; the definable forcing relation and Boolean-valued formulation supply the corresponding formal consistency arguments without asserting that a transitive ground model can be proved to exist.

The assertions are relative independence results. We construct countermodels in a background model of ZFC; if ZF is consistent, its constructible universe gives a model of ZFC as well. The original model satisfies foundation and choice, supplying the positive sides.

For failure of foundation, use a Rieger-Bernays permutation model. Let $\pi$ interchange the old $\varnothing$ and $\{\varnothing \}$ and fix every other object, and put $xEy$ iff $x\in \pi (y)$. The new extension of $y$ is the old set $\pi (y)$, so the representative of any old extension $b$ is ${\pi }^{-1}(b)$. Bijectivity gives extensionality. Translate formulas recursively by this definition of $E$. Old separation gives each new definable subset, and old replacement gives each new functional range; apply ${\pi }^{-1}$ to the resulting extension to obtain its new representative.

Explicitly, a new pair is represented by ${\pi }^{-1}(\{x,y\})$; the new union of $a$ is represented by ${\pi }^{-1}({\bigcup }_{x\in \pi (a)}\pi (x))$; and the new power set of $a$ is represented byStarting from the new empty set, iterate the new successor and collect these iterates by old replacement, obtaining an inductive set. The old axiom of choice selects a member from each nonempty extension and thereby gives new choice functions. However the old object $q=\varnothing$ now satisfies $qEq$ and has exactly itself as its member: it is a Quine atom. Its new singleton has no $E$-minimal member, so foundation fails.

To make choice fail, retain infinitely many Quine atoms $A=\{a_n:n<\omega \}$ and impose a finite-support symmetry restriction. A precise background construction uses tagged set codes. Each atom $a$ has extension $\{a\}$. For every other set of available objects $S$, introduce a unique ordinary set code with extension $S$, except that $S=\{a\}$ is already represented by $a$ itself. Iterate this construction through the ordinals. Omitting the duplicate singleton codes makes the resulting relation extensional. Every permutation of $A$ extends to an automorphism of this universe by recursion through its ordinary codes.

An object has finite support if some finite $S\subseteq A$ has the property that every permutation fixing $S$ pointwise fixes that object. Retain the hereditarily finite-supported Quine-atom model: the object and every object reached through its membership relation must have finite support. The atomic self-loops are permitted. The atom set $A$ and all pure sets are retained.

The union of supports for a domain and finitely many parameters supports any definable subset or functional image: every permutation fixing those supports preserves the defining formula. Members of the subset or image are already hereditary, so this verifies separation and replacement. The collection of all retained subsets of a retained set is invariant under any permutation fixing that set's support, and each of its members is hereditary; this verifies the Axiom of power set. Pairs and unions have the union of the requisite finite supports, and the pure natural-number hierarchy verifies infinity. The coding still verifies extensionality.

If a choice function on all two-element subsets of $A$ had finite support $S$, choose distinct $a,b\notin S$. The transposition of $a,b$ fixes $S$ and the unordered pair $\{a,b\}$, but moves either possible chosen value. This contradicts invariance of the function. Hence choice fails, while a Quine atom witnesses failure of foundation. We have a model of $\mathrm{ZF}-\mathrm{Foundation}+\neg \mathrm{Choice}$. Together with the original model this proves the required independence of choice over $\mathrm{ZF}-\mathrm{Foundation}$.

Work in a background model of ZFC. Each structure below satisfies every axiom of ZF except the indicated one and falsifies that axiom; the background model supplies the other consistency direction.

For the Axiom of union, put $\lambda ={\beth }_{\omega }$ and use the hereditarily locally small membership modelThe transitive class $B_{\lambda }$ inherits extensionality and foundation. Definable subsets of its sets remain in it. If $x\in B_{\lambda }$, a functional image of $x$ has size at most $|x|<\lambda$; every member of that image is already hereditary, so the image belongs to $B_{\lambda }$. This verifies replacement, even though its values can have unbounded ranks. Also $|\mathcal{P}(x)|=2^{|x|}<\lambda$ by the strong limit cardinal property, and each subset of $x$ is locally small. This verifies the Axiom of power set. Pairing and infinity follow similarly. Butsince $|V_{\omega +n}|={\beth }_n<\lambda$ while $|V_{\omega +\omega }|=\lambda$. This is failure of sumset without failure of replacement or power set.

For the Axiom of power set, use the hereditarily countable sets $H_{{\omega }_1}$. This is a transitive class. A countable union of countable transitive closures is countable, so it is closed under pairs, unions and functional images of its countable domains; these facts verify pairing, union and replacement. Separation, foundation and infinity also hold. Every subset of $\omega$ is hereditarily countable, but the full power set $\mathcal{P}(\omega )$ is uncountable by Cantor's theorem and cannot be a member of $H_{{\omega }_1}$. Since every candidate internal power set would have to contain all those subsets, the power-set axiom fails.

For replacement, take the rank-initial transitive set $V_{\omega +\omega }$. A limit rank contains the empty set and is closed under pairs, unions, subsets and power sets; it contains $\omega$, and foundation and extensionality are inherited. Fix the parameter $V_{\omega }$. The definable mapis internally described by a finite sequence of power-set iterations. Each finite sequence belongs to $V_{\omega +\omega }$, so the definition is genuinely functional there. Its range has rank $\omega +\omega$ and is not in the structure. Thus this instance of replacement fails.

For extensionality, use the non-extensional membership model with one atom. Start with an atom $a$ having no members and build the ordinary cumulative coded sets over it, including a separate empty set. Interpret membership by the contents of a set code and give $a$ no members. The two different objects $a,\varnothing$ have identical empty extensions. Foundation holds by the code ranks; pure finite ordinals give infinity; pairs, unions, definable subsets and functional ranges have their usual set codes. For unrestricted power set, include the atom as well as all ordinary subset codes: the atom satisfies $a\subseteq x$ vacuously for every $x$. This detail is necessary because the formulas here are the unmodified ZF formulas, without a set predicate. The collection still has a set code, so the Axiom of power set holds. All axioms except extensionality are satisfied.

Finally, $V_{\omega }$, the hereditarily finite sets, satisfies every axiom except infinity. Pairs, unions, power sets and subsets of finite sets are finite, and every functional image of a finite domain is finite, so the full replacement schema holds. An inductive set would contain every finite von Neumann ordinal and would be infinite, whereas every member of $V_{\omega }$ is finite. Hence infinity fails.