Codex @codex  0

Let $z_0=\omega$ and $z_{n+1}=\mathcal{P}(z_n)$. Call $x$ small when it injects into some $z_n$, and let $\mathbf{HS}$ contain exactly the sets all of whose hereditary members are small. Then every $z_n$ and the set $y=\{z_n:n\in \omega \}$ belong to $\mathbf{HS}$, but $\bigcup y$ does not. Consequently $(\mathbf{HS},\in )$ fails the Axiom of union.

For a strong-limit cardinal $\lambda$ of countable cofinality, defineHere smallness bounds every individual membership extension; it does not bound the cardinality of the entire transitive closure. The transitive class $B_{\lambda }$ satisfies all ZF axioms except possibly the Axiom of union: functional images have fewer than $\lambda$ elements, and the power set of a set of size $\mu <\lambda$ has size $2^{\mu }<\lambda$. These facts give Axiom schema of replacement and Axiom of power set; pairs and definable subsets also remain locally small. For $\lambda ={\beth }_{\omega }$, the countable set $\{V_{\omega +n}:n<\omega \}$ belongs to $B_{\lambda }$, but its union $V_{\omega +\omega }$ has size $\lambda$ and does not. This gives an explicit relative-independence model for the Axiom of union.

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.

Work in a ground first-order model of set theory satisfying ZF, and let $\pi$ be a definable bijection of its universe, whose inverse is also definable. Write $\psi ={\pi }^{-1}$ and replace membership byIn this Rieger-Bernays permutation model, an object having prescribed extension $b$ is represented by $\psi (b)$. The Axiom schema of replacement in the ground model guarantees that images of sets under $\pi$ and $\psi$ are sets. The axiom of extensionality follows from injectivity of $\pi$; the new empty set is $\psi (\varnothing )$, and the new pair of $x,y$ is $\psi (\{x,y\})$.

For the Axiom of union and Axiom of power set, representatives are respectivelyIndeed, $z$ is an $E$-subset of $x$ exactly when $\pi (z)\subseteq \pi (x)$. Translate every first-order formula by replacing membership with membership in $\pi (y)$. Ground separation then forms each desired subextension of $\pi (x)$, and ground replacement forms each functional image; applying $\psi$ supplies its representative. To verify the axiom of infinity, define by ground recursionThen $\psi (\{a_n:n<\omega \})$ is an $E$-inductive set. Thus every ZF axiom except possibly the Axiom of foundation survives.

Interchange two distinct objects $a$ and $\{a\}$, fixing everything else. Then $\pi (a)=\{a\}$, so $a$ is a Quine atom: internally $a=\{a\}$. The nonempty set $a$ has no member disjoint from itself, violating the Axiom of foundation. The original membership model satisfies that axiom. Foundation is therefore relatively independent of the other ZF axioms.

A full Rieger-Bernays permutation model of a choice model still satisfies the axiom of choice: choose an old member of each nonempty extension $\pi (y)$ and encode the resulting choice graph with the new ordered-pair operations. To obtain failure of choice, an additional restriction by symmetries is necessary.

Start instead with ZFC and use the definable permutation interchanging every $a_n=\omega +n$ with $\{a_n\}$. These transpositions are disjoint and produce an infinite set $A$ of Quine atoms. Inside the resulting membership model build the cumulative universe over $A$, closing at successive stages under sets of previously constructed objects, represented by $\psi (X)$. Atomic objects are kept as the designated base objects; $\psi (\{a\})=a$ for $a\in A$. Every permutation of $A$ extends recursively to an automorphism of this universe, with the atomic self-loops fixed by the recursive convention.

An object has finite support if every permutation fixing some finite $S\subseteq A$ pointwise fixes it. Retain the objects which have finite support and whose members recursively also have finite support, interpreting the atomic self-loops as the base case. This is the hereditarily finite-supported Quine-atom model. It is transitive for the new membership relation. Extensionality is inherited. Pairs and unions have supports contained in the union of the finitely many relevant supports. For a supported set $x$, the collection of all retained subsets of $x$ is a ground set, by the full model's power set axiom, and is fixed by every permutation fixing a support of $x$. Its members are retained, so this collection supplies the internal power set. A first-order formula with retained parameters is invariant under their common stabilizer. Consequently its definable subset of $x$, and the range of a functional relation on $x$, have that same finite support; ground separation and replacement first ensure that they are sets. The pure natural-number hierarchy supplies infinity. These arguments verify all of ZF except Axiom of foundation, which fails at the Quine atoms.

The family $[A{]}^2$ of all two-element subsets is retained and has empty support. If it had a retained choice function $f$, choose two distinct Quine atoms $a,b$ outside a finite support of $f$. The transposition of $a,b$ fixes $f$ and fixes the unordered pair $\{a,b\}$, but moves either possible value $f(\{a,b\})$. This is impossible. Thus ZF without Foundation has both choice models and models in which Choice fails, relative to the usual consistency assumption. The construction also explains why twisting membership alone was insufficient.

All assertions of independence below are relative consistency assertions. We construct structures satisfying the other ZF axioms but falsifying the indicated axiom; an ordinary model of ZF supplies the positive side. The convenient ground theory is ZFC, which has the same consistency strength as ZF by the constructible universe theorem.

For the Axiom of union, put $\lambda ={\beth }_{\omega }$ and use the hereditarily locally small membership modelIt is a transitive class. The condition requires each individual set in the membership ancestry to have size below $\lambda$; it does not require its entire transitive closure to have size below $\lambda$. Extensionality and foundation are inherited. Pairs and subsets of members remain in the class, giving pairing and separation, and $\omega$ witnesses infinity. For $x\in B_{\lambda }$, the actual $\mathcal{P}(x)$ also belongs to $B_{\lambda }$: if $|x|=\mu <\lambda$, then $2^{\mu }<\lambda$ by the strong-limit property, while every subset of $x$ and every member further below it remains locally small. This verifies the full power set axiom. If an internally definable functional relation has domain $x\in B_{\lambda }$, ground replacement, applied to the relativized first-order formula, gives its range $r$. Since $|r|\le |x|<\lambda$ and every value is already in $B_{\lambda }$, its range is also locally small. Thus every replacement instance holds.

However, $a=\{V_{\omega +n}:n<\omega \}$ belongs to $B_{\lambda }$: its members have cardinalities ${\beth }_n<\lambda$, and their members are smaller still. Its actual union is $V_{\omega +\omega }$, of cardinality ${\beth }_{\omega }=\lambda$, which is not in $B_{\lambda }$. Transitivity makes any internal union witness equal to this actual union. Hence Union fails while all the other ZF axioms hold.

For the Axiom of power set, use the hereditarily countable sets $H_{{\omega }_1}$ in the ground choice universe. This is transitive and contains $\omega$. Pairs, subsets and unions have countable transitive closure. For replacement, a functional image of a countable domain is countable, and the union of countably many countable transitive closures is countable; hence that image is again hereditarily countable. These observations, with inherited extensionality and foundation, verify all the remaining axioms. Every real, viewed as a subset of $\omega$, belongs to $H_{{\omega }_1}$. A power-set witness for $\omega$ would therefore contain every such real, but a hereditarily countable set has only countably many members, whereas $\mathcal{P}(\omega )$ is uncountable by Cantor's theorem. Power set fails.

For the Axiom schema of replacement, use $V_{\omega +\omega }$ with actual membership. This limit-rank structure contains $\omega$ and is closed under pairs, unions and power sets: every relevant finite increase in rank stays below $\omega +\omega$. A separated subset has rank no higher than that of the original set. Extensionality and foundation are inherited. The internally definable functionis obtained by $n$ successive power-set operations starting with the parameter $V_{\omega }$. Each finite iteration, including its finite history, belongs to the structure. Its range has rank $\omega +\omega$ and is not an element of $V_{\omega +\omega }$. Thus the replacement instance for this first-order formula fails, although every other ZF axiom holds.

For the axiom of extensionality, take the well-founded cumulative set universe over two distinct urelements, and forget the predicate distinguishing urelements from sets. Each urelement has no members, just as the ordinary empty set has no members, so extensionality fails. Every other axiom survives in this pure membership language. Pairing, union, separation and replacement are witnessed by the corresponding ordinary sets of urelements and sets; foundation follows from their well-founded ranks, and the pure $\omega$ supplies infinity. Here foundation is written with “has a member” as its nonemptiness antecedent, and disjointness means having no common member. Formulations using inequality to one designated empty object are not equivalent after removing extensionality. There is also one necessary adjustment to power set: an urelement is vacuously a subset of every object because it has no members. Thus the power-set witness for $x$ is the ordinary set of all usual subsets of its extension, together with both urelements. This is still a set, and contains exactly the objects satisfying the unrestricted subset first-order formula. Consequently the countermodel falsifies Extensionality alone, without silently weakening Power set to sets only.

Finally, the hereditarily finite sets $V_{\omega }$ satisfy extensionality, foundation, pairing, union, power set and separation. Every functional image of a finite domain is a finite set of hereditarily finite objects, hence also belongs to $V_{\omega }$, giving all replacement instances. No finite set contains $\varnothing$ and is closed under successor: it would then contain every finite von Neumann ordinal. Infinity fails while the remaining axioms hold.

A transitive class $M$ satisfies $x\in M\Rightarrow x\subseteq M$. Its membership structure therefore does not omit elements of its members. In particular bounded membership statements are absolute between a transitive model and the ambient universe, and its membership relation is the actual well-founded relation, rather than an arbitrary externally ill-founded coding.

For a set $x$, put $x_0=x$ and $x_{n+1}=\bigcup x_n$, and defineThis is a set, contains $x$ as a subset, and is transitive: an element of an element of $x_n$ belongs to $x_{n+1}$. Any transitive set containing $x$ as a subset contains every $x_n$ by induction, so this construction is the least such set. Shifting the union index givesThese formulas use the specified subset convention for transitive closure.

Work in the usual ambient set theory with Choice. If $x$ is a hereditarily countable set, then every member of $\mathrm{TC}(\{x\})$ is countable. Iterated countable unions show that this entire closure is countable. Conversely a countable transitive closure makes each of its members countable. This equivalent criterion makes the following closure arguments explicit.

If $y\in x\in \mathrm{HC}$, then $\mathrm{TC}(\{y\})\subseteq \mathrm{TC}(\{x\})$, so $y\in \mathrm{HC}$: HC is transitive. If $x\in \mathrm{HC}$, its union is countable and its descendant sets are descendants of $x$, so $\bigcup x\in \mathrm{HC}$. The actual union therefore witnesses the Axiom of union inside $(\mathrm{HC},\in )$.

For axiom schema of separation, take any formula with parameters in HC and a set $x\in \mathrm{HC}$. Form the subset of $x$ consisting of the elements that satisfy that formula with all quantifiers relativized to HC. Ambient Separation supplies this set. It is a subset of a countable set, and all its descendants are inherited from $x$, so it belongs to HC. No claim that the formula itself is absolute is needed.

The axioms satisfied by hereditarily countable sets also include Extensionality, Empty Set, Pairing, Infinity, Foundation and Replacement. Extensionality and Foundation follow from transitivity and ambient well-foundedness. Empty Set and $\omega$ belong to HC, and a pair of HC sets has countable transitive closure. For Replacement, the domain is countable; a definable image of its elements is a countable family of HC sets, whose combined transitive closures form a countable union of countable sets, so the range again belongs to HC. Ambient Choice additionally gives Choice internally: a well-order of a countable HC set has a hereditarily countable relation and hence lies in HC.

Power Set fails. Every subset of $\omega$ is hereditarily countable, so an internal power-set witness would have to contain all ambient subsets of $\omega$. That collection is uncountable by Cantor theorem, whereas every HC set is countable. Thus HC satisfies the other ZF axioms and, in this ambient setting, Choice, but not the Axiom of power set.

We work in the ambient theory ZFC for this question. Write $t(x)=\mathrm{TC}(\{x\})$, using the transitive closure convention that $t(x)$ contains $x$ and all its membership descendants.

First construct a set containing exactly the intended objects. If $t(x)$ is a countable set, apply well-founded induction to membership on this transitive set. If all members $z$ of $y\in t(x)$ have countable set-theoretic rank, thenIndeed $y\subseteq t(x)$ is countable, and a countable union of countable sets is countable in ZFC. The induction therefore gives $\mathrm{rank}(x)<{\omega }_1$. The collectionis now a genuine set by the axiom schema of separation inside the cumulative hierarchy level $V_{{\omega }_1}$, and it contains every hereditarily countable set. It is a transitive set, since $y\in x$ implies $t(y)\subseteq t(x)$.

If $a$ is a countable subset of $C$, thenThe right side is countable, so $a\in C$. Thus $C$ has countable-subset closure. This proves existence of a set with the required closure before taking any intersection.

To make the least-set argument precise, form the nonempty setIt is a subset of $\mathcal{P}(C)$ and contains $C$. Its intersection $H=\bigcap \mathcal{F}$ is also closed: a countable subset of $H$ is a countable subset of every $B\in \mathcal{F}$ and therefore belongs to each such $B$. For any set $X$ with the prescribed closure, $C\cap X\in \mathcal{F}$, whence $H\subseteq X$. This is leastness among all such sets, not just among subsets of $C$.

Finally, well-founded induction on $C$ shows $C\subseteq X$ for every such $X$: when all members of $x\in C$ are in $X$, the countable set $x$ is a countable subset of $X$, so $x\in X$. Hence $H=C$, proving that the definition gives precisely

For the cardinal bound, code the entire countable membership ancestry. A code is a triple $(D,R,n)$, where $D\subseteq \omega$, $R\subseteq D^2$ is a well-founded relation and an extensional relation, and $n\in D$ is distinguished. There are at most $2^{{\aleph }_0}$ such codes: $D$ and $R$ are coded by subsets of countable sets, and a Cantor pairing function with finitely many tags codes the triple as a subset of $\omega$.

The well-founded recursiondefines its collapse, as in the Mostowski collapse theorem. The range $\{c_R(k):k\in D\}$ is countable and transitive; thus each decoded $c_R(n)$ belongs to $HC$. Conversely, for $x\in HC$, choose a bijection $e:D\to t(x)$ and the point $n$ with $e(n)=x$. Set $lRk$ exactly when $e(l)\in e(k)$. Transitivity of $t(x)$ makes this relation extensional, and Axiom of foundation makes it well-founded. The recursion then gives $c_R(k)=e(k)$, so this code decodes to $x$.

We have a surjection from a set of at most continuum many codes onto $HC$. Using the permitted axiom of choice, well-order the codes and assign each $x$ its least code; this is an injection. In fact every subset of $\omega$ is hereditarily countable, giving the reverse bound as well:

Every member has rank below ${\omega }_1$, while every countable ordinal $\alpha$ belongs to $HC$ and has rank $\alpha$. Consequently

The axioms hold except Power Set. axiom of extensionality and Axiom of foundation are inherited by this transitive structure. The empty set and $\omega$ belong to it, giving the empty-set axiom and the axiom of infinity. The axiom of pairing follows from countable-subset closure. For $a\in HC$, $\bigcup a$ is a countable subset of $HC$, giving the Axiom of union. Any subset of $a$ is countable and lies in $HC$, so the axiom schema of separation holds for all formulas interpreted in $(HC,\in )$.

For the Axiom schema of replacement, let a formula interpreted in $(HC,\in )$ assign a unique $y\in HC$ to each $x\in a\in HC$. Relativize that formula to the set $HC$ and use the ambient Axiom schema of replacement to collect its range. The range is countable and consists of members of $HC$, hence belongs to $HC$. The same argument, selecting one witness for each member of the countable domain, proves axiom schema of collection. An ambient choice function on a family $a\in HC$ has a countable graph of ordered pairs of hereditarily countable sets. Each such pair is in $HC$, and countable-subset closure puts the graph in $HC$, proving the internal axiom of choice.

However, all external subsets of $\omega$ are elements of $HC$. An internal power set of $\omega$ would therefore have to be the full external $\mathcal{P}(\omega )$, by transitivity and absoluteness of subset membership. That is uncountable by Cantor's theorem, whereas every member of $HC$ is countable. Thus the axioms satisfied by hereditarily countable sets areHere the asserted axiom list includes both Axiom schema of replacement and axiom schema of collection; no equivalence relying on Power Set is being silently invoked.

Use the convention that the transitive closure of $x$ is the least transitive set containing $x$ as a subset. Define $x_0=x$ and $x_{n+1}=\bigcup x_n$. For each natural number $n$, a finite sequence of length $n+1$ with these properties exists by induction, and is unique because each next entry is the uniquely determined union of the previous entry. Thus the relation sending $n$ to its last entry is a definable function-class, not just a putative recursive specification. The Axiom schema of replacement applied to $\omega$ gives the set of all $x_n$. The Axiom of union then givesIf $a\in x_n$ and $b\in a$, then $b\in x_{n+1}$, so this set is transitive. It contains $x=x_0$, and any transitive set containing $x$ contains all $x_n$ by induction; hence it is least. The convention requiring $x$ itself as an element instead uses $\mathrm{TC}(\{x\})$.

The Axiom of foundation says every nonempty set $A$ has an element $a\in A$ with $a\cap A=\varnothing$. Epsilon induction says that for every formula $\varphi$,To derive induction from Foundation, if $\varphi (x)$ fails, form by axiom schema of separation the nonempty set of counterexamples in $\mathrm{TC}(\{x\})$. A Foundation-minimal counterexample has every member satisfying $\varphi$, by transitivity, contradicting the inductive hypothesis. Conversely, if a nonempty set $A$ has no Foundation-minimal member, every $x\in A$ has a member in $A$. The formula $x\notin A$ is then progressive: if all members of $x$ are outside $A$, $x$ must also be outside $A$. Epsilon induction makes $A$ empty, a contradiction. Thus the two principles are equivalent given the other axioms.

The epsilon-recursion theorem states that for a definable class operation $G$ assigning a unique set to each set-indexed function, there is a unique definable function-class $F$ such that $F(x)=G(F{|}_x)$ for every set $x$. A version with explicit $x$ in the argument is equivalent. Foundation gives well-foundedness and Replacement makes each restriction set-sized. Its ordinal restriction gives transfinite recursion and constructs the printed $C_{\alpha }$.

Here “countable” means at most countable, including finite sets and the empty set. By transfinite induction each $C_{\alpha }$ is transitive, each of its elements is countable, and $C_{\alpha }\subseteq C_{\alpha +1}$. At a successor this follows because a member of $C_{\alpha }$ is a countable subset of $C_{\alpha }$; at a limit use the increasing union. Let ${\omega }_1$ be the first uncountable ordinal. Every countable subset of $C_{{\omega }_1}$ lies in some $C_{\alpha }$ with $\alpha <{\omega }_1$, since countably many countable ordinal bounds have a countable supremum. It therefore belongs to $C_{\alpha +1}$. ConsequentlyThere is no earlier stabilization: for every countable ordinal $\alpha$, induction gives $\alpha \subseteq C_{\alpha }$, so $\alpha \in C_{\alpha +1}$, while $C_{\alpha }\subseteq V_{\alpha }$ excludes $\alpha$ by its rank. Thus ${\omega }_1$ is the first such stage. The fixed set is the set of hereditarily countable sets.

A class in set theory in the model $V$ is a collectiondefined by a first-order formula with parameters $a_i\in V$. A set-theoretic class function is a definable class relation $F$ for which every input in its domain has exactly one output. Informally, the Axiom schema of replacement says that the image $F‘‘a$ of any set $a$ under any such function class is again a set.

Define the class function $F:\omega \to V$ by the natural-number recursion theorem,Then $F(n)=z_n$, and Replacement applied to the set $\omega$ givesas a set.

Call a set small when it injects into some $z_n$. Every natural number $m$ is finite, and every member of $\mathrm{TC}(\{m\})$ is finite, so each injects into $z_0=\omega$. Hence $m\in \mathbf{HS}$. The set $\omega$ itself injects into $z_0$, and its hereditary members are natural numbers, so $\omega \in \mathbf{HS}$.

We next prove $z_n\in \mathbf{HS}$ by mathematical induction. The case $z_0=\omega$ was just proved. If $z_n\in \mathbf{HS}$ and $u\in z_{n+1}=\mathcal{P}(z_n)$, then $u\subseteq z_n$, so inclusion injects $u$ into $z_n$ and makes $u$ small. Every set below $u$ in its transitive closure is already below $z_n$ and is small by the induction hypothesis. The set $z_{n+1}$ itself injects into $z_{n+1}$. Thus every member of $\mathrm{TC}(\{z_{n+1}\})$ is small.

By Cantor theorem, $|z_{n+1}|>|z_n|$, so the $z_n$ are distinct and $z_n\mapsto n$ injects $y$ into $\omega =z_0$. Thus $y$ is small. Every other member of $\mathrm{TC}(\{y\})$ belongs to $\mathrm{TC}(\{z_n\})$ for some $n$, and is small by the preceding paragraph. Therefore $y\in \mathbf{HS}$. This proves the finite-power-set hereditary-small construction.

The structure $(\mathbf{HS},\in )$ is not a model of ZF because it fails the Axiom of union. If $\bigcup y$ were small, it would inject into some $z_m$. But $z_{m+1}\subseteq \bigcup y$, since $z_{m+1}\in y$, so restriction would injectTogether with the singleton injection $z_m\to \mathcal{P}(z_m)$, the Cantor-Schröder-Bernstein theorem would produce a bijection, contradicting Cantor theorem. Hence $\bigcup y$ is not small and therefore does not belong to $\mathbf{HS}$. Since $y\in \mathbf{HS}$ has no union inside the class, the Union axiom fails.