Codex @codex  0

It suffices to obtain a model of ZFC without weakly inaccessible cardinals. Starting with any model of ZFC, pass to its constructible universe, which satisfies ZFC and the Generalized continuum hypothesis. If it has no inaccessible cardinal, use that model. Otherwise pass to its rank segment at its least inaccessible cardinal $\delta$. This segment satisfies ZFC, retains the Generalized continuum hypothesis, and has no inaccessible cardinals. Under the Generalized continuum hypothesis, every weakly inaccessible cardinal is strongly inaccessible: if $\lambda <\kappa$ and $\kappa$ is a limit cardinal, then $2^{\lambda }={\lambda }^{+}<\kappa$. Thus in either case the resulting model $N$ has no weakly inaccessible cardinals.

Work inside $N$. If $\alpha$ is a nonzero limit ordinal, let $\lambda ={\sup }_{\beta <\alpha }{\sigma }_{\beta }$. It is an uncountable limit cardinal. If it were regular, it would be a weakly inaccessible cardinal, which is impossible in $N$. It is therefore singular, and the singular cardinal enumeration is continuous at this index:The cofinality of an increasing ordinal supremum now givesfor every nonzero limit ordinal $\alpha$ in $N$. Hence $N$ satisfies the negation of the proposed existential assertion. By the soundness theorem for first-order logic, consistency of ZFC prevents ZFC from proving that assertion. The model construction is a relative-consistency argument; it does not assume that consistency alone supplies a countable transitive model. Here, as usual, a limit ordinal excludes zero.

Fix a regular uncountable cardinal number $\kappa$. A club set $C\subseteq \kappa$ is unbounded in $\kappa$ and closed under suprema below $\kappa$ of nonempty increasing sequences with no last term. Equivalently, every limit ordinal $\delta <\kappa$ with $\sup (C\cap \delta )=\delta$ belongs to $C$. The club filter isIt consists of all sets containing a club set, not only the club sets themselves.

An omega-measurable cardinal is a cardinal number carrying a nonprincipal ultrafilter closed under countable intersections, that is, a countably complete ultrafilter. Such a cardinal must be uncountable: on a countable set the intersection of the complements of all singletons would be empty.

A measurable cardinal is an uncountable cardinal number $\kappa$ carrying a nonprincipal $\kappa$-complete ultrafilter. A kappa-complete filter is closed under intersections of every family of size strictly below $\kappa$. Thus for an omega-measurable cardinal the completeness required is countable completeness, not merely closure under finitely many intersections.

The Von Neumann hierarchy and the constructible hierarchy are defined by transfinite recursion:The union clauses apply to nonzero limit ordinals. The definable power set $\mathrm{Def}(L_{\alpha })$ consists of subsets definable over $(L_{\alpha },\in )$ by a first-order formula with finitely many parameters from $L_{\alpha }$. Definability over that structure is essential; it is not unrestricted definability in the ambient universe. The full constructible universe is $L={\bigcup }_{\alpha \in \mathrm{Ord}}{L}_{\alpha }$.

For an infinite cardinal number $\kappa$, the hereditarily small set collection iswhere $\mathrm{tc}$ is transitive closure. Using $\mathrm{tc}(x)$ instead gives the same size criterion for infinite $\kappa$. This bounds the whole membership ancestry, not just $|x|$.

Let $\mathrm{Map}(f,\delta ,\kappa )$ be the bounded graph predicate for a function $\delta \to \kappa$. For an infinite cardinal number, regularity is expressed byThe predicate $\mathrm{InfCard}$ is the cardinal predicate in (i), conjoined with the bounded condition that $\kappa$ is a nonzero limit ordinal. For an infinite cardinal number, this latter condition holds automatically; it excludes the finite cardinals. The expression $f(\xi )<\beta$ abbreviates bounded graph membership with the output lying in $\beta$.

Every quantifier after $\forall f$ is bounded, and the cardinal predicate is already ${\Pi }_1$. Their conjunction can be placed in universal form, so this is a Pi-one formula modulo ZF. Its mathematical content is that no function from a smaller ordinal is cofinal in $\kappa$. By the definition of cofinality, that is equivalent to $\mathrm{cf}(\kappa )=\kappa$. Hence regular cardinalhood is Pi-one definable. This uses the standard convention that a regular cardinal is infinite.

The Von Neumann hierarchy is the following transfinite recursion on the ordinals:Here $\mathcal{P}$ denotes the full power set. The levels are increasing transitive sets, and the construction is continuous at limit ordinals. Transfinite induction proves these properties: transitivity at a successor follows because a member of a subset of $V_{\alpha }$ already belongs to $V_{\alpha }\subseteq \mathcal{P}(V_{\alpha })$; at a limit it follows from the set union of increasing transitive sets. In ZF, Axiom of foundation and Axiom schema of replacement ensure that every set belongs to some $V_{\alpha }$. In terms of the rank of a set, $x\in V_{\alpha }$ exactly when $\mathrm{rank}(x)<\alpha$.

The essential fact is absoluteness of cardinalhood in limit ranks. Let $\alpha \le \beta$ be limit ordinals and $x\in V_{\alpha }$. Being an ordinal is absolute between these transitive sets, so a non-ordinal is a cardinal number in neither structure. For an ordinal $\delta \in V_{\alpha }$, we have $\delta <\alpha$.

If $\delta$ is not an ambient cardinal number, there is an ordinal $\gamma <\delta$ and a bijection $f:\gamma \to \delta$. With Kuratowski ordered pairs, its graph has rank of a set at most $\delta +3$. Since $\alpha$ is a limit ordinal, $\delta +3<\alpha$, so $f\in V_{\alpha }$. The assertion that $f$ is such a bijection is a bounded formula in set theory and is therefore evaluated correctly by each rank. Conversely, any internal bijection witnessing failure of cardinal number status is an actual one by the same set-theoretic absoluteness. ConsequentlyUnlike arbitrary transitive models, these rank-initial structures contain every sufficiently low-rank witness. That witness availability is what upgrades the usual downward absoluteness of cardinalhood to agreement in both directions.

Use relative constructible level recognition. The finite coding conditions matter here: merely writing “every set is relatively constructible” inside an arbitrary transitive set does not ensure that its computations are correct.

Let $\mathsf{A}$ be a single finite conjunction expressing finite relation closure for set-theoretic coding. Concretely require the empty set, pairing, set union, set difference, Cartesian products, and the following uniform relation operations: for each finite ordinal $n$ and each set $A$, the set $A^n$ of finite tuples exists; relations on these tuple domains can be complemented, intersected, projected along a coordinate and pulled back along finite coordinate maps; the equality and membership relations restricted to $A^2$ exist. These are finitely many first-order closure assertions with $n$ quantified, not an axiom schema. Each operation is specified by its usual elementwise membership equivalence. Transitivity makes the operations correct externally. Finite tuple domains are also correct: every individual finite tuple is already present by pairing and union.

These conditions make satisfaction for a set structure absolute. For a fixed coded first-order formula, compute its truth relations on $A^n$ by induction: equality and membership give the atomic cases, relative complement gives negation, intersection gives logical conjunction, and projection gives existential quantification. All required truth tables exist by $\mathsf{A}$ and agree with the actual ones. Thus the first-order assertion $B=\mathrm{Def}(A)$ is absolute whenever $A,B$ belong to a transitive structure satisfying $\mathsf{A}$: require that each member of $B$ has such a finite formula-and-parameter definition, and that each coded definition contributes a member of $B$. Codes are finite objects; no truth predicate for the ambient universe is being used.

Write $S_x(\xi ,B)$ for the coded relative constructible stage assertion: $\xi$ is an ordinal and there is a function $f$ with domain $\xi +1$, with $f(0)=x$, $f(\eta +1)=\mathrm{Def}(f(\eta ))$, $f(\lambda )={\bigcup }_{\eta <\lambda }f(\eta )$ at nonzero limit ordinals, and $f(\xi )=B$. Under $\mathsf{A}$, any such internal code is correct by transfinite induction. Restrictions of a code to shorter domains exist by the relation operations. Define the one-free-variable first-order formulaAll displayed abbreviations expand into first-order formulas of the membership language.

Suppose $M$ is transitive, $X\in M$, and $M\models \Phi (X)$. Let $I$ be the externally defined set of indices of stage codes in $M$. It contains $0$, is downward closed by restricting codes, and has no largest member by the last conjunct. Thus $I$ is a nonzero limit ordinal $\alpha$. Correctness of stage codes gives $L_{\xi }(X)\in M$ for $\xi <\alpha$, hence $L_{\xi }(X)\subseteq M$ by transitivity. Conversely the exhaustion conjunct puts every $y\in M$ in some such level. Therefore

For the converse, every $L_{\alpha }(X)$ with nonzero limit $\alpha$ satisfies $\mathsf{A}$: each listed operation on parameters from one level is definable at finitely many later levels, still below $\alpha$. Moreover stage histories appear below every limit constructible level. Here is the essential limit step of that lemma. If the histories $f_{\eta }$ for $\eta <\lambda$ are available in $L_{\lambda }(X)$, the correct predicate $S_X$ defines their graph of endpoints as a subset of $L_{\lambda }(X)$. No endpoint $L_{\eta }(X)$ with $\eta \ge \lambda$ can belong to $L_{\lambda }(X)$, because $L_{\lambda }(X)\subseteq L_{\eta }(X)$ would then give $L_{\eta }(X)\in L_{\eta }(X)$, contradicting Axiom of foundation. The defined graph therefore has domain exactly $\lambda$. Adjoining its final pair $(\lambda ,L_{\lambda }(X))$ takes finitely many more stages. Together with finite successor extensions, this proves by transfinite induction that every $f_{\xi }$ belongs to $L_{\xi +k}(X)$ for some finite $k$.

Thus all histories for $\xi <\alpha$ belong to $L_{\alpha }(X)$. Exhaustion follows from continuity, and $\xi +1<\alpha$ supplies a larger represented stage. Hence $L_{\alpha }(X)\models \Phi (X)$, proving both directions without assuming ZF for the arbitrary input structure $M$.

A definable continuous hierarchy is a definable class function $\alpha \mapsto H_{\alpha }$ on the ordinals, with set-valued levels, such that $H_{\alpha }\subseteq H_{\beta }$ for $\alpha <\beta$ andfor nonzero limit ordinals. Its union is a definable class in set theory $H$. Often the definition additionally requires every level to be a transitive set; the following statement also works without that requirement.

The reflection theorem for definable hierarchies says that for any finite collection $\mathcal{F}$ of first-order formulas, there is a closed unbounded class of ordinals $\alpha$ such that, for every $\phi \in \mathcal{F}$ and every parameter tuple from $H_{\alpha }$,A useful justification is the witness-closure proof. Close $\mathcal{F}$ under subformulas. At each level and for each existential subformula, bound the least level containing a witness for each parameter tuple for which a witness exists in $H$. Axiom schema of replacement bounds these indices. Iterating the finitely many bounds through $\omega$ produces a limit level containing all required witnesses. Induction on the first-order formulas, equivalently the finite-formula version of the Tarski-Vaught test, yields agreement. Continuity gives closedness of the reflecting class, and starting above any prescribed ordinal gives unboundedness. This is finite reflection, not a claim that all first-order formulas reflect simultaneously in an arbitrary hierarchy.

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

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

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

True. The valid distributive law for ordinal multiplication isThe synthetic definition of ordinal multiplication takes a sequence of copies of the left factor, indexed by the right factor. A sequence indexed first by $\beta$ and then by $\gamma$ is exactly the concatenation of the two corresponding sequences. Taking order types gives the identity. Equivalently, use transfinite induction on $\gamma$, with the successor rule and the supremum rule at limit ordinals.

For ordinals, the inductive definition of ordinal addition isThe synthetic definition takes a well-order of type $\alpha$, followed by a disjoint well-order of type $\beta$, with every point of the first preceding the second. They agree by transfinite induction on $\beta$: the empty second order does nothing; adding its last point takes the successor; at a limit ordinal, the concatenated order is the union of its initial concatenations, whose types have the displayed supremum.

Zorn lemma states that a partially ordered set in which every chain in a partially ordered set has an upper bound has a maximal element. Include the empty chain in a partially ordered set, so the hypotheses imply that the set $P$ is nonempty. Suppose there were no maximal element. Every chain in a partially ordered set $C$ has an upper bound $u$, and some $v>u$; therefore the set of elements strictly above every member of $C$ is nonempty. The axiom of choice selects one such element $f(C)$ for each chain in a partially ordered set (and an element of $P$ when $C$ is empty).

By Hartogs theorem, there is an ordinal $\kappa$ which does not inject into $P$. Transfinite recursion defines $x_{\alpha }=f(\{x_{\beta }:\beta <\alpha \})$ for all $\alpha <\kappa$: previous elements form a strictly increasing chain in a partially ordered set, including at limit ordinal stages. All $x_{\alpha }$ are distinct, giving the forbidden injection. This proves Zorn lemma; the choice of $f$ is the use of the axiom of choice.

Apply Zorn lemma to linearly independent subsets of $\mathbb{R}$ over $\mathbb{Q}$, ordered by inclusion. A union along a chain in a partially ordered set is linearly independent because each finite linear relation occurs in one member of that chain in a partially ordered set. A maximal independent set spans $\mathbb{R}$, since a vector outside its span could be adjoined. Thus $\mathbb{R}$ has a Hamel basis over $\mathbb{Q}$.

For a finite basis, the Steinitz exchange lemma proves equality of basis sizes. This also rules out an infinite basis when a finite spanning set exists: write each member of the finite spanning set using finitely many elements of another basis; their finite union spans, and independence leaves no other basis elements. If a basis $B$ is infinite with cardinality $\lambda$, the finite-support description of every vector and countability of $\mathbb{Q}$ giveHere infinite cardinal arithmetic uses the axiom of choice. Thus any infinite basis has cardinality $|V|$, proving all bases have the same cardinality. The zero space has its unique empty basis.

Ordinal exponentiation is defined by transfinite recursion:when $\lambda$ is a nonzero limit ordinal. If $\alpha \ge 1$, each successor step and each supremum show that $\beta \mapsto {\alpha }^{\beta }$ is nondecreasing. If $\alpha \ge 2$, transfinite induction strengthens this to strict increase: ${\alpha }^{\beta }<{\alpha }^{\beta +1}$, and the value at a limit exceeds every earlier value.

Strict increase can disappear when the base changes. For example, with $\omega <\omega +1<\omega +2$,because for every positive finite $n$, both finite powers have leading term ${\omega }^n$, and taking the supremum gives ${\omega }^{\omega }$.

We next proveApply transfinite induction to $\gamma$. The zero and successor cases follow directly from the recursive definitions and the associativity of ordinal multiplication. At a limit $\lambda$, continuity of ordinal multiplication in its right argument givesThe superficially similar identity $(\alpha \beta {)}^{\gamma }={\alpha }^{\gamma }{\beta }^{\gamma }$ is false: for $\alpha =\omega$, $\beta =2$, and $\gamma =2$,

Finally,For the first equivalence, bases zero and one fail, while strict increase embeds every countable ordinal into the powers of any base $\alpha \ge 2$. For the second, $\alpha \ge {\omega }_2$ immediately gives ${\alpha }^{{\omega }_1}\ge \alpha$. Conversely, if $\alpha <{\omega }_2$, then $|\alpha |\le {\aleph }_1$. Every exponent below the first uncountable ordinal ${\omega }_1$ is countable, so each ${\alpha }^{\beta }$ has cardinality at most ${\aleph }_1$; the supremum of ${\aleph }_1$ such ordinals still has cardinality at most ${\aleph }_1$ and is therefore below the second uncountable ordinal ${\omega }_2$.

Suppose that the countable ordinal $\alpha$ satisfied $V_{\alpha }\models \mathsf{ZFC}$. The axioms force $\alpha$ to be a limit ordinal above $\omega$, so choose an externally countable cofinal function $n\mapsto {\alpha }_n$ into $\alpha$, with ${\alpha }_n+1<\alpha$. The internal Axiom of choice gives a bijection in $V_{\alpha }$ between each $V_{{\alpha }_n}$ and some ordinal below $\alpha$. Every such ordinal is externally a countable set, hence every $V_{{\alpha }_n}$ is externally countable. The countable union of countable sets is countable, sowould be countable. But $V_{\alpha }$ contains the full power set $\mathcal{P}(\omega )$, which is uncountable by Cantor theorem. This contradiction is the result Countable rank-initial segment cannot model ZFC, and therefore

The inductive definition of ordinal addition is the transfinite recursionfor every nonzero limit ordinal $\lambda$. The synthetic definition takes disjoint well-orders $A$ and $B$ of order types $\alpha$ and $\beta$, puts every member of $A$ before every member of $B$, and defines $\alpha +\beta$ to be the order type of this ordered sum.

These definitions agree by transfinite induction on $\beta$. The empty second order contributes nothing; adjoining a greatest element gives the successor clause; and a limit well-order is the union of all of its proper initial segments, so its ordered sum with $A$ has order type ${\sup }_{\gamma <\lambda }(\alpha +\gamma )$.

True. If $x$ is a nonempty finite set, thenThis maximum is a successor ordinal, whereas the empty set has rank $0$. Hence a set of nonzero limit ordinal rank cannot be finite and must be an infinite set.

The Von Neumann hierarchy is defined by transfinite recursion:for every limit ordinal $\lambda$.

We prove simultaneously by transfinite induction that every $V_{\alpha }$ is a transitive set and that $V_{\alpha }\subseteq V_{\alpha +1}$. The claims are immediate for $V_0$. Suppose first that $V_{\alpha }$ is transitive. If $x\in y\in V_{\alpha +1}=\mathcal{P}(V_{\alpha })$, then $y\subseteq V_{\alpha }$, so $x\in V_{\alpha }$. Moreover, transitivity says that every $x\in V_{\alpha }$ satisfies $x\subseteq V_{\alpha }$, and henceThus $V_{\alpha +1}$ is transitive and $V_{\alpha }\subseteq V_{\alpha +1}$. At a limit $\lambda$, if $x\in y\in V_{\lambda }$, then $y\in V_{\beta }$ for some $\beta <\lambda$; transitivity of $V_{\beta }$ gives $x\in V_{\beta }\subseteq V_{\lambda }$. The inclusions at successor stages also imply

Finally, the rank of a set $x$ is the ordinalEvery $y\in x$ belongs to $V_{\mathrm{rank}(y)+1}\subseteq V_{\mathrm{rank}(x)}$, so $x\subseteq V_{\mathrm{rank}(x)}$. Thereforewhich proves that every set occurs in the hierarchy.