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The club filter completeness argument works for every regular uncountable $\kappa$. Let $\mu <\kappa$ and let $C_i$ be a club set for each $i<\mu$. Their intersection is closed. To prove it unbounded, start above any prescribed $\beta <\kappa$ and choose an increasing sequence $({\delta }_n{)}_{n<\omega }$ so that ${\delta }_{n+1}$ lies above a chosen point of every $C_i$ greater than ${\delta }_n$. $\kappa$ is a regular cardinal, so the supremum of these $\mu$ choices remains below $\kappa$. The resulting countable supremum $\delta ={\sup }_n{\delta }_n$ likewise remains below $\kappa$. For each $i$, the chosen $C_i$ points are cofinal in $\delta$, so closure gives $\delta \in C_i$. Thus the intersection is a club set. Intersecting fewer than $\kappa$ members of the club filter still contains such an intersection of clubs. In particular, $\boxed{{\mathcal{D}}_{{\omega }_2}\text{ is }{\aleph }_2\text{-complete}}$.

It is not an ultrafilter. For a regular infinite $\theta <\kappa$, the setis stationary. Given a club set $C$, build in $C$ a strictly increasing continuous sequence of length $\theta$ and take its supremum $\delta <\kappa$. Closure gives $\delta \in C$, and the cofinality of an increasing ordinal supremum gives $\mathrm{cf}(\delta )=\theta$. This proves the stationarity of ordinals of prescribed cofinality.

At $\kappa ={\omega }_2$, the disjoint sets $S_{\omega }^{{\omega }_2}$ and $S_{{\omega }_1}^{{\omega }_2}$ are both stationary. A club contained in either $S_{\omega }^{{\omega }_2}$ or its complement would miss one of these stationary sets. Thus neither $S_{\omega }^{{\omega }_2}$ nor its complement belongs to ${\mathcal{D}}_{{\omega }_2}$, and

Weakly inaccessible cardinals are unbounded below the given cardinal. In fact, they form a stationary set there. Let $\kappa$ be the weakly Mahlo cardinal, so it is a regular uncountable limit cardinal and the set $R=\{\alpha <\kappa :\alpha =\mathrm{cf}(\alpha )\}$ is stationary.

The set $C$ of uncountable limit cardinals below $\kappa$ is a club set. For unboundedness, above any starting point choose a strictly increasing countable sequence of cardinal numbers below $\kappa$; its supremum remains below $\kappa$ by regularity and is an uncountable limit cardinal. For closure, a limit of such limit cardinals is again a limit cardinal.

Every $\alpha \in R\cap C$ is an uncountable regular cardinal and a limit cardinal, hence a weakly inaccessible cardinal. The intersection of a stationary set with a club set is stationary, because its intersection with any further club is nonempty. ThereforeThis proves the stronger form of the requested conclusion.

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.

Use the standard convention that a strongly inaccessible cardinal is uncountable, regular and strong limit. The printed explanatory definition omits uncountability: literally it also admits $\kappa =\omega$, for which no finite cardinal is a worldly cardinal. Thus that omission makes the requested conclusion false under the literal abbreviated definition. The proof below applies to the usual, intended strongly inaccessible cardinal convention $\kappa >\omega$.

First $|V_{\gamma }|<\kappa$ for every $\gamma <\kappa$. At successor stages this follows from the strong limit cardinal property, and at limits from the regular cardinal property. Consequently $V_{\kappa }\models \mathsf{ZFC}$. In particular, for Axiom schema of replacement, a domain $a\in V_{\kappa }$ has size less than $\kappa$, so its functional image has fewer than $\kappa$ elements, so the supremum of the rank of a set over its elements is below $\kappa$ by regular cardinal structure. All other axioms, including axiom of choice, have their witnesses at bounded ranks below $\kappa$; uncountability supplies axiom of infinity.

Enumerate the first-order formulas as ${\phi }_0,{\phi }_1,\dots$. For each finite collection, reflection inside the set structure $V_{\kappa }$ gives a closed unbounded subset of $\kappa$ of agreeing ranks. To see why the witness bounds stay below $\kappa$, there are fewer than $\kappa$ parameter tuples at each $V_{\gamma }$, and the supremum of their least witness ranks remains below $\kappa$ by regular cardinal structure. Closing under these bounds and taking increasing countable limits gives the usual reflection theorem for definable hierarchies argument. The countable intersection of these closed unbounded subsets is still closed unbounded because $\kappa$ is regular and uncountable. At each resulting $\alpha$,The infinite cardinal numbers below $\kappa$ also form a closed unbounded subset: they are unbounded because ${\lambda }^{+}\le 2^{\lambda }<\kappa$ for infinite $\lambda <\kappa$, and a supremum of increasing cardinal numbers is a cardinal number. Intersect the two closed unbounded subsets. Every member of the intersection is a worldly cardinal, and an unbounded subset of a regular cardinal $\kappa$ has size $\kappa$. HenceIn fact this proves the stronger worldly cardinals below an inaccessible cardinal result that these worldly cardinals contain a closed unbounded subset of $\kappa$.

A cardinal number is identified with an initial ordinal, namely an ordinal not in bijection with any smaller ordinal. An inaccessible cardinal, in the strong sense used here, is an uncountable regular cardinal that is also a strong limit cardinal. ThusA worldly cardinal is a cardinal number whose level of the Von Neumann hierarchy satisfies all of ZFC:Worldliness is this model-theoretic condition; regularity is not part of its definition. The original PDF asks for these two notions; the extra comma in the local TeX makes “cardinal” look like a separate requested item.

N. A larger model may collapse a singular cardinal number, destroying cardinalhood, so the assertion is not described by an upward absolute formula. It may instead add a short cofinal function to a regular cardinal in the smaller model, so it is not described by a downward absolute formula. This is the nonabsoluteness of singular cardinalhood.

Let $\lambda \ge \kappa$ be a cardinal number of $M$. If forcing collapsed $\lambda$, some condition would force a surjective function $\dot{f}:\mu \to \lambda$ for an ordinal $\mu <\lambda$. Replacing $\mu$ by its ground-model cardinality lets us assume that $\mu$ is a cardinal.

For each $\xi <\mu$, choose in $M$ a maximal antichain in a forcing order deciding $\dot{f}(\xi )$. The chain condition for forcing makes its size less than $\kappa$, so the set $B_{\xi }$ of possible values of $\dot{f}(\xi )$ has size below $\kappa$. This is the possible-values lemma for chain-condition forcing. Every interpreted range is contained in $B={\bigcup }_{\xi <\mu }{B}_{\xi }$. If $\mu <\kappa$, the fact that $\kappa$ is a regular cardinal gives $|B|<\kappa \le \lambda$; if $\kappa \le \mu <\lambda$, infinite cardinal arithmetic gives $|B|\le \mu <\lambda$. Either way $B$ cannot contain the range of a surjection onto $\lambda$, a contradiction. By cardinal preservation by chain-condition forcing, $\mathbb{P}$ preserves every cardinal at least $\kappa$.

Assume that $\kappa$ is regular and uncountable in $M$, as required for this claim. If an antichain $A\in M$ had size $\kappa$, apply the delta-system lemma at a regular uncountable cardinal inside $M$ to the finite domains of its conditions. After thinning, their domains form a delta-system with finite root $R$.

There are fewer than $\kappa$ possible restrictions to $R$, because each coordinate $(\alpha ,n)\in R$ has fewer than $\kappa$ possible values. Since $\kappa$ is a regular cardinal, we can thin again to $\kappa$ conditions agreeing on $R$. Any two now have disjoint domains outside $R$ and agree on $R$, so their union is a common extension. This contradicts that $A$ is an antichain. ThereforeIf the word “regular” is allowed to include $\kappa =\omega$, the printed claim needs the additional hypothesis that $\kappa$ is uncountable; the finite-condition order can have an infinite antichain when $\kappa =\omega$.

An uncountable cardinal number $\kappa$ is weakly compact when every $\kappa$-satisfiable theory in an infinitary language $L_{\kappa ,\kappa }$ with at most $\kappa$ nonlogical symbols is satisfiable. A cardinal is inaccessible when it is uncountable, regular, and a strong limit cardinal.

Two standard results supply the proof. First, every weakly compact cardinal is inaccessible. Second, every weakly compact $\kappa$ has the Keisler extension property: there is a transitive set $X⊋V_{\kappa }$ such thatand $\kappa \in X$. Since $\kappa$ is inaccessible, the relevant downward absoluteness makes $X\models$ “$\kappa$ is inaccessible”. Hence $X$ satisfies “there is an inaccessible cardinal”. By elementarity, $V_{\kappa }$ satisfies the same sentence, so it contains some inaccessible $\lambda$. Every ordinal in $V_{\kappa }$ is below $\kappa$, and inaccessibility is absolute here, giving

Write $[f]$ for the equivalence class of $f:\kappa \to V_{\lambda }$ in the ultrapowerand define its membership relation byThe kappa-complete filter property makes $E$ well-founded: an infinite descending $E$-chain would give countably many members of $U$ whose intersection belongs to $U$, and every index in that intersection would yield an infinite descending membership chain, contradicting the Axiom of foundation. The relation is extensional by Łoś's theorem.

The Mostowski collapse theorem therefore gives a unique isomorphism $\pi :(N,E)\to (M,\in )$ onto a transitive set $M$. Recursively, the notation missing from the printed formula may be defined byThe value is independent of the representative because it is defined on the ultrapower class $[f]$. Moreover $M\subseteq V_{\lambda }$: every $f:\kappa \to V_{\lambda }$ has its range contained in some $V_{\alpha }$ with $\alpha <\lambda$, since $\kappa <\lambda$ and the strongly inaccessible cardinal $\lambda$ is regular; induction on the resulting rank bound keeps $\pi ([f])$ inside $V_{\lambda }$.

Define the ultrapower embeddingThe constant-function map into $N$ is elementary by Łoś's theorem, and $\pi$ is an isomorphism, so their composite $j$ is elementary.

A strongly inaccessible cardinal is an uncountable regular cardinal $\kappa$ that is also a strong limit cardinal:

For a regular cardinal $\kappa$,Again $q\le p$ means $q\supseteq p$, and the maximal element is the empty function. The regularity of $\kappa$ ensures that the union of a descending sequence of fewer than $\kappa$ conditions still has domain of cardinality below $\kappa$ whenever the conditions form a compatible increasing chain of partial functions.

A finite collection of first-order formulas has a closed unbounded class of ordinals whose levels in a definable continuous hierarchy agree with its union for all parameters in the level. Close the collection under subformulas, bound the least witness levels for existential instances on each set-sized stage by Axiom schema of replacement, and iterate these bounds countably. Continuity supplies the closed stages, and induction on formulas proves agreement. The Tarski-Vaught test describes the same witness criterion. For a hierarchy of length an uncountable regular cardinal, bounds stay below that cardinal when every stage has smaller cardinality.

An infinite cardinal number $\kappa$ is a regular cardinal if every function from an ordinal $\delta <\kappa$ into $\kappa$ has bounded range. Write this as $\forall f\forall \delta \in \kappa (\mathrm{Map}(f,\delta ,\kappa )\Rightarrow \exists \beta \in \kappa \forall \xi \in \delta f(\xi )<\beta )$. The graph predicates are bounded formulas in set theory, and cardinalhood is Pi-one definable, so adding infinite cardinalhood yields a Pi-one formula modulo ZF.

For every infinite regular cardinal $\kappa$, every first-order structure has a $\kappa$-saturated elementary extension. At one stage use the compactness theorem with the elementary diagram to realize all complete types over subsets of the stage having size less than $\kappa$. There is a set of such types; finite subsets of the enlarged theory are satisfiable because their finitely many witnesses are independent. Iterate for $\kappa$ stages, taking elementary unions at limits. Regularity puts any parameter set of size less than $\kappa$ inside one stage, and its type is realized at the next one. Passing to a larger regular cardinal covers singular requested degrees of saturation as well.

An infinite cardinal number $\kappa$ is singular when $\mathrm{cf}(\kappa )<\kappa$. Thus a short cofinal function reaches arbitrarily high ordinals below $\kappa$. For example, ${\aleph }_{\omega }$ has cofinality $\omega$. The complementary case is a regular cardinal.

For an uncountable regular cardinal $\kappa$, choose cofinal sequences $c_{\alpha }:\omega \to \alpha$ for the stationary set of ordinals of cofinality $\omega$. Above any bound, some fixed coordinate exceeds the bound on a stationary subset; Fodor lemma makes that coordinate constant on a stationary subset. Thus the stationary constant fibers, over all coordinates, have unboundedly many values. Regularity makes one coordinate have $\kappa$ such values. Its fibers are disjoint stationary sets; adding all leftover ordinals to one piece partitions $\kappa$.