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If $V_{\lambda }$, for an infinite cardinal number $\lambda$, satisfies Axiom schema of replacement for every external functional class, then an external cofinal function from an ordinal below $\lambda$ cannot exist: its range would have rank of a set $\lambda$ and would have to belong to $V_{\lambda }$. Similarly, a surjection $\mathcal{P}(\theta )\to \lambda$ with $\theta <\lambda$ is impossible. Infinity therefore makes $\lambda$ an uncountable regular cardinal and a strong limit cardinal. Full semantics for second-order logic, allowing arbitrary external functional relations, is essential.

A large cardinal axiom says that there is an uncountable cardinal with a strong closure, reflection, combinatorial or embedding property. Its interest is not merely the magnitude of the cardinal: it packages a universe-like portion of set theory, or a coherent way of comparing the universe with a proper inner model. Such axioms form a hierarchy extending ordinary ZFC, compared through relative consistency implications. The consistency qualifications are essential; an existence axiom is not a theorem of ZFC simply because its consequences are attractive.

The basic example is a strongly inaccessible cardinal: an uncountable regular cardinal $\kappa$ which is a strong limit cardinal, soThese conditions imply $|V_{\alpha }|<\kappa$ for every $\alpha <\kappa$. Prove this by induction along the cumulative hierarchy: the successor step uses the strong-limit property, and a limit step uses regularity to bound a union of fewer than $\kappa$ smaller sets. Thus every member of $V_{\kappa }$ has cardinality below $\kappa$.

Consequently $V_{\kappa }$ is a transitive model of ZFC. Pairing, union, power set, separation, axiom of infinity and foundation stay within the hierarchy. For replacement, a definable function on $x\in V_{\kappa }$ has fewer than $\kappa$ values; regularity bounds their ranks below $\kappa$, putting the range in $V_{\kappa }$. Ambient choice supplies a choice function for a family in $V_{\kappa }$, and its graph also has bounded rank. This verifies the substantial closure axiom rather than only saying that the rank is large. It also shows why inaccessible existence cannot be proved in a consistent ZFC: it would yield a set first-order model of ZFC and hence its consistency, contrary to the Gödel second incompleteness theorem.

An uncountable regular limit cardinal is a weakly inaccessible cardinal; requiring the strong-limit inequality is stronger without additional hypotheses. Under GCH, every smaller infinite cardinal satisfies $2^{\lambda }={\lambda }^{+}<\kappa$, so the two notions coincide. This distinction separates cardinal-arithmetic assumptions from the regularity requirement.

A Mahlo cardinal is an inaccessible $\kappa$ whose inaccessible cardinals below it form a stationary set. Thus every club set meets those inaccessibles, which strengthens mere unboundedness. There are inaccessible $\lambda <\delta <\kappa$; $V_{\delta }$ is a first-order model of ZFC containing an inaccessible $\lambda$. Mahloness therefore supplies a consistency statement stronger than the bare existence of one inaccessible. Iterating the demand that smaller cardinals have the same reflection property leads to further levels of the hierarchy.

A weakly compact cardinal is a logical and combinatorial strengthening of inaccessibility. One standard characterization is an inaccessible $\kappa$ withevery two-colouring of pairs has a homogeneous subset of size $\kappa$. Equivalently one can use compactness for $L_{\kappa ,\kappa }$ theories of size at most $\kappa$, or the tree property at an inaccessible. These viewpoints explain the terminology: a local collection of compatible requirements has a global realization, and tall narrow trees cannot avoid cofinal branches. They are characterization statements; the elementary closure and measure arguments below are separate proofs of the consequences used here.

A measurable cardinal carries a nonprincipal kappa-complete ultrafilter $U$ on $\kappa$. Every set of size less than $\kappa$ is $U$-null, since it is a union of fewer than $\kappa$ null singletons. Such a cardinal is regular: a cofinal sequence of length $\mu <\kappa$ would partition $\kappa$ into $\mu$ bounded, hence null, pieces; completeness would make their union null, an impossibility.

It is also strong limit. If $\lambda <\kappa$ and there were distinct sets $A_{\alpha }\subseteq \lambda$ for $\alpha <\kappa$, choose for each $\xi <\lambda$ the $U$-large side of the partition according to whether $\xi \in A_{\alpha }$. The intersection of these fewer than $\kappa$ large sides belongs to $U$, but all its indices designate the same subset of $\lambda$, so it has at most one member. This contradicts nonprincipality. Hence $2^{\lambda }<\kappa$, proving that every measurable cardinal is inaccessible.

The measure gives an ultrapower embedding. Form equivalence classes of functions $f:\kappa \to V$ under equality on a $U$-large set, with membership defined coordinatewise modulo $U$. Induction on first-order formulas proves elementarity: Boolean operations use the ultrafilter laws, and the existential step chooses coordinate witnesses on a large set. Countable completeness makes the ultrapower well-founded. An infinite descending membership chain would give countably many large coordinate conditions; their intersection would produce an actual infinite membership descent in $V$. Collapse the ultrapower to a transitive class $M$, obtaining $j:V\to M$.

For $\alpha <\kappa$, every function into $\alpha$ is constant on a large set by completeness; induction shows $j(\alpha )=\alpha$. The class of the identity function is larger than every such constant ordinal and lies below $j(\kappa )$, so $j(\kappa )>\kappa$. Thus the critical point of an elementary embedding is $\kappa$. The measure viewpoint has become a precise self-similarity principle for the universe.

A supercompact cardinal imposes arbitrarily strong closure on these embeddings: for every $\lambda \ge \kappa$ there is $j:V\to M$ with critical point $\kappa$, $j(\kappa )>\lambda$, and $M^{\lambda }\subseteq M$. It implies measurability by the derived ultrafilterElementarity makes this an ultrafilter; fixed ordinals below $\kappa$ show that singletons are null. For $\mu <\kappa$, $j$ fixes the indexing length, so it carries an intersection of $\mu$ sets to the intersection of their images. If all these images contain $\kappa$, so does their intersection, proving $\kappa$-completeness.

These examples illustrate three connected uses of large cardinals: obtaining set first-order models and stronger consistency statements, extending finite or countable compactness and partition phenomena to uncountable domains, and constructing elementary embeddings with controlled closure. They also interact with inner models: an inaccessible remains inaccessible in the constructible universe, since it cannot acquire a new short cofinal sequence there and there are fewer subsets of smaller ordinals. A measurable cardinal's measure need not belong to that inner model, so preservation of the stronger property is a different issue. Large cardinal axioms are thus structured additional hypotheses, with carefully distinguishable consequences and relative-consistency claims, rather than a single claim that sufficiently big cardinals exist.

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.

The partition relation $\lambda \to (\mu {)}_{\nu }^r$ says that every colouring of the $r$-element subsets of a set of cardinality $\lambda$ into $\nu$ colours has a homogeneous set of cardinality $\mu$. Define the beth numbers over an infinite cardinal $\kappa$ by ${\beth }_0(\kappa )=\kappa$ and ${\beth }_{n+1}(\kappa )=2^{{\beth }_n(\kappa )}$. The Erdős-Rado theorem for finite arities gives the precise boundThus any fixed finite arity and any fixed infinite number of colours admit arbitrarily large infinite monochromatic sets once the starting cardinal is sufficiently large. For pairs, the especially useful case is $(2^{\kappa }{)}^{+}\to ({\kappa }^{+}{)}_{\kappa }^2$. The successor in this bound matters: a binary first-difference colouring on a suitably ordered set of binary sequences supplies the familiar obstruction at $2^{\kappa }$. Also, a theorem for every fixed finite arity is not a theorem homogenizing all arities simultaneously.

The mechanism is to build an end-homogeneous routing tree: each new vertex is routed according to its colours to earlier vertices on its branch. A long branch is end-homogeneous, meaning the colour from a fixed earlier vertex is independent of the later vertex. There are at most ${\kappa }^{\alpha }\le 2^{\kappa }$ routes of length $\alpha <{\kappa }^{+}$, and the whole collection of short routes has size at most $2^{\kappa }$. Inserting $(2^{\kappa }{)}^{+}$ vertices therefore forces a branch of length ${\kappa }^{+}$. One colour occurs at ${\kappa }^{+}$ earlier vertices on that branch, giving the pair theorem. Iterating the higher-arity version of this routing construction explains the successive powers in the Erdős-Rado theorem for finite arities.

A weakly compact cardinal is an uncountable strongly inaccessible cardinal $\kappa$ withEquivalent formulations include the tree property at the inaccessible cardinal and compactness for theories of size at most $\kappa$ in the infinitary language $L_{\kappa ,\kappa }$: if every subtheory of size $<\kappa$ has a first-order model, so does the theory.

A measurable cardinal is an uncountable $\kappa$ carrying a nonprincipal ultrafilter closed under intersections of fewer than $\kappa$ members. Such an ultrafilter contains no set of size $<\kappa$, since all singletons have complements in it and it is kappa-complete. Consequently a cofinal sequence of length $<\kappa$ would partition $\kappa$ into too few small pieces, so $\kappa$ is regular cardinal. If $\lambda <\kappa$ and $\kappa$ distinct subsets of $\lambda$ existed, choose, for each coordinate of $\lambda$, the bit selected by the ultrafilter. Kappa-completeness would make all those bits simultaneously constant on a large set of indices; only one distinct subset could have that bit pattern, a contradiction. Hence $2^{\lambda }<\kappa$: $\kappa$ is also a strong limit cardinal.

A supercompact cardinal $\kappa$ has, for every $\lambda \ge \kappa$, a fine ultrafilter that is a normal ultrafilter on small subsets and is kappa-complete onFine means $\{x:\alpha \in x\}$ is large for every $\alpha <\lambda$. Normal means that a function $f$ with $f(x)\in x$ on a large set is constant on some large set. Equivalently, for every such $\lambda$ there is an elementary embedding $j:V\to M$ with critical point $\kappa$, $j(\kappa )>\lambda$, and $M$ closed under $\lambda$-sequences from the universe.

These notions form the implication chain supercompact $\Rightarrow$ measurable $\Rightarrow$ weakly compact. For the first implication, an embedding as above gives the measure $U=\{X\subseteq \kappa :\kappa \in j(X)\}$. For the second, use a normal ultrafilter on a cardinal obtained by normalizing its measure. Indeed the ultrapower embedding gives the measure $U=\{X\subseteq \kappa :\kappa \in j(X)\}$; for a regressive $f$, $j(f)(\kappa )<\kappa$ is a fixed ordinal, yielding the required constant large fibre. Given a pair-colouring, select for each $\alpha$ the measure-one colour on the tail above $\alpha$. One selected colour occurs on a measure-one set of $\alpha$'s. Normality makes the diagonal intersection of their selected tails measure one; any two members of that intersection have the selected colour. This gives the required partition relation, and the preceding regularity and strong-limit argument gives inaccessibility.

Put $\mu =\mathrm{cf}(\lambda )$ and choose an increasing cofinal sequence of infinite cardinals $⟨{\lambda }_i:i<\mu ⟩$ below $\lambda$. Since $\lambda$ is a strong limit cardinal, $2^{{\lambda }_i}<\lambda$ for every $i$. Encode a subset $A\subseteq \lambda$ by its restrictions $⟨A\cap {\lambda }_i:i<\mu ⟩$. Cofinality makes this encoding injective, givingThe second inequality chooses an injection of each restriction set into $\lambda$. Conversely,since $\mu \le \lambda$ and an infinite cardinal satisfies $\lambda \cdot \mu =\lambda$. The two inequalities prove the singular strong limit power-set identityThis does not assume the singular cardinals hypothesis or determine either side as ${\lambda }^{+}$. The strong-limit assumption is what bounds each restricted power set.

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.

The exponent-two Erdős-Rado theorem for pairs says that for every infinite cardinal $\lambda$,In particular, a finite or countable coloring of pairs from $(2^{{\aleph }_0}{)}^{+}$ has an uncountable monochromatic subset. We prove the general statement by an end-homogeneous tree construction.

Let $c:[\mu {]}^2\to \nu$ be any coloring on an ordinal $\mu$. Route a vertex $\beta$ through an increasing sequence of preceding vertices as follows. Start with the least vertex. After ancestors $a_{\eta }$, $\eta <\xi$, have been selected, choose the least remaining candidate $\gamma \le \beta$, above those ancestors, satisfyingThe terminal candidate $\beta$ always qualifies, so the procedure eventually reaches it. Candidate sets shrink and their least elements increase. If an intermediate candidate is $\gamma$, its route is precisely the earlier part of the route to $\beta$, because all earlier color tests agree. Consequently the routes define a set-theoretic tree on the vertices.

Along a branch, the color from an ancestor to every later branch vertex is constant. Also a node of height $\alpha$ is determined by its sequence of $\alpha$ colors to its ancestors: reconstruct the ancestors as successive least candidates with those prescribed colors, then reconstruct the node as the least final candidate. Thus its level has at most ${\nu }^{|\alpha |}$ nodes.

Apply this with $\mu =(2^{\lambda }{)}^{+}$ and $\nu =\lambda$. If every node had height less than ${\lambda }^{+}$, each level would have size at most ${\lambda }^{\lambda }\le 2^{\lambda }$, and there would be at most ${\lambda }^{+}$ such levels. Their total size would be at most $2^{\lambda }$, contradicting $|\mu |=(2^{\lambda }{)}^{+}$. Hence some route has at least ${\lambda }^{+}$ ancestors. For each ancestor mark its constant color to all later route points. Since there are only $\lambda$ colors, one occurs on ${\lambda }^{+}$ ancestors. Those ancestors form the required monochromatic set. This proves the theorem.

To make the ordinal function precise, one useful ordinal partition-bound function isThe target here is ordinal order type. For infinite $\alpha$, the proved theorem with $\lambda =|\alpha |$ supplies a bound $(2^{\lambda }{)}^{+}$: its homogeneous set has order type at least ${\lambda }^{+}>\alpha$, and it handles all the required color counts. Finite targets are handled by finite Ramsey theorem bounds. Thus $f$ is total, nondecreasing and $f(\alpha )\ge \alpha$.

Suppose a sequence of transfinite iterates ${\alpha }_{\xi +1}=f({\alpha }_{\xi })$, with suprema at limit stages, has supremum an uncountable cardinal $\kappa$ at a limit stage, with every earlier iterate below $\kappa$. Thenchoose an earlier iterate at least $\theta$ and use the next one as a bound. If the iteration has already stabilized, it has already found a fixed point instead.

This closure forces $\kappa$ to be a strong limit cardinal. For any infinite $\lambda <\kappa$, color pairs of distinct binary strings of length $\lambda$ by their first differing coordinate. This uses $\lambda$ colors and has no monochromatic triangle: three binary digits at one coordinate cannot be pairwise different. It gives a coloring on $2^{\lambda }$ with no homogeneous target of type $\lambda +1$. Since $\lambda$ colors are allowed when the target is the ordinal $\lambda +1$, we haveFinite exponents satisfy the strong-limit requirement automatically for uncountable $\kappa$.

The needed additional condition is the tree property: every tree of height $\kappa$, with nonempty levels of size less than $\kappa$, has a cofinal branch. In the present convention it forces $\kappa$ to be a regular cardinal. If $\kappa$ were singular, take a cofinal sequence of smaller cardinals of length $\mathrm{cf}(\kappa )$ and form the disjoint union of chains of those lengths. This tree has height and total size $\kappa$, levels of size at most $\mathrm{cf}(\kappa )<\kappa$, but no branch of length $\kappa$, a contradiction.

For any $\nu <\kappa$, apply the preceding color-routing tree to a coloring $[\kappa {]}^2\to \nu$. If a route already has length $\kappa$, use it. Otherwise its levels are indexed below $\kappa$ and each has sizefor infinite exponents, by the strong-limit property; finite cases satisfy the same needed bound directly. The tree has $\kappa$ nodes, so regularity forces its height to be $\kappa$. The tree property supplies a cofinal branch. The ancestor-color partition of that branch has one class of size $\kappa$, since $\nu <\kappa$ and $\kappa$ is regular. Thus $\kappa \to (\kappa {)}_{\nu }^2$ for every $\nu <\kappa$, andWe have proved that an uncountable function-closed cardinal with the tree property is regular and strong limit, hence strongly inaccessible, and that it is the requested fixed point.

The paper's final sentence, read as a claim about the tree property alone, is false. Already ${\aleph }_0$ has it by the König infinity lemma but is not uncountable; even with uncountability stipulated, there are relative-consistency models with the tree property at ${\aleph }_2$. The latter counterexample is documented in the research account of Mitchell forcing. The rigorous conclusion in the iterated-function context is the one just proved: function closure supplies strong limit, and the tree property supplies regularity and a branch. Strong limit and regularity follow from distinct hypotheses in this context.

First express inaccessibility by universally excluding witnesses to its failure. Besides $\kappa$ being an ordinal greater than $\omega$, require that for every $\alpha <\kappa$ there is no function from $\alpha$ cofinal in $\kappa$, and no injection from $\kappa$ into $\mathcal{P}(\alpha )$. The first condition gives regular cardinalhood; the second says that $\kappa$ is a strong limit cardinal. Under the axiom of choice these are exactly the conditions for a strongly inaccessible cardinal.

These are Pi-one formulas modulo ZF: the only unbounded quantifiers universally range over possible functions. Being a function of the specified domain, having bounded or cofinal range in $\kappa$, and being an injection whose values are subsets of $\alpha$ can all be written with bounded formulas in set theory, without using a power-set object as an unbounded existential witness. Consequently inaccessibility is downward absolute to a transitive inner model of ZFC.

Now suppose $\kappa$ is Mahlo in $V$. It is inaccessible in the constructible universe $L$, and every $V$-inaccessible ordinal below it is also inaccessible in $L$. Let $C\in L$ be any set that $L$ regards as a club set in $\kappa$. Closure and unboundedness are absolute properties of this fixed set of ordinals, so $C$ is also a club set in $V$. By Mahloness in $V$, it contains an ordinal that is inaccessible in $V$, hence in $L$. Every $L$-club therefore meets the $L$-inaccessible ordinals below $\kappa$. Thus Mahloness is downward absolute to the constructible universe:

Use the countable-condition collapseordered by reverse inclusion: an extension of a function is a stronger condition. All sizes and conditions here are computed in the ground model $M$.

This forcing is countably closed: the union of a descending countable sequence is a countable partial function. Thus it adds no countable ordinal sequences and preserves ${\omega }_1$. For each $\xi <{\omega }_1$, the conditions whose domains contain $\xi$ are dense; for each $\eta <\kappa$, those whose ranges contain $\eta$ are dense. The union of the generic filter is therefore a surjection from ${\omega }_1$ onto $\kappa$, and $|\kappa {|}^{M[G]}={\aleph }_1$.

Inaccessibility gives ${\kappa }^{{\aleph }_0}=\kappa$: every countable sequence is bounded below the regular $\kappa$, and the strong limit cardinal property bounds the number of sequences at each bound below $\kappa$. Hence $|\mathbb{P}|=\kappa$. The forcing satisfies the ${\kappa }^{+}$-chain condition for forcing, so it preserves every cardinal above $\kappa$. This gives precisely the requested collapse and preservation. The assertion that $\kappa$ is collapsed to ${\aleph }_1$ concerns its new cardinality; the ordinal $\kappa$ itself does not change.

Fix $p$ and a name $\dot{f}$ such that $p$ forces a two-coloring of $[\lambda {]}^{<\omega }$. For each ground-model finite set $a\subseteq \lambda$, record its full forcing decision patternThere are at most $3^{|\mathbb{P}|}<\lambda$ such patterns, since $\lambda$ is a strong limit cardinal and $|\mathbb{P}|<\lambda$. For finite $\mathbb{P}$ the number of patterns is finite, also below $\lambda$. Encode the patterns by an ordinal $\beta <\lambda$ and apply the assumed finite-subset partition property to obtain $H\subseteq \lambda$ of size $\lambda$ such that $F$ is constant on $[H{]}^n$ for every $n<\omega$.

For each $n$, choose one $a_n\in [H{]}^n$. Conditions below $p$ deciding $\dot{f}(a_n)$ are dense. Whenever one such condition decides the value, equality of patterns means it forces that same value at every $a\in [H{]}^n$. A generic filter containing $p$ meets this dense set for each $n$, so $f$ is constant on each $[H{]}^n$ in the extension. The constants may differ with $n$, exactly as required.

The forcing has size below the regular $\lambda$, hence satisfies the $\lambda$-chain condition for forcing and preserves its cardinality. The ground-model set $H$ consequently still has size $\lambda$. Since the argument works for every name and condition,This is small forcing preservation of finite-subset partition properties; no closure assumption on the forcing is required.

For an infinite cardinal number $\kappa$, the gimel function isThe singular cardinals hypothesis asserts that for every infinite singular cardinal $\kappa$,Here ${\kappa }^{+}$ is the successor cardinal. Equivalently, for every infinite singular cardinal,To see why the second formulation adds nothing in the other case, put $\theta =\mathrm{cf}(\kappa )$. If $\kappa \le 2^{\theta }$, then $2^{\theta }\le {\kappa }^{\theta }\le (2^{\theta }{)}^{\theta }=2^{\theta }$ by infinite cardinal arithmetic. Equality $\kappa =2^{\theta }$ is impossible here: the König theorem for cardinal numbers gives $\mathrm{cf}(2^{\theta })>\theta$, whereas $\mathrm{cf}(\kappa )=\theta$. Thus in this case $2^{\theta }\ge {\kappa }^{+}$, as required by the maximum formula. For a strong limit cardinal that is singular, the hypothesis also yields $2^{\kappa }={\kappa }^{+}$: restrictions of a subset of $\kappa$ to a cofinal sequence of smaller cardinals give $2^{\kappa }\le {\kappa }^{\theta }$, while the reverse inequality is immediate. The quantified implication above is the precise general statement.

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.

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

Let $U$ be a nonprincipal $\kappa$-complete ultrafilter on the measurable cardinal $\kappa$, and fix $\lambda <\kappa$. Suppose for a contradiction that $2^{\lambda }\ge \kappa$. Choose distinct subsets $A_{\alpha }\subseteq \lambda$ for $\alpha <\kappa$. For each $\xi <\lambda$, the ultrafilter property chooses exactly one ofas a member $H_{\xi }$ of $U$. Since $\lambda <\kappa$, $\kappa$-completeness gives $H={\bigcap }_{\xi <\lambda }{H}_{\xi }\in U$.

Any two indices in $H$ give subsets having the same membership decision at every $\xi <\lambda$, so they give the same $A_{\alpha }$. The chosen subsets were distinct, hence $|H|\le 1$. This contradicts the fact that a small set is absent from a complete nonprincipal ultrafilter. Therefore $2^{\lambda }<\kappa$ for every $\lambda <\kappa$, which is precisely the strong limit cardinal condition. This is the measurable cardinal is a strong limit cardinal argument.

Fix $i\in \{0,1\}$ and write $\kappa ={\kappa }_i$, $j=j_i$, and $M=M_i$. Every ordinal below $j(\kappa )$ is represented in the ultrapower by a function $\kappa \to \kappa$. Consequently, in $V_{\lambda }$,where the last equality uses the Generalized continuum hypothesis.

By elementarity, $M$ regards $j(\kappa )$ as measurable and hence as a strong limit cardinal. Moreover $V_{\kappa +1}\subseteq M$, so $M$ and $V_{\lambda }$ have the same subsets of $\kappa$ and the same ${\kappa }^{+}$. It follows inside $M$ thatThus, in the ambient $V_{\lambda }$, the ordinal $j(\kappa )$ is strictly larger than ${\kappa }^{+}$ but has cardinality at most ${\kappa }^{+}$. It cannot be a cardinal number. Applying this argument to both $i=0$ and $i=1$ proves that neither $j_0({\kappa }_0)$ nor $j_1({\kappa }_1)$ is a cardinal in $V_{\lambda }$, exactly as in moved critical point is not an ambient cardinal under GCH.

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

Let $U$ witness that $\kappa$ is measurable. Regularity is given, so it remains to prove the strong limit cardinal property. First, every $A\in U$ has cardinality $\kappa$: if $|A|<\kappa$, thenby nonprincipality and $\kappa$-completeness, contradicting $A\in U$.

Suppose $\lambda <\kappa$ and $2^{\lambda }\ge \kappa$. Choose an injection $f:\kappa \to \mathcal{P}(\lambda )$. For each $\xi <\lambda$, exactly one oflies in $U$. Their chosen intersection lies in $U$ by $\kappa$-completeness. On that intersection every $f(\alpha )$ is the same subset of $\lambda$, contradicting injectivity because every member of $U$ has size $\kappa$. Thus $2^{\lambda }<\kappa$, and $\kappa$ is strongly inaccessible.

For a singular strong limit cardinal $\lambda$, let $\mu =\mathrm{cf}(\lambda )$ and choose cofinal cardinals ${\lambda }_i<\lambda$. Subset restrictions inject $\mathcal{P}(\lambda )$ into ${\prod }_{i<\mu }\mathcal{P}({\lambda }_i)$, whose cardinal is at most ${\lambda }^{\mu }$ by the strong-limit property. Conversely ${\lambda }^{\mu }\le (2^{\lambda }{)}^{\mu }=2^{\lambda }$. This equality alone does not imply the singular cardinals hypothesis.

If transitive models of enough set theory contain the same $V_{\alpha +1}$, then they agree on whether $\alpha$ is a strongly inaccessible cardinal. They have the same subsets and functions on every ordinal below $\alpha$, so they agree on cardinality, regularity, and the strong limit cardinal property.