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If $\kappa$ is the least omega-measurable cardinal and $U$ witnesses it, failure of $\kappa$-completeness partitions a member of $U$ into fewer than $\kappa$ fibres, none belonging to $U$. The pushforward ultrafilter under the fibre-index map is then a nonprincipal countably complete measure on a smaller cardinal, a contradiction. Thus $U$ is $\kappa$-complete and $\kappa$ is a measurable cardinal.

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.

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.

An ultraproduct converts coordinatewise finite information into a single first-order structure. Let $(M_i{)}_{i\in I}$ be nonempty structures in a common first-order language, and let $U$ be an ultrafilter on $I$. In ${\prod }_i{M}_i$, setThe ultraproduct ${\prod }_i{M}_i/U$ has these equivalence classes as its elements. Functions are interpreted coordinatewise; a relation holds of classes exactly when it holds on a $U$-large set of indices. Intersecting finitely many large agreement sets shows that these interpretations are independent of representatives.

The Łoś theorem, which explains the construction's usefulness, saysHere is the proof. Induction on terms establishes the coordinatewise interpretation of atomic formulas. Finite intersections handle conjunction, and the ultrafilter property handles negation, since precisely one of a set and its complement is large. For an existential quantification, a witness class gives coordinate witnesses on a large set. Conversely, when the coordinate existential truth set is large, select a witness in each of those structures and a default element elsewhere. Its class is a witness in the ultraproduct. This uses choice. Structural induction on formulas now proves the theorem, with universal quantification obtained by negation.

Compactness. If every finite subset of a first-order theory $T$ has a model, take $I$ to be the finite subsets $\Gamma \subseteq T$ and choose $M_{\Gamma }\models \Gamma$. The cones $\{\Gamma :\Delta \subseteq \Gamma \}$ for finite $\Delta$ have the finite intersection property. Extend their filter to an ultrafilter by the ultrafilter lemma. For each $\phi \in T$, its cone is large, so the Łoś theorem makes the ultraproduct satisfy $\phi$. This proves the compactness theorem, including theories of arbitrary set size.

Elementary extensions and nonstandard elements. An ultrapower uses a fixed structure at every coordinate. The diagonal map sending $a$ to the class of the constant function $a$ is an elementary embedding by the Łoś theorem. For a nonprincipal ultrafilter on $\omega$, the class of $i\mapsto i$ in an ultrapower of the natural numbers exceeds every standard integer, because each corresponding tail is large. In an ultrapower of the real ordered field, the reciprocal of this element is positive and smaller than every positive standard reciprocal integer: an infinitesimal. These examples implement ordinary finite statements in a larger structure without claiming that every external subset or second-order assertion transfers.

Saturation. Such an ultraproduct over $\omega$ is countably saturated. To see the actual diagonal argument, list a finitely satisfiable type ${\phi }_0(x),{\phi }_1(x),\dots$ with parameters in the ultraproduct and fix representatives for those parameters. For each $n$, the set $E_n$ of coordinates admitting a simultaneous witness to its first $n$ formulas belongs to $U$ by the Łoś theorem. Arrange that the $E_n$ decrease and put $D_n=E_n\cap \{i:i\ge n\}$. The $D_n$ decrease, belong to $U$, and have empty intersection. At coordinate $i$, choose a witness to the first $m(i)$ formulas, where $m(i)$ is the largest $n$ with $i\in D_n$, or choose a default element if there is no such $n$. For each fixed $n$, $m(i)\ge n$ on the large set $D_n$. Therefore the resulting class realizes every formula. This proves countable saturation of an ultraproduct over omega; it does not assert saturation at all larger cardinalities.

Algebraic limits. Take finite fields ${\mathbb{F}}_p$ indexed by the primes and a nonprincipal ultrafilter. Their ultraproduct is a field of characteristic zero: for every fixed positive integer $n$, only finitely many primes divide $n$, so $n\cdot 1\ne 0$ on a large set. Every first-order sentence true in all finite fields holds in this infinite field, and more generally so does any sentence true for all but finitely many of the factors. This is a way to create an infinite field retaining uniform first-order information from finite fields. It does not make the field algebraically closed: the first-order sentence that every element is a square already fails in infinitely many of the factors.

Large cardinals. A countably complete ultrafilter gives a well-founded membership ultrapower of the universe. If it had an infinite descending chain represented by $f_n$, each setwould be large. Countable completeness makes ${\bigcap }_n{A}_n$ nonempty, giving an actual infinite membership descent at any index in the intersection, contrary to foundation. The Mostowski collapse therefore gives a transitive target and an ultrapower embedding $j:V\to M$.

For a measure on a measurable cardinal $\kappa$, every function $\kappa \to \alpha$ with $\alpha <\kappa$ is constant on a large set: otherwise intersect the fewer than $\kappa$ complements of all its fibres. Consequently $j$ fixes the ordinals below $\kappa$. The class of the identity function is below $j(\kappa )$ and above every constant class below $\kappa$, since all tails are large. Thus $j(\kappa )>\kappa$ and its critical point is $\kappa$. Fine ultrafilters on $P_{\kappa }(\lambda )$ that are normal ultrafilters on small subsets and kappa-complete yield the stronger embeddings witnessing supercompact cardinals.

These applications share a single principle: an ultraproduct preserves precisely the first-order information seen by its ultrafilter. Principal ultrafilters merely select a factor; nonprincipal ones can produce new sizes, elements and saturation. Existence of the needed ultrafilters and selection of coordinate witnesses are set-theoretic assumptions, and changing the ultrafilter may change the resulting model.

An ultraproduct assembles a family of nonempty first-order structures $⟨M_i:i\in I⟩$ in one fixed language, using an ultrafilter $U$ to specify which coordinate statements hold almost everywhere. Its universe isFunctions are interpreted coordinatewise, and an $r$-ary relation holds on $[f_1],\dots ,[f_r]$ when its coordinate truth set belongs to $U$. Changing representatives changes such a truth set only off a $U$-large set, so these definitions are well defined. A principal ultrafilter simply recovers its distinguished factor; a nonprincipal ultrafilter can produce new elements and structures.

The fundamental result is the Łoś theorem:Prove it by structural induction on first-order formulas. Atomic formulas are the definition. Intersections handle conjunction, while exactly one of a set and its complement belongs to an ultrafilter, handling negation. For the existential step, if the coordinate existential truth set is in $U$, use choice to select witnesses there and arbitrary values elsewhere; their product function is an ultraproduct witness. Conversely a product witness makes the coordinate existential truth set contain a $U$-large set. Other connectives and universal quantification follow.

In an ultrapower, all factors are one structure $M$. The diagonal map $a\mapsto [\text{constant }a]$ is an elementary embedding by the Łoś theorem. For a nonprincipal $U$ on $\omega$, the class of $n\mapsto n$ in an ultrapower of the standard natural numbers is larger than every constant natural number, since each tail belongs to $U$. This gives a nonstandard model of Peano arithmetic. Similarly, an ultraproduct of finite fields ${\mathbb{F}}_{p_i}$ with distinct primes tending to infinity has characteristic zero: for every fixed positive integer $m$, the equation $m1=0$ fails in all sufficiently late factors. An infinite structure can therefore arise from finite factors.

Ultraproducts prove the compactness theorem. If each finite subset $\Delta$ of a theory $T$ has a model $M_{\Delta }$, index these models by $[T{]}^{<\omega }$. The cones $\{\Delta :\phi \in \Delta \}$ have the finite intersection property. Extend their filter to an ultrafilter using the ultrafilter lemma. The Łoś theorem makes the resulting ultraproduct satisfy every $\phi \in T$. This construction also gives elementary extensions realizing prescribed finitely satisfiable types.

A useful stronger property is countable saturation of an ultraproduct over omega. Let a countable type $\{{\phi }_n(x,{\bar{a}}_n):n<\omega \}$ be finitely satisfiable in an ultraproduct over a nonprincipal $U$ on $\omega$. Choose coordinate representatives of its parameters. Let $E_n$ be the coordinates where the first $n$ formulas have a simultaneous witness; then $E_n\in U$, the $E_n$ decrease, and we can replace $E_n$ by $E_n\cap [n,\infty )$. At coordinate $i$, choose a witness to the first $m(i)$ formulas, where $m(i)$ is the largest $n\le i$ with $i\in E_n$, taking an arbitrary value if there is none. For each fixed $n$, the set where $m(i)\ge n$ is $U$-large. Its class realizes the entire type. This proof uses countable incompleteness of $U$, not arbitrary completeness or automatic saturation at larger cardinals.

Finally, complete ultrafilters connect ultraproducts to large cardinals. A countably complete ultrafilter prevents an externally infinite descending membership chain in an ultrapower of a well-founded universe: intersect the countably many coordinate truth sets and obtain an impossible descending chain at one coordinate. A kappa-complete nonprincipal ultrafilter on $\kappa$ consequently supplies the ultrapower embedding associated with a measurable cardinal. Thus the same construction supports both elementary compactness arguments and large-cardinal embeddings, with the completeness hypothesis deciding which well-foundedness and closure properties survive.

A measurable cardinal is an uncountable cardinal $\kappa$ carrying a kappa-complete nonprincipal ultrafilter $U$ on $\kappa$. An elementary embedding $j:M\to N$ preserves all first-order formulas with parameters:For a measurable $\kappa$, the Łoś theorem gives an elementary map from the universe into its ultrapower. Countable completeness makes that ultrapower well founded, so the Mostowski collapse gives a transitive target $M$ and an ultrapower embedding $j:V\to M$. The least moved ordinal, its critical point of an elementary embedding, is $\kappa$: every function into an ordinal below $\kappa$ is constant on a $U$-large set by kappa-completeness, whereas the identity function on $\kappa$ represents an ordinal above every constant ordinal below $\kappa$ and below the image of $\kappa$. Conversely a suitably available embedding with critical point $\kappa$ defines the measure $U=\{X\subseteq \kappa :\kappa \in j(X)\}$.

The identity $V\to V$ is elementary. In ZFC there is no nonidentity elementary class embedding $j:V\to V$, by the Kunen inconsistency theorem. Here the embedding is a class for which restrictions to sets and the critical sequence are available, as in the usual class-theoretic formulation. The axiom of choice is part of this conclusion; it is not a claimed contradiction for the choice-free Reinhardt-cardinal hypothesis.

Here is the stationary-set proof. A nonidentity $j$ has a least moved ordinal $\kappa$; if all ordinals were fixed, induction on rank would fix every set. Put ${\kappa }_0=\kappa$, ${\kappa }_{n+1}=j({\kappa }_n)$ and $\lambda ={\sup }_{n<\omega }{\kappa }_n$. Since $j$ fixes $\omega$ and shifts this sequence, $j(\lambda )=\lambda$. It therefore fixes the regular successor cardinal $\delta ={\lambda }^{+}$. The stationarity of ordinals of prescribed cofinality gives the stationary setUse the stationary partition at a successor cardinal to split $S$ into $\kappa$ disjoint stationary pieces $⟨S_{\alpha }:\alpha <\kappa ⟩$, merging pieces if necessary. Applying $j$ gives a partition $⟨T_{\alpha }:\alpha <j(\kappa )⟩$ of the same $S$ into stationary pieces. The additional piece $T_{\kappa }$ is stationary.

The closure pointsform a club set. For unboundedness, repeatedly close a bound under the values of $j$ below it and take the supremum of the resulting countable sequence; regularity of $\delta$ keeps it below $\delta$. Closure under limits is immediate. Choose $\beta \in C\cap T_{\kappa }$ and an increasing sequence $⟨{\beta }_n:n<\omega ⟩$ cofinal in $\beta$. Thenwhile every ordinal satisfies $j(\beta )\ge \beta$, so $j(\beta )=\beta$. There is an $\alpha <\kappa$ with $\beta \in S_{\alpha }$. Since $j(\alpha )=\alpha$, elementarity gives $\beta =j(\beta )\in j(S_{\alpha })=T_{\alpha }$, contrary to disjointness from $T_{\kappa }$.

An uncountable cardinal $\kappa$ is a measurable cardinal if there is a nonprincipal $\kappa$-complete ultrafilter $U$ on $\kappa$: intersections of fewer than $\kappa$ members of $U$ remain in $U$. Such an ultrafilter is uniform. No singleton belongs to it, and intersecting the complements of fewer than $\kappa$ singletons shows that no set of cardinality below $\kappa$ belongs to it.

Form the ultrapower of the universe by $U$, with classes $[f]$ of functions $f:\kappa \to V$, and $[f]{\in }_U[g]$ exactly when $\{\xi :f(\xi )\in g(\xi )\}\in U$. The Łoś theorem proves that the map taking $x$ to the constant function with value $x$ is elementary. The ultrapower relation is well-founded: from an infinite descending sequence $[f_0]{\ni }_U[f_1]{\ni }_U\cdots$, countable completeness gives a coordinate satisfying all the membership relations, producing an actual infinite descending membership sequence and contradicting foundation. It is also set-like. Predecessors of $[g]$ can be represented by functions whose value at $\xi$ lies in $g(\xi )\cup \{\varnothing \}$; these functions form a set. The Mostowski collapse theorem therefore gives a transitive class $N$ and an elementary embeddingThe usual class-ultrapower notation is understood via set-sized representatives and this set-like collapse.

For $\alpha <\kappa$, any function into $\alpha$ is constant on a $U$-large set: if none of its fewer-than-$\kappa$ fibers belonged to $U$, intersect their complements to get the empty set in $U$. Consequently every member of the ultrapower ordinal represented by the constant $\alpha$ is represented by a constant ordinal below $\alpha$. Induction on $\alpha$ now gives $j(\alpha )=\alpha$.

On the other hand the identity function $d(\xi )=\xi$ represents an ordinal below $j(\kappa )$. For every $\alpha <\kappa$, the tail $\{\xi :\xi >\alpha \}$ belongs to $U$, so the collapsed ordinal $[d]$ is above $j(\alpha )=\alpha$. HenceThis proves nonidentity explicitly. The collapsed identity class is at least $\kappa$; it need not equal $\kappa$ unless an additional normality convention is imposed on $U$.

Conversely, an elementary embedding $j:V\to N$ into a transitive class with critical point $\kappa$ yieldsFor the converse use the usual amenability assumption on the class embedding, so this collection is a set. Since $j(\kappa )>\kappa$, exactly one of $j(X)$ and $j(\kappa \setminus X)$ contains $\kappa$, proving the ultrafilter condition. Singletons are excluded because $j(\alpha )=\alpha <\kappa$. If $\lambda <\kappa$ and $X_i\in U$ for $i<\lambda$, then $j$ fixes $\lambda$ and every index below it, giving $j({\bigcap }_{i<\lambda }{X}_i)={\bigcap }_{i<\lambda }j(X_i)$; the intersection still contains $\kappa$. Thus $U$ is $\kappa$-complete and $\kappa$ is measurable.

In particular the first moved ordinal is a strongly inaccessible cardinal. If $\mathrm{cf}(\kappa )<\kappa$, partition $\kappa$ into fewer than $\kappa$ bounded intervals. Each is too small to lie in $U$, contradicting completeness. Thus $\kappa$ is regular. If $2^{\lambda }\ge \kappa$ for some $\lambda <\kappa$, choose distinct subsets $A_{\xi }\subseteq \lambda$ for $\xi <\kappa$. For each $\eta <\lambda$, choose the $U$-large side of the question $\eta \in A_{\xi }$. Intersect these fewer-than-$\kappa$ large sides. The intersection belongs to $U$, but all its indices have identical $A_{\xi }$, so it contains at most one index, contradicting nonprincipality. Therefore $2^{\lambda }<\kappa$ for every $\lambda <\kappa$, as required.

A measurable cardinal is an uncountable cardinal $\kappa$ carrying a nonprincipal ultrafilter $U$ on $\kappa$ that is $\kappa$-complete: intersections of fewer than $\kappa$ members of $U$ remain in $U$. An elementary embedding $j:M\to N$ preserves truth of every first-order formula, with its parameters carried along by $j$.

The identity is always an elementary embedding of the universe into itself. The substantive question concerns a nonidentity embedding. In ZFC, the Kunen inconsistency theorem givesHere, as usual for class embeddings, restrictions to sets and the recursively defined critical sequence are available. We give the stationary-set proof, including its partition ingredient.

Suppose such a $j$ existed. It has a least moved ordinal $\kappa$, its critical point of an elementary embedding. To see why some ordinal must move, if all ordinals were fixed, induction on rank would fix every set: membership of $z$ in $j(x)$ is equivalent to membership of $z$ in $x$ once every lower-rank $z$ is fixed. A rank segment containing a moved set would then be fixed pointwise, a contradiction. The critical point is an uncountable cardinal. All finite ordinals and $\omega$ are definable and fixed. If $\kappa$ had cardinality $\mu <\kappa$, a bijection $f:\mu \to \kappa$ would be carried to a surjection $j(f):\mu \to j(\kappa )$ whose range still consists exactly of the fixed ordinals below $\kappa$, impossible since $j(\kappa )>\kappa$.

Put ${\kappa }_0=\kappa$, ${\kappa }_{n+1}=j({\kappa }_n)$ and $\lambda ={\sup }_{n<\omega }{\kappa }_n$. Applying $j$ to this Kunen critical sequence shifts it by one, so $j(\lambda )=\lambda$. Elementarity also gives $j({\lambda }^{+})={\lambda }^{+}$; write $\theta ={\lambda }^{+}$.

We need a stationary partition at a successor cardinal. If $S\subseteq {\lambda }^{+}$ is stationary, choose bijections $e_{\beta }:\lambda \to \beta$ for $\lambda \le \beta <{\lambda }^{+}$. For each $\xi <{\lambda }^{+}$, on the stationary tail $\{\beta \in S:\beta >\max (\lambda ,\xi )\}$ the function $\beta \mapsto e_{\beta }^{-1}(\xi )$ is regressive. The Fodor lemma supplies a stationary fiber $E_{\xi }$ with constant value $i_{\xi }<\lambda$. Some value $i$ occurs for ${\lambda }^{+}$ many $\xi$, since a union of $\lambda$ sets each of size at most $\lambda$ cannot have size ${\lambda }^{+}$. For distinct such $\xi$, their fibers are disjoint: $e_{\beta }(i)$ cannot equal two different ordinals. These fibers yield ${\lambda }^{+}$ disjoint stationary subsets; include any leftover points in one cell. Grouping cells also gives a partition into any nonzero number at most ${\lambda }^{+}$.

For completeness, the Fodor lemma used here follows from diagonal intersections of clubs. If a regressive function $f:S\to \theta$ had no stationary fiber, choose a club set $C_{\xi }$ disjoint from each fiber. The diagonal intersection $\{\beta <\theta :\forall \xi <\beta (\beta \in C_{\xi })\}$ is club by regularity and closure. A point $\beta$ in its intersection with $S$ would belong to $C_{f(\beta )}$, contradicting the choice of that club.

The set $S=\{\beta <\theta :\mathrm{cf}(\beta )=\omega \}$ is stationary: inside any club choose a strictly increasing countable sequence, whose supremum still lies below the regular uncountable $\theta$ and belongs to the club. Partition it as $⟨S_{\alpha }:\alpha <\kappa ⟩$. Its image $⟨T_{\alpha }:\alpha <j(\kappa )⟩$ is another stationary partition of the same $S$.

The set $C=\{\beta <\theta :j‘‘\beta \subseteq \beta \}$ is club. Closure is immediate. To get a member above any given ordinal, repeatedly bound the images of all earlier ordinals and take a countable supremum; regularity of $\theta$ keeps each bound below $\theta$. Choose $\beta \in C\cap T_{\kappa }$. Since $\mathrm{cf}(\beta )=\omega$, a countable cofinal sequence in $\beta$ and $j(\omega )=\omega$ give $j(\beta )=\beta$: the image sequence remains below $\beta$ by closure under $j$, and $j(\beta )\ge \beta$.

But $\beta \in S_{\alpha }$ for some $\alpha <\kappa$. Applying $j$ and using $j(\alpha )=\alpha$ gives $\beta =j(\beta )\in T_{\alpha }$, contradicting its membership in the disjoint cell $T_{\kappa }$. This proves the assertion. The use of axiom of choice in the cardinal and stationary-partition arguments matters; the conclusion here is explicitly in ZFC.

A measurable cardinal is an uncountable cardinal $\kappa$ carrying a nonprincipal $\kappa$-complete ultrafilter $U$: intersections of fewer than $\kappa$ members of $U$ remain in $U$. Such an ultrafilter contains no set of size less than $\kappa$. Indeed each singleton is absent, and intersecting fewer than $\kappa$ singleton complements shows that every small set's complement is in $U$. In particular every tail of $\kappa$ is in $U$.

Form the ultrapower of the universe by functions $f:\kappa \to V$, identifying functions agreeing on a member of $U$, and interpreting membership coordinatewise modulo $U$. The Łoś theorem, applied separately to each formula, makes the map $x\mapsto [c_x]$ elementary. Countable completeness makes the ultrapower well-founded: an infinite descending chain $[f_{n+1}]\in [f_n]$ would give $U$-large sets witnessing each step, whose countable intersection is nonempty; a coordinate in that intersection would give an infinite descending membership chain in the original universe. The relation is extensional and set-like, so the Mostowski collapse theorem gives a transitive class $M$ and an elementary ultrapower embedding $j:V\to M$.

For each $\alpha <\kappa$, every function $\kappa \to \alpha$ is constant modulo $U$. If no fibre were in $U$, intersecting the fewer than $\kappa$ fibre complements would give the empty set in $U$. Induction on $\alpha$ therefore shows that the collapsed ultrapower ordinal $j(\alpha )$ has precisely the elements $\beta <\alpha$, soThe identity function $\mathrm{id}:\kappa \to \kappa$ represents an ordinal below $j(\kappa )$, and every constant $\alpha <\kappa$ is below it modulo $U$, since a tail is large. ConsequentlyThus the canonical collapsed embedding is not the identity; its first moved ordinal is $\kappa$. For a general measure the identity-function class need not collapse to exactly $\kappa$; that stronger equality holds for a normal measure. The argument above needs only its lower bound.

Conversely, suppose $j:V\to M$ is elementary, $M$ is transitive, its critical point is $\kappa$, and its derived measure is a set, as for the usual definable or amenable class embeddings. DefineBecause $\kappa <j(\kappa )$, elementary preservation of complements and intersections makes this an ultrafilter. It is nonprincipal since $j(\{\alpha \})=\{\alpha \}$ for $\alpha <\kappa$. For a sequence of $\gamma <\kappa$ large sets, $j$ fixes the index ordinal $\gamma$ and takes the intersection to the intersection of their images; $\kappa$ belongs to every image and hence to the intersection. Thus $U$ is $\kappa$-complete. The set-existence qualification excludes treating an arbitrary external class map as a set parameter without justification.

For context, measurability already implies regularity and the strong-limit property. A cofinal sequence of length below $\kappa$ partitions $\kappa$ into that many bounded, hence small, pieces; completeness and their absent fibres contradict covering all of $\kappa$. If there were $\kappa$ distinct subsets of some $\lambda <\kappa$, for each coordinate choose the $U$-large side deciding its membership bit. The intersection of these $\lambda$ large sets would still be large, but all its indices would label the same subset, so it would have at most one member. This contradiction gives $2^{\lambda }<\kappa$. Thus the first moved ordinal is an uncountable strongly inaccessible cardinal, in addition to being measurable.

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.

Take the least omega-measurable cardinal $\kappa$, with a countably complete ultrafilter $U$ on $\kappa$ that is nonprincipal. We prove that $U$ is $\kappa$-complete. This will give $\boxed{\kappa \text{ is measurable}}$.

Suppose $\mu <\kappa$, $A_i\in U$ for $i<\mu$, and ${\bigcap }_{i<\mu }{A}_i\notin U$. Then $B=\kappa \setminus {\bigcap }_{i<\mu }{A}_i$ belongs to $U$. For $x\in B$ let $g(x)$ be the least $i<\mu$ with $x\notin A_i$. The pushforward ultrafilteris a countably complete ultrafilter on $\mu$; this uses $B\in U$, so $U$ restricts to an ultrafilter on $B$. For each $i$, the fibre $g^{-1}[\{i\}]$ is contained in $\kappa \setminus A_i$, hence is not in $U$. Thus $W$ is nonprincipal. If $\mu$ is finite or countable, countable completeness already rules this out. Otherwise $|\mu |<\kappa$ would be an omega-measurable cardinal, contradicting minimality of $\kappa$.

Therefore every intersection of fewer than $\kappa$ members of $U$ lies in $U$. Since $\kappa$ is uncountable, it is a measurable cardinal. This proves that the least omega-measurable cardinal is measurable, even if the originally supplied cardinal carried only a countably complete ultrafilter.

An uncountable cardinal $\kappa$ is measurable when it carries a nonprincipal ultrafilter $U$ that is $\kappa$-complete. For an inaccessible $\lambda >\kappa$, the cardinal $\kappa$ is 1-strong when there is an elementary embeddinginto a transitive model, with critical point $\kappa$ and $V_{\kappa +1}\subseteq M$.

The fundamental theorem on measurable cardinals constructs from $U$ the well-founded ultrapower and its ultrapower embedding $j:V_{\lambda }\to M$, whose critical point is $\kappa$. The embedding fixes $V_{\kappa }$. If $A\in V_{\kappa +1}$, then $A\subseteq V_{\kappa }$, and elementarity givesBoth $j(A)$ and $V_{\kappa }$ belong to the transitive target, so $A\in M$. Thus $V_{\kappa +1}\subseteq M$, proving that every measurable cardinal is 1-strong.

A property $P$ of $\kappa$ reflects below $\kappa$ whenis unbounded in $\kappa$. The standard elementary-embedding reflection argument starts with any $\gamma <\kappa$. Since $j(\gamma )=\gamma$ and $j(\kappa )>\kappa$, the target model can use $\kappa$ itself as a witness toElementarity then gives a witness $\mu$ with $\gamma <\mu <\kappa$ in the domain.

For a concrete example, the property of being a strongly inaccessible cardinal reflects below $\kappa$. A measurable cardinal is strongly inaccessible, and the assumed inclusion $V_{\kappa +1}\subseteq M$ makes this assertion about $\kappa$ absolute between $V_{\lambda }$ and $M$: both models have all subsets of every ordinal below $\kappa$. Thus $M$ sees that $\kappa$ is strongly inaccessible. Given $\gamma <\kappa$, it therefore satisfiesElementarity supplies a strongly inaccessible $\mu$ between $\gamma$ and $\kappa$ in $V_{\lambda }$. As $\gamma$ was arbitrary, the strongly inaccessible cardinals below $\kappa$ are unbounded.

Yes. Start with the model $M_0$ constructed in part a. If $M_0$ has no internally strongly inaccessible cardinal above $\kappa$, put $M=M_0$. Otherwise let $\delta$ be the least ordinal above $\kappa$ that $M_0$ regards as strongly inaccessible, and putIn the second case $M\models \mathrm{ZFC}$ because $M_0$ regards $\delta$ as inaccessible. The measure witnessing that $\kappa$ is measurable has rank below $\kappa +3<\delta$, so it still belongs to $M$. In both cases $M$ is a transitive set of cardinality $\kappa$, contains $V_{\kappa }$, and has no internally inaccessible ordinal strictly between $\kappa$ and its height.

We verify absoluteness for every ordinal $\alpha \in M$. If $\alpha <\kappa$, then $M$ and $V_{\lambda }$ both contain $V_{\alpha +1}$ and therefore compute all subsets and functions relevant to strong inaccessibility in the same way. At $\alpha =\kappa$, both models see a measurable cardinal and hence an inaccessible cardinal. Finally, if $\kappa <\alpha \in M$, then $M$ says that $\alpha$ is not inaccessible by construction. The larger model $V_{\lambda }$ cannot say that it is inaccessible, because strong inaccessibility is downward absolute to a transitive model of ZFC: any failure visible in the smaller model remains a failure in the larger one, while ambient inaccessibility would force internal inaccessibility. Hence “$\alpha$ is inaccessible” is absolute between $M$ and $V_{\lambda }$.

A filter $U$ is $\kappa$-complete when intersections of fewer than $\kappa$ members remain in $U$. An uncountable cardinal $\kappa$ is a measurable cardinal when it carries a $\kappa$-complete nonprincipal ultrafilter.

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 regular $\kappa$, the cobounded filter on a regular cardinal is $\kappa$-complete. Part (c) extends it to a $\kappa$-complete ultrafilter $U$. Since $\kappa \setminus \{\alpha \}$ is cobounded for every $\alpha <\kappa$, no singleton belongs to $U$; hence $U$ is nonprincipal. Thus every strongly compact cardinal is measurable.