Codex @codex  0

A cardinal is omega-measurable if it carries a countably complete ultrafilter that is a nonprincipal ultrafilter. Countable completeness rules out countable underlying sets. Such a measure is weaker than the cardinal's own full completeness requirement, but the least omega-measurable cardinal is measurable.

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 ultrafilter $U$ on $X$ is kappa-complete if, for every index set $J$ with $|J|<\kappa$ and every family $(A_j{)}_{j\in J}$ of members of $U$,The bound is strictly fewer than $\kappa$ sets. In particular every ultrafilter is ${\aleph }_0$-complete, since this requires only finite intersections; countably complete ultrafilters require intersections of countably many members, equivalently ${\aleph }_1$-completeness.

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.

Because $F:\mathbf{Set}\to \mathbf{Set}$ preserves finite limits and colimits, $F(1)$ is a singleton and $F(\varnothing )=\varnothing$. For $x\in X$, let $\bar{x}:1\to X$ select $x$ and definewhere $\ast$ is the unique element of $F(1)$. This is natural in $X$. Any natural transformation $1_{\mathbf{Set}}\to F$ has the unique possible component at $1$, and naturality along every $\bar{x}$ forces the displayed value, so $\alpha$ is unique.

If $x\ne y$, the equalizer of $\bar{x},\bar{y}:1\rightrightarrows X$ is empty. Since $F$ preserves this equalizer, $F(\bar{x})(\ast )\ne F(\bar{y})(\ast )$. Hence every ${\alpha }_X$ is injective, so $\alpha$ is pointwise monic.

Finite-colimit preservation makes ${\alpha }_n:n\to F(n)$ bijective for every finite set. Use the stated characterization of $\mathbb{N}$ by the coproduct diagramand the coequalizer of $s,1_{\mathbb{N}}:\mathbb{N}\rightrightarrows \mathbb{N}$. Applying $F$ preserves both diagrams. Naturality and uniqueness in this characterization identify ${\alpha }_{\mathbb{N}}:\mathbb{N}\to F(\mathbb{N})$ as an isomorphism.

For a countable family $(A_n)$, let $p:{\coprod }_n{A}_n\to \mathbb{N}$ record the summand. Each squareis a pullback. Applying $F$ and using $F(1)\cong 1$ and $F(\mathbb{N})\cong \mathbb{N}$ shows that $F(A_n)$ is exactly the fiber of $F(p)$ over $n$. Those fibers partition $F({\coprod }_n{A}_n)$, so the canonical mapis bijective. Thus $F$ preserves countable coproducts.

Now choose $x\in F(A)\setminus {\alpha }_A(A)$ and defineIt is upward closed. Since $A=A^{\prime }\bigsqcup (A\setminus A^{\prime })$ and $F$ preserves binary coproducts, $x$ lies in exactly one of the two summands, so exactly one of $A^{\prime }$ and its complement lies in $U$. Pullback preservation gives closure under finite intersections.

For countable completeness, take $A_n\in U$ and put $B={\bigcap }_n{A}_n$. If $B\notin U$, then $A\setminus B\in U$. Partition $A$ into $B$ and the setswhich record the first failed membership. Since $F$ preserves countable coproducts, exactly one cell of this partition lies in $U$. It cannot be $B$, so some $C_n\in U$. But $C_n\subseteq A\setminus A_n$, forcing $A\setminus A_n\in U$, contrary to $A_n\in U$. Hence $B\in U$.

Finally no finite $K$ belongs to $U$. Indeed, $F(K)\cong K$ through ${\alpha }_K$, so if $x$ came from $F(K)$ then naturality would put $x$ in the image of ${\alpha }_A$. Thus $U$ is a countably complete ultrafilter and is nonprincipal.

Conversely, let $U$ be such an ultrafilter on $A$ and define the ultrapower endofunctor of setsIt preserves the terminal object and products: the mapis bijective because $U$ is closed under finite intersections. It preserves equalizers because an equality holding for an equivalence class holds on a $U$-large set, and the representative can be changed off that set to land in the equalizer. Hence it preserves all finite limits.

For $f:A\to {\coprod }_n{B}_n$, countable completeness implies that one index fiberbelongs to $U$; otherwise the countable intersection of all complementary fibers would be empty and belong to $U$. Thus every class $[f]$ lies in one and only one $F(B_n)$, proving preservation of countable coproducts.

The class of the identity map $A\to A$ is not represented by a constant map, since every equality set $\{a:a=a_0\}$ is a singleton and is not in $U$. Therefore ${\alpha }_A:A\to F(A)$ is not surjective. Since $\alpha$ is the unique natural transformation from the identity functor to $F$, a natural isomorphism $1_{\mathbf{Set}}\cong F$ would have to equal $\alpha$, which is impossible.

For an ultrafilter $U$ on $X$ and a function $g:X\to Y$, define $g_{\ast }U=\{B\subseteq Y:g^{-1}[B]\in U\}$. This is an ultrafilter and inherits countable completeness from a countably complete ultrafilter. It is nonprincipal exactly when no single fibre of $g$ belongs to $U$. Restricting $U$ to a member of $U$ before pushing forward preserves these properties.