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On a regular uncountable cardinal number $\kappa$, the club filter consists of all subsets containing a club set. It is a kappa-complete filter by club filter completeness. Its members are stationary sets, although a stationary set need not be a filter member, and a filter member need not itself be closed.

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.

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.

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.

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

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

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

The critical point of an elementary embedding $j$ is the least ordinal $\alpha$ for which $j(\alpha )\ne \alpha$.

We first prove by transfinite induction that $j(\alpha )=\alpha$ for every $\alpha <\kappa$. Suppose this is known below $\alpha <\kappa$. If $[f]E[{\mathrm{const}}_{\alpha }]$, thenThe sets $A_{\beta }=\{\xi \in A:f(\xi )=\beta \}$ for $\beta <\alpha$ partition $A$ into fewer than $\kappa$ pieces. A kappa-complete filter that is an ultrafilter must contain one cell $A_{\beta }$: otherwise all their complements would belong to $U$, and their intersection would contradict $A\in U$. Hence $[f]=[{\mathrm{const}}_{\beta }]$. The predecessors of $j(\alpha )$ are consequently exactly the already-fixed ordinals below $\alpha$, so $j(\alpha )=\alpha$.

Now let $\mathrm{id}(\xi )=\xi$. An identical argument shows that the predecessors of $[\mathrm{id}]$ in the ultrapower are exactly $[{\mathrm{const}}_{\beta }]$ for $\beta <\kappa$, soSince $\{\xi <\kappa :\xi <\kappa \}=\kappa \in U$, one has $[\mathrm{id}]E[{\mathrm{const}}_{\kappa }]$. After collapsing, $\kappa \in j(\kappa )$, and therefore $j(\kappa )>\kappa$. Thus $\mathrm{crit}(j)=\kappa$.

Since the strongly inaccessible cardinal $\lambda$ is inaccessible, $(V_{\lambda },\in )$ is a model of ZFC. The Downward Lowenheim-Skolem theorem gives an elementary substructureof cardinality $\kappa$ such thatOne may obtain $X$ concretely as the Skolem hull of this set; its cardinality remains $\kappa$ because the language of set theory is countable and $|V_{\kappa }|=\kappa$.

Apply the Mostowski collapse theorem to $X$ and write $\pi :X\to M$ for the collapse. Then $M$ is a transitive set, $|M|=\kappa$, and $\pi$ fixes $V_{\kappa }$ pointwise. It also fixes $\kappa$, because it fixes every ordinal below $\kappa$. By elementarity, $X$ satisfies ZFC and regards $U$ as a kappa-complete filter that is a nonprincipal ultrafilter on $\kappa$. Therefore, with $\bar{U}=\pi (U)$,The internal ultrafilter $\bar{U}$ need not equal the original $U$.

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.