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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 $T_{\kappa }$.

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.