Measurable cardinal

An uncountable cardinal $\kappa$ is measurable when it carries a nonprincipal $\kappa$-complete ultrafilter.

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.

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.

Let $U$ be a nonprincipal $\kappa$-complete ultrafilter on $\kappa$. If $\lambda <\kappa$ and $(A_{\alpha }{)}_{\alpha <\kappa }$ were distinct subsets of $\lambda$, then for every $\xi <\lambda$ choose the $U$-large side of the partition according to whether $\xi \in A_{\alpha }$. Their intersection is $U$-large by $\kappa$-completeness, but all its indices label the same subset of $\lambda$, so it has at most one member, contradicting nonprincipality. Thus $2^{\lambda }<\kappa$.

The ultrapower embedding $j:V_{\lambda }\to M$ associated with a measure on $\kappa <\lambda$ has critical point $\kappa$. For every $A\in V_{\kappa +1}$,so $V_{\kappa +1}\subseteq M$. Hence every measurable cardinal is 1-strong.

Let ${\kappa }_0<{\kappa }_1<\lambda$, where both ${\kappa }_i$ are measurable and $\lambda$ is inaccessible, and let $j_i:V_{\lambda }\to M_i$ be ultrapower embeddings by measures on ${\kappa }_i$. ThenThe second equality follows from the critical point. For the first, regularity gives continuity of $j_0$ at ${\kappa }_1$, while strong-limitness bounds $j_0(\beta )<{\kappa }_1$ for every $\beta <{\kappa }_1$.

For a measure ultrapower $j:V_{\lambda }\to M$ with critical point $\kappa$, there are at most $2^{\kappa }$ equivalence classes of functions $\kappa \to \kappa$, so $|j(\kappa )|\le 2^{\kappa }$ in $V_{\lambda }$. The target $M$ regards $j(\kappa )$ as measurable and contains $V_{\kappa +1}$, so it sees $(2^{\kappa }{)}^M<j(\kappa )$. Under the Generalized continuum hypothesis, this places $j(\kappa )$ strictly above the ambient ${\kappa }^{+}$ while its ambient cardinality is at most ${\kappa }^{+}$; hence it is not an ambient cardinal.