Measurable cardinal

Remarks

The cardinal $\aleph_0$ satisfies clause 2: every ultrafilter is closed under finite intersections, and nonprincipal ultrafilters on $\omega$ exist by the axiom of choice; it fails only clause 1. The cardinal $\aleph_1$ fails clause 2, because every measurable cardinal is inaccessible, and $\aleph_1$ is not. Equivalent form: $\kappa$ is measurable if and only if there is an elementary embedding $j \colon V \to M$, with $M$ a transitive class, whose critical point is $\kappa$. The critical point condition means that $j(\alpha) = \alpha$ for $\alpha < \kappa$ and $j(\kappa) > \kappa$. The embedding comes from the ultrapower of $V$ by $U$, and conversely $U = \set{X \subseteq \kappa : \kappa \in j(X)}$. Scott proved that if a measurable cardinal exists, then $V \neq L$.