Measurability, normal measures and elementary embeddings

Suppose j has critical point kappa. Its ordinal map is increasing, fixes every alpha<kappa and has j(kappa)>kappa; since M is transitive and contains j(kappa), it contains kappa. Define $W=\{X\subseteq \kappa :\kappa \in j(X)\}$. F2 and Separation make W a set. Images of kappa and empty show it proper. Elementarity for complements and finite intersections, evaluated at kappa, proves the ultrafilter laws in F3. For singleton {alpha}, j({alpha})={alpha}, which omits kappa, so W is nonprincipal. If eta<kappa and every X_xi for xi<eta belongs to W, j fixes eta and the value of the image sequence at xi is j(X_xi). Thus kappa lies in the intersection of that image sequence, which is j of the original intersection. This proves kappa-completeness, including eta=0.

If f is regressive on S in W with zero omitted, then kappa belongs to j(S) and $j(f)(\kappa )<\kappa$. Let this ordinal be beta; j fixes beta. Elementarity applied to the beta-fibre says kappa belongs to $j(\{\alpha \in S:f(\alpha )=\beta \})$. That fibre therefore belongs to W. So W is normal. A measurable kappa gives the embedding by F1 and hence a normal W by this construction; a normal measure is itself a measure, and also gives the embedding by F1. This proves all three equivalences without asserting that the original measure was already normal. F4 is inherited in F1.

Let d be the collapsed identity class for U. By F1 it is an ordinal at least kappa. If U is normal, any predecessor [f] of the identity class has, on a U-large set omitting zero, ordinal values f(alpha)<alpha. Normality makes f constant there, so its collapsed class is an ordinal beta<kappa. Every beta<kappa is already below d by F1. Thus d=kappa. Conversely suppose d=kappa and f is regressive on a U-large S omitting zero. Extend f by zero outside S. Its collapsed class belongs to d=kappa, so equals beta for some beta<kappa. F1 identifies beta with the collapsed constant-beta class. Injectivity of the collapse and Scott equality give a U-large equality fibre; intersecting it with S proves normality.

For normal U and A_xi in U for xi<kappa, let $D=\{\alpha <\kappa :(\mathrm{\forall }\xi <\alpha )\alpha \in A_{\xi }\}$. If its complement were U-large, omit zero and assign to alpha the least failed xi<alpha. Normality gives a U-large constant fibre beta, disjoint from A_beta, contrary to properness. Hence D is in U. Conversely assume diagonal closure and let f be regressive on S in U, zero omitted. If no fibre were U-large, all fibre complements A_xi would belong to U. Their diagonal intersection D belongs to U, but every alpha in S fails its membership requirement at xi=f(alpha)<alpha. Thus S and D are disjoint U-members, a contradiction. This proves the stated compatibility of normality conventions.