Representation of elements in ultrapowers

Question

Maybe this is an elementary question.

Suppose that $U$ is a non-principal $\kappa$-complete ultrafilter on $\kappa$ and consider the standard ultrapower $M\cong \textrm{Ult}_U(V)$ along with the corresponding elementary embedding $j_U : V \rightarrow M$\,.

We know that if $U$ is in addition normal, then every element $y\in M$ can be written in the form $j(f)(\kappa)$ (more precisely, if $y=[f]_U \in M$ then $y=j(f)(\kappa)$, which follows from the fact that $[id]_U=\kappa$).

Of course, we can always pick $U$ to be normal, but the question is what happens in the case that we don't. More precisely:

Assume that in $M$, $[g]=\kappa<[id]_U$, for some $g:\kappa\rightarrow \kappa$. Does it follow that for every $y\in M$, $y=j(f)(\kappa)$ for some function $f$ on $\kappa$?

My guess is that the desired would follow if can pick the function $g$ that represents $\kappa$ to be 1-1 and such that $\textrm{range}(g)\in U$. Is this true? Can we do this?

Answer 1

What you get is that if j:V to M is the ultrapower by any ultrafilter U on any set X, then every element of M has the form j(f)([id]). You can prove this by building an isomorphism from the ultrapower to the sets of this form. This way of thinking is also known as "seed theory".

Theorem. Suppose that j:V to M is an elementary embedding of the universe V into M. Then j is the ultrapower map by a measure on a set if and only if there is some s in M such that every element of M has the form j(f)(s).

That is, the ultrapower embeddings are precisely the embeddings whose target is generated by a single element.

Proof. If j is the ultrapower by U on X, then let s=[id], and argue that [f]_U is j(f)(s). Conversely, if M = { j(f)(s) | f in V }, where we assume that f is a function on some set X such that a ∈ j(X), then define the measure U by A ∈ U iff s ∈ j(A). This is a κ-complete ultrafilter on P(X). One can show that Ult(V,U) is isomorphic to M, by mapping [f]_U to j(f)(s). QED

Theorem. If U is a κ complete ultrafilter on κ with ultrapower embedding j:V to M, then every element of M has the form j(f)(κ) if and only if U is isomorphic to a normal measure.

Proof. You know the backwards implication. For the forward implication, suppose that every element of M has form j(f)(κ). In particular, β = j(f)(κ), where β = [id]_U is the seed for U. But also, κ = j(g)(β), where g(α) is the smallest ξ for which f(ξ)=α. Let μ = { X subset κ | κ ∈ j(X) } be the induced normal measure. Note that X in μ iff j(g)(β) in j(X) iff β in j(g^-1X) iff g^-1X in U. So μ is Rudin-Kiesler below U. Also, U is Rudin-Kiesler below μ since X in U iff f^-1X in μ. So μ and U are isomorphic.QED

One may illustrate the situation with product measures. Suppose that U is normal. The product measure UxU is isomorphic to the two-step iteration, where j_0:V to M is the ultrapower by U, and h:M to N is the ultrapower in M by j_0(U). Every element of M has the form j_0(f)(κ), and every element of N has the form h(g)(κ₁), where κ₁ = j_0(κ). If j is the composition of j_0 and h, then j:V to N and every element of N has the form j(f)(κ, κ₁). If one only looks at j(f)(κ) inside N, then you will only get ran(h), which is isomorphic to M, but not all of N. So this would be a counterexample to what you asked about.