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?


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.

  • $\begingroup$ You might be interested in looking at mathoverflow.net/questions/13795 for an example of how seed theory is used to analyze the nature of an ultrapower embedding. $\endgroup$ – Joel David Hamkins Mar 23 '10 at 19:40
  • $\begingroup$ Dear Prof. Hamkins, Thanks a lot for the answer and sorry for my late feedback. Could you please elaborate a little bit more on why the desired property fails when the ultrafilter is not normal? Thanks again. $\endgroup$ – kvagk Mar 26 '10 at 17:11
  • $\begingroup$ I added some further explanation. It's not correct that it fails when the ultrafilter is not normal, but rather, it fails when the ultrafilter is not isomorphic to any normal ultrafilter. This is what the second theorem says. $\endgroup$ – Joel David Hamkins Mar 26 '10 at 17:42

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