Definition: Let $V\subseteq W$ be two transitive models of $ZFC$. A pseudo-Prikry sequence, $s$, at a cardinal $\kappa$ for $(V, W)$ is an $\omega$-sequence, cofinal at $\kappa$ such that for every club $D\subseteq \kappa$ from $V$, $s\setminus D$ is finite.

In the paper "On squares, outside guessing of clubs and $I_{<f}[λ]$" by Shelah and Dzamonja and independently in the paper "Some results on the nonstationary ideal II" by Gitik, stronger versions of the following theorem are proved:

Theorem: if $V\models \kappa$ is inaccessible, and $2^\kappa = \kappa^+$ and $W\models \kappa^+ = (\kappa^+)^V,\,\text{cf }\kappa=\omega$ then there is a pseudo-Prikry sequence at $\kappa$ in $W$.

I wonder how far can a pseudo-Prikry sequence get from just being the regular Prikry sequence.

For simplicity, let's assume that $\kappa$ is measurable in $V$, $\mathcal{U}$ is a normal measure on $\kappa$ and $W = V[G]$ where $G$ is a generic filter for the Prikry forcing for $\kappa$, using the measure $\mathcal{U}$.

Question: Does $W$ contain a pseudo-Prikry sequence which is not a $\mathcal{U}$-Prikry sequence?


1 Answer 1


The answer is no, there is no such pseudo-Prikry sequence in the Prikry extension.

To see this, let's first make some observations.

Lemma 1. If $j:V\to M$ is the ultrapower by a normal measure $\mu$ on $\kappa$, then $$\bigcap\{\ j(C)\mid C\subset\kappa\text{ club }\}=\{\kappa\}.$$

Proof. Certainly $\kappa\in j(C)$ for any club $C\subset\kappa$, and by considering final segments of $\kappa$, it is clear that the intersection is contained in $[\kappa,j(\kappa))$. So consider $\kappa<\alpha<j(\kappa)$. Since $\mu$ is normal, it follows that $\alpha=j(f)(\kappa)$ for some function $f:\kappa\to\kappa$. Let $C_f\subset\kappa$ be the club of ordinals $\beta$ with $f''\beta\subset\beta$. It follows that $\alpha\notin j(C_f)$, and so any particular $\alpha$ other than $\kappa$ is cast out of that intersection. QED

Now consider the $\omega$-iteration of $\mu$ $$V\to M_1\to M_2\to\cdots\to M_\omega$$

Lemma 2. Similarly, for the $\omega$-iteration embedding $j_\omega:V\to M_\omega$, we have $$\bigcap \{\ j_\omega(C)\mid C\subset\kappa\text{ club in }V\ \}=\{\ \kappa_n\mid n\in\omega\ \}.$$

Proof. For any particular club $C\subset\kappa$, since it is in $\mu$, we have that $\kappa_n\in j_\omega(C)$ for every $n$. Now suppose that $\kappa_n<\alpha<\kappa_{n+1}$ for some $n$. By standard representation results, we know that $\alpha=j_\omega(f)(\kappa_0,\ldots,\kappa_n)$ for some function $f:\kappa^{n+1}\to \kappa$. Let $C_f$ be the set of $\beta<\kappa$ such that $f''\beta^{n+1}\subset\beta$. This is club in $\kappa$, but by design, $\alpha\notin j_\omega(C_f)$. So every ordinal not on the critical sequence is cast out of the intersection. QED

It is a standard fact that the critical sequence $s=\langle\kappa_n\mid n\in\omega\rangle$ is $M_\omega$-generic for Prikry forcing with $\mu_\omega=j_\omega(\mu)$. Suppose that $t=\langle\delta_n\mid n\in\omega\rangle$ is a pseudo-Prikry sequence in $M_\omega[s]$. In particular, if $C\subset\kappa$ is club in $V$, then $j(C)$ contains a tail of the $\delta_n$. But for any particular ordinal $\alpha$ not on the critical sequence, there is a club $C\subset\kappa$ for which $\alpha\notin j_\omega(C)$. Thus, if infinitely many $\delta_n$ are not on the critical sequence, we can intersect the corresponding clubs to find a single club $C\subset\kappa$ such that $j_\omega(C)$ does not contain a tail of $t$, which would contradict our assumption that $t$ is a pseudo-Prikry sequence. Thus, $t$ must eventually be contained in the critical sequence. It follows that $t$ is a Prikry sequence over $M_\omega$, since any almost-subsequence of a Prikry sequence is a Prikry sequence.

Thus, every pseudo-Prikry sequence in $M_\omega[s]$ is actually a Prikry sequence, and then by elementarity, the same holds in the Prikry forcing extension of $V$.

  • $\begingroup$ Thank you! I think that your arguments generalize to a rather large class of forcing notions for changing cofinalities such as the supercompact Prikry forcing and the Radin forcing. $\endgroup$
    – Yair Hayut
    Feb 5, 2015 at 14:25
  • $\begingroup$ It was my pleasure; it was a very interesting question. $\endgroup$ Feb 5, 2015 at 14:27
  • $\begingroup$ @JoelDavidHamkins A nice question and a nice answer, I enjoyed the answer, thanks. $\endgroup$ Feb 8, 2015 at 5:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .