I have some questions.

The first one is about the product of Prikry's forcing. Let $\kappa$ be a measurable cardinal, $U_1, U_2$ be normal measures on $\kappa$ and let $\mathbb{P}_{U_1}, \mathbb{P}_{U_2}$ be the corresponding Prikry forcings. Let $G\times H$ be $\mathbb{P}_{U_1}\times \mathbb{P}_{U_2}-$generic over $V$. It is easily seen that in the extension there are new subsets of $\omega$ (for example if $(x_n: n<\omega), (y_n: n<\omega)$ are the Prikry sequences added by $G, H$ respectively, then $\{ n<\omega: x_n < y_n\}$ is such a set).

Question 1.1 Is $\kappa$ preserved in the extension $V[G\times H]$? Do $V$ and $V[G\times H]$ have the same cardinals?

Remark. Though the question remained unanswered in general, but by Yair Hayut's very nice answer, given a normal measure $U$ on $\kappa, \mathbb{P}_U^2$ does not collapse cardinals. His proof extends easily to show that for any natural number $n>1, \mathbb{P}_U^n$ is forcing isomorphism to $\mathbb{P}_U\times \mathbb{C},$ hence it preserves cardinals.

Question 1.2. What is the least cardinal $\lambda$ such that $\mathbb{P}_U^\lambda$ collapses some cardinals? What is the least cardinal $\delta$ such that $\mathbb{P}_U^\delta$ collapses $\kappa$? Are $\lambda$ and $\delta$ equal?

My second question is about Cohen forcing. Let $\kappa$ be a Mahlo cardinal, let $\mathbb{P}$ be the reverse Easton iteration of $Add(\alpha,1)$ for all inaccessible cardinals $\alpha\leq \kappa,$ and let $G$ be $\mathbb{P}-$generic over $V$.

Question 2. Suppose $\alpha$ is an inaccessible cardinal $\leq \kappa.$ Is there an $H\in V[G]$ which is $Add(\alpha,1)^V-$generic over $V$? (of course the answer is yest for the least inaccessible).

  • 2
    $\begingroup$ When $U_1 = U_2$, this product is isomorphic to the product of Prikry forcing and Cohen forcing (on $\omega$) so it doesn't collapse cardinals. $\endgroup$
    – Yair Hayut
    Apr 23, 2014 at 11:21
  • 3
    $\begingroup$ For the second question - I think that you can use the arguments of Hamkin's Gap Forcing in order to show that for isolated inaccessibles this iteration doesn't add a generic for $Add(\alpha,1)^V$. $\endgroup$
    – Yair Hayut
    Apr 23, 2014 at 13:33
  • 1
    $\begingroup$ Yair, what is the source for the isomorphism argument? $\endgroup$ Apr 23, 2014 at 18:27
  • 1
    $\begingroup$ @YairHayut, I don't think your claim about the product of Prikry forcing is right. Assume GCH holds at this measurable $\kappa$. One can show that Prikry forcing with a normal ultrafilter is separative and has uniform density $\kappa^+$. Using this we can find a dense subset $D$ such that for all $p \in D$, $|\{ q \in D : p \leq q \}| \leq \kappa$. So if your claim were correct, then after the first forcing, $\kappa^+$ would be a countable union of sets of size $\kappa$, which is false as $\kappa^+$ is preserved. $\endgroup$ Apr 23, 2014 at 20:44
  • 1
    $\begingroup$ I think that you all owe me a big thank you for forcing Yair to join the site! It seems to have paid quite handsomely so far. $\endgroup$
    – Asaf Karagila
    Apr 25, 2014 at 10:19

2 Answers 2


This should be a comment - but it is too long:

Assume that $U = U_1 = U_2$. I want to show that $\mathbb{P}_U ^2 \cong \mathbb{P}_U\times \mathbb{C}$ where $\mathbb{C}$ is the Cohen forcing.

Let $\{ {\alpha^0}_i\}_{i <\omega}, \{ {\alpha^1}_i \}_{i<\omega}$ be the two Prikry sequences.

Set $\{ \gamma_n \}_{n<\omega} = \{ {\alpha^0}_i\}_{i <\omega} \cup \{ {\alpha^1}_i \}_{i<\omega}$ (the Prikry sequence) and $f:\omega \rightarrow P(2)\setminus \{\emptyset\}$, $f(n) = \{ i < 2 | \gamma_n \in \{ {\alpha^i}_m \}_{m < \omega} \}$ (the Cohen real).

This gives us an isomorphism: send conditions from the dense set $( \langle s_0, A\rangle , \langle s_1, A\rangle )$ (the same $A$, $\min A > \max s_0 \cup s_1$) in $\mathbb{P}_U^2$ to $(\langle s_0\cup s_1, A \rangle, f\restriction |s_0 \cup s_1|) \in \mathbb{P}_U \times \mathbb{C}$ (we can calculate $f\restriction |s_0 \cup s_1|$ since we can only add elements in the Prikry sequence above the $\max s_0 \cup s_1$). This is an order preserving bijection between those two posets.

Edit: The answer for Question 1.2 depends on the exact support that you're using:

For finite support - $\mathbb{P}^\omega$ trivially collapses $\kappa$ to $\omega$ - the function that assign to each $n$ the first element in the $n$-th Prikry sequence is onto $\kappa$.

For full support - $\mathbb{P}^\omega$ also collapses $\kappa$: in the generic extension there is a surjection from $(2^{\aleph_0})^V$ to $\kappa$.

Choose a $\omega$-Jonsson function on $\kappa$, i.e. function $f:[\kappa]^\omega \rightarrow \kappa$, such that for every $x\subset \kappa, |x|=\kappa$, $f^{\prime\prime}([x]^\omega) = \kappa$. Let $\{\alpha^j_i \}_{i<\omega}$ be the Prikry sequence added by the $j$ component of $\mathbb{P}^\omega$.

We define a function $g: (\omega^\omega)^V \rightarrow \kappa$ by $g(z) = f(\{\alpha_{z(n)}^n | n < \omega\})$ .

I claim that $g$ is surjective: Let $p = \langle g_i, A_i | i < \omega \rangle \in \mathbb{P}^\omega$ and $\alpha < \kappa$. WLOG, $A=A_i$ for every $i$. Choose $y\in A^\omega$ such that $f(y) = \alpha$ and extends each $g_i$ by the corresponding element of $y$. Since the sequence of lengths of $g_i$ is real from $V$ - the new condition forces $\alpha \in \text{im }g$.

Since we're dealing with Prikry forcing there is at least one more support that we should consider - the Magidor support, namely the conditions are all elements of the form $\langle g_i, A_i |i < \delta\rangle$ such that $\{i<\delta | g_i \neq \emptyset \}$ is finite. In this case (as long as $\delta < \kappa$), we can apply the same idea as above and get that $\mathbb{P}^\delta \cong \mathbb{P}\times \mathbb{D}$ where $\mathbb{D}$ is a forcing that adds a generic function from $\omega$ to the set of all finite, non empty subsets of $\delta$, so as long as $\delta < \kappa$ - $\kappa$ is not collapsed.

When $\delta = \kappa$ this argument doesn't work, so I don't know if $\kappa$ is collapsed by the Magidor power $\mathbb{P}^\kappa$ or not.


Remark 1. The answer to question 1.1 is yes, even if $U_1\neq U_2.$

Theorem. Let $U,V$ be normal measures on $\kappa.$ Then forcing with $\mathbb{P}_U\times \mathbb{P}_V$ preserves all cardinals.

Proof. It suffices to consider the case where $U$ is not equal to $V$. So let $A^*\in U$ such that $\kappa-A^*\in V.$ Let $W=\{ X\subset\kappa: X\cap A^*\in U, X\cap (\kappa-A^*)\in V \}.$ It is easily seen that $W$ is $\kappa-$complete filter on $\kappa$ which is Rowbottom (for any $f: [D]^{<\omega}\to \lambda<\kappa, D\in W$, there is $E\in W, E\subset D$ such that $card(f'' [E]^{<\omega}) \leq \omega$ ). So we can define $\mathbb{P}_W$ and by Devlin's paper "Some Remarks on Changing Cofinalities"1, forcing with $\mathbb{P}_W$ preserves cardinals. As above argument, we have a forcing isomorphism from the dense subset $\{((s, A)(t, B))\in \mathbb{P}_U\times \mathbb{P}_V: A\subset A^*, B\subset \kappa-A^*, max(s\cup t)< min(A), min(B) \}$ of $\mathbb{P}_U\times \mathbb{P}_V$ to $\mathbb{P}_W\times \mathbb{C}$ given by $((s, A)(t, B))\to ((s\cup t, A\cup B), f\restriction |s\cup t|),$ where $f$ is defined as above argument. The result follows.

  1. Keith J. Devlin, Some remarks on changing cofinalities, J. Symbolic Logic 39 (1974), 27--30.

Remark 2. $\kappa$ is collapsed by the Magidor power $\mathbb{P}^\kappa$, by the following argument:

For any $\delta<\kappa,$ we may factor $\mathbb{P}^\kappa$ as $\mathbb{P}^\delta \times \mathbb{P}^{\kappa -\delta}$, so by the above argumets we can conclude that all cardinals $<\kappa$ are collapsed, so $\kappa$ is collapsed since it is singular in the extension.

  • $\begingroup$ Very interesting. I thought you meant that the second Prikry was equivalent to Cohen, but this factors the product in a different way. $\endgroup$ Apr 24, 2014 at 15:47
  • 1
    $\begingroup$ You're right. I wonder - can we show that in the full support case, $\kappa$ is collapsed to $\omega$ (and not only to the continuum)? $\endgroup$
    – Yair Hayut
    Apr 25, 2014 at 13:47

Concerning question 2, Yair Hayut's comment is exactly right, and this kind of situation is just the kind of situation that I had aimed to analyze with those results.

The basic fact is that if one performs forcing $\mathbb{P}_0$ of size $\delta$ followed by nontrivial strategically $\leq\delta$-closed forcing $\mathbb{Q}$, then the extension $V\subset V[g][H]$ has the $\delta^+$-approximation property, which means that any set $A\in V[g][H]$ with $A\subset V$ and $A\cap a\in V$ whenever $a$ has size at most $\delta$ in $V$, then $A$ is already in $V$. Thus, there can be in the extension no fresh subsets of a larger regular cardinal $\alpha$, where $A\subset\alpha$ is fresh over $V$ if $A\notin V$ but all initial segments of $A$ below $\alpha$ are in $V$.

Your iteration admits a closure point at the least inaccessible, and so it can add no fresh subsets above the least inaccessible. In particular, it adds no $V$-generic Cohen sets using $\text{Add}(\alpha,1)^V$, since this forcing would add a fresh subset of $\alpha$.

You can find details in my paper Joel David Hamkins, Extensions with the approximation and cover properties have no new large cardinals, Fund. Math. 180 (2003), no. 3, 257--277. see also my blog.

One can use this idea to show that if you add a Cohen subset to $\kappa$ and then to $\lambda>\kappa$, then you kill all supercompact cardinals between $\kappa$ and $\lambda$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.