# Questions about Prikry forcing and Cohen forcing

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).

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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. –  Yair Hayut Apr 23 at 11:21
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 $Add(\alpha,1)^V$. –  Yair Hayut Apr 23 at 13:33
Yair, what is the source for the isomorphism argument? –  Victoria Gitman Apr 23 at 18:27
@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. –  Monroe Eskew Apr 23 at 20:44
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. –  Asaf Karagila Apr 25 at 10:19

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.

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Very interesting. I thought you meant that the second Prikry was equivalent to Cohen, but this factors the product in a different way. –  Monroe Eskew Apr 24 at 15:47
Dear Yair, thanks a lot for your very interesting answer. For $\delta=\kappa$ we may factor $\mathbb{P}^\kappa$ as $\mathbb{P}^\delta\times \mathbb{P}^{\kappa-\delta},$ for any $\delta<\kappa,$ and conclude that all cardinals $<\kappa$ are collapsed, so $\kappa$ is collapsed since it is singular in the extension. –  Mohammad Golshani Apr 25 at 13:29
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)? –  Yair Hayut Apr 25 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 $\lambda$.

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