Does this mixed-support product have the countable approximation property?

Question

Recall that a forcing order $\mathbb{P}$ has the countable approximation property if for any $\mathbb{P}$-generic filter $G$ and any $x\in V[G]$, if $x\cap y\in V$ for any countable $y\in V$, $x\in V$. In other words, $\mathbb{P}$ adds no fresh sequences.

It is a classical fact that the iteration $\operatorname{Add}(\omega)*\dot{\operatorname{Add}}(\omega_1)$ (and in fact any iteration of an $\omega_1$-square-cc. and a countably closed forcing) has the countable approximation property. Moreover, the same can be shown for any finite product $\prod_{k<n}(\operatorname{Add}(\omega)*\dot{\operatorname{Add}}(\omega_1))$. However, this is not clear when taking infinite products. In this case, to guarantee the preservation of $\omega_1$, we use the mixed support product: Let $\mathbb{P}$ consist of pairs $(p,q)$ of functions on $\omega$ such that for all $n\in\omega$, $(p(n),q(n))\in\operatorname{Add}(\omega)*\dot{\operatorname{Add}}(\omega_1)$ and for cofinitely many $n\in\omega$, $p(n)=\emptyset$. This forcing can be projected onto from the product of an $\omega_1$-Knaster and a countably closed forcing, so it preserves $\omega_1$.

Question: Does $\mathbb{P}$ have the countable approximation property?

Edit: Here is a sketch for why finite products of $\operatorname{Add}(\omega)*\operatorname{Add}(\omega_1)$ have the approximation property. This uses the following Theorem from my paper https://arxiv.org/abs/2401.13446 (which is in its proof heavily based on proofs of Unger and Mitchell):

Theorem: Let $(\mathbb{P}\times\mathbb{Q},R)$ be an iteration-like partial order such that $(\mathbb{P},b(R))$ is square-$\omega_1$-cc. and $(\mathbb{P}\times\mathbb{Q},t(R)))$ is countably closed. Then $(\mathbb{P}\times\mathbb{Q},R)$ has the countable approximation property.

Here we work with partial orders of the form $(\mathbb{P}\times\mathbb{Q},R)$ which are not necessarily product orderings (e.g. iterations $\mathbb{P}*\dot{\mathbb{Q}}$ can be viewed as orderings on the product $\mathbb{P}\times\dot{\mathbb{Q}}^{\mathbb{P}}$, where $\dot{\mathbb{Q}}^{\mathbb{P}}$ is the set of all rank-minimal $\mathbb{P}$-names for elements of $\dot{\mathbb{Q}}$). The necessary definitions are in the same paper. The most important property (and the one which distinguishes the finite case from the infinite case) is the ``mixing property'' (included in being strongly iteration-like) which states that given any $(p,q_0),(p,q_1)R(p,q)$ there are $p_0,p_1$ and $q'$ with $(p_0,q')R(p,q_0)$, $(p_1,q')R(p,q_1)$ and $(p,q')R(p,q)$ (this also occurs in Mitchells paper "On the Hamkins Approximation Property")

We view the finite product as an ordering on $\prod_{k<n}\operatorname{Add}(\omega)\times\prod_{k<n}\dot{\operatorname{Add}(\omega_1)}^{\operatorname{Add}(\omega)}$ by letting $(p',q')R(p,q)$ if and only if $(p'(k),q'(k))\leq(p(k),q(k))$ in $\operatorname{Add}(\omega)*\operatorname{Add}(\omega_1)$ for all $k<n$. Then (in the finite case) the mixing property holds because given such $p,q,q_0,q_1$, we can choose $p_0$ and $p_1$ such that $p_0(k)$ and $p_1(k)$ are incompatible for any $k<n$. Then we let $q'$ be such that $q'(k)$ is forced by $p_0(k)$ to be equal to $q_0(k)$ and by conditions incompatible with $p_0(k)$ to be equal to $q_1(k)$ (so in particular this is forced by $p_1(k)$). Then it is clear that $(p_0,q')R(p,q_0)$, $(p_1,q')R(p,q_1)$ and $(p,q')R(p,q)$ (as $q'(k)$ is forced to equal either $q_0(k)$ or $q_1(k)$, both of those are forced to be below $q(k)$). In the infinite case, this of course fails.

Answer 1

This is a nice question. I think that $\mathbb{P}$ does not have the countable approximation property. Here's a sketch of a proof. I'll try to be pretty concise; let me know if you'd like more details.

Suppose that $G$ is $\mathbb{P}$-generic over $V$. In $V[G]$, for each $n < \omega$, let $g_n : \omega_1 \rightarrow 2$ be the $\mathrm{Add}(\omega_1)$-generic function added by the $n^{\mathrm{th}}$ "column" of $\mathbb{P}$. Now define a function $h : \omega_1 \rightarrow 3$ as follows. For every $\eta < \omega_1$ and $i < 2$, let $h(\eta) = i$ if $g_n(\eta) = i$ for cofinitely many $n < \omega$; otherwise, let $h(\eta) = 2$.

We first claim that $h$ is countably approximated by $V$. To this end, let $\dot{h}$ be a $\mathbb{P}$-name for $h$, fix $x \in ([\omega_1]^\omega)^V$, and fix $(p,q) \in \mathbb{P}$. We will find $(p,q') \leq (p,q)$ and $h_x : x \rightarrow 3$ such that $(p,q') \Vdash \dot{h} \restriction x = h_x$.

By standard arguments, using the countable closure of $\mathrm{Add}(\omega_1)$ and the countability of $x$, we can find $q'$ such that $(p,q') \leq (p,q)$ and, for every $\eta \in x$ and $n < \omega$, one of the following alternatives holds:

• there is $i < 2$ such that $(p(n), q'(n)) \Vdash \dot{g}_n(\eta) = i$;
• there exist $p_0, p_1 \leq p(n)$ such that, for $i < 2$, we have $(p_i, q'(n)) \Vdash \dot{g}_n(\eta) = i$.

Now define $h_x: x \rightarrow 3$ as follows. Given $\eta \in x$ and $i < 2$, if there are cofinitely many $n < \omega$ such that $(p(n), q'(n)) \Vdash \dot{g}_n(\eta) = i$, then let $h_x(\eta) = i$; otherwise, let $h_x(\eta) = 2$. We now claim that $(p,q') \Vdash \dot{h} \restriction x = h_x$. For $\eta \in x$ such that $h_x(\eta) < 2$, this is clear. If $h_x(\eta) = 2$, it follows from a routine genericity argument using the fact that the first coordinate of $\mathbb{P}$ uses finite supports (I can give more details here if needed).

Thus, $h$ is countably approximated by $V$. However, another routine genericity argument shows that $h$ cannot be in $V$. Therefore, this provides a counterexample to the countable approximation property for $\mathbb{P}$.