Products of Cohen forcings

This result is obtained with Rahman

Theorem. Suppose there is an $\aleph_1$-complete ultrafilter $U$ over $\lambda.$ Then forcing with full product $\mathbb{P}=\prod_{i<\lambda}Add(\omega, 1)$ collapses $2^\lambda$ into $\aleph_0.$

Proof. Let $(r_i: i<\lambda)$ be the generic reals added by $\mathbb{P},$ and for each limit ordinal $\alpha,$ consider the reals $(r_{\alpha+n}: n<\omega)$ as an $\omega\times \omega$ matrix of $0, 1$'s as in Hamkins answer, and so as in Hamkins we can find for each $n<\omega$ a set $A_{\alpha, n} \subseteq [\alpha, \alpha+\omega).$ Let $A_n=\bigcup \{A_{\alpha, n}:\alpha$ is a limit ordinal $<\lambda \}.$

Working in $V$, define an equivalence relation $\sim$ on $P(\lambda)$ by $A\sim B$ iff $\{ \alpha<\lambda: lim(\alpha), A\cap [\alpha, \alpha+\omega)= B\cap [\alpha, \alpha+\omega)\} \in U.$

Claim. There are $2^\lambda$ many equivalence classes.

Now we show that for any $A \subseteq \lambda,$ there is $n_0<\omega$ such that $A \sim A_{n_0}.$

Given condition $p=(p_\alpha: \alpha<\lambda)\in \mathbb{P}$, there is $n_0 <\omega$ such that $X=\{\alpha< \lambda: dom(p_\alpha)=n_0\}\in U$ (use the $\aleph_1$-completeness of $U$). Now as in Hamkins argument, it is easily seen that we can extend $p$ to some condition $q$ forcing for all $lim(\alpha)\in X, A\cap [\alpha, \alpha+\omega)=A_{\alpha, n_0}$

The result follows immediately.

Corollary. Suppose $\lambda$ is a measurable cardinal, or is above the least strongly compact cardinal. Then forcing with full product $\mathbb{P}=\prod_{i<\lambda}Add(\omega, 1)$ collapses $2^\lambda$ into $\aleph_0.$

The answer is yes.

Lemma: There is a function $F: 2^{\lambda} \rightarrow 2^{\lambda}$ such that for every $S \in [\lambda]^{\lambda}$ and function $g: \lambda \setminus S \rightarrow 2,$ $F$ restricted to extensions of $g$ is surjective.

Fix a well-ordering of all ordered pairs of the form $(g, h),$ $g$ as above and $h \in 2^{\lambda},$ of length $2^{\lambda}.$ We construct $F$ in $2^{\lambda}$ stages, each stage determining $F$ at one input. Let $(g,h)$ be the $\alpha$th pair in the ordering. At stage $\alpha,$ we have determined $F$ at less than $2^{\lambda}$ inputs, so there is some extension $g'$ of $g$ for which $F(g')$ has not been determined. We declare $F(g')=h.$ This completes the construction.

Let's consider the case where $\text{cf}(\lambda)>\omega.$ Notice that for every $h \in 2^{\lambda}$ and $p \in \mathbb{P}=\Pi_{\alpha<\lambda} Add(\omega,1),$ there are $q \le p$ and $n<\omega$ such that $n \in \text{dom}(q_{\alpha})$ for every $\alpha,$ and $F(\alpha \mapsto q_{\alpha}(n))=h.$ Here $n$ is any number such that $|\{\alpha<\lambda: n \not \in \text{dom}(p_{\alpha})\}|=\lambda.$ Thus, in $V[G],$ there is a surjection from $\omega$ to $(2^{\lambda})^V$ definable from $F$ and $G.$

Now suppose $\text{cf}(\lambda)=\omega$ and $\lambda>\omega.$ Here we use a generalization of the matrix argument in Hamkins' answer. Let $\kappa_0=0,$ and fix a cofinal $\omega$-sequence $\langle \kappa_n: 0<n<\omega\rangle$ of uncountable regular cardinals below $\lambda.$ In $V[G],$ $G$ determines an $\omega \times \lambda$ matrix of bits, each column a Cohen real. We construct a $\lambda$-sequence $z_n$ as follows: Let $n_0=n,$ and recursively define $n_{k+1}$ to be the number of consecutive 0's above the $n_k$th entry in the $\kappa_k$th column. For $\alpha \in [\kappa_k, \kappa_{k+1}),$ let $z_n(\alpha)$ be the $n_k$th entry in the $\alpha$th column.

Fix $h \in (2^{\lambda})^V$ and $p \in \mathbb{P}.$ It suffices to show there are $q \le p$ and $n<\omega$ such that $q$ forces $h$ to equal $F(z_n).$ Let $n_k>\max(\text{dom}(p_{\kappa_k}))$ be such that $|\{\alpha \in [\kappa_k, \kappa_{k+1}): n_k \not \in \text{dom}(p_{\alpha})\}|=\kappa_{k+1},$ and let $n=n_0.$ It is easy to construct $q$ as desired.

Since the question is still not answered, let me at least explain why the full-support $\omega$-product of $\text{Add}(\omega,1)$ collapses $(2^\omega)^V$ to $\omega$ in the extension $V[G]$.

The way that I think about it is this. The generic filter $G$ fills in an $\omega\times\omega$ matrix with $0$s and $1$s. In $V[G]$, let us describe a countable sequence of real numbers as follows. For each natural number $n=n_0$, we define a binary sequence $z_n$ by the following proceure: look in the first column of the matrix $G$; starting at position $n_0$ of that column, we get the first bit of $z_n$, either $0$ or $1$, and then let $n_1$ be the number of $0$s after that bit in the first column; next, we look at position $n_1$ in the second column---this gives a second bit, and then $n_2$ is the number of $0$s following this bit in that column, and so on. Each column gives you one more bit and also the location of the next bit in the next column. For any starting number $n=n_0$, we thereby produce an $\omega$-sequence of bits, and thus a real number $z_n$.

The point is that it is dense that any given ground model real $z$ is produced via $G$ by this procedure for some starting value, since any condition $p$ in the full-support product can be extended to a condition $p^+$ on which $z$ is coded in that way: simply let $n_0$ be a value on which $p$ is not yet specified, and then extend it in such a way to $p^+$ that on the first column, the value at position $n_0$ agrees with $z$, and such that the number of following zeros $n_1$ is large enough to reach a place that is not yet specified in $p$ on the second column, so that we may put the second bit of $z$ there, and put $n_2$ many zeros there and so on. Thus, every $z$ in the ground model arises from $G$ by starting at some initial position $n$, and so $2^\omega$ of the ground model is countable in $V[G]$.

It is not clear to me how to extend this idea through limits to uncountable products.