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A real $x\in2^\omega$ is random (or Solovay-random) over a model $M$ iff it's obtained via forcing by closed sets of reals, of positive measure, coded in $M$. Similarly, a pair $(x,y)$ of reals is random over a model $M$ iff it's obtained via forcing by closed sets in $2^\omega\times2^\omega$, of positive product measure, still coded in $M$. Thus such a pair is not a product-forcing pair. (To establish this, one has to define a closed set $X\subseteq2^\omega\times2^\omega$, of positive product measure, which does not include any positive-measure rectangle, not an easy exercise!) Rather $(x,y)$ is a random pair over a model $M$ iff $x$ is random over $M$ and $y$ is random over $M[x]$ iff vice versa.

Question. In spite of not being product-forcing-generic, prove that if $(x,y)$ is a random pair over a model $M$ then $M[x]\cap M[y]=M$.

A proof of $M[x]\cap M[y]\cap2^\omega\subseteq M$ can be manufactured by reduction, via continuous reading of names, to the claim that if $X\subseteq2^\omega\times2^\omega$, is a closed set of positive product measure and $f,g:2^\omega\to2^\omega$ are continuous maps such that $f(x)=g(y)$ for almost all $(x,y)\in X$, then $X=X'\cup\bigcup_nX_n$, where $X'$ is null and each $X_n$ is Borel non-null on which both $f$ and $g$ are constants. What about the general case?

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Suppose $\kappa$ is the least ordinal such that for some $A \in 2^{\kappa} \cap M[x] \cap M[y]$, $A \notin M$. You already have that $\kappa = cf(\kappa) \geq \omega_1$. Let $\tau$ be a Random name for $A \in M[x]$. WLOG, assume that the empty condition forces $\tau \in 2^{\kappa} \setminus M$ and all initial segments are in $M$. In $M$, choose $\langle (p_i, a_i) : i < \kappa \rangle$ such that $p_i$ forces $a_i = \tau \upharpoonright i$. Since random forcing is $\kappa$-Knaster, there exists $X \in [\kappa]^{\kappa}$ such that $\{p_i : i \in X\}$ has pairwise compatible conditions so that $a = \bigcup \{a_i: i \in X\} \in 2^{\kappa}$. Using cccness, choose a condition $p$ that forces $|\{i \in X: p_i \in G\}| = \kappa$. But now $p$ forces $\tau \in M$: Contradiction.

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    $\begingroup$ I don't understand how the case $\text{cf}\kappa=\omega$ is excluded (besides $\kappa=\omega$, of course). $\endgroup$ – Vladimir Kanovei Aug 23 '16 at 21:18
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This is a response to Kanoveri's comment. If $\{\kappa_n : n < \omega\}$ is strictly increasing with limit $\kappa$, then since $A \upharpoonright \kappa_n \in M$ (by minimality of $\kappa$), using ccc, we can find a countable $X \in M$ such that $M[x] \cap M[y]$ contains a new subset of $X$ which translates to a new real.

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  • $\begingroup$ Why the downvote? This seems right - any countable set $X$ of ordinals added by a c.c.c. forcing is contained in a countable set in the ground model (look at antichains generated by questions of the form, "What is the $n$th element of $X$?"). That said, this should be a comment/edit on the original answer as opposed to an answer. $\endgroup$ – Noah Schweber Aug 23 '16 at 23:29
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This is a partial answer, but is too long for a comment; see the postscript at the end.

Suppose we have names $\nu_0$ and $\nu_1$, and a condition $p$, such that $p\Vdash \nu_0[G_0]=\nu_1[G_1]$, and that $\nu_0[G_0]\subseteq V_\alpha^M$ but $\nu_0[G_0]\not\in M$ for some $\alpha$. (That is, we're looking at an $a\in M[x]\cap M[y]\setminus M$ of minimal rank.)

I claim $p$ already decides $\nu_0[G_0]$. Why? Well, let $b\in V_\alpha^M$ and suppose $q_+\le p$ forces $b\in \nu_0[G_0]$, and $q_-\le p$ forces $b\not\in \nu_0[G_0]$.

First, note that $q_-$ forces $b\not\in\nu_1[G_1]$ as well, since it's stronger than $p$. Now look at the projections of these two conditions:

$$q_+'=\{(f, g): \exists h((f, h)\in q_+)\}\cap p,$$

$$q_-'=\{(f, g): \exists h((h, g)\in q_-)\}\cap p.$$

Now, $q_+'$ forces the same things as $q_+$ does about $\nu_0[G_0]$, and similarly for $q_-'$. So all that's left to do is to show that $q_+'\cap q_-'$ has positive measure.

The more I think about it, the less obvious this claim is. It's trivial in case $p$ contains a rectangle of positive measure, but of course that's not necessarily the case. I still think it has a snappy proof, but I don't see it.

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  • $\begingroup$ Regarding Mcnulty conclusive arguments, could it be strengthened to prove this: if $a$ is a random real over $M$ and $x\in M[a]\setminus M$ an arbitrary set then there is a real $b\in M[a]$ still random over $M$ and such that $M[x]=M[b]$? This would provide a straight reduction of the original Q to the case of reals. $\endgroup$ – Vladimir Kanovei Aug 24 '16 at 3:27

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