Suppose we have a model (of $\mathsf{ZFC}$) $M$, and that $x\in 2^\omega$ is random over $M$, and that $y\in 2^{\omega}$ is Cohen over $M$. My question is whether $y$ is also Cohen over $M[x]$. In other words, if I have a real that's Cohen over a model, is it still Cohen over the model that results from performing Random forcing?

## 1 Answer

That depends on the particular random real $x$ and Cohen real $y$. On the one hand, I could first choose $x$ random over $M$ and then choose $y$ Cohen over $M[x]$. Then $y$ is also Cohen over the submodel $M$, so it's an example where the answer to your question is yes.

On the other hand, I could first choose $y$ Cohen over $M$ and then choose $x$ random over $M[y]$. Then $x$ is also random over the submodel $M$. I'll show that $y$ is not Cohen over $M[x]$.

Partition $\omega$ into intervals $I_n$ of rapidly increasing length. (In fact, it suffices to take $I_n$ of length $n$, but "rapidly" avoids the need for arithmetic.) Define a symmetric binary relation $R$ on $2^\omega$ by putting $aRb$ iff $a$ and $b$ agree on infinitely many of these intervals. Note that, for any $a$, the set of $b$'s $R$-related to $a$ is comeager but has Lebesgue measure 0. Note also that $R$ is a low-level Borel set with code in $M$ (provided you chose the sequence of $I_n$'s in $M$). Since $x$ is random over $M[y]$, we have that $x$ and $y$ are not $R$-related. But then $y$ cannot be Cohen over $M[x]$.