This question concerns the amount of information about a model $M$ that is contained in the collection of all reals Cohen over $M$.


Specifically, let $M$ and $N$ be countable transitive models of ZFC and suppose that they have the same collection of Cohen reals, i.e. any real $c\in V$ is Cohen over one of them iff it is Cohen over the other one. Does this tell us anything about $M$ and $N$?


Naively one might hope to get that $M=N$, but this is far too much. Already if $M$ and $N$ merely agree on $\mathcal{P}(\omega)$, then they will share the same Cohen reals, since the dense subsets of the Cohen poset are effectively coded by reals. So we can get easy examples of models $M\neq N$ which share the same Cohen reals, e.g. by letting $N$ be a sufficiently closed forcing extension of $M$.


This tells us that we cannot hope to recover information about $M$ and $N$ beyond their reals. But can we recover that? If $M$ and $N$ share their Cohen reals, is $\mathcal{P}(\omega)^M=\mathcal{P}(\omega)^N$?



Is a model of set theory determined by the Cohen reals over it?
Unless I'm missing something, don't you get a counterexample if $N=M[s]$ where $s$ is a Sacks real over $M$? The point is that every dense open subset of $2^{<\omega}$ in a Sacks extension includes such a dense open set in the ground model and so a Cohen real over $M$ remains Cohen over $M[s]$.


This is a standard fusion argument. One can proceed as follows:


**Lemma:**


Suppose $p\Vdash``\text{$\dot D$ is dense open in $2^{<\omega}$}"$, $n<\omega$, and $\sigma\in 2^{<\omega}$. Then there are $q\leq p$ and $\tau\in 2^{<\omega}$ such that


* $\tau\supseteq \sigma$,
* $q\upharpoonright\vphantom{B}^n2= p \upharpoonright\vphantom{B}^{n}2$, and
* $q\Vdash \check{\tau}\in\dot D$


proof: Let $\{a\_i:i<k\}$ list $p\cap\vphantom{B}^n2$, and for each $i<k$ let $p\_i$ be those elements of $p$ that extend $a\_i$ (so $p\_i\in P$). We define objects $q\_i$ and $s\_i$ by recursion for $i<k$ so that


* $q\_i$ is an extension of $p\_i$,
* $s\_i\in 2^{<\omega}$,
* $s\_0\supseteq\sigma$,
* for $i<k$ we have $s\_{i+1}\supseteq s\_i$, and
* $q\_i\Vdash``\check{s}\_i\in \dot D".$


We can do this easily because each $p\_i$ is an extension of $p$ in $P$, and hence $p\_i$ forces $\dot D$ to be dense open. In fact, for ANY $s\in 2^{<\omega}$ and condition $r\leq p$, there are $r'\leq r$ in $P$ and $t\in 2^{<\omega}$ such that $t\supseteq s$ and $r'\Vdash ``t\in \dot D"$. (We are asking $r'$ to decide a particular value for some extension of $s$ in $\dot D$, and such conditions are dense below $p$.)


Let $\tau= s\_{k-1}$, so $\sigma\subseteq s\_0\subseteq\dots\subseteq s\_{k-1}=\tau$, and since $p$ forces $\dot D$ to be open, it follows that each $q\_i$ forces $\tau$ to be in $\dot D$.


Now paste the conditions $q\_i$ back together to build $q$. We have


* $q\leq p$ in $P$, and
* $q\upharpoonright\vphantom{B}^n2= p \upharpoonright\vphantom{B}^{n}2$.


Since any extension of $q$ is compatible with some $q\_i$, it follows that


* $q\Vdash \check{\tau}\in\dot D$,


and since $\tau$ extends $\sigma$, we are done. $\square$


We use the Lemma to build a fusion sequence. I haven't introduced a definition for $\leq\_n$ on $P$, but the above Lemma is strong enough to get what we want for any of the standard variants.


Let $\langle \sigma\_n:n<\omega\rangle$ enumerate $2^{<\omega}$, and suppose $p\Vdash``\dot{D}\text{ is dense open in }2^{<\omega}\text{''}$. We build sequences $\langle p\_n:n<\omega\rangle$ and $\langle \tau\_n:n<\omega\rangle$ such that 


* $p\_0=p$,
* $p\_{n+1}\leq\_n p\_n$, and
* $p\_{n+1}$ forces that $\tau\_n$ is an extension of $\sigma\_n$ in $\dot D$


Let $q\leq p\_n$ for all $n$ (by fusion) and let $E$ be the those elements of $2^{<\omega}$ which extend $\tau\_n$ for some $n$. Then $E$ is a dense open subset of $2^{<\omega}$ in the ground model, and 


$q\Vdash\check{E}\subseteq \dot D,$


as required.


