Forcing models satisfying the genericity property — Problem 8.1

8. Hang Zhang

The following questions are motivated by the problem whether hyper-hyperfiniteness implies hyperfiniteness.

Let $M$ be a countable model for a sufficiently large fragment of ZFC. Suppose that a partial order $P$ is in $M$. Define the space $Gen^{P}_{M}$ of $M$-generics for $P$ by

If we identify $Gen^{P}_{M}$ as a subspace of $2^{P}$ with the product topology, then $Gen^{P}_{M}$ is $G_{\delta}$ in $2^{P}$ and thus a Polish space. See [1] for details. Define $G\sim H$ if and only if $M[G]=M[H]$. Consider the equivalence relation $\sim$ on $Gen^{P}_{M}$.

Example 8.1.

(Calderoni–Sinapova [2]) Let $M$ be a countable model for large fragment ZFC. Suppose $M\models\kappa$ is measurable (witnessed by $\mathcal{U}$). If $M\models\mathbb{P}=\{(s,A):s=\langle s_{0},\dots,s_{n}\rangle,\ n\in\omega,\ A\in\mathcal{U}\}$, with order $(s,A)\leq(t,B)\iff t\subseteq s\land s\setminus t\subseteq B\land A\subseteq B$. Then for any generic sequences $\vec{\alpha},\vec{\beta}$ of $\mathbb{P}$ over $M$,

Remark 8.2 ([1, Theorem 3.1]).

If $\mathbb{P}=\mathbb{C}$ (Cohen forcing), then ”$x\thicksim y\iff M[x]=M[y]$” is hyper-hyperfinite.

Remark 8.3.

If

is Cohen, then

is also Cohen. Notice that these two Cohen reals lead to the same forcing extension. So Cohen forcing does not satisfy (*) for any $M$.

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