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Suppose $\mathbb{P}$ is a notion of forcing in the ground model $V$, and $X$ is a set which is in $V[G]$ for every $\mathbb{P}$-generic filter $G$. Then $X\in V$ already, by a fairly simple (if tedious) argument.

I'm working on a paper in which this fact is mentioned, and I would like to cite it properly; however, I can't seem to find any citation for it, and I don't recall when I learned the argument. So my question is:

What is a citation for "if $X$ is in every forcing extension by some poset $\mathbb{P}$, then $X$ is already in the ground model?"

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    $\begingroup$ I think in general if $G\times H$ is $P\times P-$generic over $V$ then $V[G]\cap V[H]=V.$ $\endgroup$ – Mohammad Golshani Jan 28 '14 at 9:18
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The result is due to Solovay, see

"A model of set theory in which every set of reals is Lebesgue measurable"

In Lemma 2.5 of the above paper, Solovay shows that if $G$ and $H$ are mutually generic filters for a forcing notion $P,$ then $V[G]\cap V[H]=V.$ From this result your statement follows immediately.

Note that Solovay's result is even more general than what I said above, and it says if $G\times H$ is $P\times Q-$generic over $V$ (where $P, Q \in V$), then $V[G]\cap V[H]=V.$

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Let me supplement Mohammad's answer by supplying a proof, since the result is not difficult, and it is extremely useful.

Theorem. (Solovay) If $G\subset\mathbb{P}$ and $H\subset\mathbb{Q}$ are mutually $V$-generic, that is, $G\times H\subset \mathbb{P}\times\mathbb{Q}$ is $V$-generic, then $V[G]\cap V[H]=V$.

Proof. Clearly $V[G]\cap V[H]\supset V$, so suppose that $x\in V[G]\cap V[H]$. Thus, $x=\sigma_G=\tau_H$ for some $\mathbb{P}$-name $\sigma$ and $\mathbb{Q}$-name $\tau$. By $\in$-induction, we may assume $x\subset V$. Since $\sigma_G=\tau_H$, there must be some condition $(p,q)$ forcing that $\sigma$ and $\tau$, viewed as $\mathbb{P}\times\mathbb{Q}$-names in the canonical way, are naming the same set in that way. But now every condition $p'$ stronger than $p$ must force $\check y\in \sigma$ just in case $y\in x$, because otherwise we could find a different generic filter $G'$, mutually generic with $H$, so that $G'\times H$ contains $(p,q)$, but such that $\sigma_{G'}\neq \tau_H$. Thus, the condition $p$ already knows exactly which $y$ are elements of $x$, and so $x\in V$, as desired. QED

See further information on this MO question about mutual genericity.

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Here is the computability-theoretic version: Suppose $G=G_0\oplus G_1$ is 1-generic and suppose $X$ is computable from both: so $X=\Phi_1^{G_0}=\Phi_2^{G_1}$. Then also $X=\Phi^{G_0}=\Phi^{G_1}$ where (Posner's trick) we define $\Phi$ by: noneffectively pick a number $n_0$ with $G_0(n_0)\ne G_1(n_0)$ and use this to decide whether to run $\Phi_1$ or $\Phi_2$.

The set $S := \{\sigma = \sigma_0\oplus\sigma_1: \Phi^{\sigma_0}(n)\downarrow \ne \Phi^{\sigma_1}(n)\,\exists n\}$ is $\Sigma^0_1$ and by assumption $G$ does not meet it, i.e., $\sigma\in S$ implies $\sigma$ is not a prefix of $G$. Then by 1-genericity there is a prefix of $G$, $\tau$, such that no extension of $\tau$ lands in $S$. Then $X$ is computable. Indeed, given $n$ we may search for any extension of $\tau$ making $\Phi^{\tau_0}(n)$ converge; then $X(n)=\Phi^{\tau_0}(n)$.

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