Genericity is still a little bit mysterious to me, although not as much as it used to be.

Here is a rough paraphrase of Theorem 3.5 of Kunen's Set Theory: an Introduction to Independence Proofs

Let $M$ be a countable transitive model, $\langle\mathbb P,\leq\rangle\in|M|$ a partial order and $G\subseteq{\mathbb P}$ $M$-generic. For any $\phi$ it is the case that $M[G]\vDash\phi$ if and only if $(\exists p\in G) p\Vdash\phi$.

I will refer to the right-to-left direction of this implication as "things which are forced are true in the extension".

Assume that $\mathbb P$ is splitting and we've chosen some $\phi$ such that $M\nvDash\phi$ and there is some $p\in\mathbb P$ such that $p\Vdash\phi$. Let $\hat p$ be the principal $\mathbb P$-filter generated by $p$. Among the $M$-generic filters $G$, those for which $M[G]\vDash\phi$ are exactly those for which $\hat p\subseteq G$.

But what about $\hat p$ itself? Well, all the principal filters of $\mathbb P$ are sets of the ground model, so $\hat p\in|M|$. So forcing with $\hat p$ will give us back an extension model which is isomorphic to $M$ -- no new sets. So $M[\hat p]\nvDash\phi$.

So, if I haven't screwed up so far, we know that for generic filters "things which are forced are true in the extension", and for filters which happen to be elements of $|M|$ it is not necessarily the case that "things which are forced are true in the extension". This leaves a gap: filters which are not elements of $|M|$, yet are not $M$-generic either.

Question: does "things which are forced are true in the extension" hold for $\{any,all\}$ filters $F\supseteq\hat p$ such that $F\notin|M|$ yet $F$ is also not $M$-generic?

Phrased another way: does the right-to-left implication rely on the filter being generic, or only on the fact that it isn't an $M$-set?

I can see quite clearly why the left-to-right implication ("things which are true in the extension are forced") relies on the fact that the filter is generic, and couldn't possibly work without "intersects every dense set". I'm having a harder time seeing why genericity (rather than simply "not in $|M|$") is necessary in the other direction, though. Kunen uses it in his proof (the case where $\phi$ is $\tau_1=\tau_2$), but it's not as easy to see why (or if) the use is essential.

This isn't a homework question (or even close to any of them). Hope you'll trust me.


There are some statements $\phi$ in the forcing language that are true in $M[G]$ iff they are forced by a condition in the filter, no matter whether $G$ is generic or not.

The simplest is $p\in\Gamma$ where $\Gamma$ is the natural name for the generic filter. (Assuming the forcing notion is separative.) But I understand that this is not what you are looking for, since you are asking about statements not true in $M$ but true in the extension.

Now, the problem with filters that are not generic and also not in the ground model is that the extension might not be a model of ZFC at all. Let me give an example:
Consider the forcing for adding a single Cohen real. Every function from $\omega$ to $2$ corresponds to a filter. $M$ is countable and transitive and the ordinals intersected with $M$ are some countable ordinal $\alpha$. Fix some function $f:\omega\to 2$ that codes a well-ordering of $\omega$ that is longer than $\alpha$.

Let $G$ be the corresponding filter. If $M[G]$ was a model of ZFC, then it would contain the ordinal coded by $f$. But for every name $\sigma$ in $M$, the evaluation of $\sigma$ at $G$ has rank at most the rank of $\sigma$, showing that $M[G]$ has the same ordinals as $M$.
This technology should be enough to give you all kinds of counter examples you might be looking for.

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    $\begingroup$ Let me add a brief comment: It is in general very hard to say something about $M[G]$ if $G$ is not generic (or, at least, I find it very hard. This is a problem I've run into a few times). Under specific circumstances, one may be able to arrange some control, but I know of no general results even for very concrete collections of posets. $\endgroup$ Nov 27 '10 at 19:51
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    $\begingroup$ Thank you! This definitely answers the "all" version of my question -- negatively. And I'll interpret Andres' comment to suggest that the answer for the "any" case is not really known (but probably isn't a significant research priority either). So, to summarize, the property "generic" is used not because it is the weakest filter property which ensures "things which are forced are true in the extension", but because it is the weakest filter property known to ensure "things which are forced are true in the extension". $\endgroup$
    – Adam
    Nov 29 '10 at 0:59

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