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Given a model $M$ of ZFC, we can define $\mathbb{P}$-names and generic extensions $M[G]$ in terms of $M$.

But the usual framework for forcing involves an outer model $M\subseteq V$ of ZFC and $M$ is required to be transitive and countable with respect to $V$. The forcing extensions then become $M\subseteq M[G] \subseteq V$.

Does anyone know why this outer model $V$ is important for forcing? Extending $M$ looks enough to prove independances of axioms such as Choice or the Continuum Hypothesis.

EDIT

Following the answer below, here is a tentative to construct a generic filter on $M$ without an outer model $V$.

We want to prove the relative consistency of a new axiom of set theory. So we start by assuming that ZFC is consistent. By the Lowenheim-Skolem theorem, there is a countable model $M$ of ZFC. Let $(\mathbb{P}, \leq)$ a partial order inside $M$ of conditions that represent all the possibilities of the forcing extension.

Because $M$ is countable, so are the $\mathbb{P}$-dense parts, that we enumerate $D_0,D_1,\dots$ Let $p_0\in D_0$. By density there is $p_1\in D_1$ such as $p_1\leq p_0$. We continue this to define a decreasing sequence $p_n\in D_n$. Then we call $G=\{p\in\mathbb{P} | \exists i\in\mathbb{N}, p_i\leq p\}$. $G$ is a filter for $\leq$ and it meets every dense part, so it is $M$-generic on $\mathbb{P}$.

Is there a problem with this construction, that does not even need that $M$ is transitive?

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  • $\begingroup$ (In response to your edit) The problem is, as Noah indicated, this $G$ is not necessarily in $M$ (and in fact will not be in all interesting cases). We only know $M$ is countable from the outside... $M$ does not know it is countable. As such, $M$ does not have access to the sequence $(D_n:n\in \mathbb N).$ $\endgroup$ – spaceisdarkgreen Oct 24 '18 at 1:40
  • $\begingroup$ @spaceisdarkgreen I thought that $G\notin M$ was the point of forcing: make a proper extension $M[G]$ of $M$. The extension has to be constructed from outside of $M$. I think this is even suggested by the word "generic" : if $G$ was in $M$ it would be a particular case, not a generic case. $\endgroup$ – V. Semeria Oct 24 '18 at 10:57
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    $\begingroup$ "Is there a problem with this construction, that does not even need that M is transitive?" There is: although $G$ is a generic filter as long as $M$ is countable, regardless of transitivity, you still need to define $M[G]$ and show that it satisfies ZFC and that the forcing theorems hold. The usual proof of this uses transitivity, so this needs some work; this is the "internal-recursion" issue I mentioned in my answer. Put another way, for countable models it's not the existence of generics that's nontrivial, it's the use of generics. $\endgroup$ – Noah Schweber Oct 24 '18 at 13:23
  • $\begingroup$ @V.Semeria If you agree about this, then I’m not sure what the argument is about. In a comment I probably shouldn’t have deleted I mentioned the outermost model $V$ is just the model we’re working in when we define $M$ and $M[G].$ You alluded to “taking $G$ from the metatheory” in a comment you have deleted, but I don’t think the distinction you’re seeing is really there and at most it’s another way of saying the same thing (and Noah already responded to that comment at length). $\endgroup$ – spaceisdarkgreen Oct 24 '18 at 14:24
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The problem comes with this step:

"Letting $G$ be a $\mathbb{P}$-generic filter over $M$, we define $M[G]$ as [stuff]."

How do we know that such a $G$ exists in the first place? Certainly it will not in general from "within $M$:" as long as $\mathbb{P}$ is nontrivial, no $\mathbb{P}$-generic filter over $M$ will exist in $M$. Generic filters over $M$ have to come from some larger model.

This is one of the reasons that the Boolean-valued models approach to forcing is enticing: everything takes place within the original model, and we can talk about forcing over the "real" universe without saying horrible nonsense.

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  • $\begingroup$ Incidentally, transitivity really plays no role in forcing: we can make sense of the poset-version of forcing over an arbitrary countable model of set theory, and the Boolean-valued models version of forcing over an arbitrary model of set theory of any cardinality. Basically, the point is that the recursions needed to define the forcing relation, and hence the forcing extension (or Boolean model), are all internal to the ground model, and every model of ZFC "thinks" that it's well-founded. $\endgroup$ – Noah Schweber Oct 23 '18 at 3:23
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    $\begingroup$ The relationship between the two approaches is roughly the following: when I have a forcing notion $\mathbb{P}$ in a model $M$, I get a Boolean-valued model $M^\mathbb{P}$ with Boolean algebra of truth values $\mathbb{B}_\mathbb{P}$. This algebra is closely tied to $\mathbb{P}$ itself, with "truer" truth values corresponding roughly to larger sets of conditions. A $\mathbb{P}$-generic filter $G$ over $M$ can be thought of as a way to "quotient out" $\mathbb{B}_\mathbb{P}$ so that we're left with the two-element Boolean algebra ("in/out of $G$") and this transforms $M^\mathbb{P}$ into $M[G]$. $\endgroup$ – Noah Schweber Oct 23 '18 at 3:50
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    $\begingroup$ First of all, I don't know what it means to get a set from a metatheory. But even leaving that aside, I think you're taking "generic filters exist" as a basic principle, and this is definitely not warranted: why should generic filters exist for arbitrary models? For example, ZFC proves that if $M$ is a transitive model of ZFC which contains every real, then there is no Cohen-generic filter of $M$. Basically, the point is that the existence of generic filters over a model should in general be surprising - the only unsurprising situation being when that model is countable. $\endgroup$ – Noah Schweber Oct 23 '18 at 14:43
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    $\begingroup$ Now you might adopt some "multiverse" principle. In particular, a principle asserting that "everything is countable in some larger universe" does imply that generic filters always exist; however, it also implies that every model lives inside a larger model! Basically, the point is that any nontrivial time you want to assert that a generic filter over a model $M$ actually exists, you're relying on the assumption that $M$ lives inside a larger universe, since that's where the generic you want has to live. $\endgroup$ – Noah Schweber Oct 23 '18 at 14:46
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    $\begingroup$ @V.Semeria Again, where does your construction take place? When you say "there is a ...," what does that mean? The purpose of the ambient $V$ is precisely to be the framework within which such constructions take place. That's the whole point of my answer: that we need a mathematical universe beyond the ground model $M$ itself if we're going to go the "actual-generic-filter" route. $\endgroup$ – Noah Schweber Oct 24 '18 at 13:20
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In the context of the most elementary version of Forcing that I first learned:

(1). We do not assume there exists a set-model for ZFC. But if you are given a finite list of some of the axioms of ZFC you can use the Lowenheim-Skolem method to prove there exists a countable transitive model (c.t.m) for them. ( Trying to apply this method to a list of all the axioms would require either an infinitely long proof or an ability to quantify over collections of sentences.) So we assume that we have a c.t.m for ZFC-minus-Comprehension-minus-Replacement, that satisfies all the instances of Comprehension and Replacement that we will need. So we speak of M as if it does satisfy all of ZFC.

(2). So if the poset $P$ belongs to $M$ then the set of dense subsets of $P$ that belong to $M$ is a countable set. It is a simple exercise to show that if $S$ is a countable family of dense subsets of a poset $P$ then there exists a filter $G$ on $P$ such that $\forall s\in S\,(s\cap G\neq \phi).$ In Forcing the filter $G$ is required only to have non-empty intersection with the dense subsets of $P$ that belong to $M.$ Since $M$ is transitive but only countable, such $G$ do exist.

(3). If $P\ne \phi$ and if $P$ is separative, that is, if every $x\in P$ has two incompatible (incomparable) extensions, then $G\not \in M.$ Otherwise $P\setminus G$ would belong to $M$ and would be a dense subset of $P$ that's disjoint from $G.$

(4) $V,$ in Set Theory, is commonly used to denote the class of all sets. If instead, we say that $(V,\in^*)$ is a (class or set) model for (enough of) ZFC but not necessarily the whole universe and with $\in^*$ not necessarily being $\in$ then we may consider points (1),(2),(3) above to be re-phrased as "all relativized to $(V,\in^*)$", with the additional requirement that $M\subset^* V$.

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