I re-read Jechs chapter about forcing, and got a question. There he characterizes a (what he calls) modern way to make the forcing argument legitimate which (I think) goes like this:

It is pointed out there that, in order to establish the consistency of a statement $\varphi$ relative to ZFC, it is sufficient to exhibit a complete Boolean algebra $B$ such that the Boolean value of $\varphi$ in the Boolean-valued model $V^B$ is not zero, i.e. $|| \varphi|| = p \ne 0$.

This fact gives us an alternative approach to forcing, which avoids the assumption of a transitive, countable model $M$ of ZFC, in order to construct the generic extension $M[G]$.

Nevertheless one hardly sees a proof where the Boolean value of an interesting $\varphi$ is really evaluated because it is more easy to pretend that for the p.o. $P \quad$there exists a generic $G$ over $V$, then build $V[G]$ and show that $V[G] \models \varphi$, which is equivalent to the existence of a $p \in G$ such that $p \Vdash \varphi$, which implies that $||\varphi|| \ne 0$.

My problem with this approach is that a generic $G$ over $V$ cannot exist whenever $P$ satisfies the following property: For every $p \in P$ there exist $q \le p, r \le p$ such that $q$ and $r$ are incompatible. (To see this it suffices to realize that for every filter $F$ on $P$ the set {$ p \in P \quad : p \notin F$} is dense, thus a generic $G$ leads to a contradiction). The existence of a generic $G$ does not change the Boolean value of $\varphi$, so I think that the argumentation in the break above remains valid although $G$ cannot exist.

However it seems strange to me that one assumes the existence of thing that must not exist, to faciliate calculation. One can say that this resembles the complex numbers but to me adding a new element to the reals is way weaker than adding an element that actually must not exist to the universe.

So my question is: is the just described strategy of showing the relative consistency of a statement really valid?

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    $\begingroup$ I'm sure that other people will give longer answers. But since $F$ is the complement of $\{p \in P : p \not \in F\}$ in $P$, if $F$ is not in $V$ then that set won't be in $V$ either. So the contradiction you mention only happens if you also assume our $V$ contains every set. In the sort of systems where you can consistently force over $V$ (e.g. the Gitman/Hamkins multiverse) that will not be the case. $\endgroup$ – Carl Mummert Mar 19 '11 at 0:39

The existence of a generic $G$ over $V$ is indeed impossible in $V$ (for nontrivial forcing notions), but it has truth value 1 in appropriate Boolean-valued models. In more detail: If $P$ is a partially ordered set (to be used as a notion of forcing) and $B$ is the complete Boolean algebra of regular open subsets of $P$, then the following paragraph is true (i.e., has truth value 1) in the Boolean-valued model $V^B$:

There is a transitive class $\check V$ that contains all the ordinals and satisfies all the sentences true in the original ground model $V$ (so it can serve as a "copy" of $V$ in $V^B$). There is a subset $G$ of $P$ generic over $\check V$. Every set is the value of some forcing name in $\check V$ with respect to $G$.

Together, these say that, with truth value 1, the universe (of $V^B$) is a $P$-generic extension $\check V[G]$ of the ground model. So when people pretend to move to a (nonexistent) $P$-generic extension of $V$, a correct interpretation of this is that they move to $V^B$ and that whatever they assert about $V[G]$ is really asserted to have truth value 1 in this Boolean-valued model.

(Technicalities: "All the sentences true in the ground model $V$" should really be specified by a theory $T$, which should include a name (or at least a definition) for $P$ so that one can talk, in $V^B$, about the $P$ that lives in $\check V$.)

  • $\begingroup$ Andreas, I think I fixed your TeX problems. $\endgroup$ – François G. Dorais Mar 19 '11 at 2:34

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