I learned from Kunen's book, besides forcing over countable transitive model (c.t.m.), there is an another way to do forcing, called the "syntactic method", i.e. forcing over V. Fixed a partial order P in the ground model. When one does forcing over a c.t.m. M, it is easy that for each p in P, there is a filter G which is Pgeneric over M and contains p. when we use the "syntactic method", can we alway assume there is a filter G which is Pgeneric over V and contains arbitrary given p in P ?
Let M be a c.t.m. for ZFC, P in M, G generic over M. Let A,B in M, r be a Pname for a function in M[G] from A into B, for each a in A, choose p(a) in P and b(a) in B s.t. p(a) r(aˇ)=b(a)ˇ, let f(a)=b(a), then can we deduce that f is a fuction in M from A into B ?

$\begingroup$ I added the settheory tag and capitalization. $\endgroup$ – Joel David Hamkins Aug 21 '10 at 12:27
One can speak about the independence results without models as follows. Let $(P,\leq)$ be a poset, with 1 as the largest element. There is a relation, called forcing, which is defined by formulainduction for all formulas, and which has the property that all axioms of ZFC are forced by 1, if some formula $\phi$ follows from ZFC then it is forced by 1. Now, if, for a particular $\phi$ we find a poset $(P,\leq)$ such that 1 does not force $\phi$, then $\phi$ is independent from ZFC (or there is a contradiction in ZFC). This does not work with models (it actually does assume the existence of models of ZFC, as that is equivalent to ZFC being noncontradictory by Godel), the argument is proof theoretic, however, all steps are essenitally the same as in the correspodning model theoretic argument.
The syntactic method is simply to work inside the forcing relation, by looking at what is forced by which conditions. That is, rather than actually building the forcing extension $V[G]$, you reason as though you were inside it, by means of the forcing relation. For example, if you have condition $p$ forcing various statements $\varphi_1$, ... $\varphi_n$, and from those statements you can deduce $\psi$, then you can conclude that $p$ also forces $\psi$. In particular, note that any condition $p$ forces that $\dot G\subset\check P$ is $\check V$generic and contains $\check p$, so syntactically, you have the filter you want in the sense I described. The point is that you don't actually need to build an actual model using an actually generic filter to gain insight into what is true there, because you have the forcing relation telling you what is forced to be true there.
If you want to build an actual model by forcing over $V$, then the thing to do is to use Booleanvalued models $V^B$, where there is Booleanvalued truth. You can quotient this $B$valued structrue by an ultrafilter $U$ (no need for any genericity) to arrive at an actual first order 2valued structure $V^B/U$, which satisfies a version of Los' theorem: $V^B/U\models\phi$ if and only if $[\phi]\in U$.
For question 2, if you make your choice of $b(a)$ $p(a)$ inside $M$, then of course $f$ is a function in $M$. But there is little reason to expect that this function $f$ has much relation to the function in $M[G]$ named by $r$, since the conditions $p(a)$ might be incompatible. If you can choose a single $p=p(a)$, then it follows that $p$ forces $r=\check f$.


$\begingroup$ Oh well, I just happened to get set up at my cafe here and checked in to MO, and what do I finda question on forcing! $\endgroup$ – Joel David Hamkins Aug 21 '10 at 12:20

$\begingroup$ Tkanks for your answer :) but i am not sure about "if you make your choice of b(a) p(a) inside M , then of course f is a function in M" , How do we know f is inside M ? $\endgroup$ – sonicyouth Aug 22 '10 at 1:29

$\begingroup$ Because you defined it inside $M$. That is, if you choose $b(a)$ and $p(a)$ by using a choice function that exists in $M$ , then the function $f$ will be in $M$. The key fact is that the forcing relation is definable inside $M$; this is one of the fundamental forcing lemmas. And so $M$ knows when a condition $p$ forces $r(\check a)=\check b$ or not. $\endgroup$ – Joel David Hamkins Aug 22 '10 at 1:48