# Problem with forcing

Recently I read throught Jech's proof of the independence of the continuum hypothesis.

However there was something that bothered me a lot, the whole idea of the generic filter. Beginning the section of forcing in Jech's set theory the author stated that Cohen's original approach was to assume the existence of a countable transitive model of $\mathsf{ZFC}$, and then get a generic filter with respect to the required forcing conditions, then the author states that this cannot be done since such model canoot be proved to exists in $\mathsf{ZFC}$ unless $\mathsf{ZFC}$ is inconsistent, later stating that the usual way of working with forcing was postulating the existence of the generic filter for the required forcing notion.

I found Jech's approach really incomplete, so I searched on the internet and saw the following theorem:

If $\Lambda\subseteq ZFC$ is finite, then $\mathsf{ZFC}\vdash\exists M[M \text{ is transitive}\wedge|M|=\aleph_0\wedge M\models ZC\cup \Lambda].$

Then I argued as follows:

Suppose $\mathsf{ZFC}\vdash\mathsf{CH}$, then there exists some finite $\Lambda\subseteq\mathsf{ZFC}$ such that $\Lambda\vdash\mathsf{CH}$.Let $P$ be the notion of forcing such that for any generic $G$ of $P$ we have $M[G]\models2^{\aleph_0}\geq\aleph_2$. Let $G$ be generic over $M$; such $G$ exists as $M$ is countable.

Reading Jech's proof of the generic model theorem we can get a finite $\Omega\subseteq\mathsf{ZFC}$ such that $\Omega\supseteq\Lambda$ and if $M\models\Omega\cup\mathsf{ZC}$, then $M[G]\models\Lambda\cup\mathsf{ZC}$; the extension is required only for the different instances of the replacement axiom in $\Lambda$. Then clearly this is a contradiction as there exists such a countable $M$. Therefore $\mathsf{ZFC}\nvdash\mathsf{CH}$.

Is my approach correct?

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Another approach, which is actually what Jech does in every book of him that I read so far (counting three by now), is using Boolean valued models of set theory. This is a definable class which is a model of set theory where truth values are not $0-1$, but rather taken from a complete Boolean algebra. Then one can show that if $\varphi$ is a statement whose truth value in some Boolean valued model is neither $0$ nor $1$, then we cannot prove it from $\sf ZFC$.
Using Jech's proof of forcing lemma, how can it be seen that if $1\neq||\phi||\neq 0$, then $\phi$ is forced by some condition? – Forcing Aug 27 '13 at 0:04
That's right. However whether or not this condition is in the generic may vary. For example, by adding a single Cohen real $c$, whose canonical name is $\dot c$, one can ask whether or not $\check n\in\dot c$. It will be forced by some conditions, but its negation may be forced by other conditions, and it depends on the generic whether or not this statement ends up being true. – Asaf Karagila Aug 27 '13 at 0:09
But I notice that in order to prove that $||\phi||\neq 0$ we have to prove that $V[G]\models\phi$, so how do we know that the statement $"V[G]\models\phi"$ is sound?, or is there a way to show that $||\phi||\neq 0$ without postulating the existence of $G$? – Camilo Arosemena Sep 4 '13 at 14:42