I'm trying to understand some concept of the forcing idea, but I'm having trouble with two questions.
assume that M is countable s.t. $M\models ZFC$, $\mathbb{P}$ is cohen forcing, $\mathbb{P}\in M $,and G is an $\mathbb{P}$-genric filter over M.
- i already know that $\bigcup G$ is a function $f:\omega \to 2$, why does $f$ cannot be the zero function?
- why does $G\notin M$?
for 1 i know that $$D=\left\{ p\in\mathbb{P}:1\in\text{rng}\left(p\right)\right\} $$ is dense in $\mathbb{P}$ but i was unable to proof that $D\in M$. $$$$ Thanks a lot
Density, density, density. Almost all forcing arguments of this sort boil down to finding the appropriate dense sets.
For example, why isn't $f$ constant? Because given any $p$, there is an extension of $p$ which takes both $0$ and $1$, which one? Well, for example, just add "on top" of $p$ these two values.
Therefore, the set $\{p\mid \operatorname{rng}(p)=\{0,1\}\}$ is dense. Since it is definable from a formula, this dense set is in the ground model, so $G$ meets it, and therefore some $p\in G$ must have both $0$ and $1$ in its range. Therefore $f$ is not constant.
Similarly, $G$ can't be in $M$ itself, because if $r\in M$, then $\{p\mid\exists n: p(n)\neq r(n)\}$ is a dense set, for the same reason as above. And it is in $M$ because it is definable from $r$, and $M$ is a model of $\sf ZF$, or at least a large enough fragment of it. So this dense set is in $M$, and therefore it meets $G$, and therefore $f$, the generic real, must be different from $r$, for any $r\in M$, and so $f\notin M$.