Say we have a model of set theory $V$ and a partially ordered structure $\mathbb{P}$, and I want to talk about a $V$-generic filter $G$. A $V$-generic filter is a filter such that for every $D\in V$ such that $D\subset \mathbb{P}$, $G\cap D \ne \emptyset$.
Now, for example, say I am talking about the partially ordered structure $(\mathbb{P}, R) = (\omega^{<\omega}, R)$ in $V$ with the upside-down partial order .
Consider
$D_n = \{ t \in ω
<ω
| \text{dom}(t) ≥ n \}$, these sets are dense, so a $V-$generic filter $G$ meets each $D_n$. This $V-$generic filter $G$, associated with $\mathbb{P}$ is not in $V$, because if it were, its complement in $\mathbb{P}$ is now recognized by $V$ and is dense so it meets its complement, a contradiction. Set $g=\bigcup G$ . It's an element of the Baire Space that wasn't there before!
My Question :
My professor basically stops here in the lecture and in the class notes. I want to make sure that I understand correctly. When my professor says $g$ is an element of the Baire Space that wasn't there before, does he mean that we can create a new model of set theory W by adjoining $g$ to $V$ and this W is still a model of set theory? And assuming I'm right, do you think he will explain how this process works and why $W$ is a model of set theory in the next lecture? Is it just somehow obvious that adjoining $g$ to $V$ to make $W$ is just fine and dandy and it requires very little explanation? I don't get it...
Yes, that is correct. We adjoin $g$ to $V$ and generate a space. Or rather, we adjoin the filter $G$, and generate a new model, and then we define $g$ from $G$. In some cases, you can define $G$ from $g$, and therefore it doesn't matter which of the two you will add to $V$.
Why is this a model of $\sf ZFC$? This is a nontrivial theorem, which you can find in any book which presents forcing seriously (Kunen, Jech, Halbeisen). But the gist is that the fact that $G$ is not just some arbitrary set, but rather a very particular type of set, a generic filter, then this ensures the resulting model is a model of $\sf ZFC$.
Of course that just adding any good old set won't necessarily allow us to extend $V$ into a model of $\sf ZFC$. Think about $g$ which encodes a bijection between $V$ and $\omega$ (let's assume that $V$ is a countable model, at least for a moment), then there is no way to add $g$ without adding ordinals, and we might not be able to do so (and we certainly can't do that with forcing alone).