# Forcing and generic extensions

The following is from Halbeisen, page 289, on generic extensions:

This leads to one of the key features of forcing: By knowing whether a certain condition $p$ belongs to $G \subseteq P$, people living in $\mathbf V$ can figure out whether a given sentence of the forcing language is true or false in $\mathbf V [G]$. Moreover, it will turn out that people living in $\mathbf V$ are able to verify that in certain models $\mathbf V [G]$ all axioms of $ZFC$ are true.

I have three questions related to this passage.

One is: Why is it desirable to verify the truth or falsity of statements in $\mathbf V[G]$ from within $\mathbf V$? Why would it not be good enough to verify them in $\mathbf V [G]$?

Two: Do I understand correctly that people in $\mathbf V$ use $P$ names to verify that $ZFC$ holds in $\mathbf V [G]$? (the passage does not talk about this)

Three: I thought that if $\mathbf V$ satisfied $ZFC$ then so would $\mathbf V [G]$ but the passage says only "certain models $\mathbf V [G]$" satisfy all axioms of $ZFC$. Which do and which don't?

-
The trouble is that $\mathbf{V}[G]$ is a mathematical fiction (at least from the perspective of $\mathbf{V}$) and so we must first define what it means for something to be true or false in $\mathbf{V}[G]$. –  Zhen Lin Jan 27 '13 at 10:11
@Zhen: But the question, as I read it, is that if we work externally to the $\bf V$ (which is really just countable), why do we need to do the verification internally to $\bf V$ and not externally as well? –  Asaf Karagila Jan 27 '13 at 10:57
You could do that if $\mathbf{V}$ is really just a countable transitive model. But we could equally well take $\mathbf{V}$ to be the actual universe of all sets. –  Zhen Lin Jan 27 '13 at 11:43

1. This is the question I am most shaky on, so forgive me if this makes no sense. The desirability of people in $\newcommand{\V}{\mathbf V}\V$ being able to verify the truth of statements holding in $\V [ G ]$ is that otherwise it may not be verifiable that $\V [G]$ satisfies ZFC for $G$ $P$-generic over $\V$. Consider Separation:

Let $X , a \in \V [G]$, and suppose $\phi ( x , y )$ is some formula. We wish to construct $Y \in \V [G]$ such that $$\V [ G ] \models ( \forall x ) ( x \in Y \leftrightarrow ( x \in X \wedge \phi ( x,a ) ).$$ We know that $X$ has a name $\dot{X}$ in $\V$, and similarly $a$ has a name $\dot a$ in $\V$. We may then consider $$\sigma = \{ \langle \tau , p \rangle \in \mathrm{dom} ( \dot{X} ) \times P : p \Vdash ( \tau \in \dot{X} \wedge \phi ( \tau , \dot a ) ) \}.$$ It should be clear $\sigma$ is a $P$-name, and that under the assumptions stated above $Y = \sigma [G]$ would have the desired property. However if $\Vdash$ were not definable in $\V$ (meaning that there is no formula $\psi$ such that $(\psi ( p , \tau , \dot a ) )^{\V} \Leftrightarrow p \Vdash ( \phi (\tau , \dot a)$), then $\sigma$ need not be in $\V$ (and hence $Y$ itself need not be in $\V [G]$). Without appealing to some property in $\V$ I honestly don't know how to in general construct the desired $P$-name in $\V$ (i.e., the desired set in $\V [G]$).

The definability of $\Vdash$ in $\V$ also allows for the likelihood of using combinatorial properties of the forcing notion $P$ (which can be verified in $\V$) to determine truth of certain statements in $\V [G]$. For example, if $P$ is ccc (meaning that people in $\V$ think $P$ is ccc), then every cardinal in $\V$ remains a cardinal in $\V [ G ]$. If there was no way to determine truth in $\V [ G ]$ from $\V$ then it would be doubtful that there would be a strong connection between the combinatorics of $P$, and truth in $\V [ G ]$. It turns out that using these combinatorial properties is central to many forcing proofs.

2. As people in $\V$ only have access to $\V [ G ]$ via the $P$-names, they use $P$-names to verify $\V [ G ] \models \phi$. In particular, the Pairing Axiom is verified in $\V$ to hold in $\V [ G ]$ as follows:

Let $\sigma , \tau$ be any two $P$-names (in $\V)$. Construct the following $P$-name: $\theta = \{ \langle \sigma , \mathbf 0 \rangle , \langle \tau , \mathbf 0 \rangle \}$. By applications of the Pairing Axiom in $\V$ we know that $\theta \in \V$, and it is easy to see that $\theta$ is a $P$-name. Furthermore, given any filter $G \subseteq P$, since $\mathbf 0 \in G$ it follows that $$\theta [ G ] = \{ \sigma [ G ] , \tau [ G ] \}.$$ As all objects in $\V [ G ]$ have $P$-names in $\V$, it follows that $\V [ G ] \models \text{Pairing}$.

As above, combinatorial properties of $P$ may also be used. But at some point someone must have verified the connection between the combinatorics of $P$ and truth in $\V [G]$ (often by using these combinatorial properties to manipulate $P$-names in an appropriate manner).

3. Note that $\V [ G ]$ need not satisfy all axioms of ZFC for arbitrary $G \subseteq P$, even arbitrary filters $G \subseteq P$. Kunen's text states that Extensionality, Foundation, Pairing and Union all hold under the assumption that $G$ is just a filter in $P$. Using the countable transitive model approach to forcing, by coding a well-ordering of $\omega$ of order-type $> o(V)$ you can construct a filter $G \subseteq P$ such that $V[G]$ does not satisfy Replacement (as noted in this previous answer).

It is true that for every $P$-generic filter $G$ over $\V$ that $\V [ G ]$ will satisfy ZFC, but this notion appears to be introduced in the next section of Halbeisen's text (even though I have mentioned it copiously above).

-
To add on the first part, when we construct the real numbers as Dedekind cuts, and we want to verify that addition and multiplication are well-defined and continuous, we have to do that in terms of sets of rational numbers because even though we know what are the objects in the new structure we don't have that many ways of knowing how they behave. Similarly when adding a generic set we don't really know what is the result except for everything which is true is forced by a condition from $\bf V$. This amounts to saying that we can usually verify only from the ground model that everything holds. –  Asaf Karagila Jan 27 '13 at 10:56
@Asaf: That's a nice analogy! –  Arthur Fischer Jan 27 '13 at 11:12
In the definition of $\sigma$, did you mean $\mathrm{dom}$ of capital $x$ dot? –  Matt N. Mar 27 '13 at 20:46
@MattN.: I think you might be right. I'll give this a good read before making any edits. –  Arthur Fischer Mar 27 '13 at 21:02