Consistency of $\lnot$GCH with ZF In Chapter 14 of Jech's Set Theory, he shows that the generalized continuum hypothesis (GCH) is independent of ZF by showing the following:

Theorem 14.32. There is a generic extension $V[G]$ that satisfies $2^{\aleph_0}>\aleph_1$.
Proof. We describe the notion of forcing that produces a generic extension with the desired property. Let P be the set of all functions $p$ such that
$$(14.33) \quad \text{(i) $\text{dom}(p)$ is a finite subset of $\omega_2\times \omega$, and (ii) $\text{ran}(p)\subseteq \{0,1\},$}$$
and let $p$ be stronger than $q$ if and only if $p\supseteq q$.
If $G$ is a generic set of conditions, we let $f=\bigcup G$. We claim that
$$(14.34) \quad \text{(i) $f$ is a function, and (ii) $\text{dom}(f)=\omega_2\times \omega$.}$$
(Of course, $\omega_2$ means $\omega_2$ in the ground model.)
Part (i) of (14.34) holds because $G$ is a filter. For part (ii), the sets $D_{\alpha,n}=\{p\in P:(\alpha,n)\in \text{dom}(p)\}$ are dense in $P$, hence $G$ meets each of them, and so $(\alpha,n)\in \text{dom}(f)$ for all $(\alpha,n)\in \omega_2\times \omega$.
Now, for each $\alpha<\omega_2$, let $f_\alpha:\omega\to \{0,1\}$ be the function defined as follows:
$$f_\alpha(n)=f(\alpha,n).$$
If $\alpha\neq \beta$, then $f_\alpha\neq f_\beta$; this is because the set
$$D=\{p\in P: p(\alpha,n)\neq p(\beta,n) \text{ for some $n$}\}$$
is dense in $P$ and hence $G\cap D\neq \varnothing$. Thus in $V[G]$ we have a one-to-one mapping $\alpha\mapsto f_\alpha$ of $\omega_2$ into $\{0,1\}^{\omega}$.
Each $f_\alpha$ is the characteristic function of a set $a_\alpha\subseteq \omega$. As in Example 14.2, we call these sets Cohen generic reals. Thus $P$ adjoins $\aleph_2$ Cohen generic reals to the ground model.

I'll take a moment to stop here and ask a few questions.

*

*I understand that if a generic set $G$ exists, then everything after it follows, but why is it possible that $G$ exists? As far as my current understanding goes, if we have a countable ground model containing $P$, then there exists a generic set over $P$. However, in this case $P$ is very much not countable, and as such I don't see why it has to be possible for $G$ to exist.


*Are we making use of any of the developed forcing machinery (Generic Model Theorem, Forcing Theorem, etc), other than the fact that $G\in V[G]$, for this part of the proof? It doesn't seem like we're using anything relating to the forcing relation here, although it's possible that it's just buried implicitly in the reasoning somewhere. (If so, could someone explain where and how?)
The proof then continues as follows:

The proof of Theorem 14.32 is almost complete, except for one detail: We don't know that the ordinal $\omega_2^V$ is the cardinal $\aleph_2$ of $V[G]$. We shall complete the proof by showing that $V[G]$ has the same cardinals as the ground model ($P$ preserves cardinals).
Definition 14.33. A forcing notion $P$ satisfies the countable chain condition (c.c.c.) if every antichain in $P$ is at most countable.
The following theorem is one of the basic tools of forcing:
Theorem 14.34. If $P$ satisfies the countable chain condition then $V$ and $V[G]$ have the same cardinals and cofinalities.
Proof. It suffices to show that if $\kappa$ is a regular cardinal then $\kappa$ remains regular in $V[G]$. Thus let $\lambda<\kappa$; we show that every function $f\in V[G]$ from $\lambda$ into $\kappa$ is bounded.
Let $\dot f$ be a name, let $p\in P$ and assume
$$(14.35)\quad p\Vdash \text{$\dot f$ is a function from $\check \lambda$ to $\check \kappa$}.$$
For every $\alpha<\lambda$ consider the set
$$A_\alpha=\{\beta<\kappa:\exists q<p \text{ } q\Vdash \dot f(\alpha)=\beta\}.$$
We claim that every $A_\alpha$ is at most countable: if $W=\{q_\beta:\beta\in A_\alpha\}$ is a set of witnesses to $\beta\in A_\alpha$ then $W$ is an antichain, and therefore countable by c.c.c. Hence $|A_\alpha|\leq \aleph_0$.
Now, because $\kappa$ is regular, the set $\bigcup_{\alpha<\lambda} A_\alpha$ is bounded, by some $\lambda<\kappa$. It follows that for each $\alpha<\lambda$, $p$ forces $\dot f(\alpha)<\gamma$.
Thus for every $\dot f\in V^P$ and every $p\in P$, if (14.35) then $p\Vdash \dot f$ is bounded below $\kappa$. It follows that in $V[G]$, every function $f:\lambda\to \kappa$ is bounded.

This is another natural stopping point, and I have a few more questions:


*When the earlier part said we are adding $\aleph_2$ Cohen reals to the ground model, is that $V$'s $\aleph_2$ or $V[G]$'s $\aleph_2$? I understand these end up being the same, but which one are we doing a priori?


*The forcing theorem makes an appearance here in the last sentence, namely linking the fact that $p$ forces "$\dot f$ is a function from $\check \lambda$ to $\check \kappa$ implies $\dot f$ is bounded in $\kappa$" to the fact that "all functions from $\lambda$ to $\kappa$ are bounded" holds in $V[G]$. Is this a correct understanding of what is going on here?


*I understand the individual steps in this proof, but I'm still not quite sure why this requires all cardinals to be preserved. My question here addresses why all cofinalities are preserved, but it's not totally clear to me why this means cardinals are preserved as well. I also don't have a particualarly great picture of how cardinals can change under different circumstances, but that may be something worth addressing at a later time.
He then finishes the proof by showing that this particular choice of $P$ satisfies c.c.c, but I don't have any questions on that part so I'll leave it at this.
 A: *

*As you have noted, formally forcing is done over a countable transitive model of ZFC, say $M$. Note that when one describes a forcing, it's meant to be computed from the point of view of $M$. So in $M$ you look at the set of all finite partial functions from $\omega_2^M\times\omega^M$ into $\{0, 1\}^M$. Now note that by absoluteness $\omega^M = \omega$ and $\{0, 1\}^M=\{0, 1\}$, but what $M$ thinks is $\omega_2$, is not the true $\omega_2$; it is just a countable ordinal. So $P^M$ is countable and by the Rasiowa-Sikorski lemma you can get generics over $M$. Another thing to note is that in practice, people usually don't care and force over $V$ and assume generics exist floating around somewhere, which is done in order to avoid talking about countable transitive models and get to the main point.

*Yes, because we want to get a model of ZFC in the end and not some random structure, so we need the fundamental theorems to guarantee that $V[G]$ is in fact a model of ZFC.

*As a priori, as you can see from the definition of the forcing, it is the $\aleph_2$ of $V$. But as you point out, later on we see that by ccc, $\aleph_2^V = \aleph_2^{V[G]}$.

*Yes, precisely. So if something is forced, it's true in the extension and vice versa.

*Well if all cofinalities are preserved, then regular cardinals are preserved because being regular means exactly $\kappa= cof(\kappa)$, but if $cof(\kappa)^V = cof(\kappa)^{V[G]}$, then $\kappa$ is still regular in $V[G]$ and so is a cardinal as well. But now all singular cardinals are preserved as well, since each singular cardinal is a limit of regular cardinals and each of those gets preserved, and so in $V[G]$ it's still a limit of regular cardinals. So it must be a cardinal in $V[G]$. This shows that all cardinals are preserved. For the last part of your question, note that being a cardinal only means not being equinumerous with a smaller ordinal, so if you manage to add such bijections to from something smaller to your current cardinal, it will fail to be a cardinal in $V[G]$. For example the forcing $P$ which has as it's elements finite partial functions from $\omega$ to $\omega_1^V$, will collapse $\aleph_1^V$ to become countable in $V[G]$.

