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It seems that Jensen's proof of the consistency of CH + SH used class forcing, but the revelant properties are not clearly verified. I haven't learnt about class forcing, so I wonder whether it is justified. It is on Devlin's The Souslin Problem, p97-99.

First, the forcing notion $\mathbb{C} = \{(\alpha, A) : \alpha < \omega_1, A \subseteq \omega_1 \text{ is a club}\}$. We assume the ground model $V \models 2^\omega = \omega_1$ and $2^{\omega_1} = \omega_2$, so $\mathbb{C}$ is $\omega_1$-closed, has $\omega_2$-cc and size $\omega_2$. For any generic filter $G$, $C = \bigcup\{\alpha \cap A : (\alpha, A) \in G\} = \bigcap\{A: (0,A) \in G\}$ is the generic subset of $\omega_1$.

Also, for any transitive set $U$ and a poset $P \in U$, a filter $G \subseteq P$ is called $(U,P)$-generic if $G$ meets all dense subset of $P$ that is first-order definable in the model $\langle U, \in, x \rangle_{x \in U}$.

Take any countable $N \prec H_{\omega_2}$ and let $\pi: N \to \bar{N}$ be the transitive collapse. So we have $\pi(\omega_1) = \alpha_N = N \cap \omega_1$ and $\pi(\alpha, A) = (\alpha, A \cap \alpha_N)$ for all $(\alpha, A) \in N\cap \mathbb{C}$. Since $|\mathbb{C}| = \omega_2$ it is a proper class of $H_{\omega_2}$, and $\pi(\mathbb{C}) = \pi[\mathbb{C} \cap N]$ is also a proper class of $\bar{N}$.

Now we take a countable transitive set $U$ with $\bar{N}, \pi(\mathbb{C}) \in U$. What happens is this:

Take any $(U,\pi(\mathbb{C}))$-generic filter $G$, there is a $p = (\alpha_N, C) \in \mathbb{C}$ that is a lower bound of $\pi^{-1}[G]$ and $C \cap \alpha_N$ is a $(U,\pi(\mathbb{C}))$-generic subset of $\alpha_N$. And this claim is made:

$\bar{N}[C \cap \alpha_N] \models \pi(\phi) \iff \exists q \in G (q \Vdash_{\pi(\mathbb{C})} \pi(\phi)) \implies \exists q \in G(\pi^{-1}(q)) \Vdash _\mathbb{C} \phi$, where $\phi$ is a sentence with parameters only in $\{\check{x} : x \in N\} \cup \{\dot{C}\}$.

Since $\pi(\mathbb{C})$ is a proper class of $\bar{N}$ I wonder whether this claim is true.

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There is no real class forcing issue here. The phrase “$\pi(\mathbb{C})$ is a proper class of $\overline{N}$” only means that $\pi(\mathbb{C})$ is not an element of $\overline{N}$ even though $\pi(C)$ is of course a genuine set in $V$.

Devlin’s setup is:

In $V$ you have a forcing notion $\mathbb{C} = \{(\alpha, A): \alpha < \omega_1 \land A \subseteq \omega_1 \text{ is club}\}$, of size $\omega_2$. So $\mathbb{C}$ is an ordinary set forcing.

You take a countable $N \prec H(\omega_2)$, collapse it to a transitive $\overline{N}$ via $\pi$. Then $\pi(C) = \pi[C \cap N]$ is a subset of $\overline{N}$ that is not itself in $\overline{N}$ so it is convenient to call it a “proper class of $\overline{N}$”.

To actually do forcing, Devlin chooses a countable transitive $U$ with $\overline{N}, \pi(\mathbb{C}) \in U$. In $U$, the same $\pi(\mathbb{C})$ is an element, so you are just doing standard set forcing with $P = π(\mathbb{C}) \in U$.

A filter $G \subset P$ is $(U,P)$-generic if it meets all dense subsets of $P$ that are first-order definable in $\langle U, \in, x \rangle_{x \in U}$. Since $U$ is countable, one can build such a $G$ in the usual way, and the forcing relation “$q$ forces $\psi$” is defined inside $U$ by the standard recursion. Thus the usual forcing theorem holds over $U$:

$$U[G] \models \psi \iff \exists q \in G (q \Vdash \psi)$$

for sentences $\psi$ with parameters from $U$.

In your situation, $\phi$ is a sentence with parameters in $N$ and the canonical name for the generic club. After collapsing, $\pi(\phi)$ is a sentence with parameters in $\overline{N}$ and $C \cap \alpha_N$. The model $\overline{N}[C \cap \alpha_N]$ sits inside $U[G]$, so

$$\overline{N}[C \cap \alpha_N] \Vdash \pi(\phi)$$

is a statement about $U[G]$, and by the forcing theorem over $U$ this is equivalent to

$$\exists q \in G (q \Vdash_{\pi(\mathbb{C})} \pi(\phi))$$

Finally, since $N \prec H(\omega_2)$ and $\pi$ is the collapse, you can pull $q$ back to $p = \pi^{-1}(q) \in \mathbb{C}$ and get

$$p \Vdash_{\mathbb{C}} \phi$$

So the argument Devlin sketches is just ordinary set forcing over a countable transitive $U$; the fact that $\pi(\mathbb{C})$ is a “proper class of $\overline{N}$” does not cause any problem and does not require genuine class forcing.

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    $\begingroup$ I converted your answer to LaTeX, I advise you use it in the future too. However, I can’t check the actual logic. $\endgroup$ Commented Nov 17 at 9:13
  • $\begingroup$ I haven't seen this definition of (U,P)-generic before, but here U is only demanded to be transitive. Would this suffice to garantee the forcing theorem? And is this U[G] defined by intepreting all its definable names? $\endgroup$ Commented Nov 17 at 10:29
  • $\begingroup$ Yes. Here it is enough that $U$ is a transitive set with $P\in U$. The forcing relation for $P$ is defined in the ambient $V$, the forcing theorem is proved there, and then applied to $\langle U,\in\rangle$ and any $(U,P)$-generic $G$. The extension $U[G]$ is defined in the usual way, by interpreting all $P$-names in $U$, not just definable ones. May I ask what your main goal is here, what you are trying to solve? $\endgroup$ Commented Nov 17 at 11:36
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    $\begingroup$ Sorry but I still have a little question. The last part I guess is like this: $$U[G] \models (\bar{N}[C \cap \alpha_N] \models \pi \phi) \implies \exists q \in G (q \Vdash_{\pi \mathbb{C}} (\pi \phi)^{\bar{N}[\dot{C} \cap \alpha_N]})^U \implies (\pi^{-1}q \Vdash_\mathbb{C} (\phi)^{N[\dot{C}]})^V \implies (\pi^{-1}q \Vdash_\mathbb{C} (\phi)^{H_{\omega_2}[\dot{G}]})^V.$$ But how can we get the second implication in this formula? Does the $\pi^{-1}: \bar{N} \prec H_{\omega_2}$ suffice? $\endgroup$ Commented Nov 18 at 8:38
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    $\begingroup$ Yes, in a sense you can relativize this forcing statement to $\bar N$. Since $\bar N$ is transitive and $\pi(\mathbb C)\in\bar N\subseteq U$, one can define an internal relation $\Vdash^{\bar N}{\pi(\mathbb C)}$ by the same recursion, and for conditions and names in $\bar N$ we have $q\Vdash^{\bar N}{\pi(\mathbb C)}\psi \iff q\Vdash^{U}{\pi(\mathbb C)}\psi$ by absoluteness of the defining formula. But in the argument I preferred to phrase the transfer via the elementary embedding $i\circ\pi^{-1}:\bar N\to H{\omega_2}$ and the fact that $\Vdash$ is first-order definable there. $\endgroup$ Commented Nov 20 at 14:34

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