# Class forcing as set forcing followed by truncation

My question arises from thinking about how we can obtain class forcing extensions from truncations of set forcing extensions in the presence of a (strongly) inaccessible cardinal in the following sense. Let $\kappa$ be an inaccessible cardinal and let $\mathbb P$ be a forcing such that $\mathbb P \subseteq V_{\kappa}$. Now we can consider $\mathbb P$ simultaneously as a set forcing (in $V$) and as a class forcing over $V_{\kappa}$. Now let $G$ be a $(V;\mathbb P)$-generic filter. Then we can ask how $V[G]$ and $V_{\kappa}[G]$ relate to each other, where in this context $V_{\kappa}[G] := \{ \sigma_G : \sigma\in V^{\mathbb P}\cap V_{\kappa} \}$.

A naive approach would be to conjecture that $V_{\kappa}[G] = (V_{\kappa})^{V[G]}$. This is of course not true in general: if $\mathbb P$ is the Levy collapse $\mathrm{Col}(\omega, <\kappa)$, then, e.g., we have $\wp^{V[G]}(\omega) \in (V_{\kappa})^{V[G]}\setminus V_\kappa[G]$, the underlying problem being that $(V_{\kappa})^{V[G]}$ is a model of the power set axiom, whereas $V_{\kappa}[G]$ is not.

We can remedy this last problem by considering $H_{\kappa}$ instead of $V_\kappa$ in the above argument (which is the same in the ground model); and indeed we get the following

Fact: Let $\kappa$ be inaccessible and $\mathbb P\subseteq H_{\kappa}^V$ a forcing preserving the regularity of $\kappa$. Let $G$ be a $(V;\mathbb P)$-generic filter. Then $H_\kappa^V[G] = (H_\kappa)^{V[G]}$.

Proof. The difficult direction is to show that $(H_\kappa)^{V[G]} \subseteq H_\kappa^V[G]$. For this let $x\in (H_\kappa)^{V[G]}$ and assume (inductively) that $x\subseteq H_\kappa^V[G]$. Then let $\sigma\in V^{\mathbb P}$ be such that $\sigma_G = x$ and $\mathrm{dom}(\sigma) \subseteq V^{\mathbb P}\cap H_{\kappa}^V$. In $V[G]$ we define a function $f: x\to \kappa$ by setting $$f(y) := \min\{ \alpha < \kappa : \exists\langle \tau, p\rangle\in \sigma\cap V_{\alpha+1} (p\in G \wedge \tau_G = y) \}$$
for any $y\in x$. Now note that $fx$ is bounded below $\kappa$, since $\kappa$ is regular in $V[G]$ and by assumption there can be no surjection from $x$ onto $\kappa$. Thus take $\alpha < \kappa$ such that $fx\subseteq \alpha$ and let $\sigma' := \sigma\cap V_{\alpha}$. Then $\sigma'_G = \sigma_G = x$ and $\sigma' \in V^{\mathbb P}\cap H^V_{\kappa}$ by inaccessibility of $\kappa$ in $V$. Therefore $x\in H_\kappa^V[G]$. $\Box$

This proof depends on the fact that $\mathbb P$ preserves the regularity of $\kappa$, but on the other hand I could not construct a counterexample for a forcing singularizing $\kappa$. This leads me to ask the following

Question: Let $\kappa$ be inaccessible. Can there be a forcing $\mathbb P\subseteq H_{\kappa}^V$ that does not collapse $\kappa$ and such that there is some $(V;\mathbb P)$-generic filter $G$ with $H_{\kappa}^V[G] \neq (H_{\kappa})^{V[G]}$ ?

• Singularizing an inaccessible cardinal requires large cardinal hypotheses well beyond the existence of a strong inaccessible. Zero sharp is certainly a lower bound, and the only way I know of doing it is with Prikry forcing (which requires a measurable). For your question, you want $\mathbb P \subseteq H^V_\kappa$, which is not true of Prikry forcing. So a sub-question of your main question that I already don't know the answer to is Can there be a forcing $\mathbb P \subseteq H^V_\kappa$ that singularizes $\kappa$ without collapsing it? – Will Brian Nov 7 '16 at 16:34
• Welcome to MathOverflow, you bespectacled redhead! – Asaf Karagila Nov 7 '16 at 17:13

## 1 Answer

No, there can be no such forcing. The point is that such a forcing should have size $\kappa$ and by my proof given in Singularizing forcing of "small" cardinality?, we have $\Vdash_{\mathbb{P}} |\kappa|=cf(\kappa)$, so $\mathbb{P}$ can not change the regularity of $\kappa$ as else it collapses $\kappa.$ Now the result follows from your argument.

• This is a nice fact. Thank you for pointing it out. – Alexander Block Nov 7 '16 at 21:59