# Levy collapse gone bad

Let $\kappa$ be strongly inaccessible, and let $\mu<\kappa$ be regular. What is the effect on $\kappa$ of forcing with the following?

(1) The product of $Col(\mu,\alpha)$ for $\alpha<\kappa$, with supports of size $<\kappa$. (bounded support)

(2) The product of $Col(\mu,\alpha)$ for $\alpha<\kappa$, with supports of size $\leq \kappa$. (full support)

The failure of $\kappa$-cc is clear, but do these collapse $\kappa$?

(1) In general, no. If $$\diamondsuit(\lambda)$$ holds stationarily often below $$\kappa$$ (e.g., if $$V=L$$ and $$\kappa$$ is Mahlo, or $$\diamondsuit(\kappa)$$ and $$\kappa$$ is weakly compact), then $$\mathbb{P}=\mathbb{P}_{\text{bounded}}$$ does not collapse $$\kappa.$$

For $$p \in \mathbb{P},$$ let $$\text{supp}(p)= \sup \{\alpha<\kappa: p_{\alpha} \neq \emptyset\}.$$

Lemma: If $$p \Vdash \dot{\gamma} < \check{\kappa},$$ there is $$p' \le p$$ and $$\beta<\kappa$$ such that $$p' \restriction \text{supp}(p)=p$$ and $$p' \Vdash \dot{\gamma}<\check{\beta}.$$

First we'll show the lemma implies the result. Suppose there is $$p$$ and $$\dot{f}$$ such that $$p \Vdash$$ "$$\dot{f}$$ is a surjection from $$\mu$$ to $$\kappa.$$" We construct a descending sequence $$\langle p_{\alpha}: \alpha \le \mu\rangle$$ by

1. $$p_0=p.$$
2. $$p_{\alpha+1}$$ is a $$p' \le p_{\alpha}$$ such that $$p' \restriction \text{supp}(p_{\alpha})=p_{\alpha}$$ and $$p'$$ bounds $$\dot{f}(\check{\alpha}).$$
3. For limit $$\alpha,$$ $$p_{\alpha}= \bigcup_{\xi<\alpha} p_{\xi}.$$

Then $$p_{\mu} \Vdash$$ "$$\dot{f}$$ is bounded," contradiction.

Now we prove the lemma. Let $$p \Vdash \dot{\gamma}<\check{\kappa}.$$ Let $$\langle r_{\alpha}: \alpha<\kappa \rangle$$ enumerate $$\mathbb{P}.$$

We define an ascending sequence $$\langle \xi_{\alpha}: \alpha<\kappa \rangle$$ by

1. $$\xi_0=0$$ and $$\xi_1=\text{supp}(p).$$
2. For $$\alpha>0,$$ $$\xi_{\alpha+1} \ge \sup\{\eta: \text{supp}(r_{\eta}) \le \xi_{\alpha}\}$$ is minimal such that, for every $$r$$ supported on $$\xi_{\alpha},$$ if there is $$r'\le r$$ with $$r' \restriction \xi_{\alpha} = r$$ such that $$r'$$ decides $$\dot{\gamma},$$ then there is such an $$r'$$ with $$\text{supp}(r') \le \xi_{\alpha+1}.$$
3. For limit $$\alpha,$$ $$\xi_{\alpha}= \sup_{\eta<\alpha} \xi_{\eta}.$$

Let $$\lambda$$ be least such that $$\diamondsuit(\lambda)$$ holds and $$\xi_{\lambda}=\lambda.$$ Fix a sequence $$\langle f_{\alpha}: \alpha<\lambda \rangle$$ of functions $$f_{\alpha}: \alpha \rightarrow \alpha$$ such that, for every $$f: \lambda \rightarrow \lambda,$$ there is $$\alpha>0$$ such that $$f \restriction \alpha = f_{\alpha}.$$

We construct a descending sequence $$\langle p_{\alpha}: \alpha \le \lambda \rangle,$$ each $$p_{\alpha}$$ supported on $$\xi_{\alpha},$$ and a sequence $$\langle \beta_{\alpha}: \alpha < \lambda \rangle$$ by

1. $$p_0=\emptyset,$$ $$\beta_0=0,$$ and $$p_1=p.$$

2. Let $$\alpha>0.$$ Suppose that for all $$\eta < \alpha,$$ $$r_{f_{\alpha}(\eta)}$$ is supported on $$[\xi_{\eta}, \xi_{\eta+1}).$$ Further suppose there is $$(q,\beta)$$ such that $$q$$ is supported on $$[\xi_{\alpha}, \xi_{\alpha+1})$$ and $$(p_{\alpha} \cup q \cup \bigcup_{\eta<\alpha} r_{f_{\alpha}(\eta)} ) \Vdash \dot{\gamma}=\check{\beta}.$$ Then let $$p_{\alpha+1}=p_{\alpha} \cup q$$ and $$\beta_{\alpha}=\beta$$ for some such pair.

Otherwise, let $$p_{\alpha+1}=p_{\alpha}$$ and $$\beta_{\alpha}=0.$$

3. For limit $$\alpha,$$ $$p_{\alpha}= \bigcup_{\xi<\alpha} p_{\xi}.$$

I claim $$p':=p_{\lambda} \Vdash \dot{\gamma}<\sup_{\alpha<\lambda} \beta_{\alpha}.$$ Suppose not. Let $$q \le p'$$ and $$\beta \ge \sup_{\alpha<\lambda} \beta_{\alpha}$$ be such that $$q \Vdash \dot{\gamma} = \check{\beta}.$$ Define $$f: \lambda \rightarrow \lambda$$ by $$r_{f(\eta)}=q \restriction [\xi_{\eta}, \xi_{\eta+1}).$$ Let $$\alpha>0$$ be such that $$f_{\alpha}=f \restriction \alpha.$$ Then $$q \le p_{\alpha+1} \cup \bigcup_{\eta<\alpha} r_{f_{\alpha}(\eta)} \Vdash \dot{\gamma}=\check{(\beta_{\alpha})},$$ contradiction.

(2) Yes, and in fact $$\mathbb{P}=\mathbb{P}_{\text{full}}$$ collapses $$(2^{\kappa})^V$$ to $$\mu.$$ In $$V[G]$$ there is a $$g$$ generic for $$\Pi_{\alpha<\kappa} Add(\mu, 1),$$ and we can construct from $$g$$ a surjection from $$\mu$$ to $$(2^{\kappa})^V$$ by a simple generalization of the "$$\text{cf}(\lambda)>\omega$$" case here: https://mathoverflow.net/a/334974/109573.