Collapsing every cardinal outside the Prikry sequence

Question

All variants of Prikry forcing with collapses that i have been able to find preserve some points outside of the generic sequence (at least the successors). This is done for two reasons, (1) to obtain a suitable guiding filter and (2) to obtain the Prikry property. I am curious if this is actually necessary for point (2). In particular, i want to know about the behaviour of the following forcing $\mathbb{P}$: Conditions are sequences $(\alpha_0,p_0,...,\alpha_{n-1},p_{n-1},F)$ such that $\alpha_0<...<\alpha_{n-1}$, $p_0\in Coll(\omega,<\alpha_0)$, $p_i\in Coll(\alpha_{i-1},<\alpha_i)$ and $F(\alpha)\in Coll(\alpha,<\kappa)$, ordered by $(\beta_0,q_0,...,\beta_{k-1},q_{k-1},G)\leq(\alpha_0,p_0,...,\alpha_{n-1},p_{n-1},F)$ if and only if $k\geq n$, $\beta_i=\alpha_i$ as well as $q_i\leq p_i$ for all $i<n$, $q_i\leq F(\beta_i)$ for all $i\geq n$, $dom(G)\subseteq dom(F)$ and $G(\alpha)\leq F(\alpha)$ for all $\alpha\in dom(G)$. We write $\leq^*$ for "$\leq$ and $k=n$".

Does $\mathbb{P}$ have the Prikry Property, i.e. given $(\alpha_0,p_0,...,\alpha_{n-1},p_{n-1},F)$ and a sentence $\phi$ in the forcing language, does there exist $(\alpha_0,q_0,...,\alpha_{n-1},q_{n-1},G)\leq^*(\alpha_0,p_0,...,\alpha_{n-1},q_{n-1},F)$ that decides $\phi$?

Answer 1

The forcing $\mathbb{P}$ collapses $\kappa$ to be countable.

Actually, in the forcing that you present, there is no restriction on adding the new $\beta_n$ so by simply splitting $\kappa$ in $\kappa$ many unbounded sets, a density argument provides a generic for $\mathrm{Col}(\omega,\kappa)$.

Let us add the normal measure $U$ and the restriction of the extensions to large sets in $U$. So a condition now is of the form: $$p = \langle \alpha_0, q_0, \dots, \alpha_{n-1},q_{n-1}, A, F\rangle,$$ with $A\in U$ and the natural ordering.

Let me show that this still doesn't work.

Let $G$ be a generic filter. Let $\langle \rho_n \mid n < \omega\rangle$ be the Prikry sequence. In the generic extension there is a sequence of generic surjections $c_n \colon \rho_n\to\rho_n^+$. Let me claim that $\langle c_n(\rho_{n-1}) \mid n < \omega\rangle$ is going to cover all of $\kappa$.

Fix $\delta < \kappa$. Let $p$ be a condition with last part $F$. Since for $\alpha \in \mathrm{dom}\, F$, $F(\alpha) \in \mathrm{Col}(\alpha,< \kappa)$, and in particular has domain bounded in $\alpha$. By pressing down, there is a large set on which the domain of the collapse is fixed to be bounded by some $\gamma$ everywhere. Now, add two Prikry points larger than both $\gamma$ and $\delta$, $\alpha_n, \alpha_{n+1}$ and set $q_{n+1}(\alpha_{n+1}^+,\alpha_n) = \delta$ (namely, we set $c_{n+1}(\alpha_n)$ to be $\delta$).