Here is a naive idea for a forcing $\mathbb A(\kappa)$, for an inaccessible cardinal $\kappa$. Conditions are pairs $(P,p)$, where $P \in V\_\kappa$ is a partial order and $p \in P$. We define the ordering $(Q,q) \leq (P,p)$ to hold when $P$ is a regular suborder of $Q$, and $q \leq\_Q p$.


The Amoeba's body grows larger and its nucleus gets wiser.


It is easy to see that whenever $G \subseteq \mathbb A(\kappa)$ is generic and $(P,p) \in G$, then $G$ induces a generic filter $G\_P \subseteq P$. Also, every cardinal below $\kappa$ is collapsed to $\omega$.


**Question:** Is $\kappa$ preserved? Does $\mathbb A(\kappa)$ add any bounded subsets of $\kappa$ that aren't added by some $G\_P$ as above?



Amoeba collapse
$\kappa$ is preserved, and moreover all reals are added by the small generics.


Let $(P\_0,p\_0)$ be a condition and let $\sigma$ be a name for a real. First, enumerate the elements of $P\_0$ below $p\_0$ as $\langle p\_i : 0<i< \lambda \rangle$. Let $(Q\_1,q\_1) \leq (P\_0,p\_1)$ decide $\sigma(0)$. Then let $(Q\_2,q\_2) \leq (Q\_1,p\_2)$ also decide $\sigma(0)$. Note that we are enlarging the partial order but going below $p\_2$ instead of $q\_1$. This makes sense because $P\_0 \lhd Q\_1$. Consider what happens at stage $\omega$. We have an increasing sequence of posets $P\_0 \lhd Q\_1 \lhd Q\_2 \lhd \dots$. If we let $Q\_\omega$ be the union, then this is a regular superorder of each $Q\_n$, since this is expressible as the first-order property, for all $q \in Q\_\omega$ (which is in some $Q\_m$ for $m \geq n$), there is $r \in Q\_n$ such that all $s \leq r$ are compatible with $q$.


So we continue transfinitely until we reach a poset $Q\_\lambda$ that is a regular superorder of $P\_0$ and each $Q\_i$, $0<i<\lambda$. It has the property that for all $p \in P\_0$ below $p\_0$, there is $r \in Q\_\lambda$ below $p$ such that $(Q\_\lambda,r)$ decides $\sigma(0)$. Now repeat process $\omega$-times until we reach some poset $R\_1$ such that for all $p \in P\_0$ below $p\_0$ and all $n<\omega$, there is $r \leq p$ in $R\_1$ such that $(R\_1,r)$ decides $\sigma(n)$.


Next, repeat this whole process with respect to $R\_1$ and iterate, reaching a closure point $R\_\omega \in V\_\kappa$. We will have that for all $r \in R\_\omega$ below $p\_0$, and all $n<\omega$, there is $r' \leq r$ in $R\_\omega$ such that $(R\_\omega,r')$ decides $\sigma(n)$. In other words, for each $n$, the set of $r \in R\_\omega$ such that $(R\_\omega,r)$ decides $\sigma(n)$ is dense below $p\_0$ in $R\_\omega$.


Now let us explain the claim in the OP that the generic for $\mathbb A$ adds a generic $G\_P$ for all posets $P$ appearing in $G$. Fix $(P,p) \in \mathbb A$, and suppose $D$ is a dense subset of $P$. For any $(Q,q) \leq (P,p)$ there is $q' \leq q$ in $Q$ such that $q' \leq d$ for some $d \in D$, since $D$ is predense in $Q$. Thus $(P,1)$ forces that the set $G\_P := \{ p \in P : (P,p) \in G \}$ is generic.


So if we force below $(R\_\omega,p\_0)$, then for each $n \in \omega$, the generic $G$ will have some element of the form $(R\_\omega,r)$ deciding $\sigma(n)$. This means that $\sigma^G$ will be an element of $V[G\_{R\_\omega}]$. By the arbitrariness of $(P\_0,p\_0)$ and $\sigma$, the desired conclusion follows.


