Amoeba collapse

Question

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?

Answer 1

$\kappa$ 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.

Answer 2

This is basically a long comment to your answer saying a bit more about the structure of $\mathbb A(\kappa)$. The forcing $\mathbb A(\kappa)$ is equivalent to $\mathrm{Add}(\kappa, 1)\ast\dot{\mathbb P}$ where $\dot{\mathbb P}$ is forced to have the following properties:

• It is $\kappa$-cc of size $\kappa$,
• collapses all cardinals below $\kappa$ and
• is not isomorphic to any forcing in $V$, in particular it is not the Levy collapse $\mathrm{Col}(\omega,{<}\kappa)$.

Let me sketch how to see this. Let $\mathbb Q$ be the forcing consisting of only the first components of $\mathbb{A}(\kappa)$ with the order inherited in the obvious way. $\mathbb Q$ adds a directed system of forcings so that the forcings appearing earlier in the system are regular subforcings of the later ones. We have that forcing with $\mathbb A(\kappa)$ is equivalent to $\mathbb Q\ast\dot{\mathbb P}$ where $\dot{\mathbb{P}}$ is a name for the direct limit of the system added by $\mathbb Q$. Further, $\mathbb Q$ is a nonatomic ${<}\kappa$-closed forcing of size $\kappa$ and hence equivalent to $\mathrm{Add}(\kappa, 1)$.

Now let $G$ be $\mathbb{Q}$-generic and $\mathbb P=\dot{\mathbb{P}}^G$. Clearly $\mathbb P$ is of size $\kappa$ and the argument in your answer shows that $\mathbb P$ is $\kappa$-cc: If $\dot A$ is a $\mathbb Q$-name for a maximal antichain in $\dot{\mathbb P}$ then there is a forcing $\mathbb R\in G$ so that $$A_0:=\{r\in\mathbb R\mid \mathbb R\Vdash_{\mathbb Q}\check r\in\dot A\}$$ is a maximal antichain in $\mathbb R$. Now $\dot A$ cannot grow any larger later, so $\dot A^G=A_0$ is small.

As you note, $\mathbb A(\kappa)$ collapses all cardinals below $\kappa$, the same must be true for $\mathbb P$ (which can also be seen in the same way directly).

Finally, suppose toward a contradiction that $\mathbb P$ is isomorphic to some forcing $\mathbb P'$ in $V$. By nature of how $\mathbb P$ arises, we can find a sequence $\vec {\mathbb P}:=\langle \mathbb P_\alpha\mid\alpha<\kappa\rangle$ which satisfies

• $\mathbb P=\bigcup_{\alpha<\kappa}\mathbb P_\alpha$,
• all $\mathbb P_\alpha$ are of size ${<}\kappa$ and
• $\mathbb P_\alpha\lessdot\mathbb P_\beta\lessdot \mathbb P$ whenever $\alpha<\beta<\kappa$ ($\lessdot$ denotes regular subforcing).

We can find (in $V$!) a sequence $\vec{\mathbb P}'=\langle\mathbb P_\alpha'\mid\alpha<\kappa\rangle$ with analogous properties relative to $\mathbb P'$. Now consider $$\Delta\left(\vec{\mathbb P}\right)=\left\{\alpha<\kappa\mid\bigcup_{\beta<\alpha}\mathbb P_{\beta}\lessdot\mathbb P\right\}.$$

This set modulo $\mathrm{NS}_\kappa$ does not depend on the particular choice of $\vec{\mathbb P}$. It thus suffices to show that $\Delta\left(\vec{\mathbb P}\right)\neq\Delta\left(\vec{\mathbb P}'\right)\mod\mathrm{NS}_\kappa$ and for this it suffices to show that $\Delta\left(\vec{\mathbb P}\right)$ splits every stationary subset of $\kappa$ in $V$ into two stationary sets. The main idea here is that whenever $(\mathbb R_\alpha)_{\alpha<\gamma}$ is a strictly decreasing sequence of complete Boolean algebras in $\mathbb Q$ (note that cBa's are dense in $\mathbb Q$) of length $\gamma<\kappa$ then both the direct limit $\mathbb R_{\mathrm{dir}}$ and the inverse limit $\mathbb R_{\mathrm{inv}}$ along this sequence produce lower bounds in $\mathbb Q$. However, we have that $\bigcup_{\alpha<\gamma}\mathbb R_\alpha$ is not a regular subforcing of $\mathbb R_{\mathrm{inv}}$ but it is of $\mathbb R_{\mathrm{dir}}$ (in fact this is $\mathbb R_{\mathrm{dir}}$). This in turn decides whether or not $\bigcup_{\alpha<\gamma}\mathbb R_\alpha\lessdot\mathbb P$.