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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?

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    $\begingroup$ Very nice. Woodin told me once that in his dissertation, he had used a "generic Boolean algebra" construction, which has a certain similarity to this (but without the inaccessible). One expands the Boolean algebra generically. But I think his forcing wasn't fully amoebic like yours. $\endgroup$ Feb 16, 2023 at 15:19
  • $\begingroup$ Is there an easy way to see that this is different from the plain Levy collapse? $\endgroup$ Feb 17, 2023 at 14:18
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    $\begingroup$ @FrançoisG.Dorais It's not $\kappa$-c.c. Let $\{ f_\alpha : \alpha < \kappa \}$ be a sequence of binary functions, where $f_\alpha$ has domain $\alpha+1$, takes value 0 below $\alpha$ and value 1 at $\alpha$. Each is a member of $Add(\lambda)$, where $\lambda$ is any regular cardinal between $\alpha$ and $\kappa$. So consider the conditions $(Add(\alpha^+),f_\alpha)$ in Amoeba. I claim these are incompatible in Amoeba. $\endgroup$ Feb 17, 2023 at 14:54
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    $\begingroup$ continued: This is because, if you take two of them for $\alpha<\beta$, then $f_\alpha,f_\beta \in Add(\beta^+)$, so a regular superorder of both would have to respect that they are incompatible in $Add(\beta^+)$. This may seem silly but it's the nature of the ordering of Amoeba which is designed to "freeze" how the smaller partial orders appear by having the larger ones be literal super-orders. Now I suspect something deeper is true; it doesn't add any generics for $\kappa$-c.c. posets. $\endgroup$ Feb 17, 2023 at 14:57
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    $\begingroup$ Rather, I conjecture it doesn't add generics for $\kappa$-c.c. posets of size $\kappa$. $\endgroup$ Feb 17, 2023 at 15:05

2 Answers 2

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$\kappa$ 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.

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  • $\begingroup$ So... this is just a "different way to amalgamate all the small forcings", as opposed to the Levy collapse where we just take the product, right? $\endgroup$
    – Asaf Karagila
    Feb 19, 2023 at 8:42
  • $\begingroup$ @AsafKaragila right, one of many. I’m hoping it has some useful kind of universal property… $\endgroup$ Feb 19, 2023 at 10:48
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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 $\bigcup_{\alpha<\gamma}\mathbb R_\alpha\lessdot\mathbb P$.

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    $\begingroup$ Nice. But could you remind me what the general definition of the inverse limit is? I only know a construction when we know we have a commuting system of projections. Do you need to pass to the Boolean completions? $\endgroup$ Feb 19, 2023 at 14:13
  • $\begingroup$ You are right, we should take the Boolean completions! Edited it now :) $\endgroup$ Feb 19, 2023 at 14:32

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