# Existence of a regular subposet which collapses everything except the top cardinal

Suppose $\delta$ is an inaccessible cardinal, and $\mathbb{P}$ is the Levy Collapse $\text{Col}(\kappa, \delta)$ which adds a surjection from $\kappa \to \delta$ (for some regular $\kappa < \delta$). It is well-known that there is a poset---namely $\text{Col}(\kappa, < \delta)$---which is absorbed by $\mathbb{P}$, and which has the same collapsing effects as $\mathbb{P}$, except that it doesn't collapse $\delta$.

Is this true in general? For simplicity, assume GCH and, if necessary, that $\delta$ is a very large cardinal. Suppose $\mathbb{P} \subset V_\delta$ is a poset which collapses $\delta$. Must there exist a poset $\mathbb{Q}$ with the following properties?

1. $\mathbb{Q}$ is absorbed by $\mathbb{P}$ as a subforcing, i.e. there is a regular embedding from $\mathbb{Q} \to \text{r.o.}(\mathbb{P})$;
2. $\mathbb{Q}$ collapses a tail end of cardinals below $\delta$, but does not collapse $\delta$.

Note: The answer is yes" in the special case that the closure of $\mathbb{P}$ matches $|\delta|^{V^{\mathbb{P}}}$, using standard absorption theory for Levy collapses. In particular, it's true if $\mathbb{P}$ makes $\delta$ countable.

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Do you have a counterexample when you drop the $\mathbb{P}\subset V_\delta$ restriction? –  Joel David Hamkins Jun 19 at 2:24
@Joel: No, I don't. –  Sean Cox Jun 19 at 3:21
Small observation: if we perform an additional small forcing $\mathbb{R}$, then we can find such a $\mathbb{Q}$ inside $\mathbb{P}\ast\mathbb{R}$. The reason is that we can let $\mathbb{R}$ collapse $|\delta|^{V^{\mathbb{P}}}$ to $\omega$, which is small relative to $\delta$ in the ground model, and so the composition will collapse $\delta$ to $\omega$, placing it into the case you mention at the end of the question. –  Joel David Hamkins Jun 19 at 5:09

Consider the forcing notion $P= Add(\omega, \delta)*\dot{Add}(\omega_1,1)$. It collapses $\delta$ to $\omega_1$. Let me show that any subforcing $Q$ of $P$ which collapses all uncountable cardinals below $\delta$ to $\omega_1,$ also collapses $\delta$ to $\omega_1.$

So let $Q$ be a subforcing of $P$ which collapses all uncountable cardinals below $\delta$ to $\omega_1.$ We show that $Q$ collapses $\delta$ to $\omega_1$ in three steps:

Step 1. $Q$ forces $2^{\omega}=\delta$: this is trivial, as otherwise $Q$ can be considered as a subforcing of $Add(\omega, \gamma)*\dot{Add}(\omega_1,1),$ for some $\gamma<\delta,$ so it will have size less than $\delta,$ which is in contradiction with our assumption on $Q$ that collapses a tail end of cardinals below $\delta$

Step 2. $Q$ is Proper: For this see Monroe's answer here "Bad subforcings of nice forcing notions"

Step 3. By an unpublished result of Todorcevic (see Is it possible for countably closed forcing to collapse $\aleph_2$ to $\aleph_1$ without collapsing the continuum?) any semi-proper forcing notion which collapses $\omega_2$ to $\omega_1,$ also collapses the continuum to $\omega_1,$ so by steps 1 and 2, the forcing $Q$ collapses $\delta$ to $\omega_1.$

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$\text{Add}(\omega_1,1)$ is always isomorphic to $\text{Coll}(\omega_1,\frak{c})$, and so the second factor in your forcing is the same as $\text{Coll}(\omega_1,\delta)$. Of course $\text{Coll}(\omega_1,\lt\delta)$ here will be a subforcing, but does it collapse $\delta$? –  Joel David Hamkins Dec 10 at 11:58
Yes, because $2^\omega=\delta.$ For example $Col(\omega_1, \omega_2)$ can be considered as a subforcing $Col(\omega_1, <\delta)$ and since $2^\omega=\delta$ it collapses $\delta$ to $\omega_1.$ –  Mohammad Golshani Dec 10 at 12:11
Ah, yes, of course. Do you have an argument in mind for the proposal in your answer? –  Joel David Hamkins Dec 10 at 12:22
I have presented my strategy for the proof in the answer. –  Mohammad Golshani Dec 14 at 11:18
Mohammad, can you give more explanation for step 1? There are two potential issues: (1) The cardinality of the continuum is not absolute even between models with the same reals, (2) The name for $Add(\omega_1)$ is given in two different forcing languages, and a generic for a model with fewer reals cannot be extended to a generic for a model with more reals. –  Monroe Eskew Dec 14 at 18:32