Suppose $\kappa$ is a weakly compact cardinal. Is there a $\kappa$-c.c. forcing $\mathbb{P}$ such that $\mathbb{P} \subseteq V_\kappa$ and $\Vdash_{\mathbb{P}} \kappa = \aleph_1$, where $\mathbb{P}$ is provably NOT equivalent to the Levy collapse $Col(\omega,<\kappa)$?

I ask this because if $\kappa$ is weakly compact and $\mathbb{P}$ has the above properties, then one can prove:

a) There are stationarily many $\alpha < \kappa$ such that $\mathbb{P} \cap V_\alpha$ is a regular suborder of $\mathbb{P}$.

b) There are unboundedly many $\alpha < \kappa$ such that $\mathbb{P} \cap V_\alpha$ is equivalent to $Col(\omega,\alpha)$.

c) If $G$ is $\mathbb{P}$-generic over $V$, then there is a further forcing $\mathbb{Q}$ such that if $H \subseteq \mathbb{Q}$ is generic, then in $V[G][H]$, there is $G' \subseteq Col(\omega,<\kappa)$ which is generic over $V$, and $\mathbb{R}^{V[G]} = \mathbb{R}^{V[G']}$.

So in some sense $\mathbb{P}$ resembles $Col(\omega,<\kappa)$.

  • $\begingroup$ It sort of reminds me Proposition 10.20 in Kanamori (Ch. 3; p. 129) that if a forcing of cardinality $|\alpha|$ adds a new surjection from $\omega$ to $\alpha$ then the forcing embeds densely the Levy collapse. Of course in our case the first condition doesn't necessarily hold. $\endgroup$
    – Asaf Karagila
    Feb 10, 2013 at 2:35
  • $\begingroup$ Asaf, yes I agree. It is the proposition you mention that allows claim (b) above. $\endgroup$ Feb 10, 2013 at 21:16

2 Answers 2


The Silver collapse forcing $S(\omega, <\kappa)$ has the required properties. The conditions are functions $f$ such that $dom(f)=n\times X,$ for some $n<\omega$ and $X\in [\kappa]^{\omega}$ and $f(i, \alpha) <\alpha$ for all $i<n, \alpha\in X.$

For more details see Cummings paper "Iterated forcing and elementary embeddings", section 20.

Added remarks:

Theroem. (1). There are no Levy generic filters over $V$ in any generic extension via the Silver collapse over $V$.

(2). There are no Silver generic filters over $V$ in any generic extension via the Levy collapse over $V$.

So in particular the Levy collapse and the Silver collapse are not forcing equivalent. The theorem follows from the following lemma, that I state without proof.

Lemma. Let $G$ be a generic filter for $Col(\omega, <\kappa)$ and $H$ be a generic filter for $S(\omega, <\kappa).$ Then:

A. There exists $A \in [\kappa]^{\kappa} \cap V[G]$ of size $\kappa$ (which is of course $\omega_1$ of $V[G]$) such that $A$ does not contain any countable ground model set.

B. For any set $A \in [\kappa]^{\kappa} \cap V[H]$ of size $\kappa$ (which is of course $\omega_1$ of $V[H]$) there exists a countable set $B\in V$ such that $B \subseteq A$.

C. For any $A \in [\kappa]^{\kappa} \cap V[G]$ there exists $X \in [\kappa]^{\kappa} \cap V$ such that for all $Y \in [X]^{\omega} \cap V, Y\cap A \neq \emptyset.$

D. There exists $A \in [\kappa]^{\kappa} \cap V[H]$ such that for all $X \in [\kappa]^{\kappa} \cap V$ there exists $Y \in [X]^{\omega} \cap V, Y\cap A = \emptyset.$

  • 1
    $\begingroup$ How do you prove the Silver collapse is inequivalent to the Levy collapse? $\endgroup$ Dec 8, 2013 at 19:22

Here is a different example: Let $\lambda<\kappa$ be regular uncountable. Then (perhaps surprisingly?) $\mathrm{Col}(\omega,{<}\kappa)$ and $\mathrm{Col}(\omega, \lambda)\times\mathrm{Col}(\lambda, {<}\kappa)$ are not forcing equivalent. This is a consequence of the following fact:

(Let us write $\mathbb B\lessdot\mathbb C$ for $\mathbb B$ is a regular subalgebra of $\mathbb C$.)

Suppose $\mathbb B$ is a $\kappa$-cc atomless cBa that can be written as $\mathbb B=\bigcup_{\alpha<\kappa}\mathbb B_\alpha$ so that

  1. $\vert \mathbb B_\alpha\vert<\kappa$ for $\alpha<\kappa$.
  2. $\mathbb B_\alpha\lessdot\mathbb B_\beta$ for $\alpha<\beta<\kappa$.
  3. Any $\mu<\kappa$ is countable in $V^{\mathbb B_\alpha}$ for some $\alpha<\kappa$.

Then $\mathbb B\cong\mathrm{RO}(\mathrm{Col}(\omega,{<}\kappa))$ iff $\{\alpha<\kappa\mid\bigcup_{\beta<\alpha}\mathbb B_\beta\lessdot\mathbb B\}$ contains a club.

We can write $\mathbb B=\mathrm{RO}(\mathrm{Col}(\omega, \lambda)\times\mathrm{Col}(\lambda, {<}\kappa))$ as $\bigcup_{\alpha<\kappa}\mathbb B_\alpha$ with $$\mathbb B_\alpha=\mathrm{RO}(\mathrm{Col}(\omega, \lambda)\times\mathrm{Col}(\lambda, {<}\alpha))$$ and this satisfies 1.-3. above. However, $$\{\alpha<\kappa\mid\bigcup_{\beta<\alpha}\mathbb B_\beta\lessdot\mathbb B\}\cap\mathrm{Lim}=\{\alpha<\kappa\mid\mathrm{cof}(\alpha)\geq\lambda\}$$ is costationary in $\kappa$ and hence $\mathbb B$ is not isomorphic to $\mathrm{RO}(\mathrm{Col}(\omega,{<}\kappa))$. The homogeneity of these two Boolean algebras implies that they are not forcing equivalent either.

This inequivalence here is an inconvenience that indeed comes up in practice. For example there is a difficult argument due to Woodin that shows, among other things, that one can force a dense ideal on $\omega_1$ from an almost huge cardinal (see Theorem 7.60 in Foreman's handbook article). The idea is to turn $\kappa$ into $\omega_1$ and $j(\kappa)$ (where $j$ is an embedding witnessing that $\kappa$ is almost huge) into $\omega_2$ and finally to show that a lift of $j$ exists after collapsing $\kappa$. To make that work, Woodin drops to a symmetric submodel of a forcing extension. In the inconsistent world where $\mathrm{Col}(\omega, {{<}j(\kappa)})\cong\mathrm{Col}(\omega, \kappa)\times\mathrm{Col}(\kappa, {<}j(\kappa))$ this would not be necessary and the easier naive approach would go through.


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