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 '13 at 2:35
  • $\begingroup$ Asaf, yes I agree. It is the proposition you mention that allows claim (b) above. $\endgroup$ – Monroe Eskew Feb 10 '13 at 21:16

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? $\endgroup$ – Monroe Eskew Dec 8 '13 at 19:22

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