5
$\begingroup$

Let $\operatorname{Col}(\omega,<\kappa)$ denote the Lévy collapse of an inaccessible cardinal $\kappa$. A variant of the Factor Lemma is as follows:

Lemma. Suppose that $\kappa$ is an inaccessible cardinal and that $\mathbb{P}$ is a poset of size $<\kappa$. Let $G$ be $\operatorname{Col}(\omega,<\kappa$)-generic over $V$. If in $V[G]$ there is a filter $h \subseteq \mathbb{P}$ that is $\mathbb{P}$-generic over $V$, then there is $G^* \in V[G]$ that is $\operatorname{Col}(\omega,<\kappa)$-generic over $V[h]$ and such that $V[h][G^*] = V[G]$.

For those interested, I took this variant from the paper Happy and mad families in $L(\mathbb{R})$, Lemma 17 of Page 10.

This lemma shows that, in particular, $h \in V[G]$. Is it true that there must exist some ordinal $\beta < \kappa$ such that $h \in V[G\upharpoonright\beta]$?

$\endgroup$

1 Answer 1

10
$\begingroup$

Note that the lemma doesn't show that $h$ is in $V[G]$, it assumes this. But yes, if $h\in V[G]$ is a subset of a set $X\in V$ such that $|X| < \kappa$, then for some $\beta < \kappa$, $h\in V[G\restriction \beta]$. The argument to follow is by no means original, but I don't remember where I saw it. Fix a name $\dot h$ such that $h = \dot h_G$, and for each $x\in X$, let $A_x\subseteq \text{Col}(\omega,{<}\kappa)$ be a maximal antichain consisting of conditions that force either $\check{x}\in \dot h$ or its negation. Recall that $\text{Col}(\omega,{<}\kappa)$ has the $\kappa$-cc, so $|A_x| < \kappa$. Let $A = \bigcup_{x\in X} A_x$. Since $|X| < \kappa$, $|A| < \kappa$, and therefore there is some $\beta < \kappa$ be such that $A\subseteq \text{Col}(\omega,{<}\beta)$. We have $h\in V[G\restriction \beta]$ since $h = \{x\in X : \exists q\in G\restriction \beta\, (q\Vdash \check{x} \in \dot h)\}$.

$\endgroup$
2
  • 1
    $\begingroup$ I think that, in the definition of $A_x$, $\check p\in\dot h$ should be $\check x\in\dot h$, and two lines later "Since $|\mathbb P|<\kappa$" should be "Since $|X|<\kappa$". $\endgroup$ Commented Jan 26, 2022 at 0:45
  • $\begingroup$ Agh, thanks for pointing that out! $\endgroup$ Commented Jan 26, 2022 at 2:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .