5
$\begingroup$

1-Let $P=Add(\omega_1, \kappa)$, and let $D$ be a dense open subset of $P$. Then there is a dense subset $S$ of $D$ such that for every $f \in S$ and any $g \in P$, if $domf=domg$ and the set $\{ \beta: f(\beta) \neq g(\beta) \}$ is finite then $g \in D$.

Why is this true?

2-Assume $0^{\sharp}$ exists, and let $\lambda \geq \aleph_2^{L[0^{\sharp}]}.$ In $L$ let $P$ be the Easton support product of the forcing notions $Add(\delta^{++}, 1)$ where $\delta < \lambda$ is a limit cardinal in $L$, and for $p \in Add(\delta^{++}, 1)$ we require $domp$ be a subset of $(\delta, \delta^{++})$. Let $F_{0}$ from the union of the intervals $(\delta, (\delta^{++})^{L})$ where $\delta$ is a limit cardinal in $L$ into $2$ be the resulting generic function. Let $I$ be the class of silver indiscernibles and for $\delta \in I$ let $\langle \alpha_{n}^{\delta}: n < \omega \rangle \in L[0^{\sharp}]$ be a cofinal sequence through $(\delta^{++})^{L}.$ Also let $\langle r_{\alpha}:\alpha < \lambda \rangle$ be a sequence of Cohen reals generic over $L[0^{\sharp}]$. Define $F_{0}^{*}$ by the same domain as $F_{0}$ by:

$F_{0}^{*}(\beta)=r_{\delta}(n)$ , if $\beta=\alpha_{n}^{\delta}, \delta \in I$

$F_{0}^{*}(\beta)=F_{0}(\beta)$ otherwise

Show that $F_{0}^{*}$ is $P-$generic over $L$.

$\endgroup$
0

1 Answer 1

8
$\begingroup$

Let me answer question 1. Conditions in $P$ are partial functions $p$ from $\omega_1\times\kappa\to 2$, with countable domain, ordered by inclusion. For any condition $p$, since $\text{dom}(p)$ is countable, there are only countably many finite modifications of $p$ on this domain. For each such finite modification $p^\ast$, there is a stronger condition $q^\ast\leq p^\ast$ with $q^\ast\in D$, and hence a corresponding $q\leq p$ such that the same modification to $q\mapsto q^\ast$ places it into $D$. So we may build a descending sequence of conditions $p_0\geq p_1\geq p_2\geq\cdots$ such that at each step, moving from $p_n$ to $p_{n+1}$, we handle one finite modification of $p_n$ to $p_n^\ast$, and extend $p_n$ to $p_{n+1}$ such that the corresponding finite modification $p_{n+1}^\ast$ is in $D$. Since the domains are countable, we can arrange by suitable bookkeeping to handle all the finite modifications that arise (details: for example, we could handle the $k$-th modification to $p_r$ at stage $\langle k,r\rangle$, using a pairing function). By the countable closure of the forcing, the union $p_\omega=\bigcup_n p_n$ is a condition, and this condition has the property that for every finite modification $p_\omega^\ast$ of it, there is some $p_n$ supporting this modification, and we already arranged that $p_{n+1}^\ast\in D$, and since $D$ is open this means $p_\omega^\ast\in D$ as well. Thus, all finite modifications of $p_\omega$ land into $D$. So the set $S$ of all such $p_\omega$ is a dense subset of $P$ with the desired property.

For question 2, perhaps you can provide more details. I don't have the paper here.

$\endgroup$

You must log in to answer this question.

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