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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.