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Suppose $p_{n}$, $n\in \omega$ is a sequence in a partial order $P$ and $Q_{n}$ is a dense open subset of $P$ for each $n\in\omega$ such that $p_{n}\in Q_{n}$ and $\bigcap_{n}Q_{n}=\emptyset$. Is it true that no generic $G$ can contain all $p_{n}$?

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1 Answer 1

No, it is not true. In fact, if the $Q_n$ are all of the dense subsets of $P$, the filter generated by $\{p_n:n\in\omega\}$ is $P$-generic.

Added: Let $P$ be the set of finite partial functions from $\omega$ to $\omega_1$, partially ordered by $\supseteq$. For $\alpha<\omega_1$ let $D_\alpha=\left\{p\in P:\exists n\in\operatorname{dom}p\big(p(n)=\alpha\big)\right\}$; it’s easy to see that $D_\alpha$ is dense in $P$. Suppose that $G\subseteq P$ and $G\cap D_\alpha\ne\varnothing$ for each $\alpha<\omega_1$. For $\alpha<\omega_1$ fix $p_\alpha\in G\cap D_\alpha$; there is an $n_\alpha\in\operatorname{dom}p_\alpha$ such that $p_\alpha(n_\alpha)=\alpha$. For each $n\in\omega$ let $A_n=\{\alpha<\omega_1:n_\alpha=n\}$; clearly there must be some $k\in\omega$ such that $A_k$ is uncountable. But then $p_\alpha$ and $p_\beta$ are incompatible (i.e., have no common extension) whenever $\alpha$ and $\beta$ are distinct elements of $A_k$. In particular, $G$ cannot be generic.

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Suppose in addition $P$ has uncountably many dense subsets. Is that true? Generally, in which cases can we deduce that? –  sonicyouth Jul 19 '12 at 7:19
    
@sonicyouth: It’s been almost 40 years since I worked with this stuff, so I can’t give you a really complete answer off the top of my head, but I have extended my answer to give a specific negative example. –  Brian M. Scott Jul 19 '12 at 7:48
    
Thank you very much :) –  sonicyouth Jul 20 '12 at 4:30

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