# If $\mathbb{P}$ is a separative poset that doesn't add $\theta$-sequences then every intersection of $\theta$ dense open sets is dense in $\mathbb{P}$

I am looking for a hint (not a solution) to exercise IV.7.28 of Kunen's Set Theory book (2013). Recall that a poset $\mathbb{P}$ is separative if for every $p,q\in \mathbb{P}$, $p\nleq q$ implies $\exists s\in \mathbb{P}$ such that $s\leq p$ and $s\perp q$. The exercise is as follows:

Let $M$ be a ctm for ZFC, let $\mathbb{P}$ be a poset in $M$ and $\theta$ a cardinal in $M$. Prove that (1)$\implies$(2) and (2)$\implies$(1) if $P$ is separative:

(1) In $\mathbb{P}$, every intersection of $\theta$ dense open sets is dense in $\mathbb{P}$.

(2) $\mathbb{P}$ doesn't add $\theta$-sequences.

The exercise asks the reader to interpret these statements in the ctm approach. The way I see it is that $\mathbb{P}$ doesn't add $\theta$-sequences if for any $G$ that is $\mathbb{P}$-generic over $M$ and every $f\in M[G]$ with dom($f$)=$\theta$ and ran($f$)$\subseteq M$ it holds that $f\in M$.

Now, my attempt at (1)$\implies$(2), inspired by the proof of lemma IV.7.15, is the following: Let $f$ be as above and pick $\dot{f}$ a name for $f$, we wish to see that $f\in M$. For every $\alpha<\theta$, let $e_\alpha\in M$ be such that $M[G]\models \dot{f}(\alpha)=e_\alpha$ and let $A_\alpha=\{p\in\mathbb{P} : p\Vdash \dot{f}(\alpha)=e_\alpha\}$. The problem, of course, is that the $A_\alpha$ need not be dense in $\mathbb{P}$ (indeed, there may be a $q\in \mathbb{P}$ such that $q\Vdash \dot{f}(\alpha)\neq e_\alpha$). But I think this problem could be solved if I take

$A_\alpha=\{p\in\mathbb{P} : p\Vdash \dot{f}(\alpha)=e_\alpha\}\cup\{p\in\mathbb{P} :\exists q\in\mathbb{P}(p\perp q\land q\Vdash \dot{f}(\alpha)=e_\alpha) \}$

and then use (1) to find a element of the intersection that is also in $G$.

However, I have no clue what to do about (2)$\implies$(1) if $\mathbb{P}$ is separative. My initial approach was to take $\{A_\alpha : \alpha < \theta\}$ a family of dense open sets and then assume that there is a $p\in\mathbb{P}$ such that for no element $q\in\bigcap_{\alpha<\theta}A_\alpha$ do we have $q\leq p$. For every $\alpha<\theta$ I would then take $b_\alpha\in A_\alpha$ such that $b_\alpha\leq p$. But I'm not sure how to proceed from there, or where to use the fact that $\mathbb{P}$ is separative.

• The separativity is subtle, and it is mostly for avoiding silly counterexamples (e.g. replace each point with a decreasing $\omega$-sequence, and generate a poset which has a countable sequence of dense set with an empty intersection). – Asaf Karagila Nov 18 '14 at 4:43
• Here are two hints that may or may not be very obtuse: (1) it often helps to pass from open dense sets to maximal antichains because those are much "thinner" objects; (2) separativity usually comes into play via the following: if $P$ is separative then $p\Vdash \check{q}\in G$ iff $p\leq q$ (you should prove this if it is unfamiliar). – Miha Habič Nov 18 '14 at 5:01
• Please check my edit. – Camilo Arosemena-Serrato Nov 19 '14 at 13:28
• I've managed to do everything except showing that $q\in\bigcup_{\alpha<\theta}U_\alpha$, but I think this shouldn't be too difficult to achieve. – Miguel De Ávila Nov 19 '14 at 17:28
• @MiguelDeÁvila, we have that $q\Vdash \dot f=\check h$, so in particular for all $\alpha<\theta, q\Vdash h(\alpha)\in U_\alpha$, so $q\leq h(\alpha)$. – Camilo Arosemena-Serrato Nov 20 '14 at 2:50

With respect to the first implication, the family $\langle A_\alpha:\alpha<\theta\rangle$ you defined does not necessarily belong to $M$ because you're using $f$ to define it. Why don't you try instead $A_\alpha:=\{p\in\Bbb P:$ for some $e\in M, p\Vdash \dot f(\check\alpha)=\check e\}$. Prove that $\langle A_\alpha:\alpha<\theta\rangle\in M$.
Now, for the second implication, if $U_\alpha$ is an open dense subset of $\mathbb P$ for each $\alpha<\theta$, choose for all $\alpha<\theta$ an antichain $A_\alpha\subseteq U_\alpha$ and consider the name $\dot f=\{((\check \alpha,\check r),r):\alpha<\theta,r\in A_\alpha\}$.Check that $\Bbb 1\Vdash\dot f$ is a function from $\check\theta$ into $\check{\Bbb P}"$. Then if you pick an arbitrary $p\in\Bbb P$, prove that there is some $q\leq p$ and $h:\theta\rightarrow\Bbb P$ in $M$ such that $q\Vdash \dot f=\check h$. Then ask yourself why the separability of $\Bbb P$ implies $q\in\bigcap_{\alpha<\theta} U_\alpha$.
• Let me remark that the second implication is using the axiom of choice (for the existence of antichains, and for choosing them) and this use is essential. It might be that there is a countable sequence of open dense sets with an empty intersection, but no new $\omega$-sequences are added. – Asaf Karagila Nov 19 '14 at 6:22