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Suppose $\mathbb{P}$ is a separative partial order of uniform density $\kappa$, i.e. for all $p \in \mathbb{P}$, the least size of dense set below $p$ is $\kappa$. Does forcing with $\mathbb{P}$ add an unbounded $A \subseteq \kappa$ such that for all unbounded $X \subseteq \kappa$ in the ground model, $X \cap A \not= \emptyset$ and $X \setminus A \not= \emptyset$?

In my answer here, I showed one half of this: There is an unbounded $A$ such that $X \setminus A \not= \emptyset$ for all ground model unbounded $X$. However, the method of construction does not yield the stronger "splitting" property.

Answers with additional assumptions such as $\kappa$ is regular and/or $\mathbb{P}$ is $\kappa$-c.c. would still be useful. But we already know the case $\kappa = \omega$ is true!

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  • $\begingroup$ Probably you mean to assume that $\mathbb{P}$ is splitting or separative, since otherwise there is a counterexample: the order $\langle\kappa,\gt\rangle$, which is trivial as a forcing notion, adds no new sets, but is uniformly $\kappa$-dense, if $\kappa$ is regular. $\endgroup$ May 28 '14 at 19:55
  • $\begingroup$ You're right, I will add that. $\endgroup$ May 28 '14 at 19:58
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    $\begingroup$ @MonroeEskew. Let us call such a set $A$ an independent (or splitting) subset of $\kappa.$ The following is proved in Independence of Boolean algebras and forcing: $\endgroup$ Jul 27 '14 at 9:42
  • $\begingroup$ @MonroeEskew Theorem. Suppose $B$ is a complete Boolean algebra. The following are equivalent: 1) Forcing with $B$ does not adds an independent subset of $\kappa.$ 2) $B$ is $\kappa-$dependent i.e. for all $⟨b_\beta: \beta<\kappa⟩$ of elements of $B$, there exists a partition $P$ of unity such that for all $p∈P$, $p≤b_\beta$ or $p≤−b_\beta$ for $\kappa-$many $\beta<\kappa.$ $\endgroup$ Jul 27 '14 at 9:42
  • $\begingroup$ @MonroeEskew So your question reduces to: Suppose $B$ is of uniform density $\kappa.$ Is $B$ $\kappa-$independent? $\endgroup$ Jul 27 '14 at 9:43
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Edit: The "proof" of (1) as given below is flawed as Monroe pointed out in the comments, but it seems that the answer to Monroe's question is yes under GCH. The idea is to use a result of K. Bozeman connecting weak density and density of boolean algebras (see the reference) which in our context implies that for a dense set of $p \in B$, (1) holds with $B$ replaced by $B \upharpoonright p = \{a \in B : a \leq p\}$. See proposition 6.53 here. A consistent counterexample is still possible when $\kappa = \omega_2$. We will have to look at boolean algebras in which weak density is strictly less than density.

Let $B$ be a complete boolean algebra with density $\kappa$ everywhere. The following hold:

(1) For every $F \in [B]^{< \kappa}$, there exists a $p \in B$ that splits every (nonzero) condition in $F$ - i.e., for every $q \in F$, both $p \cap q$ and $\overline{p} \cap q$ are nonzero.

Proof: List $F = \{ p_i : i < \lambda \}$, $\lambda < \kappa$. Using the fact that $B$ has density $\kappa$ below $p_0$, get $q_0 \leq p_0$ such that each $p_i$ satisfies $p_i \cap q_0 = 0$ or $q_0 \subseteq p_i$ for $i \geq 1$. Split $q_0$, remove each $p_i$ in $F$ above $q_0$ and continue.

(2) There is a sequence $\langle p_i : i < \kappa \rangle$ such that for every $p \in B$, there exists $j < \kappa$ such that $p_i$ splits $p$ for every $i > j$.

Proof: Use (1) to construct such a sequence.

(3) Let $X$ be $B$-name for a subset of $\kappa$ defined by letting the boolean value of $i \in X$ to be $p_i$. Then $X$ splits every unbounded subset of $\kappa$ in $V$.

Proof: Let $Y$ be an unbounded subset of $\kappa$ in $V$. Suppose some $p \in B$ forces $Y$ to be disjoint with $X$. Get $i \in Y$ large enough such that $p_i$ splits $p$. Then $p \cap p_i$ forces $i \in X$ - contradiction. Similarly no $p \in B$ can force $Y$ to be contained in $X$.

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  • $\begingroup$ I don't understand that argument for (1). $\endgroup$ Jul 8 '14 at 15:51
  • $\begingroup$ I see how the induction proceeds assuming the first step can always be done, but how do you get $q_0$ with that property? A priori, we just have "$p_i \nleq q$ for all $i$," so how do you get some $q$ that is also either disjoint or comparable with all $p_i$? $\endgroup$ Jul 8 '14 at 16:14
  • $\begingroup$ For example, consider $\mathrm{Add}(\omega,2)$, and look at the set of conditions $\{ (p,1) : p \in 2^{<\omega} \}$. These are not dense in the forcing, but anything witnessing that, say $(p_0,q_0)$ for some nontrivial $q_0$, is not disjoint with $(p,1)$ for $p \leq p_0$. But for $p$ strictly below $p_0$, $(p_0,q_0)$ is not below $(p,1)$. $\endgroup$ Jul 8 '14 at 16:23
  • $\begingroup$ You are right. I guess this works only for $\kappa = \omega_1$ by thinning down every condition finitely often. I will have to think about the general situation again. $\endgroup$
    – Ashutosh
    Jul 8 '14 at 16:24
  • $\begingroup$ Indeed. I formulated this question in an attempt to attack Monk's problem from a forcing properties angle. $\endgroup$ Aug 11 '14 at 22:03

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