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Assume GCH and that $\kappa$ is a regular uncountable cardinal. Let $\mathbb{P}$ be a separative, $<\kappa$-directed closed, nowhere trivial, $\kappa^+$-cc poset of size $\kappa^+$. Must $\mathbb{P}$ be forcing equivalent to $\text{Col}(\kappa, < \kappa^+)$? (which I believe is equivalent to adding $\kappa^+$ many Cohen subsets of $\kappa$). If not, then what if we also assume that the $<\kappa$ directed closure of $\mathbb{P}$ is witnessed by greatest lower bounds? (i.e. that $\mathbb{P}$ satisfies a strong version of the ``well-met" requirement).

This question is somewhat related to the following earlier question on MO:"Resembling the Levy collapse". There it was shown that for $\delta > \kappa$ where $\delta$ is inaccessible, the Silver collapse to turn $\delta$ into $\kappa^+$ has all the relevant properties but is not equivalent to Levy collapse $\text{Col}(\kappa, < \delta)$. I don't think the analogue of the Silver collapse between $\kappa$ and $\kappa^+$ is $\kappa^+$-cc, however, so I don't think that provides a counterexample to this question.

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  • $\begingroup$ You probably should require also that $\mathbb{P}$ is separative or at least splitting, since otherwise we can make a trivial example by using $\text{Add}(\kappa,1)$, but putting a linear chain of size $\kappa^+$ on top to pump up the size to $\kappa^+$. $\endgroup$ – Joel David Hamkins Oct 9 '15 at 19:12
  • $\begingroup$ We're looking forward to seeing you soon in New York! $\endgroup$ – Joel David Hamkins Oct 10 '15 at 0:12
  • $\begingroup$ I am looking forward to it too! $\endgroup$ – Sean Cox Oct 10 '15 at 15:55
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First note that in the special case $\kappa = \omega_1$, countably closed implies countably directed-closed.

Answer 1: No. In any model of CH, there is a countably closed, $\omega_2$-c.c. forcing that is inequivalent to $\mathbb P = \text{Add}(\omega_1,\omega_2)$.

Proof: Let $\mathbb Q$ be Jensen's partial order for adding a Kurepa tree. Conditions are $(t,b)$, where $t$ is a countable tree of successor height $\alpha$, $b$ is a countable subset of $\mathcal P(\omega_1)$, and for all $x \in b$, $x \cap \alpha \subseteq t$. Ordering is $(t_1,b_1) \leq (t_0,b_0)$ when $t_1$ end-extends $t_0$ and $b_0 \subseteq b_1$. It is not hard to show that $\mathbb Q$ is countably closed and $\omega_2$-c.c. under CH. Any two conditions with the same first coordinate are compatible. The union of the first coordinates in any generic is a Kurepa tree $T$ by a density argument.

Note that every subset of $\omega_1$ added by $\mathbb P$ is added by some $\omega_1$-sized regular suborder. If $\mathbb Q$ were forcing equivalent to $\mathbb P$, then there would be some $\omega_1$-sized regular subalgebra $\mathbb R$ of $ro(\mathbb Q)$ that adds the generic Kurepa tree $T$. $\mathbb Q$ forces an $\omega_2$-size subset of $\mathcal P(\omega_1)^V$ to be contained in the set of branches of $T$. Thus suppose $G * H$ is $\mathbb R * \mathbb Q / \mathbb R$-generic. But in $V[G]$ we can compute whether $(p,q) \in H$, since we just determine whether $p \subseteq T$ and whether all $x \in q$ are branches of $T$. Thus $V[G * H] = V[G]$, and $\mathbb Q$ must have a dense set of size $\omega_1$. But this is false: If $\{ (t_i,b_i) : i < \omega_1 \} \subseteq \mathbb Q$, then pick $x \notin \bigcup b_i$. No $(t_i,b_i)$ can force $x$ to be a branch of $T$ since we can extend $t$ to rule out $x$ as a branch.

Note that $\mathbb Q$ doesn't have infima to countable chains since we have a lot of choices for the "top level" of the countable tree. But...

Answer 2: Not necessarily, even for $\omega_1$-closed with infima. Let $V$ be a model of GCH, and let $G \subseteq \text{Col}(\omega_1,\omega_2)$ be generic. $\omega_3^V = \omega_2^{V[G]}$. I claim that in $V[G]$, $\mathbb P = \text{Col}(\omega_1,<\omega_2)^{V[G]}$ and $\mathbb Q = \text{Col}(\omega_2,<\omega_3)^{V}$ are both countably closed with infima and both $\omega_2$-c.c., but are not forcing equivalent.

The closure follows from the fact that $V$ and $V[G]$ have the same countable sequences. The $\omega_2$-c.c. for $\mathbb Q$ follows from the fact that $\mathbb Q \times \text{Col}(\omega_1,\omega_2)$ is $\omega_3$-c.c. in $V$.

Next, one can prove a similar fact to Mohammad's answer. In $V[G]$,

(1) $\Vdash_{\mathbb P} \exists A \in [\omega_2]^{\omega_2}$ such that $A$ does not contain any $\omega_1$-sized set from $V[G]$.

(2) $\Vdash_{\mathbb Q} \exists A \in [\omega_2]^{\omega_2}$ such that for all $X \in [\omega_2]^{\omega_2} \cap V[G]$, there is $y \in [X]^{\omega_1} \cap V[G]$ such that $y \cap A = \emptyset$.

The justification for (1) is the same as for Mohammad's answer. To see (2), let $\dot A$ be a name for a code of the generic $H$ for $\mathbb Q$. In $V$, let $\dot B$ be a name for a code for $G \times H$. If $X \in ([\omega_2]^{\omega_2})^{V[G]}$, let $\dot X$ be a name for it in $V$. Let $p \in \text{Col}(\omega_1,\omega_2)$ be arbitrary, and let $\{ p_i : i < \omega_3 \} \subseteq \text{Col}(\omega_1,\omega_2) \restriction p$ be such that $p_i$ decides the $i^{th}$ member of $\dot X$. Let $p'$ be such that $p_i = p'$ for $\omega_3$ many $i$, giving a large set $Y$ such that $p' \Vdash \check Y \subseteq \dot X$. Let $q \in \mathbb Q$ be arbitrary. Interpreting $Y$ as a subset of $\mathbb Q$, find $Y' \subseteq Y$ of size $\omega_3$ such that $q \nleq y$ for all $y \in Y'$. For each such $y$, choose $q_y \leq q$ such that $q_y \perp y$. Now form a $\Delta$-system among the $q_y$, and let $z$ be an infimum of a set $S$ of $\omega_1$ many of them. Then $z$ forces that the generic $H$ misses $y$ for $q_y \in S$. As $p$ and $q$ were arbitrary, we are done.

Likewise, we can show that $\mathbb P$ forces the negation of the the property under the forcing sign in (2), and vice versa. Some details are in section 2.1.2 of my thesis. I may post more later.

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