We say that a poset $\mathbb{P}$ is absolutely Knaster if, for every $ccc$ poset $\mathbb{Q}$, $1 \Vdash_{\mathbb{Q}} \text{$\mathbb{P}$is Kanster''}$.
In general, we say that a poset $\mathbb{P}$ is absolutely $X$ if $\mathbb{P}$ remains have the property $X$ after any $ccc$ forcing, where $X$ is some property, $ccc$, Knaster, etc.

We know that:

1. Every $\sigma$-linked poset is absolutely Knaster because every $\sigma$-linked poset is absolutely $\sigma$-linked.

2. $\mathrm{MA}_{\aleph_1}$ implies that every Knaster poset is absolutely Knaster, because every $ccc$ poset is $\sigma$-linked under such an assumption.

So, the following implications hold:
$\text{$\sigma$-linked} \Rightarrow \text{absolutely Knaster} \Rightarrow \text{Knaster} \Rightarrow \text{absolutely ccc}$

My question is the following:
Is it consistent relative to $\mathsf{ZFC}$ that the above middle arrow is not reversible?
Or, can we prove in $\mathsf{ZFC}$ that every Knaster poset is absolutely Knaster?

Every Knaster poset is absolutely Knaster.

Proof: First let's prove a lemma (probably well known) that $\kappa$-c.c. implies "generically $\kappa$-Knaster" for regular $\kappa$. More precisely, if $\{ p_\alpha : \alpha < \kappa \} \subseteq \mathbb{P}$, then there is $q \in \mathbb{P}$ such that $q$ forces $\{ \alpha : p_\alpha \in G \}$ has size $\kappa$. Otherwise by the $\kappa$-c.c., 1 forces that for some $\beta< \kappa$, $\{ \alpha : p_\alpha \in G \} \subseteq \beta$. But $p_{\beta +1}$ forces the opposite.

Now suppose $\mathbb{P}$ is Knaster and $\mathbb{Q}$ is ccc. Suppose $1_\mathbb{Q}$ forces that $\{ \tau_\alpha : \alpha < \omega_1 \} \subseteq \mathbb{P}$. Let $q \in \mathbb{Q}$ be arbitrary. For each term $\tau_\alpha$, pick $q_\alpha \leq q$ deciding $\tau_\alpha = \check{p}_\alpha$ for some $p_\alpha \in \mathbb{P}$. There is some uncountable $A \subseteq \omega_1$ (in the ground model) such that $\{ p_\alpha : \alpha \in A \}$ is pairwise compatible. Now by the lemma, there is some $r \leq q$ such that $r$ forces $\{ \alpha \in A : q_\alpha \in G \}$ is uncountable. Forcing below $r$ gets that $\{ p_\alpha : \alpha \in A$ and $q_\alpha \in G \} \subseteq \{ \tau_\alpha^G : \alpha < \omega_1 \}$. As $q$ was arbitrary, it's forced that there is an uncountable pairwise compatible subset of $\{ \tau_\alpha : \alpha < \omega_1 \}$.

• Great argument! In your lemma, do you also need that cofinality of $\kappa$ is uncountable (which is true when $\kappa = \omega_1$)? – hot_queen Feb 11 '14 at 3:33
• I meant to consider only regular $\kappa$. – mbsq Feb 11 '14 at 6:28
• @mbsq Thanks for your answer! – srb Feb 11 '14 at 10:27

If you interpret "$\Vdash_{\mathbb{Q}} \mathbb{P}$ is Knaster" as $"V^{\mathbb{Q}} \models \mathbb{P}^{V^{\mathbb{Q}}}$ is Knaster", then the answer is yes, it's consistent that there is a Knaster forcing notion that is not absolutely Knaster.

The following fact is well known: If $MA_{\aleph_1}$ holds, then every ccc forcing notion satisfies Knaster's condition.

So suppose $V\models MA_{\aleph_1}$. The following result is due to Judah, and can be found in the book of Bartoszynski and Judah in pages 179-185: There is a ccc forcing notion $\mathbb{Q}$ and (a definition of) a Suslin ccc forcing notion $\mathbb{P}$ such that $V^{\mathbb{Q}} \models "\mathbb{P}$ does not satisfy the Knaster condition". Hence, $\mathbb{P}^V$ is Knaster (in $V$) but $\mathbb{P}^{V^{\mathbb{Q}}}$ is not Knaster in $V^{\mathbb{Q}}$.

• Thanks for the answer. By "$\Vdash_{\mathbb{Q}} \mathbb{P}$ is Knaster", I have intended to mean "$V^{\mathbb{Q}}\vDash \mathbb{P}^V$ is Knaster". How is this case...? – srb Feb 10 '14 at 8:34