TITLE: $\kappa$-c.c. vs. $\kappa$-Knaster
QUESTION [8 upvotes]: For an infinite cardinal $\kappa$ and a partial order $\mathbb{P}$, we say:
(a) $\mathbb{P}$ has the $\kappa$ chain condition ($\kappa$-c.c.) iff there is no subset of $\mathbb{P}$ of size $\kappa$ consisting of pairwise incompatible elements.
(b) $\mathbb{P}$ has the $\kappa$-Knaster property iff for every $A \subseteq \mathbb{P}$ of size $\kappa$ there is $B \subseteq A$ of size $\kappa$ consisting of pairwise compatible elements.
Clearly $\kappa$-Knaster implies $\kappa$-c.c.  Also, it is trivial that the $\kappa$-c.c. implies the $\delta$-c.c. for every $\delta \geq \kappa$.
Question 1: If $\mathbb{P}$ is $\kappa$-Knaster, is it $\delta$-Knaster for every $\delta \geq \kappa$?
Question 2: If $\mathbb{P}$ is $\kappa$-c.c., is it necessarily $\kappa^+$-Knaster?

REPLY [5 votes]: An answer to Question 1 (Thanks to hot_queen for the suggestion):
Consider the partial order $Fn(\aleph_\omega,2)$ of finite partial functions from $\aleph_\omega$ to 2, ordered by subset.  Here is a large subset of $Fn(\aleph_\omega,2)$ without a large pairwise-compatible subset:
For each $\alpha \in [\aleph_0, \aleph_\omega)$, let $p_\alpha$ be a function with domain $[0,n] \cup \{ \alpha \}$, where $n$ is such that $\alpha \in [\aleph_n,\aleph_{n+1})$.  For $\alpha \in [\aleph_n,\aleph_{n+1})$, let $p_\alpha(k) = 0$ for $k<n$, $p_\alpha(n) = 1$, and $p_\alpha(\alpha)=0$.  There is no subset of size $\aleph_\omega$ of pairwise compatible elements, because if $\alpha < \beta$ are of distinct cardinalities, then their lower parts are disagree, and we must have such $\alpha, \beta$ in any $\aleph_\omega$-sized subset.