Poset where no infinite descending chain has a lower bound

Question

Suppose $(\mathbb{P},\leq)$ is a partial order that is (i) separative, i.e. if $x\nleq y$ then $\exists ~z\leq x$ s.t. no $w$ satisfies $w\leq z$ and $w\leq y$ (ii) every strictly descending chain does not have a lower bound (so this is the exact opposite of $\sigma$-closedness). Can we say anything about the structure of $\mathbb{P}$? Does it contain a tree of height $\omega$ as a dense subset? A naive attempt is to take a maximal antichain $A_0$, and a maximal antichain $A_1$ strictly below $A_0$, etc. But I couldn't show that $\bigcup_i A_i$ is dense.

Answer 1

For $X$ infinite and $|Y| \geq 2$, define $Fn(X, Y)$ to be the set of all finite partial functions from $X$ to $Y$ ordered by $p \leq q \iff q \subseteq p$. All of these satisfy

($\star$) No strictly descending sequence of conditions has a lower bound.

If $|Y| \leq \omega < |X|$, then $Fn(X, Y)$ is ccc forcing of uncountable density ($ = |X|$) and hence a countable union of antichains in $Fn(X, Y)$ cannot be dense. There is no upper limit on either the chain-condition (increase $|Y|$) or the density (increase $|X|$) of these posets. Moreover, every separative sub-poset of these posets also satisfies $(\star)$. So I doubt that there is a useful characterization for this class.