A poset $\mathbb{P}$ is called *well-met* iff every pair of compatible conditions in $\mathbb{P}$ has a greatest lower bound. 


Question: Suppose $\mathbb{P}$ is a separative partial order which is $\lambda$-directed closed (for some regular infinite cardinal $\lambda$). Can we always view $\mathbb{P}$ as a dense suborder of a well-met poset which is still $\lambda$-directed closed?


I'm vaguely aware that Boolean completions can screw up properties like directed closure, but I'm only asking for finite infima, not arbitrary infima. It's not clear to me that the obvious "well-met closure" of $\mathbb{P}$ is still $\lambda$-directed closed, but I also don't have a counterexample.



Introducing meets while preserving directed closure
$\newcommand\P{\mathbb{P}}$The answer is no. For a counterexample, consider the following
partial order $\P$. On the bottom layer, we have countably many
incompatible atoms $a\_n$ for $n<\omega$. On a second layer, we have a
collection of pairwise-incomparable elements $b\_k$, with $a\_n<b\_k$
just in case $k\neq n$. So each $b\_k$ is above all $a\_n$ except
$a\_k$. In this sense, $b\_k$ is like $\neg a\_k$ in the Boolean algebra. 


This partial order is $\lambda$-directed closed for any $\lambda$,
because any directed set can contain at most one atom $a\_n$, and
if it contains two different $b\_k$'s then it must contain at least
one $a\_n$ below both of them. And in this case that $a\_n$ will be
a lower bound.


Also, it is easy to check that $\P$ is separative.


But I claim that $\P$ is not a dense suborder of any directed closed well-met partial order $\bar\P$. If it were, then consider the elements of $\bar\P$ given by $b\_0$, $b\_0\wedge b\_1$, $b\_0\wedge b\_1\wedge b\_2$ and so on. This
is a descending sequence in $\bar\P$, but it can have no lower
bound in $\bar\P$, since the atoms of $\P$ must be dense in $\bar\P$, but no $a\_n$ is below all those finite meets, as $a\_n$ is excluded once $b\_n$ is included.


One can make a non-atomic counterexample by replacing each atom
with some $\lambda$-directed partial order and using the same
argument otherwise. 


