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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.

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    $\begingroup$ It sounds weird that Boolean completions can screw up closure properties. $\endgroup$
    – Asaf Karagila ♦
    Commented May 28, 2015 at 20:55
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    $\begingroup$ @NoahSchweber No infinite complete Boolean algebra is even countably closed, since there must be countably infinite antichains, and you can make an $\omega$-descending sequence by joining the tail starting further and further out. These meet to zero, so the algebra is not countably closed. $\endgroup$
    – Joel David Hamkins
    Commented May 28, 2015 at 21:11
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    $\begingroup$ @Joel: Is it possible to characterize these sort of failures? Namely, "if $A\subseteq\mathcal B(\Bbb P)$ is a counterexample to some closure property of $\Bbb P$, then ..." or something like that? $\endgroup$
    – Asaf Karagila ♦
    Commented May 28, 2015 at 21:36
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    $\begingroup$ These could be relevant: Sh1036 and Assaf's blogpost blog.assafrinot.com/?p=3841 $\endgroup$
    – Ashutosh
    Commented May 28, 2015 at 21:39
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    $\begingroup$ On Sh1036: I proof-read the current arxiv version (part of my job), and requested Shelah to make some changes. They should appear soon. I forgot to say that well-met condition does matter (forcing-wise). Assaf explains this well on his blogpost. $\endgroup$
    – Ashutosh
    Commented May 28, 2015 at 22:00

1 Answer 1

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$\newcommand\P{\mathbb{P}}$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.

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    $\begingroup$ The partial order $\P$ is just the atoms and co-atoms in a power set. $\endgroup$
    – Joel David Hamkins
    Commented May 29, 2015 at 0:20
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    $\begingroup$ Very nice. So not even countable closure is preserved, much less directed closure. $\endgroup$
    – Sean Cox
    Commented May 29, 2015 at 0:35

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