## Free stars and filtrators

Below is a special case of a general research problem I try to attack.

Let $\mathfrak{A}$ be a poset. Free stars on $\mathfrak{A}$ are such $S\in\mathscr{P} \mathfrak{A}$ that the least element (if it exists) is not in $S$ and for every $X,Y\in\mathfrak{A}$

$$\forall Z\in\mathfrak{A}:(X\supseteq Y\wedge Z\supseteq Y\Rightarrow Z\in S)\Leftrightarrow X\in S\vee Y\in S.$$

This can be simplified for join-semilattices:

Let $\mathfrak{A}$ be a join-semilattice. Then $S\in\mathscr{P}\mathfrak{A}$ is a free star iff the least element (if it exists) is not in $S$ and for every $X,Y\in\mathfrak{A}$

$$X\cup Y\in S\Leftrightarrow X\in S\vee Y\in S.$$

Let $\mathfrak{Z}$ is a sub-poset of $\mathfrak{A}$.

By definition $\mathrm{up}\,\, L = \{X\in\mathfrak{Z} | X\supseteq L\}$.

Let $f$ is a free star. Under which additional conditions we can prove $\mathrm{up}\, L\subseteq f\Rightarrow L\in f$ for every $L\in\mathfrak{A}$?

One condition we may naturally require is $\bigcap^{\mathfrak{A}} \mathrm{up}\,L=L$ for every $L\in\mathfrak{A}$.

But this seems not enough. What condition we may add to make the questioned statement true?

To make my question more formal: Consider a special case: Let $\mathfrak{A}$ be the set of filters on some poset, and $\mathfrak{Z}$ be the set of principal filters on that poset, both ordered reverse to set theoretic inclusion of filters. I don't know whether that are sufficient conditions for my question to be true.

-
 Counter-examples are also welcome. – porton Oct 4 2011 at 20:02 A counter-example is mathoverflow.net/questions/77354/… – porton Oct 6 2011 at 20:27