Attempting to generalize properties of filter bases in my research:
Let $\mathfrak{A}$ and $\mathfrak{Z}\subseteq\mathfrak{A}$ be two complete lattices (with $\bigcup$ and $\bigcap$ supremum and infimum), order on which agrees.
I will denote $\operatorname{up} a = \{ x\in\mathfrak{Z} \mid x\geq a \}$ for every $a\in\mathfrak{A}$.
Let the filtrator $(\mathfrak{A},\mathfrak{Z})$ be filtered, that is the following two equivalent (see my free book) conditions hold:
- $\forall a,b\in\mathfrak{A}: (\operatorname{up}a \supseteq \operatorname{up}b \Rightarrow a\leq b)$;
- $a = \bigcap^{\mathfrak{A}}\operatorname{up}a$ for every $a\in\mathfrak{A}$.
Property 1 $S$ is nonempty subset of $\mathfrak{Z}$ and there exists $f\in\mathfrak{A}$ such that $f$ is a lower bound for $S$ and $\forall X\in\operatorname{up}f\exists T\in S:T\leq X$.
Property 2 $S$ is nonempty subset of $\mathfrak{Z}$ and for every $X,Y\in S$ and $Z\in\operatorname{up}(X\cap^{\mathfrak{A}} Y)$ there is $T\in S$ such that $T\leq Z$.
Note: It is easy to show if $\mathfrak{A}$ is a poset of filters ordered reverse to set theoretic inclusion and $\mathfrak{Z}$ is the subset of principal filters, then property 2 is a characterization of $S$ being a filter base.
Conjecture 1 Let $S$ conforms to property 2. If $Z\in\operatorname{up}\bigcap^{\mathfrak{A}}S$ then $Z\in\operatorname{up}(X_0\cap^{\mathfrak{A}}\dots\cap^{\mathfrak{A}}X_n)$ for some $n\in\mathbb{N}$, $X_0,\dots,X_n\in S$.
It is easy to show that property 1 implies property 2.
Conjecture 2 Property 2 implies property 1.
Note: I formulated conjecture 1 to serve as a lemma while attempting to prove conjecture 2.
If the conjectures do not hold, they may hold under some additional conditions. I am interested in these conditions (if any).
One condition we may add that $\mathfrak{Z}$ is closed (as a subset of $\mathfrak{A}$) under arbitrary suprema.
I am thinking about these problems. Your help may be useful.
All the above appeared trying to solve this open problem (which you may not understand unless you read my book).