In any partial order $(P,\leq)$ it is easy to see that every chain generates (i.e., by taking the upwards closure) a filter, and any filter built as a result of the Rasiowa-Sikorski lemma in forcing is of this form. When can we reverse this?


More specifically, I am interested in the partial order of all closed subsets, with nonempty interior, of a metric space, ordered by inclusion. (I suppose the nonempty regular open sets would work, too.) Given a filter, can we say that it is generated by a countable chain?



When is a filter generated by a (countable) chain?
$\mathbf{Theorem}$ Let $X$ be a $T\_{1}$-space. Let $\mathcal{Z}$ denote the lattice of all closed subsets of $X$. Let $M\subseteq\mathcal{Z}$ be a maximal filter. Then 
$M=\{C\in\mathcal{Z}|x\in C\}$ for some $x\in X$ or $M$ is not countably generated.


$\mathbf{Proof}$ Assume that $M$ is countably generated, and $M\neq\{C\in\mathcal{Z}|x\in C\}$ for all $x\in X$. Then $\bigcap M=\emptyset$. Otherwise, if $x\in\bigcap M$, then $M\subseteq\{C\in\mathcal{Z}|x\in C\}$, so $M=\{C\in\mathcal{Z}|x\in C\}$ by maximality. Assume $M$ is countably generated by some decreasing sequence $(C\_{n})\_{n}$ of closed sets. Without loss of generality, we may assume that the sequence $(C\_{n})\_{n}$ is strictly decreasing. Let $x\_{n}\in C\_{n}\setminus C\_{n-1}$ for all $n$. Let $U\_{n}=C\_{n}^{c}$ for all $n$. Then $(U\_{n})\_{n}$ is a countable open cover of $X$. However, if $A\subseteq\{x\_{n}|n\in\mathbb{N}\}$, then $A\cap U\_{n}$ is finite for all n, hence $A\cap U\_{n}$ is closed in $U\_{n}$ for all $n$. Therefore, since $(U\_{n})\_{n}$ covers $X$, the set $A$ is closed. Therefore, the set $C=\{x\_{2n}|n\in\mathbb{N}\}$ is a closed set in $X$, but $C\_{n}\not\subseteq C$ for all $n$ contradicting the maximality of $M$. QED


