All the filters are generated by a chain of elements? If we consider a chain of elements in a lattice, then this chain is able to generate a filter. My question is whether every filter is generated by a suitable chain
 A: Let $\mathbf L$ be the lattice with underlying set
$$\{\varnothing\} \cup \{X\subseteq \mathbb R: |\mathbb R\setminus X| \in \mathbb N\},$$
ordered by inclusion. 
So $\mathbf L$ is the set of cofinite subsets of $\mathbb R$ together with the empty set. 
It is easy to see this forms a lattice.
Now let $F=\{X\subseteq \mathbb R: |\mathbb R\setminus X| \in \mathbb N\}$ (the cofinite subsets of $\mathbb R$), and let $C$ be a chain such that $C \subseteq F$.
For each element $c$ of $C$ , 
$$c= \mathbb R \setminus A,$$
where $A$ is a finite set.
Thus
$$C=\{ \mathbb R \setminus A_i : i \in I \},$$
where $A_i$ is a finite set.
Since $C$ is a chain, so is $(A_i)_{i \in I}$ (because $\mathbb R \setminus A_i \subseteq \mathbb R\setminus A_j$ implies that $A_j \subseteq A_i$), and indeed, it is a well-ordered set. 
It follows that $\bigcup_{i\in I} A_i$ is countable, whence
$$\bigcap C = \bigcap\{ \mathbb R \setminus A_i : i \in I \} = \mathbb R \setminus \bigcup_{i\in I}A_i \neq \varnothing.$$
Let $x_0 \in \bigcap C$.
Then $x_o \in c$ for all $c \in C$.
For each $x \in \mathbb R$, let $A_x = \mathbb R\setminus\{x\}$.
Clearly, $A_x \in F$, for every $x \in \mathbb R$.
In particular, $A_{x_0} \in F$.
If $c\in C$ is such that $c \subseteq A_{x_0}$, then $x_0 \notin c$, which cannot happen as we saw.
So there exists no $c \in C$ such that $c \subseteq A_{x_0}$, and therefore $A_{x_0}$ doesn't belong to the filter generated by $C$.
