I am interested in finitely generated closure algebras (as a special case of Heyting algebras), and in decreasing sequences of elements within such an algebra that have no lower bound.
Call two decreasing sequences equivalent if any member of one sequence includes some member of the other sequence as a subset, and vice versa (informally, they have the same limit point in some completion).
After factoring out equivalence, can there be an uncountable number of such decreasing sequences?
This is not an answer, but a long comment.
Nik asks:
Wouldn't any finitely generated closed set system also be finite?
Let me describe an infinite $1$-generated closure algebra.
Start with the Boolean algebra ${\mathcal P}(\omega)$ and take as the closed subsets of a closure algebra structure on ${\mathcal P}(\omega)$ the sets $\emptyset$ and the final segments $[n,\infty)$ (including the whole set $\omega = [0,\infty)$). The infinite $1$-generated subalgebra I will describe is generated by the set $X = \{1,3,5,\ldots\}$ = odds. We are allowed to generate with the Boolean operations and the closure operation $C$. We can get these things:
$\omega = \{0,1,2,\ldots\}= [0,\infty)$.
$X = \{1, 3, 5, \ldots\}$.
$C(X) = \{1,2,3,\ldots\} = [1,\infty)$.
$C(X)-X = \{2,4,6,\ldots\}$.
$C(C(X)-X) = \{2,3,4,\ldots\} = [2,\infty)$.
$C(C(X)-X) - (C(X)-X) = \{3,5,7,\ldots\}$.
$C(C(C(X)-X) - (C(X)-X)) = \{3,4,5,\ldots\} = [3,\infty)$.
Etc. (If you look at every other set on this list you see that all nonempty closed sets appear.)
