Let $\mathrm{r}\mathscr{O}$ be the family of open domains (regular open sets) of a topological space $\langle X,\mathscr{O}\rangle$, that is:
$$A\in\mathrm{r}\mathscr{O}\iff A=\mathrm{int(\mathrm{cl(A)})}.$$


Define $A$ to be *well inside* $B$:
$$\tag{df $\Subset$}
A\Subset B\iff \mathrm{cl}(A)\subseteq B,
$$


Call a set $X$ of regular open sets *contracting* iff $\forall\_{A\in X}\exists\_{B\in X}\,B\Subset A$. Let $C$ be a maximal contracting chain with respect to well inside relation and consider the filter $\mathscr{F}$ generated from $C$. $\mathscr{F}$ is obviously contracting. My question is: 



> 
> Does $\mathscr{F}$ have to be a maximal contracting filter?
> 



Special filters in the algebra of regular open sets of a topological space
The answer is no. Consider the case of a discrete space $X$, where every set is open and hence also regular open. In this case, the well-inside relation coincides with the ordinary inclusion relation. Your question asks whether the filter generated by an inclusion chain in $P(X)$ is maximal, and the answer is that this can fail. 


For example, if $X$ is infinite, then there is a maximal inclusion chain containing no finite sets, and so the filter is not principal. Let $A\_\alpha$ be a sequence that is coinitial (downward cofinal) in the chain, with $A\_0=X$ the whole space, and let $E=\bigsqcup\_{\alpha\text{ even}}A\_\alpha-A\_{\alpha+1}$ and $O=\bigsqcup\_{\alpha\text{ odd}}A\_\alpha-A\_{\alpha+1}$. That is, you chop the chain into pieces, and let $E$ consist of the union of the even differences, and $O$ the odd differences. So $E\sqcup O=X$, but neither contains an element of the chain. So the filter generated by the chain is not an ultrafilter, and so it isn't a maximal filter. 


Essentially the same argument shows that a nonprincipal ultrafilter in a complete Boolean algebra is never generated by a linearly ordered set, and so one can make such examples even in any non-discrete spaces. (See lemma 43 in my paper [Well-founded Boolean ultrapowers as large cardinal embeddings](http://jdh.hamkins.org/boolean-ultrapowers/).) This result shows that in any complete Boolean algebra (which can always be viewed as the regular open sets of some topological space), a nonprincipal ultrafilter is never generated by a linearly ordered set.


