I know that if a topological structure $\Omega$ on a set $X$ is linearly ordered by inclusion such that $(X,\Omega)$ contains no strictly increasing infinite chain of open sets has exactly one base, where by definition a base $B$ is a collection of nonempty open sets such that any open set in $\Omega$ is a union of sets in $B$.
My question is, 1) what about the converse: if a topological space has exactly one base, does it follow that its topological structure is linearly ordered by inclusion such that it contains no strictly increasing infinite chain of open sets?
2) Altering the definition of base and defining it to be a collection of nonempty open sets excluding the whole space, what would be the case?
This question is a bit different from What topological structures have exactly one base? so do not refer me to this link.
As to question 1, the answer is positive: It cannot contain any strictly increasing infinite chain of open sets as (an amendment of) it would yeild an open set other than those appearing in the chain.
As to question 2, the answer is “not necessarily”.