Can a basis ever be a semiring? I have a question that I've been wondering about the past day or so while trying to relate measure theory back to some general topology.
Is it ever possible for some family of (open) sets in $\mathbb{R}^2$ to be both a base for the usual topology on $\mathbb{R}^2$, as well as a semiring?
To avoid confusion, I mean a semiring of sets. So by semiring, I mean a collection of subsets of $\mathbb{R}^2$ which has $\emptyset$ as an element, is closed under finite intersections, and for any $A,B$ in the semiring, $A\setminus B=\bigcup_{i=1}^n C_i$ for disjoint $C_i$ in the semiring.
 A: In a non-trivial $T_1$ connected, locally connected space this can never happen, I think: as soon as we can find non-empty proper connected open subsets $U \subset V$ in a base/semiring, where the inclusion is proper, then $V \setminus U$ cannot be open (so in particular cannot be a disjoint union of base elements) because it's already closed in $U$. But a base could consist of non-connected subsets, so the argument is not quite clear. But I think something like this should be provable.
The situation can occur for other topologies, like the discrete one, or more interestingly, the Sorgenfrey plane (so $\mathbb{R}^2$ with sets of the form $[a,b) \times [c,d) $ as base),
which has the standard semiring for the Lebesgue measure as a topological base.
Note that every semiring of sets is a base for some topology (as it's closed under intersections).
A: I believe here's a proof for $\mathbb{R}^2$, but I'm not positive how general this can be; as Henno Brandsma mentions, connectivity seems to be key.
Assume you have a base for $\mathbb{R}^2$ that is closed under finite intersections. Let $A$ be a nonempty basic open set, and let $B$ be a connected component of $A$. Let $C$ be a nonempty open set of the basis that is properly contained in $B$ (always possible in $\mathbb{R}^2$). Since $\mathbb{R}^2$ is equal to the disjoint union of $C$, $\partial C$, and the interior of the complement of $C$, and $B$ is connected, then $B$ must intersect $\partial C$. Let $x\in B\cap\partial C$. Then $x\in A-C$; but there can be no open set that contains $x$ and is contained in $A-C$, since every open neighborhood of $x$ intersects $C$ nontrivially. Therefore, $A-C$ cannot be open, which shows that the base cannot be a semiring.
