A semiring (of sets) is a nonempty class $\mathcal{P}$ of subsets of the whole space $X$ that is closed under intersections and is such that any difference of two sets in $\mathcal{P}$ can be expressed as a finite disjoint union of sets in $\mathcal{P}$.
Motivation for this question: for any class $\mathcal{E}$ of subsets of the whole space $X$, does there exist a unique smallest semiring containing $\mathcal{E}$? In other words, does there exist such a thing as the semiring generated by $\mathcal{E}$? I believe the answer is "no," but I don't really have a good reason why.
The actual question: Normally, to prove the existence of rings (or fields, I'll just stick to rings for simplicity) generated by a set $\mathcal{E}$, we first show that the intersection of an arbitrary collection of rings is again a ring. My hunch is that this result does not hold for semirings, and is the reason why there (possibly) does not exist such a thing as a semiring generated by $\mathcal{E}$.
Let $\mathcal{P}_{\gamma}$ be a semiring for every $\gamma$ in some index set $\Gamma$. Define $\mathcal{P} = \cap \{\mathcal{P}_{\gamma}: \gamma \in \Gamma\}$. Then if $E, F \in \mathcal{P}$, we have that $E,F \in \mathcal{P}_{\gamma}$ for every $\gamma$. Then by the definition of semiring for each $\gamma$ there exists a disjoint collection of sets $\{E_{n_{\gamma}}\} \in \mathcal{P}_{\gamma}$ such that $E-F = \cup E_{n_{\gamma}}$ (for each $\gamma$ this sequence may be different).
But for some reason I'm having trouble seeing why I can't take this disjoint sequence $\{E_{n_{\gamma}}\}$ to be common across $\gamma$. If I can, then it follows that an intersection of an arbitrary collection of semirings is a semiring, and then it follows that generated semirings exist. But I'm becoming convinced that they don't exist. Can anyone point out where I'm tripping up?
Best Answer
Consider $X=\{1,2,3,4\}$. Then we let $S_1=\{\emptyset, \{1\}, \{1,2,3,4\}, \{2,3\}, \{4\}\}, S_2=\{\emptyset, \{1\}, \{1,2,3,4\}, \{3,4\}, \{2\}\}$. Both are semirings but their intersection is not.