Let $X$ be a topological space, let $R$ a commutative ring. All algebras are required to be commutative unital associative, and algebra morphisms are required to be unital.
Let us say a sheaf $\mathcal{O}$ of $R$-algebras on $X$ is "continuous" if for every open $U \subset X$, for every $f \in \mathcal{O}(U)$, for every $x \in U$, there is a unique element $\lambda \in R$ such that $f-\lambda$ is non-invertible in every open neighborhood of $x$ (contained in $U$).
Would this definition be a sensible one? (My apologies if there is already a better notion of a "continuity" property for sheaves, or if this definition is flawed.)
For examples:
I think I've managed to show every sheaf of local $k$-algebras, for $k$ an algebraically closed field, is continuous.
Also, the sheaf of all continuous $F$-valued functions on $X$, where $F$ is any Hausdorff topological field, appears to be continuous; and I think any $F$-subalgebra-sub-sheaf of this would also be continuous.
(Are these correct? Could anyone weigh in on these?)
I was wondering if anyone could give an example of an "interesting" sheaf which is not continuous? (For example, I believe the sheaf of all real-valued, functions not necessarily continuous, would fail to be continuous; but is there a more "restrictive" sheaf which fails to be continuous?)
Edit: added above a requirement for $F$ to be Hausdorff; I think we need at least its points to be closed.