In Brian Conrad's notes 
[here](http://math.stanford.edu/~conrad/papers/aws.pdf) for the 2007 Arizona winter school, bottom of p18, he says that there is an affinoid rigid-analytic space and a sheaf of abelian groups on it equipped with a non-zero section such that all stalks vanish (at all the "usual" points corresponding to maximal ideals in the affinoid algebra). He uses this to motivate Berkovich spaces etc, and explains why the existence of such a section does not contradict anything (the resulting open cover on which the section vanishes might not be an admissible cover) but does not give an explicit example of such an affinoid/sheaf/section. 


What is an explicit example?



Why do rigid spaces have "not enough points"?
If you are familiar with Berkovich spaces, you can do the following construction. Let $X$ be an affinoid space of positive dimension and pick a point $x$ in $X$ that is not a rigid point. Consider the inclusion map $i\colon x \to X$. Then the sheaf $F = i\_\*\mathbb{Z}$ does the job. Since the space $X$ is Hausdorff, the point $x$ is closed and the sheaf $F$ has no section on the open set $X\setminus \{x\}$.


Let me try to rewrite this example purely in terms of rigid geometry in a simple case: $X$ is the closed unit disc over an algebraically closed field $k$ and $x$ is the point at its boundary (the Gauss point). The previous sheaf may then be described as follows: for an affinoid domain $V$ of $X$, we have $F(V)=0$ if every connected component of $V$ is contained in an open unit disc (with any center) and $F(V)=\mathbb{Z}$ otherwise. 


