Suppose I have a rigid-analytic $K$-space $X$ and a cover $(U_{i})_{i \in I}$ of $X$ consisting of admissible open subsets, such that each $U_{i}$ satisfies a certain property $P$. Is there any strategy of obtaining from this an admissible cover $(V_{j})_{j \in J}$ of $X$ such that each $V_{j}$ satisfies $P$?
Let me give an example:
By a vector bundle on a $G$-ringed space $(X, \mathcal{O}_{X})$, I mean an $\mathcal{O}_{X}$-module $\mathcal{F}$ such that there is an admissible cover $(U_{i})_{i \in I}$ of $X$ with $\mathcal{F}|_{U_{i}}$ free of finite rank. In algebraic geometry, we know that a coherent $\mathcal{O}_{X}$-module $\mathcal{F}$ on a scheme $X$ is a vector bundle if and only if all stalks $\mathcal{F}_{x}$ are free $\mathcal{O}_{X,x}$-modules.
In rigid-analytic geometry the same assertion seems to hold (c.f. Remark at the end of Definition 4.5.1 in "Rigid Analytic Geometry and Its Applications" by Fresnel and van der Put). The proof from algebraic geometry, that I know, can be carried over to obtain a cover $(U_{i})_{i \in I}$ of $X$ by admissble open subsets such that $\mathcal{F}|_{U_{i}}$ is free of finite rank. However, I don't see how to obtain from that an admissible cover in order to satisfy the definition of vector bundle.
I have tried to use the special properties of the $G$-topology on rigid-analytic spaces (often called (G0), (G1) and (G2)) but that didn't help me. Actually, it seems like it shouldn't be possible at all to find a strategy as mentioned above, since the only way I see this happening is by somehow creating an admissible refinement of the original cover. But then because of (G2), the original cover itself must have been admissible already.
I would be grateful for any help!