Does every vector bundle on a Stein space have a finite local trivialisation?


Definitions:


* *Stein space* means either a complex analytic Stein space or a nonarchimedean Stein space in the sense of Kiehl.
* *Vector bundle* means either holomorphic vector bundle or rigid analytic vector bundle (= locally free sheaf). Edit: Say, of constant rank $r$.
* *finite local trivialisation* = local trivialisation where the covering consists of finitely many sets.


In the nonarchimedean case, assume moreover that the base field is spherically complete or that the Stein space is unbounded.


My thoughts so far: Usually, the space is noncompact, so the answer is probably "no". On the other hand, I could not come up with an example yet. If the Stein space is the analytification of an algebraic variety, then at least those vector bundles that are analytifications of algebraic bundles will have a finite trivialisation. Every vector bundle on a Stein space is determined by finitely many global sections. Could that help? For an example, one should probably use the non-finitely generated ideals in the algebra of global functions of the Stein space. I cannot write down any of them either.


As said, I suspect the answer to be "no". It would be great if somebody knew an example or a reference.



Trivialisation of vector bundles on Stein spaces
In the complex analytic world\*, every vector bundle which is globally generated by a finite set of global sections has a finite trivialising cover:


Let $X$ be a complex space and let $E$ be a locally free (hence coherent) $\mathcal{O}\_X$-module of constant rank $r$, generated by the global sections $s\_1,s\_2,\dots s\_d$. For each subset $I=\{i\_1,\dots i\_r\}$ of $\{1,\dots d\}$, consider the morphism $\varphi\_I\colon\mathcal{O}\_X^r\to E$ given by the sections $s\_{i\_1},\dots s\_{i\_r}$.


I claim that the subsets $U\_I\subset X$ consisting of the points where $\varphi\_I$ is an isomorphism are open and cover $X$. Note that if the map on germs $(\varphi\_I)\_x\colon \mathcal{O}\_{X,x}^r\to E\_x$ is surjective, then it's already an isomorphism, hence $X-U\_I$ is just the support of the cokernel of $\varphi\_I$. This is generated by the residue classes of the finitely many sections generating $E$ and so $U\_I$ is open for every $I$.


For every $x\in X$, by assumption, the $\mathcal{O}\_{X,x}$-module $E\_{x}$ is generated by $(s\_1)\_x,\dots (s\_d)\_x$. Since it is also free of rank $r$, a subset of $r$ germs is enough to generate $E\_x$, and so $x$ is contained in some $U\_I$. Thus, the $U\_I$ are covering $X$.
This proves everything I claimed. 




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* The argument works for locally ringed spaces ~~with coherent structure sheaf~~ I guess.


