Let $K$ be a local field of characteristic zero and $X$ an affinoid rigid space over $K$. Let $U\subset X$ be an affinoid subdomain, and consider a finite family of points $\{p\_{1},\cdots, p\_{n}\}\subset U$. Is it true that there is a rational subdomain $V\subset X$ such that $\{p\_{1},\cdots, p\_{n}\}\subset V\subset U$? The Theorem of Gerritzen-Grauert, Theorem 7.3.5 in the reference, implies that this is true for a single point. I think a result like this would make sense, but I don't see a straightforward way of proving it just from the statement of the theorem.
Is there a known answer, or at least any literature related to this question?


*Bosch, S.; Güntzer, Ulrich; Remmert, Reinhold*, Non-Archimedean analysis. A systematic approach to rigid analytic geometry, Grundlehren der Mathematischen Wissenschaften, 261. Berlin etc.: Springer Verlag. XII, 436 p. DM 168.00 (1984). [ZBL0539.14017](https://zbmath.org/?q=an:0539.14017).



On a consequence of the Gerritzen-Grauert Theorem
Yes, this is true . You can proceed as follows. First pick finite many functions $f\_1,\dots,f\_m$ on $U$ such that $V(f\_1,\dots,f\_m)$ consists of finitely many points, including the ones you want. (For instance, if you embed $U$ into a polydisk, you can take a non-constant polynomial $P\_i$ that vanishes on the $i$-th projections of your points. The family of the $P\_i$'s would work.) Now, for $r >0$, consider $V\_r = \{ |f\_1|\le r,\dots,|f\_r|\le r\}$. It is a rational (even Weierstrass) domain and, for $r$ small enough, the connected component of $V\_r$ containing any $p\_i$ will be contained in $U$. It remains to get rid of finitely many connected components, which you can do by intersecting with domains of the form $\{|g|\ge s\}$. The resulting domain is a rational (even Laurent) domain with the properties you want.


By the way, in the case you have only one point, the result follows from the fact that any point has a basis of neighborhoods made of Laurent domains. No need for Gerritzen-Grauert here.


