# On a consequence of the Gerritzen-Grauert Theorem

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.

• May I ask why you are interested in this? Jun 22, 2022 at 16:37

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.