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Let $K$ be a $p$-adic field and let $X$ be the rigid space $ \operatorname{Max} K\langle T_1, T_2 \rangle$, i.e. the 2-dimensional closed unit ball.

Consider the sets $U := \{ |T_1| < 1\}$ and $V := \{ |T_2| = 1\}$. It's standard that $U$ and $V$ are both admissible open subsets of $X$ (for the strong $G$-topology); e.g. this follows from Propositions 9.1.4/4 and 9.1.4/5 of Bosch--Guentzer--Remmert. However, it's not part of the axioms for a G-topology that the union of two admissible open sets be admissible open.

  • Is $U \cup V$ an admissible open set for the $G$-topology of $X$?
  • If so, is $\{ U, V\}$ an admissible covering of $U \cup V$?

EDIT: I just realised these things can't both be true. Since admissible coverings are by definition stable under pullback by morphisms of affinoids, if $U \cup V$ is admissible and admissibly covered by $U$ and $V$, then the same has to be true after intersecting with the diagonal $Z = \{ T_1 = T_2\} \cong \operatorname{Max} K\langle T \rangle$. But intersecting with $Z$ gives precisely the canonical example of a non-admissible covering, $\{ |T| = 1\} \cup \{ |T| < 1\}$.

So either $U \cup V$ is not an admissible open, or it is admissible but the covering $\{U, V\}$ of it is not an admissible covering. I'd still like to know which of these is true.

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    $\begingroup$ Isn't this also part of the statement of 9.1.4/5 of BGR? $\endgroup$
    – Jakob Werner
    Commented Aug 16, 2019 at 9:39
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    $\begingroup$ No, it is not part of that statement. BGR prop 9.1.4/5 only covers sets defined by strict inequalities, and its conclusions are clearly false if you mix strict and non-strict inequalities, e.g. the covering of the closed disc given by $\{ |T| < 1\} \cup \{ |T| \ge 1\}$ is the canonical example of a non-admissible covering. $\endgroup$
    – David Loeffler
    Commented Aug 16, 2019 at 10:13
  • $\begingroup$ I think that the affinoid subdomains $|T_1|^n \leq |T_2| \leq 1$ for all $n \in \mathbb N^+$ and $|T_1| \leq (1-\epsilon)$ for all $\epsilon>0$ might have the desired finiteness property for $U \cup V$, rendering it admissible, but I wasn't able to check this. $\endgroup$
    – Will Sawin
    Commented Aug 18, 2019 at 15:19

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