Is this subset of a rigid space an admissible open?

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.

• Isn't this also part of the statement of 9.1.4/5 of BGR? – Jakob Werner Aug 16 at 9:39
• 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. – David Loeffler Aug 16 at 10:13
• 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. – Will Sawin Aug 18 at 15:19