Rational subsets in rigid analytic and adic spaces I'm confused about rational subsets of rigid analytic spaces and their corresponding adic spaces.
Suppose $A$ is an affinoid $K$-algebra, $X = \operatorname{Sp}A$ and $f \in A$.
Then $X$ can be covered as a set by the admissible open subsets
$$U = \operatorname{Sp}A\langle f \rangle = \{x \in X\, |\, \vert f(x)\vert \leq 1\}
$$
and
$$
V = \{x \in X\, |\, 1 < \vert f(x)\vert\}.
$$
But this cover is not admissible in general.
However, $V$ can be covered admissibly by $V_n = \{x \in X\, |\, \vert\varepsilon_n\vert \leq \vert f(x)\vert \}$ where $\vert \varepsilon_n \vert \in \vert K \vert$ converges to $1$ from above.
So far so good.
Now Huber associates to $X$ an adic space $r(X)$ (see e.g. section (1.1.11) in his book Étale Cohomology of Rigid Analytic Varieties and Adic Spaces) and he says that a family $X_i, i \in I$, of admissible open subsets of $X$ cover $X$ admissibly if and only if the $r(X_i)$ cover $r(X)$ as sets.
Using this we get that $r(V) = \bigcup_n r(V_n)$.
But
$$r(V_n) = \operatorname{Spa}(A\langle \frac{\varepsilon_n}{f} \rangle, A\langle \frac{f}{\varepsilon_n} \rangle^\circ) = \{x \in r(X)\, |\, \vert \varepsilon_n\vert \leq \vert f(x)\vert \},
$$
so $r(V) = \{x \in r(X)\, |\, 1 < \vert f(x)\vert\}$.
But then $r(X)$ is the union of $r(U) = \{x \in r(X)\, |\, \vert f(x)\vert \leq 1\}$ and  $r(V) = \{x \in r(X)\, |\, 1 < \vert f(x)\vert\}$ which would mean that $X = U \cup V$ is an admissible cover.
Clearly, this isn't true so I wonder where I went wrong.
It's probably a very silly mistake but it would be great if someone could clear this up for me.
 A: Your equality
$$\bigcup_n r(V_n)=\{x\in r(X): 1<|f(x)|\}$$
is incorrect in general. The reason basically is that you can (and will) have higher rank points $x\in r(X)$ with the property that $1<|f(x)|\leqslant \varepsilon_n$ for every $n\geqslant 0$ -- points with $|f(x)|$ 'infinitesimally bigger than $1$'. For instance, see Example 2.20 of [Scholze] for a recollection of how such higher rank points with this property might look.
That said, I claim that
$$\bigcup_n r(V_n)=\left\{x\in r(X): 1<|f(x)|\right\}^\circ.$$
To see this, note that
$$\{x\in r(X): 1<|f(x)|\}=r(X)-\{x\in r(X)| :|f(x)|\leqslant 1\},$$
and $\{x\in r(X): |f(x)|\leqslant 1\}$ is a rational domain, so a quasi-compact open (and thus retro-compact since $r(X)$ is quasi-separated). Then, by [FK, Chapter II, 4.2.6] we may describe
$$\{x\in r(X):1<|f(x)|\}^\circ = \{x\in r(X): 1<|f(x^\mathrm{max})|\}\qquad (\ast).$$
Here for a point $x$ of $r(X)$ we are denoting by $x^\mathrm{max}$ its maximal generization${}^{\color{red}{(1)}}$. I then claim that
$$\{x\in r(X): 1<|f(x^\mathrm{max})|\}=\bigcup_{n\geqslant 1} \{x\in r(X): \varepsilon_n \leqslant |f(x)|\}=\bigcup_n r(V_n).$$
To see this, suppose that $1<|f(x^\mathrm{max})|$. Since $x^\mathrm{max}$ is a rank $1$ point, then we actually do have that $x^\mathrm{max}$ belongs to $\{y\in r(X): \varepsilon_n <|f(y)|\}$ for some $n$, and as this set is closed (and thus closed under specialization) we have that $x$ also belongs to $\{y\in r(X): \varepsilon_n<|f(y)|\}$. Thus, the left-hand side is contained in the right-hand side. Conversely, if $x$ belongs to the right-hand side then $x$ belongs to some $\{y\in r(X): \varepsilon_n\leqslant |f(x)|\}$. But, this set is open and so closed under generization, so it contains $x^\mathrm{max}$ and so $1<\varepsilon_n<|f(x^\mathrm{max})|$. Thus, right right-hand side is contained in the left-hand side.
In summary, you see from $(\ast)$ that the difference of closed sets (of the form you were considering) and their interiors is entirely a function of higher rank points -- they have the same rank $1$ points.
$\color{red}{(1)}$ : Unlike the case of locally topologically Noetherian schemes (i.e. schemes we are used to) the underlying topological space of an analytic adic spaces (at least those considered by Huber in his book on etale cohomology) is valuative meaning that the set of generizations of a given point $x$ is a chain and, in particular, has a maximal element (which we denote by $x^\mathrm{max}$). By [Huber, Lemma 1.1.10] one may describe $x^\mathrm{max}$ as the unique rank $1$ generization of $x$.
References:
[Huber] Huber, R., 2013. Étale cohomology of rigid analytic varieties and adic spaces (Vol. 30). Springer.
[FK] Fujiwara, K. and Kato, F., 2018. Foundations of rigid geometry. European Mathematical Society.
[Scholze] Scholze, P., 2012. Perfectoid spaces. Publications mathématiques de l'IHÉS, 116(1), pp.245-313.
