sorry if this is a too vague. For $K$ some nonarchimedean discretely valued field the rigid disc of radius 1 $\mathrm{Sp} \, K\langle T \rangle$ has a formal model $\mathrm{Spf} \, K^{\circ} \{T\}$. I just read a bit about Berkovich spaces and his theory allows discs of any radius $r$: $K\langle r^{1} T \rangle$. Is there some theory of "formal models" in this context?

1$\begingroup$ Potentially related: M. Temkin "On local properties of nonArchimedean analytic spaces. II." Israel J. Math. 140 (2004), 1–27. From the MR by GrosseKlönne: "The key discovery is an efficient notion of reduction for general kanalytic spaces. For example, the reduction algebra $\tilde A$ of an algebra $A$ as above is defined here as the $\mathbf{R}_+^\times$graded algebra associated to the spectral seminorm filtration on $A$. " Link: mathscinet.ams.org/mathscinetgetitem?mr=2054837 $\endgroup$– Piotr AchingerMar 6 '21 at 10:39

$\begingroup$ Also, I don't know if for $r\notin \overline{K}^\times$ it is safe to identify the closed disc $\{T\leq r\}$ with the open disc $\{T<r\}$ (I guess not, since the former should be qc). But of course $\{T<r\}$, being quasiparacompact, admits a (somewhat complicated) formal model. $\endgroup$– Piotr AchingerMar 6 '21 at 10:42

$\begingroup$ Thanks Piotr! I wasn't aware of Temkin's 2. paper of this name. I will check it out. $\endgroup$– KonstantinMar 6 '21 at 17:02

2$\begingroup$ @PiotrAchinger: If you want your theory to allow extensions of scalars, you'd better not identify $\{T\le r\}$ with $\{T< r\}$. Otherwise, after base changing to a field $L$ with $r\inL^*$ (which exists) and changing the variable, you will identify $\{T'\le 1\}$ with $\{T'< 1\}$. $\endgroup$– Jérôme PoineauMar 6 '21 at 21:50
The formal models used in Berkovich geometry are the same that are used in rigid geometry. In particular, their generic fibers are built from spectra of quotients of Tate algebras of the form $K\langle T_1,\dotsc,T_n\rangle$, called strictly $K$affinoid in Berkovich's terminology.
To answer your question, one checks that, for $r\in\mathbf{R}_{>0}$, the algebra $K\langle r^{1}T\rangle$ is strictly $K$affinoid if, and only if, $r\in \lvert K^*\rvert^\mathbf{Q}$. (For the if direction, use the trick in A.B.'s answer. For the only if direction, use that the supremum of a function on a strictly $K$affinoid space belongs to $\lvert K^*\rvert^\mathbf{Q}$.) So, you will not get any formal model for $K\langle r^{1}T\rangle$ if $r\notin \lvert K^*\rvert^\mathbf{Q}$.
On the other hand, as pointed out by Piotr Achinger in his comments, Michael Temkin worked out a theory of reduction for all $K$affinoids, strict or not. It has properties very similar to the usual reduction of affinoids, with objects only slightly more complicated, actually graded versions of the usual objects. For $r\notin \lvert K^*\rvert^\mathbf{Q}$, the reduction of $K\langle r^{1}T\rangle$ is $\tilde{K}[r^{1}\tilde{T}]$, where the notation means that $\tilde{T}$ has degree (with respect to the graduation) $r$. It follows that $\tilde{K}[r^{1}\tilde{T}]$ has few homogeneous elements, only the monomials. As a result, its graded spectrum, which is Temkin's reduction of $K\langle r^{1}T\rangle$, has only two points: the generic point $(0)$ and the closed point $(\tilde T)$.
Not really an answer as I am not familiar with Berkovich stuff. But, ... in classical rigid geometry, one also has a disc of any rational radius $r$. If $r=\frac{a}{b}$ and $\pi$ is a uniformizer of $\mathcal{O}_K=K^{\circ}$, then $\textrm{Spf}(K^{\circ}\{T, U\}/(T^b\pi^{a}U))$ is a formal model (Raynaud style) of the subdisc $K\langle r^{1}T\rangle$. However, it may not be as nice as you may want. For example, its special fiber is not reduced. It becomes so after a sufficiently large extension of $K$. Since the formal models are basically the same in both theories, the discs in Berkovich with radii in $K$ should have these kinds of models.

1$\begingroup$ The "sufficiently large extension" in this case is simply $K(\sqrt[b]{\pi})$. $\endgroup$ Mar 6 '21 at 11:31

$\begingroup$ Thank you. I know about this formal model, but as you said it's much less nice than I would like it to be. E.g. $\mathrm{Spf} K^{\circ} \{T,U\}/(T^b\pi^aU) \rightarrow \mathrm{Spf} K^{\circ} \{T\}$ is not an open immersion etc... $\endgroup$ Mar 6 '21 at 17:04