Fix $R$ a complete DVR. Recall from Berkovich's Vanishing Cycles for Formal Schemes II paper that we have a class of *special formal schemes* which are not topologically of finite type over $\operatorname{spf} R$, but modeled on *special $R$-algebras* which are controlled enough to work with. An example is the completion of $R[x]$ along the ideal $(\pi,x)$ where $\pi$ is a uniformizer of $R$. It turns out special $R$-algebras are always quotients for some $n,m \geq 0$ of the adic ring

$$R^{n,m} := R[[T_1,\ldots,T_n]]\{ S_1,\ldots,S_m \}.$$

To construct the generic fiber of a special $R$-algebra $A$, we write it as a quotient

$$ I \subset R^{n,m} \to R^{n,m}/I \simeq A $$

Now to construct the generic fiber of $\operatorname{spf}A$, we consider $R^{n,m}$ to be a formal model for a partially-closed polydisc $E^m \times D^n$, and we *define* the generic fiber of $A$ to be the closed subspace of $E^m \times D^n$ determined by the coherent ideal $I \cdot \mathscr{O}_{E^m\times D^n}$

My question is how does the generic fiber constructed above compare to the naive generic fiber $\operatorname{sp} A \otimes_R K$? If $A$ is topologically of finite type they agree. In general the rings $\mathscr{O}_{E^m\times D^n} / I \cdot \mathscr{O}_{E^m\times D^n}$ and $A \otimes_R K$ are different, but is the former the completion of the latter with respect to some norm? Are the Berkovich spectra isomorphic?

It seems to be the case that $\mathscr{O}_{E^m\times D^n} / I \cdot \mathscr{O}_{E^m\times D^n}$ is a decreasing intersection of Banach rings $C_n$, and further each $C_n$ is a $A \otimes_R K$ algebra which further satisfies that the image of $A \otimes_R K \to C_n$ is topologically everywhere dense. Does this imply that $A\otimes_R K \to \mathscr{O}_{E^m\times D^n} / I \cdot \mathscr{O}_{E^m\times D^n}$ is topologically everywhere dense?