A question on correct interpreting of following remark in Alex Youcis' survey on underlying topological picture of adic spaces:
The universal separated quotient. The underlying topological space of a rigid $K$-space $X$ is valuative in the sense of [12, Chapter 0, Definition 2.3.1]. This means that while the topology of X is locally spectral, and thus is scheme-theoretic in nature, the generizations of any point of $X$ form a totally ordered set. In particular, unlike the case of schemes locally of finite type over a field, rigid $K$-spaces admit non-trivial continuous maps to separated topological spaces.
So far I understand the upshot is that the underlying topological space of an adic space $X$ is valuative (...consequently the set of valuations $X_v$ inside this space generalize a fixed valuation $v$ (ie those $w$ with $v \in \overline{\{w\}}$) form a total order, ie any two memebers are comparable in sense one is generalization or specialization of the other) and locally spectral (every "point" $v$ admits an open neighborhood $U_v$ which is spectral, ie homeomorphic to prime spectrum of a ring.
Questions:
(1) But wouldn't this imply that any such open neighborhood $U_v$ on $v$ which is spectral, is topologically always $1$-dimensional, ie prime spectrum of open curve? Note that two points $x,y \in U_v$ are "comparable" (=one lies in topological closure of the other, equivalently one generalize/specialize the other) in $U_v$ iff comparable in $X$.
Asking differently, are there local rings of dimension bigger $1$ such that for each two picked primes there one neccessarily includes the other? (note that spectrum of local ring corresponds precisely to prime generalizing to unique maximal prime). Appears cumbersome, except it's a local ring ring of a curve.
So I'm not sure if I understood correctly in which sense such spaces are locally spectral, except such spectral neighborhoods are neccessarily spectra of curves.
(2) In contrast to spectral spaces valuative spaces admit nontrivial maps to Hausdorff spaces. Prototype candicates are maps of type $m_f:X \to \Bbb R_{\ge 0}:x \mapsto \vert f \vert_{v_x}$ where $v_x$ corresponds to valuation represented byy $x$ and $f$ some element of affinoid algebra belonging to $X$.
But so far I understand $X$ as topological space of adic space itself is not Hausdorff. And that's because even though there still exist nontrivial maps to Hausdorff spaces, but just "not enough" to distinguish all points in $X$, right?
More precisely, I guess that there no problem with distinction of points on "Type 1" (=classical points" coming from maximal ideals) by means of maps of kind $m_f$, but the real "culprits" preventing $X$ to be Hausdorff are valuations generalizing/extending valuations infinitesimally, right?
(because clearly, if there is a continuous map from $X$ to a Hausdorff space $Y$, then any two points maped to different images in $Y$ are obviously also separable by disjoint open neighborhoods in $X$ inself)