While trying to prove some properties of a subset of a prime spectrum I arrived at the following question: Let $R$ be a commutative ring and let $P \in \mathrm{Spec}(R)$. We can consider $\mathrm{Spec}(R\_P)$ as a subset of $\mathrm{Spec}(R)$. My question is: What kind of subset (subspace) is this? I guess it is not true that $\mathrm{Spec}(R\_P)$ is open, right? Is it true that a neighborhood of the generic point in $\mathrm{Spec}(R\_P)$ is a neighborhood of $P$ in $\mathrm{Spec}(R)$?



What kind of subset is Spec(R_P) in Spec(R)?
As already said, in general the image is not open and even not constructible. But it is pro-constructible (that is, locally an intersection of locally constructible subsets, see EGA IV.1.9.4, and 1.9.5(ix) because $X:=\mathrm{Spec}(R)$ is affine hence quasi-compact). 


Denote this image by $S$. This is a subset of $X$ having exactly one closed point and which is stable by generization as pointed out by Martin. I claim that these two properties characterize all possibles images $S$ when $X$ is noetherian (this hypothesis could be weakened if we notice that $S$ is always quasi-compact, but I don't know how exactly). Here $X$ needs not be affine. 


Indeed, let $s$ be the unique closed point of $S$. Let us show $S$ is the intersection of all open neighborhoods of $s$ in $X$ (then it is easy to show $S$ is the image of $\mathrm{Spec}(O\_{X,s})$). Let $U$ be any open neighborhood of $s$ in $X$. Then $S\cap (X\setminus U)$ is noetherian hence admits a closed point if non-empty. But this would be a closed point of $S$ different from $s$. So $S\cap (X\setminus U)=\emptyset$ and $S\subseteq U$. Conversely, any point $x\in X$ in the intersecion of all $U\ni s$ is a generization of $s$ (otherwise the complementary of $\overline{\{ x\}}$ is an open neighborhood of $s$ not containing $x$). 


