What is $\overline{Y}$ in $\text{Spec}A$? 
Consider a subset $Y$ of $\text{Spec}(A)$. (Here $A$ is a commutative ring.) What is the closure of $Y$ (or $\overline{Y}$)? 

I have been under the impression that $\overline{Y}$ is the set of prime ideals containing all prime ideals in $Y$. In other words, $\overline{Y}=V(Y)$. However, I'm not quite sure.
Any help would be greatly appreciated.
 A: If $Y$ is a subset of $\mathrm{Spec}(A)$, we may define the ideal $I(Y) \subseteq A$ of regular functions which vanish on $Y$, i.e. just $I(Y) = \bigcap_{\mathfrak{p} \in Y} \mathfrak{p}$. Conversely, if $I \subseteq A$ is an ideal, we may associate the zero set $V(I) = \{\mathfrak{p} : I \subseteq \mathfrak{p}\}$. We obtain a Galois connection between the subsets of $\mathrm{Spec}(A)$ and the ideals of $A$. If $I$ is an ideal, we have $\sqrt{I}=I(V(I))$ (this is the usual formula for the radical of an ideal as an intersection of prime ideals). If $Y \subseteq \mathrm{Spec}(A)$ is a subset, we have $\overline{Y} = V(I(Y))$ (since one easily checks that $V(I(Y))$ is the smallest closed subset containing $Y$). Hence, the Galois connection restricts to a duality between the closed subsets of $\mathrm{Spec}(A)$ and the radical ideals of $A$.
This answers in particular the question: We have $\overline{Y} = \{\mathfrak{p} : \bigcap_{\mathfrak{q} \in Y} \mathfrak{q} \subseteq \mathfrak{p}\}$.
