Irreducible closed subsets of a scheme corresponds to points (Jim's solution) I have a question concerning Jim's answer to the 1-1 correspondence ofirreducible closed subsets of a scheme and points.
Specifically, I don't know how to verify the following statement:

Now $U \cap Z$ has a unique generic point when $U$ is affine.  Show that if $V$ is also affine and $V \cap Z$ is nonempty then the generic point of $U \cap Z$ lies in $V$ and hence equals the unique generic point of $V \cap Z$.

especially why "the generic point of $U \cap Z$ lies in $V$" implies that the generic point of $U\cap Z$ "hence equals the unique generic point of $V \cap Z$" 
 A: Recall the definition of a generic point: a point $\eta$ of a scheme $X$ is generic if $\overline{\eta}=X$.
The scenario here is that $Z$ is an irreducible closed subscheme of a scheme $X$, and $U,V$ are affine open subschemes of $Z$.
Recall that from topology, a subset $A\subset B$ is dense if $\overline{A}=B$ or equivalently $A$ intersects every nonempty open subset of $B$.
Let $\eta_U$ be the unique generic point of $U\cap Z$. The closure of $\eta_U$ in $U$ is $U\cap Z$. The the closure of $\eta_U$ in $X$ must contain $U\cap Z$, which means that it must be all of $Z$. This is true for the following reason: suppose it wasn't all of $Z$ - then $Z\setminus (U\cap Z)$ and $\overline{\eta_U}$ would be a decomposition of $Z$ into two proper nontrivial closed subsets, which would give that $Z$ is reducible.
Since $\overline{\eta_U}=Z$, it is a dense subset of $Z$, and thus must be contained in every nonempty open subset of $Z$. In particular, $\eta_U\in V\cap Z$. The closure of $\eta_U$ in $V\cap Z$ is $Z$, so it is also a generic point for $V\cap Z$, and thus must equal the generic point $\eta_V$. This is true because we've shown (or Jim has suggested) that for any affine scheme, there is a unique generic point.
