The product of two spectral spaces Notice: the following statements about the product topologies are all Cartesian product topology, we are in the category of topology not the category of schemes.
In this page of sober space, it said any product of sober spaces is sober. What does it mean by "any"? Any index or any finite products?

Fact 1. Could anyone give a proof about this fact that the product $X\times Y$ of any two sober spaces $X,Y$ is sober?

I am considering the product topology of two spectral spaces. Let $X=\operatorname{Spec} A$, $Y=\operatorname{Spec} B$ where $A,B$ commutative rings. Then I wonder to know

Does there exist a canonical choice of a ring $C$ such that $\operatorname{Spec} C$ is cannonically isomorphic to the Cartesian topology of $X\times Y$?

The topologies of $X,Y$ are quasicompact, and they have bases consisting of quasicompact opens and the intersection of any two quasicompact open is quasicompact open. Those above properties are preserved in $X\times Y$ (if I am right). So by Fact 1, the product topology of $X\times Y$ is a spectral space, thus it can be realized as a spectrum of a commutative ring (see the same wiki article).
Moreover, if we are considering the product topology $X\times_Z Y$ (the induced topology of the product topology $X\times Y$), where $X,Y,Z$ are affine schemes the maps $X\to Z$, $Y\to Z$ are induced by the ring maps, what will happen in this case, is $X\times_Z Y$ a spectral space, etc?
Thanks!
 A: Proof of Fact 1 (for any product). 
Let $\{ X_i \}_{i\in I}$ be a family of non-empty sober spaces. Let $F$ be a closed irreducible subset of $X:=\prod_i X_i$. By replacing $X_i$ with the closure of the projection of $F$ in $X_i$, we can suppose $F\to X_i$ has dense image for all $i$. 
I claim that $F=X$. If $\eta_i$ is the generic point of $X_i$, then it is clear that $(\eta_i)_i$ is the generic point of $X$. So let's prove the claim. Suppose the open subset $X\setminus F$ is non-empty. Then it contains a product 
$$X\setminus F \supseteq U_{i_1}\times \cdots \times U_{i_n} \times \prod_{i\ne i_1,..., i_n} X_i$$
with non-empty open subsets $U_{i_j}\subseteq X_{i_j}$. 
So $F$ is covered by finitely many closed subsets of $X$:
$$ F\subseteq \cup_{1\le j\le n} Z_{i_j}\times \prod_{i\ne i_j} X_i$$
where $Z_{i_j}=X_{i_j}\setminus U_{i_j}$. As $F$ is irreducible, it is contained in one of them, say 
$$ F\subseteq Z_{i_1}\times \prod_{i\ne i_1} X_i.$$
But then the projection of $F$ to $X_{i_1}$ is not dense. Contradiction.
