I am studying an article about the space of valuation rings by Koji Sekiguchi and while trying to establish a morphism of ringed spaces, he used the fact that the space $Zar(K\mid A) $ of valuation rings contained in a fixed field K and containing a ring A, equipped with the basis of topology:
$$
\Sigma = \{Zar(K\mid E) \mid E \ \text{is a finite subset of} \ K\} \quad \quad (1)
$$
is irreducible.
The justification he used is a result about singletons in this space:
$$
\overline{\{R\}} = \{R^\prime \in Zar(K \mid A) \mid R^\prime \subset R \} \quad \quad (2)
$$
I cannot see how $(2)$ would lead to the fact that $Zar(K\mid A)$ is irreducible.
(a topological space is said to be irreducible if it cannot be expressed as a union of two proper closed subets.)
Here's a citation for the article:
Sekiguchi, Koji, Prüfer domain and affine scheme, Tokyo J. Math. 13, No. 2, 259-275 (1990). ZBL0726.14001.
Equation (2) implies that the closure of the point $K\in Zar(K\mid A)$ is the whole space, since $K$ contains every other valuation ring. A space which is the closure of a single point is irreducible, since if you decompose it as a union of two closed subsets then one of them must contain that point and thus be the whole space.