**Proof.**
Proof of (1). If $X$ is Kolmogorov then so is $X_ i$ for every $i\in I$ by Lemma 5.8.5. Suppose $X_ i$ is Kolmogorov for every $i\in I$. Let $x,y\in X$ with $\overline{\{ x\} }=\overline{\{ y\} }$. There exists $i\in I$ with $x\in X_ i$. There exists an open subset $U\subset X$ such that $X_ i$ is a closed subset of $U$. If $y\notin U$ we get the contradiction $x\in \overline{\{ x\} }\cap U=\overline{\{ y\} }\cap U=\emptyset $. Hence $y\in U$. It follows $y\in \overline{\{ y\} }\cap U=\overline{\{ x\} }\cap U\subset X_ i$. This shows $y\in X_ i$. It follows $\overline{\{ x\} }\cap X_ i=\overline{\{ y\} }\cap X_ i$. Since $X_ i$ is Kolmogorov we get $x=y$. This shows $X$ is Kolmogorov.

Proof of (2). If $X$ is quasi-sober then so is $X_ i$ for every $i\in I$ by Lemma 5.8.5. Suppose $X_ i$ is quasi-sober for every $i\in I$. Let $Y$ be an irreducible closed subset of $X$. As $Y\neq \emptyset $ there exists $i\in I$ with $X_ i\cap Y\neq \emptyset $. As $X_ i$ is open in $X$ it follows $X_ i\cap Y$ is non-empty and open in $Y$, hence irreducible and dense in $Y$. Thus $X_ i\cap Y$ is an irreducible closed subset of $X_ i$. As $X_ i$ is quasi-sober there exists $x\in X_ i\cap Y$ with $X_ i\cap Y=\overline{\{ x\} }\cap X_ i\subset \overline{\{ x\} }$. Since $X_ i\cap Y$ is dense in $Y$ and $Y$ is closed in $X$ it follows $Y=\overline{X_ i\cap Y}\cap Y\subset \overline{X_ i\cap Y}\subset \overline{\{ x\} }\subset Y$. Therefore $Y=\overline{\{ x\} }$. This shows $X$ is quasi-sober.

Proof of (3). Immediately from (1) and (2).
$\square$

## Comments (0)

There are also: