**Proof.**
Proof of (1). Suppose $X$ is Kolmogorov. Let $x,y\in Y$ with $x\neq y$. Then $\overline{\overline{\{ x\} }\cap Y}=\overline{\{ x\} }\neq \overline{\{ y\} }= \overline{\overline{\{ y\} }\cap Y}$. Hence $\overline{\{ x\} }\cap Y\neq \overline{\{ y\} }\cap Y$. This shows that $Y$ is Kolmogorov.

Proof of (2). Suppose $X$ is quasi-sober. It suffices to consider the cases $Y$ is closed and $Y$ is open. First, suppose $Y$ is closed. Let $Z$ be an irreducible closed subset of $Y$. Then $Z$ is an irreducible closed subset of $X$. Hence there exists $x \in Z$ with $\overline{\{ x\} } = Z$. It follows $\overline{\{ x\} } \cap Y = Z$. This shows $Y$ is quasi-sober. Second, suppose $Y$ is open. Let $Z$ be an irreducible closed subset of $Y$. Then $\overline{Z}$ is an irreducible closed subset of $X$. Hence there exists $x \in \overline{Z}$ with $\overline{\{ x\} }=\overline{Z}$. If $x\notin Y$ we get the contradiction $Z=Z\cap Y\subset \overline{Z}\cap Y=\overline{\{ x\} }\cap Y=\emptyset $. Therefore $x\in Y$. It follows $Z=\overline{Z}\cap Y=\overline{\{ x\} }\cap Y$. This shows $Y$ is quasi-sober.

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

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