Specialization preorder

Remarks

The relation is a preorder: $x \in \overline{\set{x}}$, and if $x \in \overline{\set{y}}$ and $y \in \overline{\set{z}}$ then $\overline{\set{y}} \subseteq \overline{\set{z}}$, so $x \in \overline{\set{z}}$. Equivalently, $x \le y$ if and only if every open set that contains $x$ also contains $y$.

The preorder is antisymmetric, hence a partial order, if and only if $X$ is a T0 space. It is the equality relation if and only if every point of $X$ is closed, that is, $X$ is T1. Every continuous map is monotone for the specialization preorders.

Example: in the spectrum of a commutative ring with the Zariski topology, the closure of a prime $\mathfrak q$ is the set of primes that contain $\mathfrak q$. So $\mathfrak p \le \mathfrak q$ holds exactly when $\mathfrak q \subseteq \mathfrak p$.

Non-example: on $X = \set{a, b}$ with the indiscrete topology, $a \le b$ and $b \le a$ but $a \ne b$. So the specialization preorder is not a partial order, and indeed $X$ is not T0.

Some authors use the opposite convention, with $x \le y$ when $y \in \overline{\set{x}}$.