Let $X$ be a T0 space. For any $x,y\in X$ , we define a binary relation $\le$ on $X$ as follows: $$x\le y\mbox{ iff }x \in \overline{\lbrace y\rbrace}.$$

Proposition. The binary relation just defined is a partial order.

Proof. Clearly $x\le x$ . Suppose next that $x\le y$ and $y\le x$ . If $x\ne y$ , then there is an open set $A$ such that $x\in A$ and $y\notin A$ . So $y\in A^c$ , the complement of $A$ , which is a closed set. But then $x\in A^c$ since it is in the closure of $\lbrace y\rbrace$ . So $x\in A\cap A^c=\varnothing$ , a contradition. Thus $x=y$ . Finally, suppose $x\le y$ and $y\le z$ . Let $C$ be a closed set containing $z$ . Since $y$ is in the closure of $\lbrace z\rbrace$ , $y\in C$ . Since $x$ is in the closure of $\lbrace y\rbrace$ , $x\in C$ also. So $x\le z$ .

This turns the topological space $X$ into a poset.

$\le$ is called the specialization order of $X$ . We have the following

$x\le y$ iff $x\in U$ implies $y\in U$ for any open set $U$ in $X$

Proof. $(\Rightarrow):$ if $x\in U$ and $y\notin U$ , then $y\in U^c$ . Since $x\le y$ , we have $x\in U^c$ , a contradiction. $(\Leftarrow) :$ if $x\notin \overline{\lbrace y\rbrace}$ , then for some closed set $C$ , we have $y\in C$ and $x\notin C$ . But then $x\in C^c$ , so that $y\in C^c$ , a contradiction.

Remarks.

• If $X$ is any topological space, then $\le$ is merely a preorder.
• $\overline{\lbrace x\rbrace}=\downarrow x$ , the lower set of $x$ . ($z\in \downarrow x$ iff $z\le x$ iff $z\in\overline{\lbrace x\rbrace}$ ).
• But if $X$ is a Hausdorff space, then the partial ordering just defined is trivial (the diagonal set), since singletons in $X$ are closed (for verification, just modify the antisymmetry portion of the above proof).