partial ordering in a topological space

Let $X$ be a topological space. For any $x,y\in X$, we define a binary relation $\le$ on $X$ as follows:

$$x\le y\mathit{\hspace{1em}\hspace{1em}}\text{ iff }\mathit{\hspace{1em}\hspace{1em}}x\in \overline{\{y\}}.$$

Proposition 1.

$\le$ is a preorder.

Proof.

Clearly $x\le x$. Next, suppose $x\le y$ and $y\le z$. Let $C$ be a closed set containing $z$. Since $y$ is in the closure of $\{z\}$, $y\in C$. Since $x$ is in the closure of $\{y\}$, $x\in C$ also. So $x\le z$. ∎

We call $\le$ the specialization preorder on $X$. If $x\le y$, then $x$ is called a specialization point of $y$, and $y$ a generization point of $x$. For any set $A\subseteq X$,

• the set of all specialization points of points of $A$ is called the specialization of $A$, and is denoted by $\mathrm{Sp}(A)$;

• the set of all generization points of points of $A$ is called the generization of $A$, and is denoted by $\mathrm{Gen}(A)$.

Proposition 2.

. If $X$ is $T_{\mathrm{0}}$ (http://planetmath.org/T0), then $\mathrm{\le }$ is a partial order.

Proof.

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 $\{y\}$. So $x\in A\cap A^c=\mathrm{\varnothing }$, a contradition. Thus $x=y$. ∎

This turns a $T_0$ topological space into a poset, where $\le$ here is called the specialization order of the space.

Given a $T_0$ space, we have the following:

Proposition 3.

$x\le y$ iff $x\mathrm{\in }U$ implies $y\mathrm{\in }U$ 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 $x\notin \overline{\{y\}}$, 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.

• $\overline{\{x\}}=\downarrow x$, the lower set of $x$. ($z\in \downarrow x$ iff $z\le x$ iff $z\in \overline{\{x\}}$).

• But if $X$ is $T_1$ (http://planetmath.org/T1), then the partial ordering just defined is trivial (the diagonal set), since every point is a closed point (for verification, just modify the antisymmetry portion of the above proof).