# partial ordering in a topological space

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

 $x\leq y\qquad\mbox{ iff }\qquad x\in\overline{\{y\}}.$
###### Proposition 1.

$\leq$

###### Proof.

Clearly $x\leq x$. Next, suppose $x\leq y$ and $y\leq z$. 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 $x\leq z$. ∎

We call $\leq$ the specialization preorder on $X$. If $x\leq y$, 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 $T_{0}$ (http://planetmath.org/T0), then $\leq$ is a partial order  .

###### Proof.

Suppose next that $x\leq y$ and $y\leq x$. If $x\neq y$, 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 $x\in A\cap A^{c}=\varnothing$, a contradition. Thus $x=y$. ∎

This turns a $T_{0}$ topological space into a poset, where $\leq$ here is called the specialization order of the space.

Given a $T_{0}$ space, we have the following:

###### Proposition 3.

$x\leq y$ 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 $x\leq y$, 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 $z\leq x$ 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).

Title partial ordering in a topological space PartialOrderingInATopologicalSpace 2013-03-22 16:35:02 2013-03-22 16:35:02 CWoo (3771) CWoo (3771) 10 CWoo (3771) Definition msc 54F99 specialization order specialization preorder specialization generization