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 $A\subseteq X$,

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

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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.

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

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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 

Canonical name  PartialOrderingInATopologicalSpace 
Date of creation  20130322 16:35:02 
Last modified on  20130322 16:35:02 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  10 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 54F99 
Defines  specialization order 
Defines  specialization preorder 
Defines  specialization 
Defines  generization 