You are here
Homepartial ordering in a topological space
Primary tabs
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.
$\leq$ is a preorder.
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 $T_{0}$, 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.

But if $X$ is $T_{1}$, 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).
Mathematics Subject Classification
54F99 no label found Forums
 Planetary Bugs
 HS/Secondary
 University/Tertiary
 Graduate/Advanced
 Industry/Practice
 Research Topics
 LaTeX help
 Math Comptetitions
 Math History
 Math Humor
 PlanetMath Comments
 PlanetMath System Updates and News
 PlanetMath help
 PlanetMath.ORG
 Strategic Communications Development
 The Math Pub
 Testing messages (ignore)
 Other useful stuff
 Corrections
Comments
Is this specialization order?
Isn't this the same thing which is usually called specialization order?
http://planetmath.org/encyclopedia/Specialization.html
http://en.wikipedia.org/wiki/Specialization_order
If yes, it should be added into the entry as an alternative name.
Re: Is this specialization order?
That's right. I will make the necessary link. Thanks!