partial ordering in a topological spacePartialOrderingInATopologicalSpace2007-01-16 13:27:582007-01-18 21:44:12Definitionspecialization order\usepackage{amssymb,amscd}
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Let $X$ be a T0 space. For any $x,y\in X$, we define a binary relation $\le$ on $X$ as follows: $$x\le y\mbox{ iff }x \in \overline{\lbrace y\rbrace}.$$
\textbf{Proposition}. The binary relation just defined is a partial order.
\begin{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 $x\le z$.
\end{proof}
This turns the topological space $X$ into a poset.
$\le$ is called the \emph{specialization order} of $X$. We have the following \begin{quote} $x\le y$ iff $x\in U$ implies $y\in U$ for any open set $U$ in $X$\end{quote}
\begin{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. \end{proof}
\textbf{Remarks}.
\begin{itemize}
\item If $X$ is any topological space, then $\le$ is merely a preorder.
\item $\overline{\lbrace x\rbrace}=\downarrow x$, the lower set of $x$. ($z\in \downarrow x$ iff $z\le x$ iff $z\in\overline{\lbrace x\rbrace}$).
\item 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).
\end{itemize}