separated uniform space
SeparatedUniformSpace
2007-02-18 12:01:52
2008-06-02 20:28:24
Definition
uniform space
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\PMlinkescapeword{separated}
\PMlinkescapeword{separation axiom}
Let $X$ be a uniform space with uniformity $\mathcal{U}$. $X$ is said to be \emph{separated} or \emph{Hausdorff} if it satisfies the following \emph{separation axiom}:
$$\bigcap \mathcal{U}=\Delta,$$
where $\Delta$ is the diagonal relation on $X$ and $\bigcap \mathcal{U}$ is the intersection of all elements (entourages) in $\mathcal{U}$. Since $\Delta\subseteq \bigcap \mathcal{U}$, the separation axiom says that the only elements that belong to every entourage of $\mathcal{U}$ are precisely the diagonal elements $(x,x)$. Equivalently, if $x\ne y$, then there is an entourage $U$ such that $(x,y)\notin U$.
The reason for calling $X$ separated has to do with the following assertion:
\begin{quote}
$X$ is separated iff $X$ is a Hausdorff space under the topology $T_{\mathcal{U}}$ \PMlinkname{induced by}{TopologyInducedByAUniformStructure} $\mathcal{U}$.
\end{quote}
Recall that $T_{\mathcal{U}}=\lbrace A\subseteq X\mid \mbox{for each }x\in A\mbox{, there is }U\in \mathcal{U}\mbox{, such that }U[x]\subseteq A\rbrace$, where $U[x]$ is some uniform neighborhood of $x$ where, under $T_{\mathcal{U}}$, $U[x]$ is also a neighborhood of $x$. To say that $X$ is Hausdorff under $T_{\mathcal{U}}$ is the same as saying every pair of distinct points in $X$ have disjoint uniform neighborhoods.
\begin{proof}
$(\Rightarrow)$. Suppose $X$ is separated and $x,y\in X$ are distinct. Then $(x,y)\notin U$ for some $U\in \mathcal{U}$. Pick $V\in \mathcal{U}$ with $V\circ V\subseteq U$. Set $W=V\cap V^{-1}$, then $W$ is symmetric and $W\subseteq V$. Furthermore, $W\circ W\subseteq V\circ V\subseteq U$. If $z\in W[x]\cap W[y]$, then $(x,z),(y,z)\in W$. Since $W$ is symmetric, $(z,y)\in W$, so $(x,y)=(x,z)\circ (z,y)\in W\circ W\subseteq U$, which is a contradiction.
$(\Leftarrow)$. Suppose $X$ is Hausdorff under $T_{\mathcal{U}}$ and $(x,y)\in U$ for every $U\in \mathcal{U}$ for some $x,y\in X$. If $x\ne y$, then there are $V[x]\cap W[y]=\varnothing$ for some $V,W\in \mathcal{U}$. Since $(x,y)\in V$ by assumption, $y\in V[x]$. But $y\in W[y]$, contradicting the disjointness of $V[x]$ and $W[y]$. Therefore $x=y$.
\end{proof}