uniform neighborhoodUniformNeighborhood2007-02-18 11:48:112007-04-21 00:36:07Definitionuniform space\usepackage{amssymb,amscd}
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Let $X$ be a uniform space with uniformity $\mathcal{U}$. For each $x\in X$ and $U\in \mathcal{U}$, define the following items
\begin{itemize}
\item $U[x]:=\lbrace y\mid (x,y)\in U\rbrace$, and
\item $\mathfrak{N}_x:=\lbrace (x,U[x])\mid U\in \mathcal{U}\rbrace$
\item $\mathfrak{N}=\bigcup_{x\in X} \mathfrak{N}_x$.
\end{itemize}
\textbf{Proposition}. $\mathfrak{N}_x$ is the abstract neighborhood system around $x$, hence $\mathfrak{N}$ is the abstract neighborhood system of $X$.
\begin{proof} We show that all five defining conditions of a neighborhood system on a set are met:
\begin{enumerate}
\item
For each $(x,U[x])\in \mathfrak{N}$, $x\in U[x]$, since every entourage contains the diagonal relation.
\item
Every $x\in X$ and every entourage $U\in \mathcal{U}$, $U[x]\subseteq X$ with $(x,U[x])\in \mathfrak{N}$
\item
Suppose $(x,U[x])\in \mathfrak{N}$ and $U[x]\subseteq Y\subseteq X$. Showing that $(x,Y)\in\mathfrak{N}$ amounts to showing $Y=V[x]$ for some $V\in \mathcal{U}$. First, note that each entourage $U$ can be decomposed into disjoint union of sets ``slices'' of the form $\lbrace a\rbrace \times U[a]$. We replace the ``slice'' $\lbrace x\rbrace \times U[x]$ by $\lbrace x\rbrace \times Y$. The resulting disjoint union is a set $V$, which is a superset of $U$. Since $\mathcal{U}$ is a filter, $V\in \mathcal{U}$. Furthermore, $V[x]=Y$.
\item $a\in U[x]\cap V[x]$ iff $(x,a)\in U\cap V$ iff $a\in (U\cap V)[x]$. This implies that if $(x,U[x]),(x,V[x])\in \mathfrak{N}$, then $(x,U[x]\cap V[x]) = (x,(U\cap V)[x])\in \mathfrak{N}$.
\item Suppose $(x,U[x])\in \mathfrak{N}$. There is $V\in\mathcal{U}$ such that $(V\circ V)[x]\subseteq U[x]$. We show that $V[x]\subseteq X$ is what we want. Clearly, $x\in V[x]$. For any $y\in V[x]$, and any $a\in V[y]$, we have $(x,a)=(x,y)\circ (y,a)\in V\circ V$, or $a\in (V\circ V)[x]\subseteq U[x]$. So $V[y]\subseteq U[x]$ for any $y\in V[x]$. In order to show that $(y,U[x])\in \mathfrak{N}$, we must find $W \in \mathcal{U}$ such that $U[x]=W[y]$. By the third step above, since $V[y]\subseteq U[x]$, there is $W\in \mathcal{U}$ with $W[y]=U[x]$. Thus $(y,U[x])=(y,W[y])\in \mathfrak{N}$.
\end{enumerate}
\end{proof}
\textbf{Definition}. For each $x$ in a uniform space $X$ with uniformity $\mathcal{U}$, a \emph{uniform neighborhood} of $x$ is a set $U[x]$ for some entourage $U\in\mathcal{U}$. In general, for any $A\subseteq X$, the set $$U[A]:=\lbrace y \in X \mid (x,y)\in U\mbox{ for some }x\in A\rbrace $$ is called a \emph{uniform neighborhood} of $A$.
Two immediate properties that we have already seen in the proof above are: (1). for each $U\in\mathcal{U}$, $x\in U[x]$; and (2). $U[x]\cap V[x]=(U\cap V)[x]$. More generally, $\bigcap U_i[x]=(\bigcap U_i)[x]$.
\textbf{Remark}. If we define $T_{\mathcal{U}}:=\lbrace A\subseteq X\mid \forall x\in A, \exists U\in \mathcal{U}\mbox{ such that }U[x]\subseteq A\rbrace$, then $T_{\mathcal{U}}$ is a \PMlinkname{topology induced by the uniform structure}{TopologyInducedByAUniformStructure} $\mathcal{U}$. Under this topology, uniform neighborhoods are synonymous with neighborhoods.