Let $X$ be a uniform space with uniformity $\mathcal{U}$ . For each $x\in X$ and $U\in \mathcal{U}$ , define the following items

• $U[x]:=\lbrace y\mid (x,y)\in U\rbrace$ , and
• $\mathfrak{N}_x:=\lbrace (x,U[x])\mid U\in \mathcal{U}\rbrace$
• $\mathfrak{N}=\bigcup_{x\in X} \mathfrak{N}_x$ .

Proof. We show that all five defining conditions of a neighborhood system on a set are met:

1. For each $(x,U[x])\in \mathfrak{N}$ , $x\in U[x]$ , since every entourage contains the diagonal relation.
2. Every $x\in X$ and every entourage $U\in \mathcal{U}$ , $U[x]\subseteq X$ with $(x,U[x])\in \mathfrak{N}$
3. 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, $V[x]=Y$ .
4. $a\in U[x]\cap V[x]$ 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 $(x,U[x]\cap V[x]) = (x,(U\cap V)[x])\in \mathfrak{N}$ .
5. 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 $(y,U[x])=(y,W[y])\in \mathfrak{N}$ .

Definition. For each $x$ in a uniform space $X$ with uniformity $\mathcal{U}$ , a 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 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, $\bigcap U_i[x]=(\bigcap U_i)[x]$ .

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 topology induced by the uniform structure $\mathcal{U}$ . Under this topology, uniform neighborhoods are synonymous with neighborhoods.