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

$U[x]:=\{y\mid(x,y)\in U\}$, and

$\mathfrak{N}_{x}:=\{(x,U[x])\mid U\in\mathcal{U}\}$

$\mathfrak{N}=\bigcup_{{x\in X}}\mathfrak{N}_{x}$.
Proposition. $\mathfrak{N}_{x}$ is the abstract neighborhood system around $x$, hence $\mathfrak{N}$ is the abstract neighborhood system of $X$.
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 $\{a\}\times U[a]$. We replace the “slice” $\{x\}\times U[x]$ by $\{x\}\times Y$. 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. 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
$U[A]:=\{y\in X\mid(x,y)\in U\mbox{ for some }x\in A\}$ 
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 $T_{{\mathcal{U}}}:=\{A\subseteq X\mid\forall x\in A,\exists U\in\mathcal{U}% \mbox{ such that }U[x]\subseteq A\}$, then $T_{{\mathcal{U}}}$ is a topology induced by the uniform structure $\mathcal{U}$. Under this topology, uniform neighborhoods are synonymous with neighborhoods.
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