PlanetMath: uniform neighborhood

Let

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

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

For each $(x,U[x])\in \mathfrak{N}$ , $x\in U[x]$ , since every entourage contains the diagonal relation.
Every $x\in X$ and every entourage $U\in \mathcal{U}$ , $U[x]\subseteq X$ with $(x,U[x])\in \mathfrak{N}$
Suppose $(x,U[x])\in \mathfrak{N}$ and $U[x]\subseteq Y\subseteq X$ . Showing that $(x,Y)\in\mathfrak{N}$ amounts to showing for some $V\in \mathcal{U}$ . First, note that each entourage 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 , which is a superset of . Since $\mathcal{U}$ is a filter, $V\in \mathcal{U}$ . Furthermore, .
$a\in U[x]\cap V[x]$ 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 $(x,U[x]\cap V[x]) = (x,(U\cap V)[x])\in \mathfrak{N}$ .
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 . By the third step above, since $V[y]\subseteq U[x]$ , there is $W\in \mathcal{U}$ with . Thus $(y,U[x])=(y,W[y])\in \mathfrak{N}$ .

$\qedsymbol$

Definition. For each

in a uniform space

with uniformity $\mathcal{U}$ , a uniform neighborhood of

is a set

for some entourage $U\in\mathcal{U}$ . In general, for any $A\subseteq X$ , the set

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 $T_{\mathcal{U}}:=\lbrace A\subseteq X\mid \forall x\in A, \exists U\in \mathcal{U}$ 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.