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.

$a\in U[x]\cap V[x]$ 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]:=\{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 (http://planetmath.org/TopologyInducedByAUniformStructure) $\mathcal{U}$ . Under this topology, uniform neighborhoods are synonymous with neighborhoods.

Title	uniform neighborhood
Canonical name	UniformNeighborhood
Date of creation	2013-03-22 16:42:31
Last modified on	2013-03-22 16:42:31
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	6
Author	CWoo (3771)
Entry type	Definition
Classification	msc 54E15
Related topic	TopologyInducedByAUniformStructure