uniform neighborhood

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

• $U[x]:=\{y\mid (x,y)\in U\}$, and

• ${𝔑}_x:=\{(x,U[x])\mid U\in 𝒰\}$

• $𝔑={\bigcup }_{x\in X}{𝔑}_x$.

Proposition. ${𝔑}_x$ is the abstract neighborhood system around $x$, hence $𝔑$ 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 $(x,U[x])\in 𝔑$, $x\in U[x]$, since every entourage contains the diagonal relation.

2. Every $x\in X$ and every entourage $U\in 𝒰$, $U[x]\subseteq X$ with $(x,U[x])\in 𝔑$

3. Suppose $(x,U[x])\in 𝔑$ and $U[x]\subseteq Y\subseteq X$. Showing that $(x,Y)\in 𝔑$ amounts to showing $Y=V[x]$ for some $V\in 𝒰$. 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 $𝒰$ is a filter, $V\in 𝒰$. 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 $(x,U[x]),(x,V[x])\in 𝔑$, then $(x,U[x]\cap V[x])=(x,(U\cap V)[x])\in 𝔑$.

5. Suppose $(x,U[x])\in 𝔑$. There is $V\in 𝒰$ 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 $(y,U[x])\in 𝔑$, we must find $W\in 𝒰$ such that $U[x]=W[y]$. By the third step above, since $V[y]\subseteq U[x]$, there is $W\in 𝒰$ with $W[y]=U[x]$. Thus $(y,U[x])=(y,W[y])\in 𝔑$.

∎

Definition. For each $x$ in a uniform space $X$ with uniformity $𝒰$, a uniform neighborhood of $x$ is a set $U[x]$ for some entourage $U\in 𝒰$. In general, for any $A\subseteq X$, the set

$$U[A]:=\{y\in X\mid (x,y)\in U\text{ for some }x\in A\}$$

is called a uniform neighborhood of $A$.

Two immediate properties that we have already seen in the proof above are: (1). for each $U\in 𝒰$, $x\in U[x]$; 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 $T_{𝒰}:=\{A\subseteq X\mid \forall x\in A,\exists U\in 𝒰\text{ such that }U[x]\subseteq A\}$, then $T_{𝒰}$ is a topology induced by the uniform structure (http://planetmath.org/TopologyInducedByAUniformStructure) $𝒰$. Under this topology, uniform neighborhoods are synonymous with neighborhoods.