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

1. 1.

   For each $(x,U[x])\in\mathfrak{N}$, $x\in U[x]$, since every entourage contains the diagonal relation.

2. 2.

   Every $x\in X$ and every entourage $U\in\mathcal{U}$, $U[x]\subseteq X$ with $(x,U[x])\in\mathfrak{N}$

3. 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. 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. 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 $(y,U[x])=(y,W[y])\in\mathfrak{N}$.

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