Every Cauchy filter canonically determines a unique minimal Cauchy filter coarser than it

Let $\mathcal B$ consist of all $E[A]$ with $A\in\mathcal F$ and symmetric entourage $E$. Every such set contains the nonempty set $A$. Given $E[A],D[B]\in\mathcal B$, the symmetric entourage $E\cap D$ and the member $A\cap B\in\mathcal F$ give $(E\cap D)[A\cap B]\subseteq E[A]\cap D[B].$ Thus $\mathcal B$ is a proper downward-directed filter base. Let $m(\mathcal F)$ be the filter it generates.

Since $A\subseteq E[A]$, every member of $\mathcal B$ belongs to $\mathcal F$, so $m(\mathcal F)\subseteq\mathcal F$. To prove it Cauchy, let $U$ be an entourage and choose a symmetric $E$ with $E^{\circ3}\subseteq U$. Choose $A\in\mathcal F$ with $A\times A\subseteq E$. If $y,z\in E[A]$, take $a,b\in A$ with $aEy$ and $bEz$; symmetry gives $yEaEbEz$, so $(y,z)\in E^{\circ3}\subseteq U$. Hence $E[A]\in m(\mathcal F)$ is $U$-small.

Let $\mathcal G\subseteq\mathcal F$ be Cauchy, and fix $E[A]\in\mathcal B$. Choose a symmetric $D$ with $D\subseteq E$, and a $D$-small $B\in\mathcal G$. Since $A,B\in\mathcal F$, choose $c\in A\cap B$. Then $B\subseteq D[c]\subseteq E[A]$, so $E[A]\in\mathcal G$. Thus every Cauchy filter coarser than $\mathcal F$ contains $m(\mathcal F)$.

If a Cauchy filter is coarser than $m(\mathcal F)$, step 2.2 places $m(\mathcal F)$ inside it, so equality holds; hence $m(\mathcal F)$ is minimal. Any minimal Cauchy filter coarser than $\mathcal F$ contains $m(\mathcal F)$ by step 2.2 and must equal it by minimality. This proves uniqueness.

For $x\in X$, the set $\{x\}$ belongs to $\mathcal P_x$, and $\{x\}\times\{x\}\subseteq\Delta_X\subseteq E$ for every entourage $E$. Thus $\mathcal P_x$ is Cauchy by [L1], and step 3.1 supplies its associated minimal Cauchy filter.