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 $(E\cap D)[A\cap B]\subseteq E[A]\cap D[B]\mathrm{.}$ 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 $E^{\circ 3}\subseteq U$. 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 $(y,z)\in E^{\circ 3}\subseteq U$. 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 $\{x\}\times \{x\}\subseteq {\mathrm{\Delta }}_X\subseteq E$ for every entourage $E$. Thus ${\mathcal{P}}_x$ is Cauchy by [L1], and step 3.1 supplies its associated minimal Cauchy filter.