The uniform space of minimal Cauchy filters is complete

For $A\subseteq X$, put $A^{\mathrm{\#}}:=\{\mathcal{M}\in \widehat{X}:A\in \mathcal{M}\},$ and define $\mathcal{F}:=\{A\subseteq X:A^{\mathrm{\#}}\in \mathrm{\Phi }\}$. Since $X^{\mathrm{\#}}=\widehat{X}$, ${\mathrm{\varnothing }}^{\mathrm{\#}}=\mathrm{\varnothing }$, $(A\cap B{)}^{\mathrm{\#}}=A^{\mathrm{\#}}\cap B^{\mathrm{\#}}$, and $A^{\mathrm{\#}}\subseteq B^{\mathrm{\#}}$ whenever $A\subseteq B$, [L4] shows that $\mathcal{F}$ is a filter on $X$.

The filter $\mathcal{F}$ is Cauchy. Given an entourage $U$, choose a symmetric $D$ with $D^{\circ 3}\subseteq U$. Choose a $\widehat{D}$-small $S\in \mathrm{\Phi }$, a filter ${\mathcal{M}}_0\in S$, and a $D$-small $C\in {\mathcal{M}}_0$. For every $\mathcal{N}\in S$, the relation ${\mathcal{M}}_0\widehat{D}\mathcal{N}$ has witnesses $P\in {\mathcal{M}}_0$ and $Q\in \mathcal{N}$ with $P\times Q\subseteq D$. A point of $C\cap P$ shows $Q\subseteq D[C]$, hence $D[C]\in \mathcal{N}$. Thus $S\subseteq (D[C]{)}^{\mathrm{\#}}$, so $D[C]\in \mathcal{F}$. Moreover $D[C]\times D[C]\subseteq D^{\circ 3}\subseteq U$, making this a $U$-small member of $\mathcal{F}$.

Let $\mathcal{M}=m(\mathcal{F})$. Given an entourage $E$, choose a symmetric $D$ with $D^{\circ 2}\subseteq E$, and choose a $D$-small $A\in \mathcal{F}$. Then $A^{\mathrm{\#}}\in \mathrm{\Phi }$. If $\mathcal{N}\in A^{\mathrm{\#}}$, the sets $D[A]\in \mathcal{M}$ and $A\in \mathcal{N}$ satisfy $D[A]\times A\subseteq D^{\circ 2}\subseteq E$, so $\mathcal{N}\in \widehat{E}[\mathcal{M}]$. Hence $A^{\mathrm{\#}}\subseteq \widehat{E}[\mathcal{M}]$, and the ball $\widehat{E}[\mathcal{M}]$ belongs to $\mathrm{\Phi }$. Therefore $\mathrm{\Phi }\to \mathcal{M}$.

Since every Cauchy filter $\mathrm{\Phi }$ converges, $\widehat{X}$ is complete by [L3].