The uniform space of minimal Cauchy filters is complete

For $A\subseteq X$, put $A^\#:=\{\,\mathcal M\in\widehat X:A\in\mathcal M\,\},$ and define $\mathcal F:=\{A\subseteq X:A^\#\in\Phi\}$. Since $X^\#=\widehat X$, $\varnothing^\#=\varnothing$, $(A\cap B)^\#=A^\#\cap B^\#$, and $A^\#\subseteq B^\#$ 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 $D^{\circ3}\subseteq U$. Choose a $\widehat D$-small $S\in\Phi$, 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 $S\subseteq(D[C])^\#$, so $D[C]\in\mathcal F$. Moreover $D[C]\times D[C]\subseteq D^{\circ3}\subseteq U$, making this a $U$-small member of $\mathcal F$.

Let $\mathcal M=m(\mathcal F)$. Given an entourage $E$, choose a symmetric $D$ with $D^{\circ2}\subseteq E$, and choose a $D$-small $A\in\mathcal F$. Then $A^\#\in\Phi$. If $\mathcal N\in A^\#$, the sets $D[A]\in\mathcal M$ and $A\in\mathcal N$ satisfy $D[A]\times A\subseteq D^{\circ2}\subseteq E$, so $\mathcal N\in\widehat E[\mathcal M]$. Hence $A^\#\subseteq\widehat E[\mathcal M]$, and the ball $\widehat E[\mathcal M]$ belongs to $\Phi$. Therefore $\Phi\to\mathcal M$.

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