Proof. Since $\fV$ and $\fB$ are both non-empty, $\mathfrak{M}$ is non-empty as well. Let $V_{1}, V_{2} \in \fV$ and $M_{1}, M_{2} \in \fB$. By (F2) of $\fV$ and $\fB$, there exists $V_{0} \in \fV$ and $M_{0} \in \fB$ such that $V_{0} \subset V_{1} \cap V_{2}$ and $M_{0} \subset M_{1} \cap M_{2}$. So $V_{0}(M_{0}) \subset V_{1}(M_{1}) \cap V_{2}(M_{2})$, and $\mathfrak{M}$ satisfies (FB1). Given that $\emptyset \not\in \fV, \fB$, $\emptyset \not\in \mathfrak{M}$, so $\mathfrak{M}$ satisfies (FB2). By Proposition 4.2.3, $\mathfrak{M}$ is a filter base for

To see that $\mathfrak{M}$ is a Cauchy filter base, let $V \in \fV$. By (U2), there exists $W \in \fV$ such that $W \circ W \circ W \subset V$. Since $\fB$ is a Cauchy filter base, there exists $M \in \fB$ such that $M \times M \subset W$. Let $(x, y) \in W(M) \times W(M)$, then there exists $(p, q) \in M \times M$ such that $(p, x), (q, y) \in W$. As $M \times M$ $W$, $(x, y) \in W \circ W \circ W$ by symmetry of $W$ and Lemma 5.1.11. Therefore $W(M) \times W(M) \subset V$ and $\mathfrak{M}$ is a Cauchy filter base.

Finally, let $\fF' \subset \fF$ be a Cauchy filter. Let $M \in \fB$ and $V \in \fV$, then there exists $N \in \fF'$ such that $N \times N \subset V$. Since $N \in \fF$, $M \cap N \ne \emptyset$, and there exists $y \in M \cap N$. For any $x \in N$, $N \times N \subset V$ implies that $x \in V(y)$. Hence $N \subset V(M)$ and $\fF' \supset \fF_{0}$.$\square$