Lemma 4.2.8. Let $X$ be a set, $\fF, \mathfrak{G}\subset 2^{X}$ be filter subbases. If for any $E \in \fF$ and $F \in \mathfrak{G}$, $E \cap F \ne \emptyset$, then there exists a filter $\fU \supset \fF \cup \mathfrak{G}$.
Proof. Let $\seqf{E_j}\subset \fF$ and $\seqf[m]{F_j}\subset \mathfrak{G}$, then $\bigcap_{j = 1}^{n} E_{j} \in \fF$ and $\bigcap_{j = 1}^{m} F_{j} \in \mathfrak{G}$. Thus their intersection is non-empty, and $\fF \cup \mathfrak{G}$ is a filter subbase. By Definition 4.2.7, there exists a filter containing $\fF \cup \mathfrak{G}$.$\square$