Proposition 11

If a filter $\mathfrak{F}$ has a countable base, it is the intersection of the elementary filters which are finer than $\mathfrak{F}$.

Let us arrange the countable base of $\mathfrak{F}$ as a sequence $(A_n)_{n \in \mathbf{N}}$; if we put

then the $B_n$ again form a base of $\mathfrak{F}$ (no. 3, Proposition 3) and we have $B_{n+1} \subset B_n$ for each $n$. Let $a_n$ be any element of $B_n$ for each $n \in \mathbf{N}$; then it is clear that $\mathfrak{F}$ is coarser than the elementary filter associated with $(a_n)$. Hence the intersection $\mathfrak{J}$ of the elementary filters which are finer than $\mathfrak{F}$ exists and is finer than $\mathfrak{F}$; if $\mathfrak{J}$ is strictly finer than $\mathfrak{F}$ there exists a set $M \in \mathfrak{J}$ such that $B_n \cap \complement M \neq \varnothing$ for each $n$; if $b_n \in B_n \cap \complement M$, the elementary filter associated with the sequence $(b_n)$ is finer than $\mathfrak{F}$ and does not contain $M$. This contradicts the definition of $\mathfrak{J}$.