Proposition 3

A subset $\mathfrak{B}$ of a filter $\mathfrak{F}$ on $X$ is a base of $\mathfrak{F}$ if and only if every set of $\mathfrak{F}$ contains a set of $\mathfrak{B}$.

If $\mathfrak{B}$ is a base of $\mathfrak{F}$, then clearly every set of $\mathfrak{F}$ contains a set of $\mathfrak{B}$; conversely, if every set of $\mathfrak{F}$ contains a set of $\mathfrak{B}$, then the set of subsets of $X$ containing a set of $\mathfrak{B}$ coincides with $\mathfrak{F}$ by reason of $(\mathrm{F}_1)$.