Let $X$ be a set and $(x_i)_{i\in D}$ a non-empty net in $X$ For each $j\in D$ define $S(j):=\lbrace x_i\mid i\le j\rbrace$ Then the set $$S:=\lbrace S(j)\mid j\in D\rbrace$$ is a filter basis: $S$ is non-empty because $(x_i)\neq \varnothing$ and for any $j,k\in D$ there is a $\ell$ such that $j\le \ell$ and $k\le \ell$ so that $S(\ell) \subseteq S(j)\cap S(k)$

Let $\mathcal{A}$ be the family of all filters containing $S$ $\mathcal{A}$ is non-empty since the filter generated by $S$ is in $\mathcal{A}$ Order $\mathcal{A}$ by inclusion so that $\mathcal{A}$ is a poset. Any chain $\mathcal{F}_1\subseteq \mathcal{F}_2\subseteq\cdots $ has an upper bound, namely, $$\mathcal{F}:=\bigcup_{i=1}^{\infty} \mathcal{F}_i.$$ By Zorn's lemma, $\mathcal{A}$ has a maximal element $\mathcal{X}$

Definition. $\mathcal{X}$ defined above is called the section filter of the net $(x_i)$ in $X$

Remark. A section filter is obviously a filter. The name ``section'' comes from the elements $S(j)$ of $S$ which are sometimes known as ``sections'' of the net $(x_i)$