From Planetmath:
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 $\mathcal{X}$. $\mathcal{X}$ defined above is called the section filter of the net $(x_i)$ in $X$.
- I was wondering if the sentence in bold is contrary to the definition of $S(j)$? Under that definition, it should be $S(\ell) \supseteq S(j)\cap S(k)$ instead, which means $S$ cannot be a filter basis.
- Isn't the maximal element $\mathcal{X}$ always the power set of $X$?
How is the "section filter" of a net related to the filter generated by the net? The filter generated by a net $(x_i)_{i\in D}$ that I found out elsewhere in the internet is defined as
For each $j \in D$, define $x_j := \{x_i : i ≥ j\}$. The collection of tails $\{x_j : j ∈ D\}$ is a filterbase on $X$. The filter generated by this filterbase is called the filter generated by the net.
Thanks and regards!