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Homesection filter

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# section filter

Let $X$ be a set and $(x_{i})_{{i\in D}}$ a non-empty net in $X$. For each $j\in D$, define $S(j):=\{x_{i}\mid i\leq j\}$. Then the set

$S:=\{S(j)\mid j\in D\}$ |

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 $j\leq\ell$ and $k\leq\ell$, 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 $(x_{i})$.

## Mathematics Subject Classification

54A99*no label found*03E99

*no label found*

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