# 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 $S(j):=\{{x}_{i}\mid i\le j\}$. Then the set

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

is a filter basis: $S$ is non-empty because $({x}_{i})\ne \mathrm{\varnothing}$, and for any $j,k\in D$, there is a $\mathrm{\ell}$ such that $j\le \mathrm{\ell}$ and $k\le \mathrm{\ell}$, so that $S(\mathrm{\ell})\subseteq S(j)\cap S(k)$.

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 ${\mathcal{F}}_{1}\subseteq {\mathcal{F}}_{2}\subseteq \mathrm{\cdots}$ has an upper bound, namely,

$$\mathcal{F}:=\bigcup _{i=1}^{\mathrm{\infty}}{\mathcal{F}}_{i}.$$ |

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 $({x}_{i})$.

Title | section filter |
---|---|

Canonical name | SectionFilter |

Date of creation | 2013-03-22 16:41:37 |

Last modified on | 2013-03-22 16:41:37 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 4 |

Author | CWoo (3771) |

Entry type | Definition |

Classification | msc 54A99 |

Classification | msc 03E99 |