请输入您要查询的字词:

 

单词 ProofThatEveryFilterIsContainedInAnUltrafilter
释义

proof that every filter is contained in an ultrafilter


Let Y be the set of all non-empty subsets of X which are not contained in . By Zermelo’s well-orderding theoremMathworldPlanetmath, there exists a relationMathworldPlanetmath’ which well-orders Y. Define Y={0}Y and extend the relation ‘’ to Y by decreeing that 0y for all yY.

We shall construct a family of filters Si indexed by Y using transfinite inductionMathworldPlanetmath. First, set S0=. Next, suppose that, for some jY, Si has already been defined when ij. Consider the set ijSi; if A and B are elements of this set, there must exist an ij such that ASi and BSi; hence, AB cannot be empty. If, for some ij there exists an element fSi such that fj is empty, let Sj be the filter generated by the filter subbasis ijSi. Otherwise {j}ijSi is a filter subbasis; let Sj be the filter it generates.

Note that, by this definition, whenever ij, it follows that SiSj; in particular, for all iY we have Si. Let 𝒰=ijSi. It is clear that and that 𝒰.

It is easy to see that 𝒰 is a filter. Suppose that AB𝒰. Then there must exist an iY such that ABSi. Since Si is a filter, ASi and BSi, hence A𝒰 and B𝒰. Conversely, if A𝒰 and B𝒰, then there exists an iY such that ASi and BSi. Since Si is a filter, ABSi, hence AB𝒰. By the alternative characterizationMathworldPlanetmath of a filter, 𝒰 is a filter.

Moreover, 𝒰 is an ultrafilterMathworldPlanetmath. Suppose that A𝒰 and B𝒰 are disjoint and AB=X. If either A or B, then either A𝒰 or B𝒰 because 𝒰. If AY and ASA, then A𝒰 because SA𝒰. If AY and ASA, there must exist x𝒰 such that Ax is empty. Because B is the complementPlanetmathPlanetmath of A, this means that xB and, hence B𝒰.

This completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof that 𝒰 is an ultrafilter — we have shown that 𝒰 meets the criteria given in the alternative characterization of ultrafilters.

随便看

 

数学辞典收录了18232条数学词条,基本涵盖了常用数学知识及数学英语单词词组的翻译及用法,是数学学习的有利工具。

 

Copyright © 2000-2023 Newdu.com.com All Rights Reserved
更新时间:2025/5/4 13:58:57