proof of Tukey’s lemma
Let be a set and a set of subsets of such that is of finite character. By Zorn’s lemma, it is enough to show that is inductive. For that, it will be enough to show that if is a family of elements of which is totally ordered by inclusion, then the union of the is an element of as well (since is an upper bound on the family ). So, let be a finite subset of . Each element of is in for some . Since is finite and the are totally ordered by inclusion, there is some such that all elements of are in . That is, . Since is of finite character, we get , QED.
|Title||proof of Tukey’s lemma|
|Date of creation||2013-03-22 13:54:58|
|Last modified on||2013-03-22 13:54:58|
|Last modified by||Koro (127)|