proof of Tukey’s lemma
Let S be a set and F a set of subsets of S such that F is
of finite character. By Zorn’s lemma, it is enough to show that
F is inductive. For that, it will be enough to show that if
(Fi)i∈I is a family of elements of F which is totally ordered
by inclusion, then the union U of the Fi is an element of F
as well (since U is an upper bound on the family (Fi)).
So, let K be a finite subset of U. Each element of
U is in Fi for some i∈I. Since K is finite and
the Fi are totally ordered by inclusion, there is some j∈I
such that all elements of K are in Fj. That is, K⊂Fj.
Since F is of finite character, we get K∈F, QED.
Title | proof of Tukey’s lemma |
---|---|
Canonical name | ProofOfTukeysLemma |
Date of creation | 2013-03-22 13:54:58 |
Last modified on | 2013-03-22 13:54:58 |
Owner | Koro (127) |
Last modified by | Koro (127) |
Numerical id | 4 |
Author | Koro (127) |
Entry type | Proof |
Classification | msc 03E25 |