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 |
---|---|
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 |