# 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
${({F}_{i})}_{i\in I}$ is a family of elements of $F$ which is totally ordered^{}
by inclusion, then the union $U$ of the ${F}_{i}$ is an element of $F$
as well (since $U$ is an upper bound on the family $({F}_{i})$).
So, let $K$ be a finite subset of $U$. Each element of
$U$ is in ${F}_{i}$ for some $i\in I$. Since $K$ is finite and
the ${F}_{i}$ are totally ordered by inclusion, there is some $j\in I$
such that all elements of $K$ are in ${F}_{j}$. That is, $K\subset {F}_{j}$.
Since $F$ is of finite character, we get $K\in F$, 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 |