# 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 ProofOfTukeysLemma 2013-03-22 13:54:58 2013-03-22 13:54:58 Koro (127) Koro (127) 4 Koro (127) Proof msc 03E25