Kuratowski's Lemma

Theorem
Let $\struct {S, \preceq}, S \ne \O$ be a non-empty ordered set.

Then every chain in $S$ is the subset of some maximal chain.

Proof
Let $S$ be a (non-empty) ordered set, and $C$ be a chain in $S$. Let $P$ be the set of all chains that are supersets of $C$. Now, here comes the clever part, where we construct chains of chains.

Let $\mathcal{C}$ be a chain in $\mathcal{P}(P)$ (partially ordered by set-inclusion), and define $C' = \bigcup \mathcal{C}$. Note that the elements of $P$ are chains on $(S)$, so the elements of $\mathcal{C}$ are also chains in $S$ ($\mathcal{C}$ is a subset of $P$). Then, that means, $\bigcup \mathcal{C}$ contains elements in $S$, so $C' \subseteq S$.

First, note that $C'$ is a chain in $S$. Let $x,y \in C'$, which means $x \in X$ and $y \in Y$ for some $X,Y \in \mathcal{C}$. However, as $\mathcal{C}$ is an chain in $\mathcal{P}(P)$, that means either $X \subseteq Y$ or $Y \subseteq X$ - so $x,y$ belong to the same chain in $S$, and thus either $x \leq y$ or $y \leq x$. Thus $C'$ is a chain on $S$.

Now, let $x\in C$. Then $\forall A \in P(x\in A)$. Then, because $\mathcal{C} \subseteq P$, we have $\forall A \in \mathcal{C}(x\in A)$. So, we will have $x\in \bigcup\mathcal{C}$, so $C\subseteq C'$. Thus, $C' \in P$.

Now, note $C'$ is an upper bound on $\mathcal{C}$. To prove this consider $x\in D \in \mathcal{C}$, which means $x\in \bigcup\mathcal{C} = C'$, so $D\subseteq C'$. Note that the chain in $P$ was arbitrary, so every chain in $P$ has an upper bound. Thus, by Zorn's Lemma, $P$ has a maximal element, which must be a maximal chain containing $C$.

Note
One can also prove that Zorn's lemma follows from Kuratowski's Lemma, which shows that they are equivalent statements. Thus, this is another statement equivalent to the Axiom of Choice.

Also see

 * Zorn's Lemma
 * Kneser's Lemma
 * Tukey's Lemma
 * Hausdorff Maximal Principle