Hausdorff's Maximal Principle

Theorem
Each chain in the partial order $\mathcal {P}$ is included in a maximal chain in $\mathcal {P}$.

Proof
Let $\le$ be a partial order on the set $\mathcal {P}$ and let $X$ be a chain in $\mathcal {P}$.

A maximal chain in $\mathcal {P}$ that includes $X$ is a chain $Y$ in $\mathcal {P}$ such that $X \subseteq Y$ and there is no chain $Z$ in $\mathcal {P}$ with $X \subseteq Z$ and $Y \subsetneq Z$.

Let $\mathcal {C} = \lbrace Y \vert Y$ is a chain in $\mathcal {P}$ and $X \subseteq Y \rbrace$. Then a maximal chain in $\mathcal {P}$ that includes $X$ and a maximal element of $\mathcal {C}$ under the partial order induced on $\mathcal {C}$ by inclusion are one and the same. It thus suffices to show that $\mathcal {C}$ contains such a maximal element.

According to Zorn's Lemma, $\mathcal {C}$ contains a maximal element if each chain in $\mathcal {C}$ has an upper bound in $\mathcal {C}$.

Let $W$ be a chain in $\mathcal {C}$, and let $Z = \bigcup { W }$. Suppose that $a$ and $b$ belong to $Z$, then there are sets $A$ and $B$ in $W$ such that $a \in A$ and $b \in B$. Since $W$ is a chain in $\mathcal {C}$, one of $A$ and $B$ includes the other; say $A \subseteq B$. Then both $a$ and $b$ belong to $B$. Since $B$ is a chain in $\mathcal {P}$, either $a \le b$ or $b \le a$. Therefore, $Z$ is a chain in $\mathcal {P} $. For each $Y \in W$, $X \subseteq Y \subseteq Z$. Thus $Z$ is an upper bound for $W$ under inclusion. Finally, since $X \subseteq Z$, $Z \in \mathcal {C}$.