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 $$X \subseteq Z$$, $$Z \in \mathcal {C}$$.