Hausdorff Maximal Principle/Proof 2

Proof
Let $\preceq$ be an ordering on the set $\PP$.

Let $X$ be a chain in $\left({\PP, \preceq}\right)$.

By definition, a maximal chain in $\PP$ that includes $X$ is a chain $Y$ in $\PP$ such that $X \subseteq Y$ and there is no chain $Z$ in $\PP$ with $X \subseteq Z$ and $Y \subsetneq Z$.

Let us define $\CC$ as:
 * $\CC = \leftset {Y : Y}$ is a chain in $\PP$ and $\rightset {X \subseteq Y}$

Then:
 * a maximal chain in $\PP$ that includes $X$

and:
 * a maximal element of $\CC$ under the partial order induced on $\CC$ by inclusion

are one and the same.

It thus suffices to show that $\CC$ contains such a maximal element.

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

Let $W$ be an arbitrary chain in $\CC$.

Let $Z = \bigcup W$.

It will be shown that $Z$ is an upper bound for $W$ in $\CC$.

Let $a, b \in Z$.

Then:
 * $\exists A, B \in W: a \in A, b \in B$

Since $W$ is a chain in $\CC$, one of $A$ and $B$ includes the other.

, suppose $A \subseteq B$.

Then $a, b \in B$.

Since $B$ is a chain in $\PP$, either $a \preceq b$ or $b \preceq a$.

Therefore $Z$ is a chain in $\PP$.

By definition of $Z$, we have that:
 * $\forall A \in W: A \subseteq Z$

Thus $Z$ is an upper bound for $W$ under inclusion.

Finally we have that:
 * $\forall A \in W: X \subseteq A$

and so $X \subseteq Z$.

Thus:
 * $Z \in \CC$

We have shown that for an arbitrary chain $W$ in $\CC$, $W$ has an upper bound $Z$ in $\CC$.

The result follows.