# Proving that every poset has a maximal well ordered subset

This has been given as an guided exercise in Pinter's set theory textbook. I'm unable to complete the problem.

Let $$(A, \le )$$ be a partially ordered set and let $$\mathcal{A}$$ be the set of all well ordered subsets of $$A$$. For $$C \in \mathcal{A}$$ and $$D \in \mathcal{A}$$, define $$C \preccurlyeq D$$ iff $$C$$ is a section of $$D$$. (Here's the definition of section: Let $$A$$ be a partially ordered set. A subset $$B$$ of $$A$$ is called a section if for all $$x \in A$$ and all $$b \in B$$, if $$x\le b$$ then $$x\in B$$.)

1. Show that $$\mathcal{A}, \preccurlyeq$$ is a partially ordered set.
2. Show that every chain in $$\mathcal{A}$$ has a upperbound in $$\mathcal{A}$$.
3. Show by Zorn's Lemma that $$A$$ has a maximal well ordered set.

I see how parts 1 through 3 work. Let $$\mathcal{C}$$ be a chain of $$\mathcal{A}$$. I did manage to show that $$\bigcup \mathcal{C}$$ is totally ordered and that it is an upperbound for $$\mathcal{C}$$. I'm unable to show that $$\bigcup \mathcal{C}$$ is well ordered. Any hints would be appreciated.

Here's my attempt: Let $$D$$ be nonempty subset of $$\bigcup \mathcal{C}$$. If $$\mathcal{C}$$ has a greatest element $$C$$, then $$\bigcup \mathcal{C} = C$$ and we are done. Suppose it does not. So for all $$C \in \mathcal{C}$$, there must be $$C' \in \mathcal{C}$$ such that $$C \prec C'$$ and it would follow that $$C=\{ x \in C' \, : \, x < c' \}$$ for some $$c' \in C'$$. Now for each $$d \in D$$, there is a $$C_d \in \mathcal{C}$$ such $$d\in C_d$$. So, we obtain $$\{ C_d : d \in D \}$$.If I could find a $$C' \in \mathcal{C}$$ such that $$C_d \subseteq C'$$ for all $$d\in D$$. We'll be done. But I'm not sure how to finish this.

If $$D$$ is a non-empty subset of $$\bigcup\cal C$$, then take $$d\in D$$ to be some arbitrary point, and consider $$C\in\cal C$$ such that $$d\in C$$. Consider now $$D\cap C$$, it is non-empty. So it has a minimal element, $$m$$.
Now, argue that every $$C'\in\cal C$$ such that $$C\subseteq C'$$, we have that $$D\cap C$$ is an initial segment of $$D\cap C'$$, and therefore $$m$$ is still minimal there. Finally, use this to conclude that $$m$$ is in fact the minimal element of $$D$$ in $$\bigcup\cal C$$.