Proving that every poset has a maximal well ordered subset

elementary-set-theorywell-orders

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.

Best Answer

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 $\bigcup\cal C$.

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