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 $\bigcup\cal C$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.