I am dealing with a problem in Pinter set theory text. This is the statement
Let A be a partially ordered set and let $\mathcal{A}$ be te set of all the well ordered subsets of A. For $C \in \mathcal{A}$ and $D \in \mathcal{A}$, defince $C<D$ if and only if C is a section of D.
The first problem is to show that $<$ is a partial order relation in $\mathcal{A}$ and I was successful with this.
However, the second one is proving that $\mathcal{A}$, ordered by $<$ is inductive, and I'm stuck on this.
I must show that every chain of A has an upper bound in $\mathcal{A}$. For any chain $\mathcal{C} \subseteq \mathcal{A}$, I showed that $\cup \mathcal{C}$ is a least upper bound of $\mathcal{C}$, but I couldn't show that that this is an element of $\mathcal{A}$.
I believe that it is sufficient to show that every nonempty subset of $\cup \mathcal{C}$ has a least element, which implies that $\cup \mathcal{C}$ is well ordered so that this is an element of $\mathcal{A}$.
However, I couldn't prove that this has a least element. It just makes a vicious circle. How should I deal with this?
To show that every subset of $\cup \mathcal{C}$ has a least element, we would like to work with the elements of $\mathcal{C}$ instead of $\cup \mathcal{C}$ itself, since we know that the elements of $\mathcal{C}$ are well-ordered. Let $S$ be a nonempty subset of $\cup \mathcal{C}$, and let $a \in S$. Then $a$ is an element of $C$ for some $C$ in $\mathcal{C}$. We know that $S \cap C$ is an initial segment of $S$, and so they share the same minimum element, if one exists. But since $C$ is well-ordered, this minimum element does exist!
The key part of this argument was being able to 'jump down' from $\cup \mathcal{C}$ to an element of $\mathcal{C}$.