Particular union of well ordered sets that is well ordered

Question

Suppose $X$ has a partial ordering $\leq$. If $A$ and $B$ are subsets of $X$, we say that $A\leq B$ if and only if $A \subseteq B$ and for $b \in B$, $b \leq a \in A$ implies $b \in A$. Now, if $F$ is a collection of well ordered subsets (by the ordering on $X$) of $X$ such that for $C, D \in F$ either $C\leq D$ or $ D\leq C$, then prove that the union $\bigcup_{C \in F} C$ is well ordered. I can prove that the union is totally ordered, but I cannot find a least element for a subset of the union. How would I go about that?

Answer 1

Okay, I misread the problem as saying that $F$ was a well-ordered collection (under the order defined on $\mathcal{P}(X)$) of well-ordered subsets of $X$.

Hint. Let $R$ be a nonempty subset of $\cup_{C\in F} C$. Let $I=\{D\in F\mid D\cap R\neq \varnothing\}$.

Since $I$ is nonempty, and is a subset of $F$, it has a least element $D_0$. And $R\cap D_0\neq\varnothing$. Try the least element of $R\cap D_0$.

---

This is a bit subtler. If $C$ and $D$ are well-ordered subsets of $X$, and $C\leq D$, then $C$ is what is called an initial segment of $D$. What you are asked to prove is that if you have a collection of well-ordered subsets of a poset $X$, with the property that given any two sets in the collection one is an initial segment of the other, then the union is well-ordered.

We may assume that the union is nonempty (for if it is empty, it is trivially well-ordered). Let $R$ be a nonempty subset of $\cup_{C\in F}C$. Let $r\in R$. Then there exists $C_r\in F$ such that $r\in C_r$, and so in particular, $R\cap C_r\neq\varnothing$. Let $r_0$ be the least element of $R\cap C_r$, which exists since $C_r$ is well-ordered.

I claim that $r_0$ is the least element of $R$. Indeed, let $s\in R$; then there exists $C_s\in F$ such that $s\in C_s$. Now, either $C_r$ is an initial segment of $C_s$ (that is, $C_r\leq C_s)$ or $C_s$ is an initial segment of $C_r$ (that is, $C_s\leq C_r$).

If $C_s\leq C_r$, then $C_s\subseteq C_r$, and hence $s\in C_s\cap R \subseteq C_r\cap R$. Since $r_0$ is the least element of $C_r\cap R$, then $r_0\leq s$.

If $C_r\leq C_s$, then $s\leq r_0$ implies that $s\in C_r$ (since $r_0\in C_r$ and $s\leq r_0$), and hence $s\in C_r\cap R$, and thus $r_0\leq s$, so in fact we have $r_0=s$.

Either way, $r_0\leq s$. This proves that $r_0$ is the least element of $R$.

Thus, $\cup_{C\in F} C$ is well-ordered.