This is the question I am struggling with. I understand what well-ordering is – a totally ordered set with a smallest element. I just cannot see how to actually show these two properties.
Q: Let $F$ is a set of well-ordered subsets of some partially ordered set, such that for every pair
$C, C’ \in F$ either $C$ is an initial segment of $C’$, or $C’$ is an initial segment of $C$
If I let $U:= \cup_{C \in F} C$
How do I show $U$ is in fact well-ordered and also that $C$ is an initial segment of $U$?
I have no idea how to approach this so a direction of where to go is what I ask.
Best Answer
A well-ordered set $X$ is a set where every non-empty $S\subset X$ has a minimal element. The well-ordering axiom says that given any set $X$, there exists a total ordering on the set which is a well-ordering.
Suppose $U=\bigcup F$ isn't well ordered. That implies that there exists some $V\subset U$ such that $x\in V$ but $V$ has no minimal element.
Note that $x\in C$ for some $C\in F$ by the definition of $U$. So, consider $C\cap V\subset C$. Since this set is non-empty and $C$ is well-ordered, we know that there exists $y$ minimal element in $C\cap V$.
Since $V$ is not well-ordered, there exists $z\in V$ such that $z<y$. So, $z\notin C$. Let $C'\in F$ such that $z\in C'$. Note that $C$ is an initial segment for $C'$ since $C'\not\subset C$.
This is a contradiction since $C$ contains $y$, but doesn't contain $z<y$. That means $C$ can't be an initial segment of $C'$.
Hence, $U$ is well-ordered. The remainder of the proof is quite trivial.