Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable

Theorem
Let $N$ be a class.

Let $N$ be slowly well-ordered under the subset relation.

Then $\ds \bigcup N$ is well-orderable.

Proof
Let $N$ be slowly well-ordered under $\subseteq$.

Let $A = \ds \bigcup N$.

For $a \in A$, let $\map F a$ denote the smallest element of $N$ that contains $a$.

Then for $a, b \in A$, we define $a \preccurlyeq b \iff \map F a \subseteq \map F b$.

It will be shown that $\preccurlyeq$ is a well-ordering on $A$.

By Class which has Injection to Subclass of Well-Orderable Class is Well-Orderable, it is sufficient to show that $F$ is an injection.

So, let us assume that $\map F a = \map F b$ for arbitrary $a, b \in A$.

By condition $S_1$ of the definition of slowly well-ordered under $\subseteq$, $\O \in N$ is the smallest element of $N$.

We have that $a \in \map F a$ by definition.

Hence $\map F a$ cannot be the smallest element of $N$.

Suppose $\map F a$ were a limit element of $N$.

Then by condition $S_3$ of the definition of slowly well-ordered under $\subseteq$, $\map F a$ would be the union of its lower section $\ds \map \bigcup {\paren {\map F a}^\subset}$

But no element of the lower section of $\map F a$ can contain $a$.

Hence $\ds \map \bigcup {\paren {\map F a}^\subset}$ cannot contain $a$.

Hence $\map F a$ cannot be a limit element of $N$.

By Categories of Elements under Well-Ordering, $\map F a$ must therefore be the immediate successor element of some $x \in N$.

As $a$ is the immediate predecessor of $x$, it follows by the definition of $\map F a$ that $a \notin x$.

However, by condition $S_2$ of the definition of slowly well-ordered under $\subseteq$, $\map F a$ contains exactly $1$ more element than $x$.

Hence as $a \in \map F a$ but $a \notin x$ it must be the case that $a$ must be that $1$ more element.

Hence:
 * $\map F a = x \cup \set a$

As $\map F a = \map F b$ :
 * $\map F b = x \cup \set a$

As $\map F b$ is the smallest element of $N$ that contains $b$, we have:
 * $b \ne x$

Hence:
 * $b \in \set a$

which means:
 * $b = a$

We have shown that:
 * $\map F a = \map F b \implies a = b$

and so by definition $F$ is an injection.

Thus, by Class which has Injection to Subclass of Well-Orderable Class is Well-Orderable, the result follows.

The following results also hold: