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

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Theorem

Let $N$ be a class.

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


Then $\bigcup N$ is well-orderable.


Proof

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

Let $A = \bigcup N$.


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 $\map \bigcup {\paren {\map F a}^\subset}$

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

Hence $\map \bigcup {\paren {\map F a}^\subset}$ 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 $\map F a = \map F b$ by hypothesis:

$\map F b = x \cup \set a$

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.

$\blacksquare$


The following results also hold:


Corollary 1

$a \preccurlyeq b \iff \forall x \in N: b \subseteq x \implies a \subseteq x$


Corollary 2

$\forall x \in N, a \in \bigcup N: x \in \map F a \implies x \preccurlyeq a$


Corollary 3

$\forall a \in \bigcup N:$ if $a$ is not the greatest element of $\bigcup N$, then the immediate successor of $a$ is the smallest element of $\bigcup N \setminus \map F a$.


Corollary 4

Every immediate successor element of $N$ is $\map F a$ for some $a \in N$.


Sources

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