G-Tower is Well-Ordered under Subset Relation

Theorem
Let $M$ be a class.

Let $g: M \to M$ be a progressing mapping on $M$.

Let $M$ be a $g$-tower.

Then $M$ is well-ordered under the subset relation such that:

Proof
Let $M$ be a $g$-tower.

By $g$-Tower is Nest, $M$ is a nest.

Hence $\subseteq$ is a total ordering on $M$.

It remains to be shown that $\subseteq$ is a well-ordering, by showing the following:

Let $A \subseteq M$ be an arbitrary non-empty subclass of $M$.

Let $L$ be the set of all elements $x$ which are proper subsets of all elements of $A$:
 * $L = \set {x \in M: \forall z \in A: x \subsetneqq z}$

Since $L \subseteq \powerset z$ for all $z \in A$, it follows by Subclass of Set is Set that $L$ is in fact a set.

It is to be shown that the following conditions hold:


 * $(1): \quad$ If $L$ is empty, then $\O$ is the smallest element of $A$.


 * $(2): \quad$ If $L$ is non-empty and contains a greatest element $x$, then $\map g x$ is the smallest element of $A$.


 * $(3): \quad$ If $L$ is non-empty and contains no greatest element, then its union $\bigcup L$ is the smallest element of $A$.


 * Case $(1)$ -- $L$ is empty

As $L$ is empty:
 * $\O \notin L$

Hence $\O$ is not a proper subset of every element of $A$.

However, $\O$ is a subset (although not necessarily proper) of every element of $A$.

Hence $\O$ must be one of the elements of $A$.

Hence $\O$ is by definition the smallest element of $A$.


 * Case $(2)$ -- $L$ is non-empty and contains a greatest element $x$

Let $L$ be non-empty and contain a greatest element $x$.

We have that:
 * from $g$-Tower is Closed under Mapping that $M$ is closed under $g$
 * $g$ is a progressing mapping on $M$.

From $g$-Tower is Nest: Lemma 2 we have:
 * $\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$

From Fixed Point of $g$-Tower is Greatest Element:
 * if $x = \map g x$, then $x$ is the greatest element (under the subset relation) of $M$.

Hence the conditions are fulfilled for Closed Class under Progressing Mapping Lemma be applied, which gives us that:


 * if:
 * $x$ is a proper subset of all elements of $A$ and the greatest element of $A$ with that property,
 * then:
 * $\map g x \in A$ and is the smallest element of $A$.

This is statement $(2)$.


 * Case $(3)$ -- $L$ is non-empty and contains no greatest element

Suppose $\bigcup L$ were an element of $L$.

Then $\bigcup L$ would be the greatest element of $L$.

But $L$ contains no greatest element.

Hence $\bigcup L$ is not an element of $L$.

From Restriction of Total Ordering is Total Ordering, $\subseteq$ is a total ordering on $L$.

This makes $L$ a nest.

As $L$ is a set, $L$ is a chain.

By $g$-Tower is Closed under Chain Unions:
 * $\bigcup L \in M$

Let $y \in A$.

We have by definition of $L$ that each element of $L$ is a subset of $y$.

Hence:
 * $\forall y \in A: \bigcup L \subseteq y$

But :
 * $\bigcup L \notin L$

Hence there exists at least one $z \in A$ such that $\bigcup L$ is not a proper subset of $z$.

Hence for such a $z$ it must be that:
 * $\bigcup L = z$

That is:
 * $\bigcup L \in A$

and because:
 * $\forall y \in A: \bigcup L \subseteq y$

it follows by definition that $\bigcup L$ is the smallest element of $A$.

Thus it has been shown that every non-empty subclass of $M$ has a smallest element.

Hence by definition $M$ is well-ordered under the subset relation.