Closed Class under Progressing Mapping Lemma

Theorem
Let $N$ be a class which is closed under a progressing mapping $g$.

Let $g$ be such that:
 * $\forall x, y \in N: \map g x \subseteq y \lor y \subseteq x$
 * if $\map g x = x$, then $x$ is the greatest element of $N$.

Let the following hold:
 * $A \subseteq N$ is a subclass of $N$
 * $x \in N$ is an element of $N$

Let $x$ be:
 * 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 $A$.

Proof
Let the hypothesis be assumed.

Let $A$ be an arbitrary non-empty subclass of $N$.

Let $L$ be the class of all elements $y$ of $N$ such that $y$ is a proper subset of all elements of $A$.

Let $x$ be the greatest element of $L$.

It is to be shown that $\map g x$ is the smallest element of $A$.

We have that $x$ is a proper subset of all elements of $A$.

Then by Corollary 1 to the Sandwich Principle:
 * $\map g x$ is a subset of all elements of $A$.

It remains to be shown that $\map g x$ is itself an element of $A$.

We have that $x$ is a proper subset of all elements of $A$.

We also have that $x$ contains at least one element.

Therefore $x$ cannot be the greatest element of $N$.

Therefore, by hypothesis:
 * $x \ne \map g x$

Thus $x$ is a proper subset of $\map g x$.

Because $x$ is the greatest element of $L$, $\map g x$ is not therefore an element of $L$.

Therefore there exists at least $1$ element $y$ of $A$ such that $\map g x$ is not a proper subset of $y$.

But we have shown that $\map g x \subseteq y$.

Hence:
 * $\map g x = y$

and so:
 * $\map g x \in A$

Hence the result.