Fixed Point of g-Tower is Greatest Element

Theorem
Let $M$ be a class.

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

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

Let $x = \map g x$.

Then $x$ is the greatest element (under the subset relation) of $M$.

Proof
Let $x$ be an element of $M$ such that $\map g x = x$.

We will demonstrate by the Principle of Superinduction on $y$ that:
 * $\forall y \in M: y \subseteq x$

The proof proceeds by superinduction.

For all $y \in M$, let $\map P y$ be the proposition:
 * $y \subseteq x$

Basis for the Induction
From Empty Set is Subset of All Sets, we have that:
 * $\O \subseteq x$

Thus $\map P \O$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that if $\map P y$ is true, where $y \in M$, then it logically follows that $\map P {\map g y}$ is true.

So this is the induction hypothesis:
 * $y \subseteq x$

from which it is to be shown that:
 * $\map g y \subseteq x$

Induction Step
This is the induction step:

Suppose that $y \subseteq x$.

We note that by the

Then by the Sandwich Principle for G-Towers: Corollary 2:
 * $\map g y \subseteq \map g x$

But we have that $\map g x = x$.

Thus:
 * $\map g y \subseteq x$

So $\map P y \implies \map P {\map g y}$.

Closure under Chain Unions
It remains to demonstrate closure under chain unions.

Let $\map P y$ be true for all $y \in C$, where $C$ is an arbitrary chain of elements of $M$.

That is:
 * $\forall y \in C: y \subseteq x$

Then it follows by Union of Subsets is Subset that:
 * $\ds \bigcup C \subseteq x$

Thus:
 * $\forall C: \forall y \in C: \map P y \implies \map P {\ds \bigcup C}$

The result follows by the Principle of Superinduction.

Therefore:
 * $\forall y \in M: y \subseteq x$

and so $x$ is the greatest element of $M$.

Also see

 * Fixed Point of Progressing Mapping on Minimally Inductive Class is Greatest Element