Sandwich Principle for G-Towers

Theorem
Let $M$ be a class.

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

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

Let $x, y \in M$ be arbitrary.

Then it cannot be the case that:
 * $x \subsetneqq y \subsetneqq \map g x$

Proof
From Lemma $2$ of $g$-Tower is Nest we have that:


 * $\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$

From the Sandwich Principle:


 * $\forall x, y \in M: x \subseteq y \subseteq \map g x \implies x = y \lor y = \map g x$

That is, there is no element $y$ of $M$ such that:
 * $x \subsetneqq y \subsetneqq \map g x$

Also see

 * Sandwich Principle