Sandwich Principle/Corollary 2

Theorem

Let $A$ be a class.

Let $g: A \to A$ be a mapping on $A$ such that:

for all $x, y \in A$, either $\map g x \subseteq y$ or $y \subseteq x$.

Let $g$ be a progressing mapping.

Let $x \subseteq y$.

Then:

$\map g x \subseteq \map g y$

Proof

Let $x \subseteq y$.

Suppose $x = y$.

Then $\map g x \subseteq \map g y$ and the result holds.

$\Box$

Suppose that $x \ne y$.

Then $x \subset y$

It follows from Corollary 1 that:

$\map g x \subseteq y$

As $g$ is a progressing mapping on $A$:

$y \subseteq \map g y$

Hence by Subset Relation is Transitive:

$\map g x \subseteq \map g y$

$\Box$

So in either case:

$\map g x \subseteq \map g y$

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Lemma $4.9 \ (3)$