Minimally Closed Class under Progressing Mapping induces Nest/Proof

Proof
Let $\RR$ be the relation on $N$ defined as:


 * $\forall x, y \in N: \map \RR {x, y} \iff \map g x \subseteq y \lor y \subseteq x$

We are given that $g$ is a progressing mapping.

From the Progressing Function Lemma, we have that:

In what follows, $(1)$ is not needed.

From the Double Induction Principle for Minimally Closed Class, if $\RR$ is a relation on $N$ which satisfies:

then $\map \RR {x, y}$ holds for all $x, y \in N$.

As $\Dom g = N$ it is seen that $(\text D 2)$ is fulfilled by $(2)$ directly.

It remains to be demonstrated that $(\text D 1)$ holds.

Indeed, from Smallest Element of Minimally Closed Class under Progressing Mapping, we have that:


 * $\forall x \in N: b \subseteq x$

By the Rule of Addition:
 * $\forall x \in N: \map g x \subseteq b \lor b \subseteq x$

That is:
 * $\forall x \in N: \map \RR {x, b}$

Thus by the Double Induction Principle for Minimally Closed Class:
 * $\forall x, y \in N: \map \RR {x, y}$

That is:
 * $\forall x, y \in N: \map g x \subseteq y \lor y \subseteq x$

As $g$ is a progressing mapping, we have:
 * $x \subseteq \map g x$

Hence from:
 * $\map g x \subseteq y \lor y \subseteq x$

we have:
 * $x \subseteq y \lor y \subseteq x$

Hence the result by definition of a nest.