Set is G-Set iff Element of G-Ordered Set

Theorem
Let $g$ be a progressing mapping.

Let $x$ be a set.

Then:
 * $x$ is a $g$-set


 * $X$ is an element of a $g$-ordered set.
 * $X$ is an element of a $g$-ordered set.

Proof
Let $M$ be the class of all $g$-sets.

Then $M$ is a $g$-tower.

Sufficient Condition
Let $x$ be a $g$-set.

Hence by $g$-tower is $g$-ordered, $M$ is $g$-ordered.

Hence for $x \in M$, the lower section $x^\subseteq$ of $x$ is also $g$-ordered.

As $x \in x^\subseteq$ it follows that $x$ is an element of a $g$-ordered set.

Necessary Condition
Let $y$ be a $g$-ordered set.

Then by Transfinite Induction on the $g$-ordering of $y$, every element of $y$ must be an element of $M$.

Indeed:


 * $\O \in M$

Let $x \in y$ be an element of $M$ other than the greatest element of $y$

Then the immediate successor of $x$ is $\map g x$.

But $\map g x$ is an element of $M$.

Let $z$ be a limit element $y$.

Let each element of $z^\subset$ be an element of $M$.

Then:
 * because $z = \ds \bigcup x^\subset$
 * $M$ is closed under chain unions.

$z$ is an element of $M$.

Thus every element of $y$ is an element of $M$.