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

if and only if:

$x$ 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.

$\Box$

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 $z = \bigcup x^\subset$

$M$ is closed under chain unions.

$z$ is an element of $M$.

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

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {III}$ -- The existence of minimally superinductive classes: $\S 8$ Another characterization of $g$-sets