Cowen's Theorem/Proof

Theorem

Let $g$ be a progressing mapping.

Let $x$ be a set.

Let $\powerset x$ denote the power set of $x$.

Let $M_x$ denote the intersection of the $x$-special subsets of $\powerset x$ with respect to $g$.

Let $M$ be the class of all $x$ such that $x \in M_x$.

Then $M$ is minimally superinductive under $g$.

Proof

Lemma $1$

$\powerset x$ is $x$-special with respect to $g$.

$\Box$

Lemma $2$

$\O \in M$

$\Box$

Lemma $3$

$x \subseteq y \implies M_x \subseteq M_y$

$\Box$

Lemma $4$

$M$ is closed under chain unions.

$\Box$

Lemma $5$

$x \subseteq y \implies M_y \subseteq M_x \cup \paren {\powerset y \setminus \powerset x}$

$\Box$

Lemma $6$

$\forall z: M_z \subseteq M$

$\Box$

Lemma $7$

$M$ is closed under $g$ relative to $x$.

$\Box$

Recall:

From Lemma $2$:

$\O \in M$

From Lemma $7$:

$M$ is closed under $g$ relative to $x$

From Lemma $4$:

$M$ is closed under chain unions.

Hence by definition:

$M$ is superinductive under $g$.

Lemma $8$

Let $A$ be a class which is superinductive under $g$.

Then:

$M_x \subseteq A$

$\Box$

We have a priori that $M$ is superinductive under $g$.

Let $A$ be superinductive under $g$.

Let $x \in M$.

Then:

$x \in M_x$

and from Lemma $8$:

$M_x \subseteq A$

Hence:

$x \in A$

As $x$ is arbitrary, it follows by definition of subclass that:

$M \subseteq A$.

Hence by definition $M$ is minimally superinductive under $g$.

$\blacksquare$

Source of Name

This entry was named for Robert H. Cowen.

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: $\S 7$ Cowen's theorem: Proposition $7.12$