Cowen's Theorem/Lemma 6

Lemma for Cowen's 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 $x \in M_x$.

We have that:

$\forall z: M_z \subseteq M$

Proof

Let $x \in M_z$.

Let $y = x \cup z$.

From Set is Subset of Union:

$z \subseteq y$

Hence by Lemma $3$:

$M_z \subseteq M_y$

Hence:

$x \in M_y$

Also, we have:

$x \subseteq y$

and so from Lemma $5$:

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

Hence:

$x \in M_x \cup \paren {\powerset y \setminus \powerset x}$

But we have that:

$x \notin \powerset y \setminus \powerset x$

Hence:

$x \in M_x$

Thus:

$x \in M$

and so:

$M_z \subseteq M$

and the result is shown to hold.

$\blacksquare$

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 $7.9$