Cowen's Theorem/Lemma 8

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

Then:

$M_x \subseteq A$

Proof

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

We show that $A \cap \powerset x$ is $x$-special with respect to $g$.

Let us recall the definition of $x$-special with respect to $g$.

$S$ is $x$-special (with respect to $g$)

if and only if:

\((1)\)	$:$							$\O \in S$
\((2)\)	$:$							$S$ is closed under $g$ relative to $x$
\((3)\)	$:$							$S$ is closed under chain unions

We take the criteria one by one:

$(1)

\quad \O \in A \cap \powerset x$

We have by definition that $\O \in A$ and $\O \in \powerset x$.

Hence:

$\O \in A \cap \powerset x$

$\Box$

$(2)

\quad A \cap \powerset x$ is closed under $g$ relative to $x$

Let $y \in A \cap \powerset x$ and $\map g y \in \powerset x$.

Because $y \in A$ we have:

$\map g y \in A$

Hence:

$\map g y \in A \cap \powerset x$

That is, $A \cap \powerset x$ is closed under $g$ relative to $x$.

$\Box$

$(3)

\quad A \cap \powerset x$ is closed under chain unions

Because $A$ and $\powerset x$ are both closed under chain unions, then so is $A \cap \powerset x$.

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$\Box$

We have demonstrated that $A \cap \powerset x$ is $x$-special with respect to $g$.

We have that:

$A \cap \powerset x \subseteq \powerset x$

Hence:

$M_a \subseteq A \cap \powerset x$

Hence it follows that:

$M_y \subseteq A$

$\blacksquare$

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: Lemma $7.11$