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 $\powerset x$ $g$.

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 $g$.

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


 * $S$ is $x$-special ( $g$)



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


 * $(2): \quad A \cap \powerset x$ 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 $x$.


 * $(3): \quad A \cap \powerset x$ is closed under chain unions

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

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

We have that:
 * $A \cap \powerset x \subseteq \powerset x$

Hence:
 * $M_a \subseteq A \cap \powerset x$

Hence it follows that:
 * $M_y \subseteq A$