Cowen's Theorem/Lemma 3

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 $\powerset x$ $g$.

We have that:
 * $x \subseteq y \implies M_x \subseteq M_y$

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


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



Let $x \subseteq y$.

First we show that $M_y \cap \powerset x$ is $x$-special $g$.

We take the criteria one by one:


 * $(1): \quad \O \in M_y \cap \powerset x$

From Lemma $1$ we have that $\powerset x$ is $x$-special $g$.

Thus $\O \in \powerset x$.

We have by definition that $\O \in S$ for all $y$-special $S$ $g$.

Hence $\O$ is an element of the intersection of all $y$-special sets $g$.

That is:
 * $\O \in M_y$

Hence:
 * $\O \in M_y \cap \powerset x$


 * $(2): \quad \O \in M_y \cap \powerset x$ is closed under $g$ relative to $x$

Let $z \in M_y \cap \powerset x$ be arbitrary.

Suppose $\map g z \subseteq x$.

Because $x \subseteq y$ we have by Subset Relation is Transitive that $\map g z \subseteq y$.

Because $z \in M_y$ and $\map g z \subseteq y$, we have that $\map g z \in M_y$.

Thus:
 * $\map g z \in \powerset x$

and:
 * $\map g z \in M_y$

Hence:
 * $\map g z \in M_y \cap \powerset x$

and so $M_y \cap \powerset x$ is closed under $g$ relative to $x$.


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

We have by definition that $M_y$ and $\powerset x$ are both closed under chain unions.

From Closure under Chain Unions is Preserved by Intersection:
 * $M_y \cap \powerset x$ is closed under chain unions.

All the criteria $(1)$, $(2)$ and $(3)$ are fulfilled by $M_y \cap \powerset x$.

Hence $M_y \cap \powerset x$ is $x$-special $g$.

By Intersection is Subset:
 * $M_y \cap \powerset x \subseteq \powerset x$

and so:
 * $M_x \subseteq M_y \cap \powerset x$

That is:
 * $M_x \subseteq M_y$