Cowen's Theorem/Lemma 1

Lemma for Cowen's Theorem
Let $g$ be a progressing mapping.

Let $x$ be a set.

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

We have that:
 * $\powerset x$ is $x$-special $g$.

Proof
By definiton of $x$-special:


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



In this context:
 * $S = \powerset x$

We have from Empty Set is Element of Power Set:
 * $\O \in \powerset x$

Then by definition of closed under $g$ relative to $x$:
 * $y$ is closed under $g$ relative to $x$


 * $\forall z \in y \cap \powerset x: \map g z \in y$
 * $\forall z \in y \cap \powerset x: \map g z \in y$

In this context we have:
 * $\forall z \powerset x: \map g z \in \powerset x$

This is trivially true, as $\Img g \subseteq \powerset x$ by definition.

Then by definition of closed under chain unions:
 * $A$ is closed under chain unions


 * for every chain $C$ of elements of $A$, $\ds \bigcup C$ is also in $A$
 * for every chain $C$ of elements of $A$, $\ds \bigcup C$ is also in $A$

where $\ds \bigcup C$ denotes the union of $C$

Replacing $A$ with $\powerset x$, this is also trivially true.

Hence the result.