Cowen's Theorem/Lemma 5

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 $M$ be the class of all $x$ such that $x \in M_x$.

We have that:
 * $x \subseteq y \implies M_y \subseteq M_x \cup \paren {\powerset y \setminus \powerset x}$

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


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



Let $x \subseteq y$.

Let $T = M_x \cup \paren {\powerset y \setminus \powerset x}$

First we show that $T$ is $y$-special $g$.

We take the criteria one by one:


 * $(1): \quad \O \in T$

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

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

That is:
 * $\O \in M_x$

Hence:
 * $\O \in M_x \cup \paren {\powerset y \setminus \powerset x}$


 * $(2): \quad T$ is closed under $g$ relative to $y$

Let $z \in T$ and $\map g z \subseteq y$.

We are to show that $\map g z \in T$.

Suppose $\map g z \notin \powerset x$.

Then because $\map g z \subseteq y$ it must follow that $\map g z \in \powerset y \setminus \powerset x$.

Suppose $\map g z \in \powerset x$.

That is:
 * $\map g z \subseteq x$

Because $g$ is progressing:
 * $z \in \map g z$

Hence:
 * $z \subseteq x$

and so:
 * $z \subseteq \powerset x$

Hence it is not the case that $z \in \powerset y \setminus \powerset x$.

But we have that $z \in T$.

So:
 * $z \in M_x$

But then because $\map g z \subseteq x$, we have:L
 * $\map g z \in M_x$

Hence:
 * $\map g z \in T$

Thus:
 * $\paren {z \in T \land \map g z \subseteq y} \implies \map g z \in T$

and so $T$ is closed under $g$ relative to $y$.


 * $(3): \quad T$ is closed under chain unions

Suppose $C$ is a chain of elements of $T$.

We have that all elements of $M_x$ are subsets of $x$.

Hence by Subset Relation is Transitive, all elements of $M_x$ are subsets of $y$.

Also, by definition of power set, all elements of $\powerset y \setminus \powerset x$ are also subsets of $y$.

Hence all elements of $T$ are subsets of $y$.

Because all elements of $C$ are subsets of $y$:
 * $\ds \bigcup C \subseteq y$

Hence:
 * $\ds \bigcup C \in \powerset y$

Suppose $\ds \bigcup C \notin \powerset x$.

Then:
 * $\ds \bigcup C \in \powerset y \setminus \powerset x$

Hence:
 * $\ds \bigcup C \in T$

Suppose $\ds \bigcup C \in \powerset x$.

Then:
 * $\forall z \in C: z \subseteq \ds \bigcup C \subseteq x$

and so:
 * $z \subseteq x$

Then:
 * $z \notin \powerset y \setminus \powerset x$

and so:
 * $z \in M_x$

So:
 * $\ds \bigcup C \in \powerset x \implies C \subseteq M_x$

and so:
 * $\ds \bigcup C \in M_x$

and it follows that:
 * $\ds \bigcup C \in T$

Thus, by Proof by Cases:
 * $\ds \bigcup C \in T$

That is, $T$ is closed under chain unions.

We have demonstrated that $T$ is $y$-special $g$.

We have that every element of $M_x$ is a subset of $x$.

Hence every element of $M_x$ is a subset of $y$.

Also:
 * $\powerset y \setminus \powerset x \subseteq \powerset y$

and it follows that:
 * $T \subseteq \powerset y$

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

and the result follows.