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 $\powerset x$ with respect to $g$.

We have that:

$x \subseteq y \implies M_x \subseteq M_y$

Proof

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

Let $x \subseteq y$.

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

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 with respect to $g$.

Thus $\O \in \powerset x$.

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

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

That is:

$\O \in M_y$

Hence:

$\O \in M_y \cap \powerset x$

$\Box$

$(2)

\quad 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 $x$.

$\Box$

$(3)

\quad 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.

$\Box$

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

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

By Intersection is Subset:

$M_y \cap \powerset x \subseteq \powerset x$

and so:

$M_x \subseteq M_y \cap \powerset x$

This article, or a section of it, needs explaining. In particular: Not quite sure why the above $M_x \subseteq M_y \cap \powerset x$ follows, and what exactly from You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code.

That is:

$M_x \subseteq M_y$

$\blacksquare$

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 $7.6$