Cowen's Theorem/Lemma 6
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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 $g$.
Let $M$ be the class of all $x$ such that $x \in M_x$.
We have that:
- $\forall z: M_z \subseteq M$
Proof
Let $x \in M_z$.
Let $y = x \cup z$.
From Set is Subset of Union:
- $z \subseteq y$
Hence by Lemma $3$:
- $M_z \subseteq M_y$
Hence:
- $x \in M_y$
Also, we have:
- $x \subseteq y$
and so from Lemma $5$:
- $M_y \subseteq M_x \cup \paren {\powerset y \setminus \powerset x}$
Hence:
- $x \in M_x \cup \paren {\powerset y \setminus \powerset x}$
But we have that:
- $x \notin \powerset y \setminus \powerset x$
Hence:
- $x \in M_x$
Thus:
- $x \in M$
and so:
- $M_z \subseteq M$
and the result is shown to hold.
$\blacksquare$
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: Proposition $7.9$