Class Mapping has Minimally Superinductive Class
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Theorem
Let $g$ be a (class) mapping.
Then there exists a class $M$ that is minimally superinductive under $g$.
Proof
This theorem requires a proof. In particular: According to the rubric of the exercise, this is to be proved using one of the versions of the Transfinite Recursion Theorem that appears in S&F You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. 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 {{ProofWanted}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
Sources
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $6$: Order Isomorphism and Transfinite Recursion: $\S 5$ Transfinite recursion theorems: Exercise $5.2$