Order Isomorphism between Ordinals and Proper Class/Lemma
Order Isomorphism between Ordinals and Proper Class/Lemma 0 17766 335895 297355 2018-01-11T20:45:11Z Barto 3079 notation for image clarified wikitext text/x-wiki {{MissingLinks|one or two}} == Lemma for Order Isomorphism between Ordinals and Proper Class ==  Suppose the following conditions are met: Let $A$ be a class. We allow $A$ to be a proper class or a set. Let $\left({A, \prec}\right)$ be a strict well-ordering. Let every $\prec$-initial segment be a set, not a proper class. Let $\operatorname{Im} \left({x}\right)$ denote the image of a subclass $x$. Let $G$ equal the class of all ordered pairs $\left({x, y}\right)$ satisfying: ::$y \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right)$ ::The initial segment $A_y$ of $\left({A, \prec}\right)$ is a subset of $\operatorname{Im} \left({x}\right)$ Let $F$ be a mapping with a domain of $\operatorname{On}$. Let $F$ also satisfy: :$F \left({x}\right) = G \left({F \restriction x}\right)$ Then: : $G$ is a mapping : $G \left({x}\right) \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right) \iff \left({A \setminus \operatorname{Im} \left({x}\right)}\right) \ne \varnothing$  Note that only the first four conditions need hold: we may construct classes $F$ and $G$ satisfying the other conditions using transfinite recursion. {{explain|Far too woolly. If it is important to the statement of the theorem, then it needs to be expounded properly. If it is not, then it needs to be put somewhere else, either in a separate section, or as a separate proof, or be deleted.}} == Proof == {{begin-eqn}} {{eqn | o =        | r = \left({\left({x, y}\right) \in G \land \left({x, z}\right) \in G}\right)       | c =  }} {{eqn | o = \implies        | r = \left({y \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right) \land z \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right)}\right)       | c = Definition of $G$ }} {{eqn | o = \implies       | r = \left({y \notin A_z \land y \notin A_y}\right)       | c = $A_y$ is disjoint with $\left({A \setminus \operatorname{Im} \left({x}\right)}\right)$.  Same with $A_z$. }} {{eqn | o = \implies       | r = \left({y \not \prec z \land z \not \prec y}\right)       | c = Definition of initial segment }} {{eqn | o = \implies       | r = y = z       | c = $\prec$ is a strict well-ordering }} {{end-eqn}} Therefore, we may conclude, that $G$ is a single-valued relation and therefore a mapping. For the second part: {{begin-eqn}} {{eqn | o =        | r = \left({A \setminus \operatorname{Im} \left({x}\right)}\right) \ne \varnothing       | c =  }} {{eqn | n = 1       | o = \implies       | r = \exists y \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right): \left({A \cap A_y}\right) \setminus \operatorname{Im} \left({x}\right) = \varnothing       | c = Proper Well-Ordering Determines Smallest Elements }} {{eqn | o = \implies       | r = G \left({x}\right) = y       | c = Conditions are satisfied for $\left({x, y}\right) \in G$.  Follows from first part. }} {{eqn | o = \implies       | r = G \left({x}\right) \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right)       | c = equation $(1)$, $y \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right)$ }} {{end-eqn}} Furthermore: :$G \left({x}\right) \in \left({A \setminus \operatorname{Im} \left({x}\right)}\right) \implies \left({A \setminus \operatorname{Im} \left({x}\right)}\right) \ne \varnothing$ by the definition of nonempty.  == Also see == * Transfinite Recursion * Condition for Injective Mapping on Ordinals * Maximal Injective Mapping from Ordinals to a Set == Sources == * {{BookReference|Introduction to Axiomatic Set Theory|1971|Gaisi Takeuti|author2 = Wilson M. Zaring}}: $\S 7.48$