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Topic: $\Omega$$Replies: 0  Search Thread: Advanced Search  ruppVzorn Posts: 4 From: Russia Registered: 5/25/11 \Omega$$ Posted: May 31, 2011 6:09 AM  Plain Text Reply With Von Neumann's definition of$\Omega$, the ordinal set , we have$[0,\
omega)=\omega \forall \omega \in \Omega$. In fact with Von Neumann approach ,$\Omega$cannot be considred as a set ( since$\\Omega$is itself well- ordered and belong to itself . With the "modern" definition of$\Omega\$ that consist in defining the
ordinals as the equivalence classes of well-ordered sets ( two well ordered
sets being equivalent if there exist a monotonous bijective mapping between
them ) , do we still have this property ? Apparently yes because the
demopnstration by transfinite induction is still valid .