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Topic: $\Omega$$
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ruppVzorn

Posts: 4
From: Russia
Registered: 5/25/11
$\Omega$$
Posted: May 31, 2011 6:09 AM
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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 .



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