Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


ruppVzorn
Posts:
4
From:
Russia
Registered:
5/25/11


$\Omega$$
Posted:
May 31, 2011 6:09 AM


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 wellordered 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 .



