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I mean for von Neumann ordinal a $\in$-transitive set well-ordered by $\in$, I proved that the class of such ordinals is well ordered by $\in$ (or equally by inclusion), and so different von Neumann ordinals are one element of the other, moreover they are one initial segment of the other. I should prove that order-isomorphism classes (order-isomorphism is a bijective monotone function) are different.

I need it to prove that any set of ordinals (I mean isomorphism classes of well-ordered sets) is also well-ordered, like von Neumann ordinals (identifying a von Neumann ordinal with the relative isomorphism class).

Thanks in advance!

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  • $\begingroup$ Neither $[0,1)$ nor $[0,\infty)$ is well-ordered in the usual order. $\endgroup$ – Ross Millikan Sep 15 '18 at 15:23
  • $\begingroup$ To the last question: any set of von Neumann ordinals is well-ordered (being a subset of the successor to its union, which is a von Neumann ordinal again). Does this help you? $\endgroup$ – metamorphy Sep 15 '18 at 15:26
  • $\begingroup$ @Ross Millikan Oh thanks, you're right, i'll edit $\endgroup$ – nicola Sep 15 '18 at 15:27
  • $\begingroup$ @metamorphy Thanks, for real I proved yet that von Neumann ordinals are well ordered, i should prove that ordinals, intended as isomorphism classes of well-ordered sets, are well ordered, with ordered given by being isomorphism class of an initial segment. $\endgroup$ – nicola Sep 15 '18 at 15:32
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Let $<$ be a well-order on a set $Y$ and let $x\in Y.$ Let $pred_<(x)=\{y\in Y:y<x\}.\quad$ ("pred" for predecessors). Suppose there exists a bijection $f:Y\to pred_<(x).$ Then $f$ cannot preserve the $<$ order.

Proof: The set $\{y\in Y:f(y)\ne y\}$ is not empty because $x$ belongs to it , so it has a $<$-least member $y_0.$ Now $y_0\ne f(y_0),$ and for any $y<y_0$ we have $y=f(y)\ne f(y_0).$

So $y_0<f(y_0).$ And we have $f(y_0)\in pred_<(x) $ so $f(y_0)<x.$

So $y_0<f(y_0)<x.$ So $y_0\in pred_<x=\{f(y):y\in Y\}.$ So there exists $z_0$ with $f(z_0)=y_0.$

Now $z_0>y_0$ because $z<y_0\implies f(z)=z\ne y_0,$ and because $z=y_0\implies f(z)=f(y_0)\ne y_0.$

We now have $z_0>y_0$ and $f(z_0)=y_0<f(y_0). $

Remark: We could also suppose, by contradiction, that $f$ preserves order. Then let $x=x_0$ and let $x_n=f(x_{n-1})$ for $n\in \Bbb Z^+.$ Then $\forall n\in \Bbb Z^+\;(x_n<x_{n-1})$ so the non-empty set $\{x_{n-1}:n\in\Bbb Z^+\}$ has no $<$-least member. But I prefer not to use the axiom of Infinity.

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