Proper initial segment $X$ of a von Neumann ordinal $Y$ has different order-isomorphism class from $Y$ 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!
 A: 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. 
