Is the comparison of well-orders theorem true for classes

set-theorywell-orders

The classical comparison of well-orders theorem states that for every two wosets (well-ordered sets), one woset is an initial segment of the other one. I am wondering whether it is possible to use a similar theorem for the isomorphism of $\mathbf{ON}\cong \mathbf{ON} \times \mathbf{ON}$.

Is the following also true?

Theorem. Let $\langle X,< \rangle $ and $\langle Y, \prec \rangle $ be two well-ordered classes; that is, $<$ and $\prec$ are two transitive, assymetric, linear, well-founded class relations. Then there exists an initial segment $\langle C, <\rangle $ of $\langle X,< \rangle $ that is isomorphic to $\langle Y, \prec \rangle $, or vice versa.

Attempt of proof. Denote $X_{<x} = \{ y\in X \mid y< x \}$. In an analogous way to the classical proof, define
$$
F = \{ \langle x,y \rangle \mid X_{<x} \cong Y_{\prec y} \}.
$$

  1. We claim that $F$ is a class function. If both $\langle x,y \rangle$ and $\langle x,y' \rangle$ are members of $F$, we have $Y_{\prec y} \cong Y_{\prec y'}$ and we can deduce that $y=y'$.

  2. Now prove that $F$ is a homomorphism of orders. If $y=F(x)$, there exist an isomorphism $g: X_{<x} \to Y_{\prec y} $. So if $x' < x$ we have $y' = F(x') $ for some $y'\prec F(x)$, since $g\restriction X_{<x'}$ is an isomorphism from $X_{<x'}$ to $Y_{\prec y'}$. Rephrasing, if $x<x'$ than also $F(x) \prec F(x')$. This proves that $F$ is a homomorphism, and that $C=\operatorname{dom}(F)$ is an initial segment.

  3. Since $\cong$ is a symmetric relation, everything we proved so far is also true regarding $F^{-1}$. In particular, $F$ is an invertible homomorphism (since $F^{-1}$ is a function) and $D=\operatorname{ran}(F)$ is an initial segment in $Y$, s.t. $C\cong D$.

  4. Assume to the negation that both $C$ and $D$ are proper initial segments in $X$ and $Y$ correspondingly. So we can write $C = X_{<x}$ and $D = Y_{\prec y}$ for some $x\in X$ and $y\in Y$. So, by definition of $F$, $\langle x,y\rangle \in F$, contradicting the definition of $C = \operatorname{dom}(F)$. $\square$

If this indeed proves the comparison theorem of class well-orders, is there also a uniqueness theorem stating the every proper well-ordered class is isomorphic to $\mathbf{ON}$?

Best Answer

Your proof does not work in NBG. It does, however, go through in the stronger MK.

It is valid to discuss, for each $x \in X$, the well-ordered class $X_{< x}$. However, this class could be proper; a simple example can be found in the set $X = \mathbf{ON} \cup \{*\}$, where $*$ is some non-ordinal defined to be larger than all ordinals; then $X_{<*} = \mathbf{ON}$. Incidentally, this answers your other question; $X$ is a larger well-ordered set than $\mathbf{ON}$.

And it is valid to discuss the contention that $X_{< x} \cong X_{< y}$. However, doing so requires a quantification over classes. Therefore, one would not be able to conclude that the class $F$ exists from NBG’s class comprehension scheme, since this scheme only works with formulas that quantify only over sets. However, $F$ can be constructed from MK’s class comprehension scheme.

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