This is a follow-up to my question here. Ordinals are order types of well-ordered sets. Proper classes can be well-ordered too, the most famous example being the class of all ordinals under the standard ordering. Now the set of all order types of well-orderings of a given set is always too big to be put in one to one correspondence with that set. For instance the set of countable ordinals is too big to be put in one to one correspondence with $\mathbb{N}$. I’m wondering if the same is true of proper classes.
To be precise, I am trying to ascertain the truth value of a particular sentence in the language of NBG/MK set theory: there exists a two-place class relation $R$ and a three-place class relation $S$ such that for any fixed $x$, the two-place relation $S(x,y,z)$ is a well-ordering on the class of all $y$ such that $R(x,y)$, and such that for any class $C$ and any class well-ordering $T$ on $C$, there exists an $x$ such that the two-place relation $S(x,y,z)$ is order-isomorphic to $T$. My question is, can this sentence be proven or disproven in NBG or MK? Or is it independent of both.
I’m guessing that this sentence is false, in analogy to the set case, but I’d like to confirm it.
Yes, this is impossible.
Suppose (changing notation slightly) that $(S_i)_{i\in V}$ is a "class of class-well-orderings" with $S_i$ having class domain $D_i$. For $i\in V, j\in D_i$, let $b_{i,j}$ be the set of all $\langle c, x\rangle$ such that
$x\in D_c$,
$S_c$ up to $x$ is order-isomorphic to $S_i$ up to $j$, and
$rk(c)$ is the smallest ordinal $\alpha$ such that $c,x$ with the above properties exist. (Basically, we're using Scott's trick here.)
Let $\mathcal{B}$ be the class of all sets of the form $b_{i,j}$ for $i\in V$ and $j\in D_i$. $\mathcal{B}$ has a natural well-ordering, namely by setting $\langle c,x\rangle\trianglelefteq\langle d,y\rangle$ iff $S_c$ up to $x$ is order-isomorphic to an initial segment of $S_d$ up to $y$, and with this ordering is clearly longer than any $S_i$.