$\text{Ordinal Replacement:}$ if $\phi(x,y)$ is a formula in two free variables $x,y$, then:

$\forall x \ [ordinal(x) \to \exists! y \ (ordinal (y) \wedge \phi(x,y)) ] \to \forall A \ (\forall x \in A (ordinal(x)) \to \exists B \ \forall y \ (y \in B \leftrightarrow \exists x \in A \ \phi(x,y))) $

is an axiom.

In English for every set $A$ of von Neumann ordinals and a function $F$ from von Neumann ordinals to von Neumann ordinals, then we can get a set $B$ whose elements are all the $F$-images of elements of $A$.

i.e. we can replace elements of a set of ordinal by ordinals.

Now is full Replacement provable in $\text{Z + Ordinal Replacement}$?


No. Consider the model $\langle V_{\omega_1},\in\rangle$. This is a model of Zermelo's theory, but all ordinals in it are countable. Thus, it satisfies the ordinal-replacement axiom, since the image of a countable ordinal under any function will be countable, and $V_{\omega_1}$ contains all its countable subsets, since $\omega_1$ is regular.

In fact, in this case, we needn't restrict $B$ to consist of ordinals. Rather, we get the general version of replacement for functions whose domain is a set contained in the ordinals.

Since this model does not satisfy full replacement, it shows that full replacement is not provable from ordinal replacement over Z.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.