The following scheme can be understood as a form of replacement. Axiomatizing $\sf ZF$ with it instead of the usual replacement schema renders it immune to removal of extensionality; see here.

In an article by Gitman/Hamkins/Johnstone, it was shown that the usual version of replacement is in some sense problematic if power set axiom is removed, while at the same time having the well-ordering principle.

Would the following version of replacement work as collection does if powerset is removed while having the well-ordering principle?

Replacement$^*$: if $\varphi$ is a formula in which $B$ is not free, then:

$$\forall A \exists B \forall y \, (y \in B \leftrightarrow \exists x \in A: \forall z (z \in y \leftrightarrow \varphi(x,z)))$$

If we have extensionality, then this is:

$$\forall A \exists B \forall y \, (y \in B \leftrightarrow \exists x \in A: y= \{z \mid \varphi(x,z)\})$$


1 Answer 1


The answer is no.

Your version of replacement is a weakening of ordinary replacement. To see this assume that replacement holds (but perhaps not power set or collection), and then observe that for any set $A$ and any formula $\varphi$, we can first restrict to the subset of $x\in A$ for which $\varphi(x,z)$ defines a set $y$, and then apply replacement to gather these $y$s together into a set.

Since replacement does not imply collection, by the results in the paper you mention, it follows that your principle Replacement* also does not imply collection.

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    $\begingroup$ In the presence of Extensionality, this would be equivalent to replacement, but in absence of Extensionality, this principle is stronger than replacement. But, for the power set issue, it is not stronger. I see. Thanks! $\endgroup$ Feb 23 at 18:13

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