A set $M$ is called amenable if it is transitive and satisfies the following conditions:

  1. For all $x,y\in M$, $\{x,y\}\in M$
  2. For all $x\in M$, $\bigcup x \in M$
  3. $\omega \in M$
  4. For all $x,y \in M$, $x\times y \in M$
  5. ($\Sigma_0$ comprehension) Whenever $\Phi$ is a $\Sigma_0$ formula of one free variable with parameters from $M$, then for all $x\in M$, $\{z\in x | \Phi(z)\}\in M$

Although the definition of an amenable set does not include replacement, some very limited amount of replacement follows from the axioms given. For example, for all $x,y\in M$, it must be that $\{\{z,w\}|z\in x, w\in y\}\in M$ and $\{\{z\}|z\in x\}\in M$. So just how limited is the replacement in amenable sets? In particular,

If $M$ is an amenable set and $x\in M$, does it follow that $\{\bigcup z | z \in x\} \in M$?


I think the following is a counterexample to your specific question. Let AH be the set of those $x$ such that (1) each element of $TC\{x\}$ has cardinality at most $\aleph_\omega$ and (2) all but finitely many elements of $TC\{x\}$have cardinality strictly smaller than $\aleph_\omega$. (By $TC\{x\}$, I mean the transitive closure of the singleton, so it contains $x$, all its members, all their members, etc.) If $y\in x$ then $TC\{y\}$ is a subset of $TC\{x\}$, so AH is transitive. AH contains $\omega$ and is easily seen to be closed under pairing, union, and (binary) Cartesian product. Furthermore, it satisfies not only $\Sigma_0$-comprehension but full comprehension, because if $y$ is a subset of $x$ then $TC\{y\}$ is a subset of $TC\{x\}\cup\{y\}$. So AH is amenable.

For each natural number $n$, let $A_n=\{\{n\}\times\aleph_k:k\in\omega\}$, and notice that AH contains not only each of the $A_n$'s but also the set $X$ consisting of all the $A_n$'s. The union of any particular $A_n$ is $\{n\}\times\aleph_\omega$, which is in AH and has cardinality $\aleph_\omega$, but the set of all these unions is not in AH because it has these infinitely many elements of size $\aleph_\omega$. Summary: $X$ is in AH but $\{\bigcup z:z\in X\}$ is not.

Comment: If one modifies the definition of AH by requiring all elements of $TC\{x\}$ to have size strictly below $\aleph_\omega$, one gets the standard example of a model of all the ZFC axioms except the axiom of union. By allowing, in the definition of AH, finitely many exceptions of size $\aleph_\omega$ one revives the axiom of union and in particular one lets each of the sets $\bigcup A_n$ into AH (but just barely) but not the collection of all of them.


I agree with Andreas Blass's solution. The problem, or difficulty, with the definition of amenable is highlighted with this example: $\Sigma_0$-comprehension in this questioner's scheme is not really adequate.

For this reason it is sometimes replaced with $\Sigma_0$- (or `rudimentary') -closure for $\Delta_0$ formulae $\varphi$: $$\forall x \exists w \forall \vec v \in x \exists t\in w \forall u (u \in t \leftrightarrow u \in x \wedge \varphi[u, \vec v]) $$

This is more useful, implies $\Sigma_0$-Comprehension, and rules out the undesirable effect of the example.

  • 1
    $\begingroup$ Philip, welcome to MathOverflow! $\endgroup$ – Joel David Hamkins Jun 21 '10 at 10:26

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.