TITLE: When does replacement (accidentally) hold in amenable sets?
QUESTION [9 upvotes]: A set $M$ is called amenable if it is transitive and satisfies the following conditions:

For all $x,y\in M$,  $\{x,y\}\in M$
For all $x\in M$, $\bigcup x \in M$
$\omega \in M$
For all $x,y \in M$, $x\times y \in M$
($\Sigma_0$ 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 $\{\bigcup z | z \in x\} \in M$?

REPLY [5 votes]: 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.