About the unionset axiom I found this formulation of the unionset axiom: 

For each set $\mathcal E$ there exists a set $B$ whose members are the members of the members of $\mathcal E$, i.e., so that it satisfies the equivalence $$t\in B\iff (\exists X\in\mathcal E)(t\in X)$$

An then: 

The unionset operation is obviously most useful when $\mathcal E$ is a family of sets, i.e., a set all of whose members are also sets.

So the last statement implies that the elements of $\mathcal E$ are not necessarily sets. But doesn't the first quotation imply that all elements of $\mathcal E$ are sets? Otherwise how can $t$ be a member of an element of $\mathcal E$?
 A: I don't have Moschovakis on hand, but if I recall correctly he's working in (essentially) a set theory with urelements (EDIT: also called atoms); this is a set theory where not everything is a set, but instead we have a collection of "urelements" which can be elements of sets but are not themselves sets. 


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*In some sense, in a model of set theory with urelements we build a set-theoretic universe on top of the collection of urelements. There is a lot of freedom here (e.g. should there be a "set of all urelements"?). The usual set theory ZFC (and ZF) does not allow urelements. However, we can whip up a "ZF(C) with urelements" without serious difficulty. Conversely, set theory with urelements isn't really very different (in most contexts, anyways) than set theory without urelements, so not much is lost by omitting them.


In a set theory with urelements, we can have a set some of whose elements are sets and others of which aren't. For example, suppose $a$ and $b$ are urelements. Then $$\mathcal{E}=\{a, \{b\}, \{\{a,b\}\}, \{\{\}\},\{\}\}$$ is a perfectly valid set. The union construction can be applied to such an $\mathcal{E}$, giving $$\bigcup\mathcal{E}=\{b, \{a,b\}, \{\}\}.$$ Note that the element $a$ of $\mathcal{E}$ doesn't contribute anything to $\bigcup\mathcal{E}$: by definition $\bigcup\mathcal{E}$ is the set of all elements of elements of $\mathcal{E}$, and $a$ - while an element of $\mathcal{E}$ - has no elements of its own. More trivially, "$\{\}$" is an element of $\mathcal{E}$ which doesn't contribute anything to the union.


*

*Note that there is no "typing" here: even if $a$ is an urelement, an expression like "$t\in a$" makes perfect grammatical sense (it's just false).


Note that if $\mathcal{E}$ is any set then $\bigcup\mathcal{E}=\bigcup\mathcal{E}'$ where $\mathcal{E}'$ is the subset of $\mathcal{E}$ gotten by removing all urelements (= all non-sets); so there's no real reason to consider taking the union of a set which isn't a family of sets. And in any reasonable set theory with urelements, being an urelement is a definable property (so we can form $\mathcal{E}'$ from $\mathcal{E}$ via the separation/subset axiom). There are two obvious ways to guarantee this:


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*We could have a predicate naming the urelements.

*We could have a constant symbol $\emptyset$ naming the emptyset; then an object $x$ of our universe is an urelement iff $x\not=\emptyset\wedge\forall y(y\not\in x)$.
Note that these are ultimately equivalent.
