I want to be clear that I am not asking about the axiom of union. I understand that for an infinite set $A$, $\bigcup A$ exists. My question is specifically about the more widely used (as far as I have seen) version of the notation:
$\bigcup_{b\in B} S_b$.
The technical issue that seems to appear to me is that applying the axiom of union requires first the construction of a set $A$ such that $x\in A \iff \exists b\in B(x= S_b)$.
Clearly in the case that the indexing set $B$ is finite, iterated application of the axiom of pairing with axiom of union can give the required set, but when $B$ is infinite I cannot see any way to justify $A$'s existence, and thus existence of the union in general.
Best Answer
An indexed family $S_b, b \in B$ is a set $I$ already: namely consisting of ordered pairs (pairing axiom) $(b,x)$ where $b \in B$, and which is "functional":
$$\forall z \in I: \exists x,y: (z=(x,y) \land x \in B) \text{ and } \forall b \in B: \exists x: (b,x) \in I \text{ and } \forall b \in B: \forall x,x': ((b,x) \in I \land (b,x') \in I) \to x=x'$$
So the second part of the pair $(b,x) \in I$ is the set we denote $S_b$, but the total collection (assignment) $I$ must already be a set in the universe.
And it's straightforward to define the union $\bigcup_{b \in B} S_b$ from $I$: define the range of $I$ first, and apply the union axiom to that.
If however $S_b$ is defined by some predicate, we can apply the replacement axiom instance for that predicate to get the range as a set in our universe as well.