Axiom Of Union in ZFC

Question

In the Wikipedia article on ZF set theory it says the following about the axiom of union:

Formally, the axiom of union states that for any set of sets $\mathcal{F}$ there is a set $A$ containing every element that is a member of some member of $\mathcal{F}$: $$\forall \mathcal{F} \,\exists A \, \forall Y\, \forall x [(x \in Y \land Y \in \mathcal{F}) \Rightarrow x \in A].$$ While this doesn't directly assert the existence of $\bigcup\mathcal F$, it can be constructed from $A$ in the above using the axiom schema of specification.

Then why use the axiom of union? Is not it redundant?

Answer 1

Most axioms of ZF can be phrased in one of two ways:

1. For every $x$, there is some $y$, such that $y$ is exactly the desired set.

2. For every $x$, there is some $y$, such that $y$ contains all the elements of the desired set.

These, a priori are not equivalent. But using specification and extensionality, we can actually prove that they are. Because "the desired set" is something which we can define in the language of set theory, so once we have assured there is an even larger set, we can carve it out using specification.

But the second version itself is not somehow provable from specification. You still need to say something about the existence of such a set.

(And this can be applied to union, power set, replacement, and pairing.)

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And while we are at it, there's nothing wrong with having a few redundant axioms.

Using the stronger formulation of the axioms (1 in the above part), we can omit specification entirely. Pairing is generally a consequence of a few other axioms. We can include them for simplicity in some cases, or remove them in others.