Axiomatic Set Theory: Why do we need the "Axiom of Union"?

Question

I've been reviewing the axioms of ZFC, and I'm trying to make sense of why the "Axiom of Union" was put in place. While the existence of the intersection (of two) sets seems to be a "Theorem" we can prove from the Axiom of Separation, at first I had trouble seeing the gap between the Axiom of Separation and the construction of the union.

Is the idea that we have no way to express the construction of an arbitrary union (of subsets) by appropriate formula $F(x)$ involving a general set of variables, as we do in the construction of the intersection?

I can see how the Axiom of Separation gives the existence of the union for a finite number of (chosen) subsets, but I guess that using the Axiom of Separation is the strongest construction?

Answer 1

No, the problem is that union gives a bigger set. Given $A$ and $B$ what set contains $A\cup B$ ?

On the other hand $A$ contains $A \cap B$.

Answer 2

First let's take a look at the entire formal statement which gives the Axiom Schema of Separation:

$$\forall w_1, \dots ,w_n \forall X \ \exists Y \ \forall x(x\in Y \iff [x \in X \ \land \ \phi(x, X,w_1, \dots ,w_n)])$$

What Thomas Andrews has written in the comment below the previous answer is that there can be no such formulation of $\phi$ where $(x\in A\lor x\in B)$ because there would have to be a set $X$ that already contained $A \cup B$. Note that the Axiom Schema of Separation implies that $Y$ be a subset of $X$ but we can't say that the existence of such a $X$ is assured and therefore $\phi$ cannot be formulated to satisfy the statement of the axiom schema.

Now we can prove that the unions of finite sets exist without the union axiom but that requires the use of the Power Set Axiom. Please see Can we prove the existence of $A\cup B$ without the union axiom? if the questions aren't marked as a doubles yet.