# Is the union axiom really needed to prove existence of intersections?

I once asked a while ago how to prove that the intersection $$\bigcap S$$ of any non-empty set $$S$$ exists, using the ZFC axioms. The answer involved the union axiom. However, I don't think it is necessary. Since the intersection of $$S$$ is a subset of an element $$s$$ of $$S$$, I think all that is needed is the axiom schema of separation. Is this true? Or is the union axiom essential?

If $$\mathcal{C}$$ is a non-empty collection of sets we can pick $$C_0 \in \mathcal{C}$$ (by that non-emptyness) and define

$$\bigcap \mathcal{C} := \{x \in C_0\mid \forall C \in \mathcal{C}: x \in C\}$$

purely from separation. You might want to show too that the resulting set does not depend on the set $$C_0$$ we choose first. The union axiom is only used to avoid that last extra proof I think.

Yes, you can do this. However, there is a nice uniformity aspect to the union-based argument which the union-free argument lacks, namely that it does not involve a choice at any step. In the union-free argument we need to pick some element of $$S$$ to "frame" our intersection. Of course the choice doesn't matter, and first-order logic lets us perform such a choice in the first place, but for some people there's still something annoying about this.

This can be brought out more clearly by an "algebraic" interpretation of the situation. Imagine an expansion of the language of set theory by some new operations corresponding to the various axioms, including:

• a unary function for union $$f_{union}$$ with the intended meaning that $$f_{union}(x)=\bigcup x,$$ and

• for each formula $$\varphi(x_1,...,x_n,y)$$ in the language of set theory with only the displayed free variables, an $$(n+1)$$-ary separator function $$g_{sep, \varphi}$$ with the intended meaning that $$g_{sep,\varphi}(a_1,...,a_n,b)=\{c\in b: \varphi(a_1,...,a_n,c)\}.$$

The union-based argument lets us construe the intersection map $$x\mapsto \bigcap x$$ as a term $$\tau$$ in this language: it's just $$x\mapsto g_{sep,\psi}(x, f_{union}(x))$$ where $$\psi(x,y)\equiv\forall z\in x(y\in z)$$.

By contrast, I believe (the argument is pretty tedious sadly) that the union-free argument does not yield such a term, and indeed if we drop the function corresponding to union we can no longer represent intersection as a term. in the resulting more limited language.

So the union-based argument has genuine content that the union-free argument lacks, but this content only shows up when we consider expanded contexts (= set theory + some particular additional operations).

Incidentally, note that $$\tau(x)$$ is defined even when $$x$$ is empty: we have $$\tau(\emptyset)=\emptyset$$. We don't run into the "$$\bigcap \emptyset=V$$"-issue since we've trapped everything inside the specific set $$f_{union}(\emptyset)$$. This is kind of cute!