# Defining arbitrary intersection as a subset of arbitrary union? (ZFC)

I've seen it mentioned a few times that there is an issue with taking the arbitrary intersection over the empty set -- namely, that every $x$ vacuously satisfies the definition of the arbitrary intersection (and thus generates the problematic universal set), when it is defined as[1] $$\cap M = \{ x : \forall A \in M, x \in A\},$$ whose existence is justified when $M$ is non-empty by imposing the condition that x be a member of some set in $M$ and applying the subset axiom scheme.

Why would it not be preferable to use the following definition and note that it is equivalent to the above definition in the case that $M$ is non-empty (since the condition for set membership implies $x\in \cup M$ in this case)? $$\cap M = \{ x \in \cup M : \forall A \in M, x \in A\}$$ It seems that you could justify this definition for any set $M$ (including the empty set) by the union axiom and the subset axiom scheme. Also, you would have that $\cap\varnothing=\varnothing$, since $x\not\in\cup\varnothing$ for all $x$.

I think that "$\cap\emptyset=\emptyset$" is undesirable, because in many instances we want the intersection of a empty family of subsets of an "ambient space" to be equal to that ambient space.
For example, given a commutative ring with unity $R$ and an ideal $I$ in $R$, we define the radical of the ideal $I$ as the set $\sqrt I:=\{r\in R: r^n\in I\ \text{for some}\ n\geq1\}$. It can be shown, using $\mathsf{AC}$, that for a proper ideal $I$ (that is $I\subsetneqq R$) we have $\sqrt I=\cap\mathcal F$, where $\mathcal F$ is the family of prime ideals in $R$ containing $I$; actually the nonemptyness of this family $\mathcal F$ is also proved using $\mathsf{AC}$. But what about $\sqrt R$? evidently we have $\sqrt R=R$ according to the (primary) definition, and in this case $\mathcal F=\emptyset$ because prime ideals are proper by definition, so it is desirable to have $\cap\emptyset=R$ in this case.
This would be perfectly fine. The only issue is a minor one, namely that (working in the framework of $\mathsf{ZF}$ set theory) the union axiom is needed to ensure the existence of $\bigcup M$, while it is not required to show the existence of $\bigcap M$, for which separation and extensionality suffice.