Suppose $I$ is a set and $X_i$ is a set $\forall i\in I$. Is there a set $X$ s.t. $\forall i\in I:X_i\subset X$? The question arose when I looked at the most general definition of cartesian products on Wikipedia. There it is assumed that we are given "indexed family of sets" $(X_i)_{i\in I}$, which seems to be a primitive notion in addition to sets and functions. So I was wondering if this is really necessary or if we can assume that there is a set $X$ such that $\forall i\in I:X_i\subset X$, meaning that we are given a function from a set $I$ to the powerset of $X$.
For example, let us consider the Zermelo-Fraenkel set theory. As far as I understand the Wikipedia article on the axiom of union, we have to show that there exists a set $A$ with $X_i\in A$ for all $i\in I$. Can this be rigorously proven? If there are different answers depending on the used axiomatic system, I am also interested to hear that (but please keep in mind that I am a beginner in set theory when writing your answer).
 A: There are at least a couple ways to interpret the notion of an indexed family of sets within axiomatic set theory:

*

*You can consider the indexed family as a function $X$ with domain $I$, and whose values are allowed to be arbitrary sets.  In other words, $(\forall a)(\forall b) ((a, b) \in X \rightarrow a \in I)$ and $(\forall i)(i \in I \rightarrow (\exists! X_i) ((i, X_i) \in X))$.  In this case, the range of the function $X$ can be constructed using the replacement axiom, and then using the union axiom on this range will give a set containing all $X_i$.

*In the case that the $X$ you have in mind is some explicit formula, then typically you can consider $X$ as being given by a first-order formula $\varphi$ with $i$ and $X_i$ as free variables, such that you can prove that this first-order formula gives a functional relation when restricted to the case $i \in I$.  Again in this situation, you can construct the image of this functional relation when applied to $I$ using the replacement axiom, and then use the union axiom to construct the containing set $X$.

*You can consider it as being given by a set $X$ and an "index" function $index : X \to I$; in this formulation, $X_i$ is interpreted as $\{ x\in X \mid index(x) = i \}$.  In this case, the containing set $X$ is built into the definition.  (This interpretation of an indexed family happens to be especially common if you study topos theory, or related topics in algebraic geometry or category theory.)

