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I tried this: $x\in \cap_{i\in \emptyset}A_i\iff x\in A_i\forall i\in \emptyset$ and the right hand side is vacuously ture--right? So this means it is equivalent to $x$ being an element of ... any set?

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Yes.$~~~~~~~~~$ – Michael Greinecker Oct 16 '12 at 22:43

In ZF set theory, which is the most commonly-used formal basis for set theory, you can't construct such a set. There are only a few ways to construct sets. One way is to take the union of previously constructed sets, but that obviously doesn't work here. To make an intersection, you have to use the "specification axiom schema", in which you take some previously-constructed set $S$ and some property $\phi(x)$ and you construct the set of all the elements $s\in S$ for which $\phi(s)$ holds, written $$\{x\in S : \phi(x)\}.$$ For example, you could define $A\cap B$ as $$ \{x\in A : x\in B \};$$

that is, as the set of all elements of $A$ that are also in $B$. (Here $S$ is $A$ and $\phi(x)$ is “$x\in B$”.)

If you have a nonempty family of sets $\{A_1,\ldots,A_n\}$ (or even an infinite family) you could define $\bigcap_{i=1}^n A_i$ similarly as $$ \{x\in A_1 : \forall i\in\{1,\ldots,n\}. x\in A_i\} .$$

But that doesn't work for your nullary intersection, because there is no $A_1$ you can use and there is no previously constructed set $S$ that is appropriate.

In set theories with a universal set, $\bigcap_{S\in\emptyset} S$ is the universal set. In certain contexts it may be useful to restrict one's attention to a limited "universe" $U$, where all sets are subsets of $U$, and in those contexts the empty intersection can be considered to be $U$.

Wikipedia has a discussion of this issue in their article on intersection. I also discussed it at some length in a recent blog article.

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