# I have a question on the concept of arbitrary intersection of elements in a set.

Given a collection $\mathcal A$ of sets, the intersection of elements of $\mathcal A$ is defined by the equation:

$$\bigcap_{A\in \mathcal A}A=\{x\mid x\in A \text{ for every } A\in \mathcal A\}.$$

If one allows $\mathcal A$ to be 'empty collection' (that is, all the sets in the collection are empty), what then does the definition mean?

The book I'm reading calls the arbitrary intersection in the particular case to be equal to $U$ (the universal set). I'm not quite sure how that can happen.

The author explains how every $x$ vacuously satisfies the given condition but I don't really follow the reasoning he's giving. Help?

Add: Answers received thus far make sense. I'd like to add that the author in the end mentions that not all mathematicians feel it's reasonable to let $\bigcap_{A \in \mathcal A} A = U$ and he too decides to leaves the intersection undefined when $\mathcal A$ is empty in order 'to avoid difficulty.' I wonder why do some have qualms with allowing $U$ to stand for the intersection? Is there another equally reasonable, but contradicting answer to the problem?

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I think I answered this before. – Asaf Karagila Aug 28 '12 at 17:57

In order not to be in the intersection $\bigcap_{A \in \mathcal{A}} A$, a point $x$ needs to fail to be in one of the sets in the intersection. But if we are taking the empty intersection, so the are no sets in the intersection, then how can you fail to be in one? So every point $x \in U$ satisfies the condition, and is in the intersection.

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Accepting this answer because well this answers the main question asked. Thanks to everyone else for responding. – Acid2 Aug 30 '12 at 19:51

The trouble with letting it mean $U$ is that suddenly the valuye of $\bigcap_{A\in\mathcal A} A$ depends not only on which set $\mathcal A$ is, but also on what you take $U$ to be.

This is not a major problem in most everyday mathematics settings where things are (somewhat informally) typed. For example, if you know that all of the $A$'s, if there had been any, would range over subsets of a particular vector space $V$, then it makes excellent sense to decide to let $\bigcap_{A\in\varnothing} A$ mean all of $V$ in that context.

It still feels a bit untidy to let the types of $A$ and $\mathcal A$ influence the result of the notation so blatantly, though. Usually their effects are more subtle. And it's not quite unambiguous either -- it might be that you also knew that all of the $A$s would have been subsets of some particular subspace of $W\subsetneq V$, and should $\bigcap_{A\in\varnothing} A$ then mean $V$ or $W$? It's often better to be explicit about the $\mathcal A=\varnothing$ case being special than to rely on the notation implicitly meaning the right thing. After all, clarity trumps everything in mathematical writing.

And it breaks down entirely when we're working in the raw untyped set theory, where everything is a set and no sets are more equal than others. Then the only reasonable value for $\bigcap_{A\in\varnothing} A$ would be the "set of all sets", but such a set does not exist in ordinary set theory!

We might decide that $\bigcap \mathcal A$ is sometimes a proper class and sometimes a set, but that creates more irritating nitty-gritty exceptions than it prevents.

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Good answer, but I think there are shades of grey. The complementation symbol "$^C$", has a similar natur, but seldom leads to confusion. – Michael Greinecker Aug 28 '12 at 19:01
@MichaelGreinecker: I agree that there is a large continuum of greys here. That was my point with using fuzzy phrasings such as "a bit untidy" and "often better". – Henning Makholm Aug 28 '12 at 19:04

Ask yourself what it would mean for an object $x\in U$ not to belong to $\bigcap\varnothing$: that would mean that there was an $A\in\varnothing$ with $x\notin A$. But there is no $A\in\varnothing$, so there is certainly no $A\in\varnothing$ such that $x\notin A$. Thus, $x\in\bigcap\varnothing$ by default (or, to use the usual terminology, vacuously).

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To answer the addendum: The problem is that there is no universal set. If you assume one exists, you run into Russell's paradox (see also this question). Therefore, it is better to leave $\bigcap_{A\in\emptyset}A$ undefined.

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Nuance: there is no universal set in ZF (which the OP is probably using/used to). Some other set theories do have a universal set, see e.g. math.stackexchange.com/a/411992/11994. See also en.wikipedia.org/wiki/Universal_set. – Marnix Klooster Sep 7 '14 at 5:55