In books on set theory $\bigcap S$ is defined for $S \neq \emptyset$ by taking some arbitrary member $x \in S$ and separating those elements from $x$ that belong to every memebr of $S$. It leads to a problem of what to do if $S = \emptyset$. This case is handled separately by defining $\bigcap S = \emptyset$.

Why don't they just define $\bigcap S := \{x \in \bigcup S |$ $x$ belongs to every member of $S\}$. Then if $S$ is empty, $\bigcup S$ will be empty and there is nothing to separate from it. If $S$ is not empty, then the result is the same as if we used the previous definition.

The only drawback of the new definition is that we need the axiom of union to get $\bigcup S$. But it is a standard axiom.

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    $\begingroup$ The problem is that the simple characterization $x\in \bigcap S\iff \forall s\in S\colon x\in s$ does not hold any more; it is replaced by the less handy $\iff (\exists s\in S\colon x\in s)\land (\forall s\in S\colon x\in s)$. Btw, I usually see $\bigcap \emptyset $ left undefined instead of specifically defined to be empty. - The latter has ugly consequences such as $S\subseteq S'\not\rightarrow \bigcap S'\subseteq \bigcap S$. $\endgroup$ – Hagen von Eitzen Aug 12 '14 at 17:01
  • $\begingroup$ Usually the sets being considered live in some universe $X$. It's natural to have „empty intersection“ = „everything (considered)“. So $\bigcap ∅ = X$. If there is no other universe, then $\bigcap ∅$ is universal class which is not a set. $\endgroup$ – user87690 Aug 12 '14 at 17:05
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    $\begingroup$ Somewhat related: math.stackexchange.com/questions/6613/…, math.stackexchange.com/questions/370188/…, math.stackexchange.com/questions/151924/… and other posts shown there among linked questions. $\endgroup$ – Martin Sleziak Aug 12 '14 at 18:18
  • $\begingroup$ Another problem with $\bigcap \emptyset = \emptyset$ is we would no longer have $\bigcap (A \cup B) = \bigcap A \cap \bigcap B$. $\endgroup$ – Mathemanic Nov 2 '16 at 2:37

I don't think it's ever correct to define $\bigcap \emptyset=\emptyset$. One always expects that taking an intersection of more things should get you a smaller result. (The set of big red ugly expensive things is a subset of the set of big things.) That is, $$\def\S{\mathscr S}\text{if $\S\subset \mathscr T$ then $\bigcap \mathscr T\subset \bigcap \S$}.$$ But having $\bigcap\emptyset = \emptyset$ would spoil that. So that answers one of your questions, which is why we don't do that. I find it hard to believe that any serious text for set theory would do as you say and make $\bigcap \emptyset = \emptyset$.

In set theories with a universal set $V$, one takes $\bigcap \emptyset= V$ for that reason; then we have $\emptyset\subset \S$ for all $\S$ and also $\bigcap \S \subset \bigcap \emptyset = V$, as we would hope. In set theories without a universal set, most notably ZF, one leaves $\bigcap \emptyset$ undefined.

In ZF there is no axiom of intersection that corresponds to the axiom of union; one uses the specification axioms to define intersections. Recall that this means that for any predicate $\Phi(x)$ and any set $X$ we can construct $\{x\in X\mid \Phi(x)\}$, the subset of $X$ for which $\Phi$ holds. Let's see how this goes in this case. We want to take the set of all $s$ that are in every $S$ that is an element of $\S$. But to use the specification axiom schema, we can't take this condition by itself; we have to attach it to some set $X$ and use it to select the desired subset of $X$.

So in ZF the best we can do is $$\{x\in X\mid \forall S\in \S . x\in S\}$$ for some set $X$, which we might agree to abbreviate as $$\def\S{\mathscr S}\bigcap_{(X)}\S.$$ This is the subset of $X$ whose elements are in every element of $\S$, or we might call it the intersection of $S$ "taken relative to $X$". If there were a universal set $V$ then we could take the intersection relative to $V$ and be done (and note that then $\bigcap_{(V)}\emptyset = V$) but ZF has no universal set.

But it should be clear (or you can take it as an exercise) that no matter what $X$ is, if $S\in \S$, we have $$\bigcap_{(X)}\S \subset \bigcap_{(S)} \S.$$ From this it immediately follows that for any $S, S'\in \S$, then $$\bigcap_{(S)}\S = \bigcap_{(S')}\S$$ so it doesn't matter which element of $\S$ we take the intersection relative to.

This shows that when $\S$ is nonempty, there is a unique maximal relative intersection of the elements of $S$, which we can abbreviate as simply $\bigcap \S$. And since $$\bigcap_{(X)}\S = X\cap \bigcap \S$$ the relative intersections weren't getting us anything interesting, and we lose nothing to forget about them. In particular it doesn't matter if we take $X = \bigcup \S$ as you suggest, because the intersection is the same regardless. So we may as well take the simpler definition.

On the other hand, when $\S=\emptyset$, we have $\bigcap_{(X)} \S = X$, which is rather useless, so we lose nothing to forget about it.

  • $\begingroup$ Carl Mummert and Arturo Magidin agree in math.stackexchange.com/questions/6613/… that some authors do make $\bigcap\emptyset=\emptyset $, to my amazement. $\endgroup$ – MJD Aug 13 '14 at 11:54
  • $\begingroup$ Another potential problem with defining $\bigcap \emptyset=\emptyset$ would be no longer having $\bigcap (A \cup B) = \bigcap A \cap \bigcap B$. $\endgroup$ – Mathemanic Nov 26 '16 at 7:50

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