# Prove $\bigcap S$ exists for all $S \ne \emptyset$. Where is the assumption $S \ne \emptyset$ used in the proof?

I know that

$\bigcap S = \{x : \forall A \in S, x \in A \}$

Here if $S = \emptyset$ then there is no $A$ which satisfies the property. However, why is it then that it defaults to every possible $x$, that is the $\bigcap S$ becomes the set of all sets, as opposed to just saying the property no longer holds?

• Which set do you suggest then that $\bigcap \emptyset$ would be? – Ittay Weiss May 4 '15 at 9:55
• Writing $\{x : \Phi(x)\}$ for some formula $\Phi$ is dangerous (as per Russel's paradox). This is unrestricted set comprehension and is not allowed in ZFC. – kahen May 4 '15 at 9:55
• For example if you instead wrote $\cap S = \{x \in \cup S \mathrel\colon A \in S \implies x \in A\}$, then there would be no problems and you'd have $\cap \emptyset = \emptyset$. – kahen May 4 '15 at 10:04

The intersection over an empty collection of sets does not make sense (to me). The reason is this:

Suppose $S = \emptyset$. Now let $x$ be any comprehensible object (set, element whatever). Now I ask you why is $x$ not in $\bigcap S$. The only reason for $x$ to not be in $\bigcap S$ is if $(\forall A \in S, x \in A)$ is false. For this formula to be false there must exist a set $A'\in S$ such that $x \not \in A'$ i.e. the negation of the "set-builder" formula is $(\exists A' \in S \in , x \not \in A')$. But there can be no such $A'$ in $S$ since $S$ is empty.

Hence, anything belongs to $\bigcap S$. Whether this is a problem is for you to decide for yourself. It depends on the level of set-theoretic formlisation you wish to incorporate into your mathematics. If you believe in a "Universal Set" then no problem! But most would have seen big fat issues like Russell's Paradox resulting due to the thought of inclusive sets.

Formally speaking, note that $x\in\bigcap S$ if and only if for every $X\in S$ we have that $x\in X$. But this is equivalent to saying that $x\notin\bigcap S$ if and only if there exists $X\in S$ such that $x\notin X$.

So in order to rule out $x\notin\bigcap S$ we need to have some witness. And in other words, we need that $S\neq\varnothing$.

It's similar to the convention that the sum of $\emptyset$ is 0 and the product of $\emptyset$ is 1. This is because $x + 0 = x$ and $x \times 1 = x$. Similarly, for which $y$ is $x \cap y = x \,\forall x$? Only $y$ the set of everything.

This is OK if all your intersections are bounded over some small universe set.