I have an exercise which says

Prove $\bigcap S$ exists for all $S\neq\emptyset$. Where is the asssumption that $S\neq\emptyset$ used in the proof?

The definition of intersection is

$x\in\bigcap S\iff x\in A$ for every $A\in S$

I HAVE read some of the other discussions where it is explained that $\bigcap\emptyset$ is the set of all sets. What I want to ask about is why my "proof" fails, because it seems like I can say $\bigcap\emptyset$ exists, which shouldn't be the case. Something should be stopping me?


My axiom schema of comprehension says that for any property $P(x)$ and for any set $A$, there exists a set $B$ such that $x\in B \iff [x\in A$ and $P(x)]$. Well given any set $S$, the union $U=\bigcup S$ exists (axiom of union). Let $P(x,S)$ be the property "$x\in A$ for every $A\in S$". Then there exists a set $I$ such that $x\in I\iff [x\in U\text{ and }P(x,S)]$.

I've constructed a set $I$, now I just have to argue that it has the property of $\bigcap S$. If $x\in I$, then $P(x,S)$ holds, so $x\in A$ for every $A\in S$. If $x\in A$ for every $A\in S$, then $x\in A$ for some $A\in S$ (so $x\in U$). So $x\in U$ and $P(x,S) \Rightarrow$ $x\in I$. This proves $x\in I \iff x\in A$ for every $A\in S$. Therefore there exists a set satisfying the definition given above. Therefore $I=\bigcap S$ exists for any set $S$.

  • $\begingroup$ Your mistake is when you say "then $x\in A$ for some $A\in S$". There is no such $A$ if $S=\emptyset$. In fact, when $S$ is the empty set, then $\forall A\in S\colon x\in A$ is true for every set $x$. A good way to check for mistakes is by going through your proof and substituting $S$ for $\emptyset$ all the time and see what happens. $\endgroup$ – Jesko Hüttenhain Apr 13 '16 at 14:09

Union Axiom: For any set $S$, there exists a set $U$ whose elements are exactly the members of the members of $S$.

If $S= \emptyset$, there are no members of $S$ and thus no memebers of members of $S$.

Conclusion: $U = \bigcup S = \emptyset$.

Thus, by Comprehension, there is the set $I$ such that:

$x∈I \leftrightarrow [x∈U \land P(x,S)]$

But $U=\emptyset$ and thus also its subset $I=\emptyset$.


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