# Talk:Intersection of Empty Set

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I'm having some trouble with this one. First, I think all the $\bigcup$s should be $\bigcap$s and that we should have $\bigcap \mathbb S = \left\{{x: \forall X \in \mathbb S: x \in X}\right\}$.

It also seems like this asserts that $x \in \varnothing$, which isn't true, is it? --Alec (talk) 02:17, 17 August 2010 (UTC)

- D'oh. Corrected symbols on proof. Must have been half asleep.

- Okay: so $\left\{{x: \forall X \in \mathbb S: x \in X}\right\}$ means:
- "All the elements in the universe which are also in (all of the sets in $\mathbb S$)", or:
- But all the elements in the universe are
*not*in (all of the sets in $\mathbb S$).

- It's an example of a vacuous truth. --prime mover 05:24, 17 August 2010 (UTC)

- I think my real question is whether $\mathbb S = \{\varnothing\}$ or $\mathbb S = \varnothing$. That is, is it the empty set or the set containing the empty set? --Alec (talk) 01:32, 18 August 2010 (UTC)
- $\text{D}'\text{oh}^2$.--prime mover 05:28, 18 August 2010 (UTC)

- I think my real question is whether $\mathbb S = \{\varnothing\}$ or $\mathbb S = \varnothing$. That is, is it the empty set or the set containing the empty set? --Alec (talk) 01:32, 18 August 2010 (UTC)

It was my understanding that $\bigcap\mathbb C := \{x\in\bigcup\mathbb C : \forall S\in\mathbb C(x\in S)\}$. This results in the intersection of the empty set being the empty set. And the definition given in the page only makes sense in the presence of the axiom of unrestricted comprehension. --Robertbiggs34 (talk) 05:38, 24 June 2013 (UTC)

- Can you cite a source for your assertion? --prime mover (talk) 06:10, 24 June 2013 (UTC)
- Here's a discussion page about the subject. http://math.stackexchange.com/questions/6613/unary-intersection-of-the-empty-set However, I don't have a specific book to cite. You can prove, though, that both definitions result in the same set for non-empty families of sets. The definition using union, however, is just one that naturally extends to the empty set which remains consistent with ZF. In ZF, the axioms of specification and existence prove that there is no universe. Of course, if your background theory accepts classes, then this conversation is moot.--Robertbiggs34 (talk) 18:02, 24 June 2013 (UTC)