Talk:Axiom of union

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I added a brief note about the reason that the "intersection of the empty set" makes no sense in ZF; but I'm no set theorist, so please feel free to revert if I'm talking crap. Dmharvey 20:28, 21 April 2006 (UTC)

Your statement seems correct, assuming an axiom of intersection which goes like: the intersection of a set is the set consisting of all sets which are contained in members of A, or \forall A, \exist B, \forall C: C \in B \iff (\forall D: (C \in D \and D \in A) \or D \notin A). But you could patch up the axiom to exclude the empty set: \forall A, \exist B, \forall C: C \in B \iff (\forall D: (C \in D \and D \in A) \or D \notin A) \and \exist E: C \in E \and E \in A or equivalently \forall A, \exist B, \forall C: C \in B \iff (\forall D: (C \in D \and D \in A) \or D \notin A) \and \exist E: E \in A which makes the intersectionof the empty set the empty set. Of course this should probably be interpreted as that there is no "intersection of the empty set", but the article now reads as if that is the reason that there cannot be an axiom of intersection, which I think is not right. --MarSch 09:19, 24 April 2006 (UTC)
I've made a few changes to address your concerns... is that an improvement? I think it's an important point to mention, because anyone with any familiarity with "naive set theory" who is reading about ZF for the first time will see the axiom of union and wonder "why isn't there an axiom of intersection", and pretty much the reasons are: (1) ZF is designed so you don't need it, and (2) if you have an unrestricted axiom of intersection, you run into all the sort of "universe" problems that ZF is specifically trying to avoid. Dmharvey 11:01, 24 April 2006 (UTC)