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$\exists X (\forall Y (\neg(Y \in X)))$

is whats given in my lecture, but I was wondering, is it the same as

$\exists X (\forall Y (Y \notin X))$

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Isn't this an axiom of zf-settheory ? – Andre Sep 30 '12 at 13:13
Not in most standard axiomatisations, for it would be redundant in the presence of other axioms. (The existence of an empty set is typically proved using separation, given the existence of some other, non-empty, set -- and we know some such set exists by the axiom of infinity.) – Peter Smith Sep 30 '12 at 13:33
@Peter: It's sort of cheating to do it that way: why not combine the axiom of the power set and the axiom of the sum set into a single axiom while you're at it? It's pedagogically useful to state the axiom of the empty set anyways -- and it also makes it easier to state the axiom of infinity. Most of the people I've seen insistent on omitting the axiom of the empty set insist on logic where $\exists x$ is tautological, and that's how they get a set to apply separation to. – Hurkyl Sep 30 '12 at 13:50
I think the empty set axiom is also included so that if one removes, say separation, then a model of the remaining axioms still has the empty set. – Quinn Culver Sep 30 '12 at 14:13
@Hurkyl If it is cheating, then canonical texts like Kunen and the Jech Bible cheat! :-) And actually, I think there is a (small) point to doing things their way. Recall that even Cantor himself didn't believe in the empty set (when describing a supposed set that turns out not to have any members, he says that strictly speaking it does not exist). Beginning students have a precedent for being suspicious. So it worth noting that you will have to fiddle with other attractive axioms if you want a set theory without the empty set. – Peter Smith Sep 30 '12 at 14:36

1 Answer 1

up vote 9 down vote accepted

Yes, this is what $\notin$ means. $A \notin B$ is a shorthand for $\neg(A \in B)$.

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