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$\begingroup$

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

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 ?2012-09-30
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    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.)2012-09-30
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    @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.2012-09-30
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    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.2012-09-30
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    @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.2012-09-30
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    @PeterSmith In a set theory I have developed, I don't assume the existence of the empty set or any other sets. If I postulate the existence of any sets, then, using my equivalent of the ZF axiom of specification (or separation), I can prove the existence of a unique empty set.2012-10-01
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    @DanChristensen You might be interested in the development of the Scott-Potter axiomatization in Potter's *Set Theory and its Philosophy* which also develops quite a bit before introducing the existential axiom that there *is* a level (i.e. a set which is a partial universe in the hierarchy).2012-10-01

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