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Has there ever been a set theory without an empty set? Is this possible?

I ask because we usually take the empty set to exist axiomatically or obtain it through separation and a nonempty set together with the standard parameter-free predicate $X\neq X$, but it seems possible to have a 'set theory' without an axiom asserting the existence of an empty set or an axiom of separation.

I put 'set theory' in quotations because such a nonstandard axiomatization might not really deserve to be called a set theory per-se (it wouldn't prove the existence of intersections of disjoint sets), but more formally I mean

Has a theory in the language of set theory whose axioms do not prove the existence of an empty set ever been explored?

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    $\begingroup$ Any reason for the downvote? $\endgroup$
    – Alec Rhea
    Commented May 28, 2022 at 17:05
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    $\begingroup$ I think it depends on what you mean by a "set theory". There is one such theory that has been explored for over 100 years in disguise: Take the usual axioms of ZFC and flip the $\in$ symbol in all instances. Then you get a "set theory". It proves the same theorems as ZFC with $\in$ flipped. Therefore this theory does not prove that there is an empty set. The "sets" here do not align with our intuition of a "set" though: For example, given sets x,y, there is a set z which is contained in both x and y, but nothing else. (I couldn't help but post this "troll" answer, sorry :) $\endgroup$
    – Burak
    Commented May 30, 2022 at 20:27

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For a good discussion of this matter, see: Kanamori, Akihiro The empty set, the singleton, and the ordered pair. Bull. Symbolic Logic 9 (2003), no. 3, 273–298.

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I’ve explored a “set theory whose axioms do not prove the existence of an empty set,” the Incomprehensive Set Theory in my “Naive View of the Russell Paradox” (https://arxiv.org/abs/2103.00090), but for axiomatic parsimony, not with the motives I suspect you’re looking for. I note that “Any assumptions about set existence, even that any sets at all exist, will be noted explicitly. What follows will be trivial unless a lower or an upper exists, and the main interest is when an empty set and a universal set exist.” I imagine Incomprehensive Set Theory plus the existence of the Universal Set is consistent and does not prove the existence of the empty set, but, as noted, that is not necessarily of much interest.

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  • $\begingroup$ Much-improved version, thanks to deservedly harsh comments from the second anonymous referee, to appear in Logique et Analyse. Preprint at arxiv.org/abs/2103.00090v5. $\endgroup$
    – Flash Sheridan
    Commented Aug 20 at 23:14

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