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Axiom of regularity (foundation). Every non-empty set $X$ has an element $y$ that is disjoint from $X$. In other words, $$\forall X\,\left(X\neq \varnothing \Longrightarrow\exists y\in X\,\left(y\bigcap X = \varnothing\right)\right) \tag{R}$$

What would happen if the axiom of regularity is replaced with the axiom $$\forall Y\,(Y\notin Y),\tag{Nei}$$

i.e. "no set $Y$ can be an element of itself"? Clearly the axiom of foundation implies $(\text{Nei})$. On the other hand, in ZFC the axiom regularity is equivalent to there being no countable descending $\in$-chains of any kind whereas $(\text{Nei})$ says essentially that there are no $\in$-chains of the constant kind. While this is not a proof, I would intuitively assume that $(\text{Nei}) \Longrightarrow (\text R)$ does not hold in general in ZFC. If this is true, one would forfeit some results if $(\text R)$ is replaced by $(\text{Nei})$ already in ZFC. So

  1. What are some of the more stark consequences of replacing $(\text R)$ with $(\text{Nei})$ in ZFC? In other words, are there important or intuitive/desirable theorems which become unprovable in ZFC – $(\text R)$ + $(\text{Nei})$?
  2. How about in ZF?

I am especially interested if there are results that are lost which could be described as "intuitive" or "elementary". By browsing through Halmos's Naive set theory, which does not have the axiom of regularity, a newbie like myself would gather that there are no profound consequences for the puropses of elementary set theory. And if the purpose of regularity in elementary set theory is to simplify the proofs of some results (seemingly exclusively of the $x\notin x$ type), it also seems that assuming $\text{(Nei)}$ instead of $\text{(R)}$ would further simplify the situation by virtue of being a more natural axiom in its syntacitc form and intuitive content.

Or is this replacement a bad idea in the long run?

Linear Christmas
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    What kind of consequences would it have? Any non-set theory stuff can be taken to live inside the von Neumann universe anyway. And indeed, $\forall x(x\notin x)$ is not enough to prove Regularity. – Asaf Karagila Jan 22 '21 at 15:07
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    Related: https://math.stackexchange.com/questions/213639/where-is-axiom-of-regularity-actually-used – Asaf Karagila Jan 22 '21 at 15:12
  • @AsafKaragila Thank you for confirming that $\forall x (x \notin x)$ is not enough to prove regularity. As for your second sentence, I have almost no clue what that means :). Probably a rough rephrasing of my question is... If I replace (R) with (Nei) in ZF, do some important statements become undecidable? Will I eventually get into trouble for doing so? Did you mean that the answer is no, you will not get into trouble? – Linear Christmas Jan 22 '21 at 15:33
  • Well, given any universe of ZF-Reg, you can still construct the von Neumann universe inside of it: that is, iterate power sets from the empty set. It will be a model of ZF. So any non set theory stuff you can just do in there. Moreover, the thing about weakening axioms is that we simply prove less. So certainly we can't prove something "new" fails. But even if you add the negation of Reg, then you just can't prove Reg (and its equivalents). – Asaf Karagila Jan 22 '21 at 15:37
  • @AsafKaragila What do you mean by "non-set theory stuff"? I agree, of course, that there can't be anything "new" that fails. By results lost, I indeed have in mind that some results in ZF could become unprovable from the weaker axioms of ZF – Reg + Nei. (Surely this is the same as undecidable/independence in the axiomatic sense?). And some results do indeed become unprovable, as you confirmed, even if just the previous axiom of regularity. But there could be other more significant statements which were provable in ZF but are not in ZF – Reg + Nei. Are there? (1/2) – Linear Christmas Jan 22 '21 at 15:55
  • Non-set theory stuff is things like analysis, or things like algebra. You can even say more, since $\omega$ is inherently well-founded, CH is a statement about well-founded sets. Similarly, almost all the other things we are interested in. – Asaf Karagila Jan 22 '21 at 15:56
  • @AsafKaragila You might have already answered that question in a previous comment. I might have been too inexperienced to understand, and if so, I apologise. I will try to get a yes/no type answer from you however reluctant you might be to give it to me :) (2/2) – Linear Christmas Jan 22 '21 at 15:56
  • I doubt that in ZF - Reg + Nei you could disprove existence of sets such that $x\in y \in x$. – Dark Malthorp Jan 22 '21 at 15:58
  • @AsafKaragila Can one define cardinal numbers as sets in the context of ZF $-$ Reg + Nei? I read recently that it is possible in ZF (via Scott's trick) but this supposedly relies heavily on Reg. Does the trick survive weaking to Nei? – Linear Christmas Jan 29 '21 at 15:03
  • I'm guessing not. Suppose that you have a collection of distinct pairs: $x={y}={{x}}$. Treat each pair as an "atom", then the usual arguments in ZFA should still work, including the failure of defining cardinals. This can get more complicated, not just with these small loops, I suppose. – Asaf Karagila Jan 29 '21 at 15:07

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