Thanks Dan. It seems that when we say sets cannot contain themselves all the problems go away. I wonder how mathematicians are able to imagine sets that contain themselves when a set is defined as a container distinct from its contents. I can make no sense of the idea, and this is what intrigues me. Why does anyone assume a set can contain itself? Is there ever a need for this assumption?

@PeterJ "when we say sets cannot contain themselves all the problems go away" have you read anything that's been written in this thread? Regularity has nothing to do with Russell's paradox.

@PeterJ: "It seems that when we say sets cannot contain themselves all the problems go away." This is 100% wrong. The axioms of ZF happen to imply that no set can contain itself, but this is completely unrelated to how ZF avoids Russell's paradox.

@PeterJ I myself have never found it necessary to invoke the Axiom of Regularity to ban set-self-membership. Then again, I haven't done any formal proofs about ordinals. With the above a restricted axiom for subsets, however, it becomes impossible to derive Russell's Paradox.

@David - Nevertheless, if sets cannot contain themselves then R's paradox does not arise. This was my simple point.

@EricWofsey - I'm asking what problems remain when we state sets cannot contain themselves. The answer seems to be none.

@PeterJ You've made it very clear that that's your simple point. Your simple point is false.\

@PeterJ Because saying $S=\{x:x\notin x\}$ leads to a contradiction regardless of whether a set can contain itself. Assume for minute that no set can contain itself. Then the notation $S=\{x:x\notin x\}$ means $\forall_x(x\in S\iff x\notin x)$. That holds for all $x$; letting$x=S$ gives $S\in S\iff S\notin S$, a contradiction. A contradiction, even though we said sets cannot contain themselves.

@DavidC.Ullrich - I suspect you've just given me a useful answer but I can't tell because I can't read the symbols. If you're saying that (regardless of whether sets are allowed to contain themselves) the set of all sets is a paradoxical idea, then I get this. It seems a more important issue than R's paradox. (To me my 'simple point' above still seems to stand).

@DanChristensen - Does it not remain the case that when we disallow sets from containing themselves R's paradox cannot arise? Then a paradox arises for the set-of-all sets, but it is a different one. No?

"Does it not remain the case that when we disallow sets from containing themselves R's paradox cannot arise?" WHY do you continue to ask the same question, no matter how many times it's been answered??????????? I just demonstrated RP for you, after assuming that sets cannot contain themselves.

@PeterJ RP would also not arise if proof by contradiction was disallowed. But whether you disallowed proof by contradiction or set-self-membership or both, you would still need a subset axiom as I have described here. But that axiom itself will prevent RP. So, to avoid RP, you do not have to ban either proof by contradiction OR set-self-membership.

@PeterJ "If you're saying that (regardless of whether sets are allowed to contain themselves) the set of all sets is a paradoxical idea, then I get this. It seems a more important issue than R's paradox. (": In fact R's paradox is the simplest way to prove that there is no set of all sets (assuming the rest of standard set theory). Because if $U$ is the set of all sets then saying $S=\{x\in U:x\notin x\}$ is legal, and it's the same as the $S$ above. (When you say this is more important than R you must have a different proof in mind: how do you show there is no set of all sets?)

@DanChristensen That sounds possibly misleading. Because in fact disallowing self-membership does not have any effect on RP (in fact self-membership is disallowed in standard set theory plus unrestricted comprehension, which was the whole point to RP in the first place)

@DavidC.Ullrich If expressions of the form $x\in x$ are disallowed, then so is the expression $\exists a: \forall b: [b\in a \iff \neg b \in b]$ and no more RP, right?

@DanChristensen I don't have any idea, because in fact such expressions are not disallowed! $x\in x$ is a wff in the relevant language

@DanChristensen Just to perhaps clarify: none of the standard axioms of set theory say anything about what expressions are allowed. That's settled before we start talking about the axioms, when we say the language is the language of FOL with equality and one infix binary predicate $\in$. You seem to be considering a context where $\in$ is an infix binary predicate but the expression $x\in x$ is not part of the language - hence I honestly have no idea what you're doing, because whatever it is it's not FOL.

@DavidC.Ullrich What I am using is not standard FOL or ZFC, but AFAICT it is the logic and set theory implicitly used in most math textbooks. It has no restrictions on set-self-membership and yet it avoids RP, i.e. it would seem to be impossible to prove $\exists a: \forall b: [b\in a \iff b \notin b]$.

It is possible, however, to prove $\neg\exists a: \forall b: [b\in a \iff b \notin b]$.

@DanChristensen "it would seem to be impossible to prove..." what seems possible to prove or not has no bearing on anything. In any case, what can or cannot be proved is not at all the same as what exprressions are disallowed! My comment was aboutyour commentabout expressions, not about proofs. (Assuming ZFC is consistent) it's impossible to prove that formula in ZFC, although $x\in x$ iscertainly a legal expression. (The whole point to RP is that in set theory with unrestricted comprehension it is possible to prove that formula, hence set theory with uc is inconsistent.)

@DavidC.Ullrich What then is meant by "we cannot have $A\in A$" at en.wikipedia.org/wiki/…

What is meant by that is that there is no set $A$ such that $A\in A$. What is not meant is anything about the expression "$A\in A$". For example, $\lnot\exists A(A\in A)$ is a theorem in the system in question, that contains the expression $A\in A$.

@DavidC.Ullrich So, we can obtain $\neg \exists a: \forall b: [b\in a \iff b\notin b]$ whether we have $\exists A: A\in A$ or its negation or neither. Set-self-membership is not the issue here.

@DavidC.Ullrich It also works for every other binary relation. The notion that various paradoxes arise due to "self-reference" is widespread.