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Suppose the Set theory in question followed all axioms of ZF or ZFC, except for the axiom of regularity. Additionally the axiom schema of specification was altered from

$$\forall z\forall w_1 \forall w_2 \dots \forall w_n\exists y\forall x[x \in y \iff ((x\in z)\wedge\phi(x))]$$

to unrestricted comprehension

$$\forall w_1 \forall w_2 \dots \forall w_n\exists y\forall x[x \in y \iff ((x\neq y)\wedge\phi(x))]$$

for the purpose of constructing sets beyond just subsets of some $z$.

Without regularity and standard specification, this immediately opens the doors to a possibility of Russel's Paradox. However, a set $R$ cannot be built (this way) to have $R\in R$, as it would have to follow $R \in R \iff ((R\neq R)\wedge\phi(R))$.

Would this avoid Russel's Paradox? Even if so, would it lead to another paradox?

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    $\begingroup$ Regularity seems irrelevant here... It’s a common misconception that regularity has something to do with preventing Russel’s paradox. It doesn’t. Adding an axiom can’t make you any safer from a contradiction, only less safe. $\endgroup$ Commented Jul 13, 2020 at 23:02
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    $\begingroup$ @spaceisdarkgreen Very true, needs to be mentioned. I specified such only because I happened to be working with models without regularity. $\endgroup$
    – Graviton
    Commented Jul 14, 2020 at 6:29

2 Answers 2

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No, this form of comprehension is inconsistent with even the ability to form singletons and unions of two sets.

Suppose $A=\{x: x\not\in x\wedge x\not=A\}$. Then I claim $A\cup\{A\}$ is the usual Russell set.

  • If $x\in A\cup \{A\}$ then $x\not\in x$. This is because for such an $x$, either $x\in A$ or $x=A$, and no element of $A$ contains itself by the "$x\not\in x$"-clause of the definition of $A$ and $A\not\in A$ by the "$x\not=A$"-clause of the definition of $A$.

  • If $x\not\in x$ then $x\in A\cup\{A\}$. Again, we reason by cases. If $x\not\in x$ then either $x\not=A$ (in which case $x\in A$) or $x=A$ (in which case $x\in\{A\}$), and either way $x\in A\cup\{A\}$.

And now we ask whether $A\cup\{A\}$ is an element of itself.

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  • $\begingroup$ Wonderful thought and time out into this. I recall that the axioms in ZFC that allow the existence of $x\cup\{x\}$ (union, pairing, specification) were put in place to make up for what unrestricted comprehensuon lost when it was replaced with specification. Many naive set theories relied solely on unrestricted comprehension. Hence, with that as the only way to build sets, I wonder if $x\cup\{x\}$ is buildable. $\endgroup$
    – Graviton
    Commented Jul 13, 2020 at 22:25
  • $\begingroup$ The first statement in the answer,"No, this form of comprehension is inconsistent with even the ability to form unions of two sets." is not correct. This form of comprehension and the ability to form unions of two sets holds in the one element domain where the one element is empty. $\endgroup$ Commented Jul 18, 2020 at 19:44
  • $\begingroup$ @GregKirmayer Good point, I also need the ability to form singletons. Thanks! $\endgroup$ Commented Jul 18, 2020 at 20:08
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The above "unrestricted comprehension" schema is equivalent to the statement that there is a single element and that element is empty.

Note that the schema holds in the one element domain where the one element is empty.

There is an empty set. Proof:There is a Y such that x∈Y⟺𝑥≠Y∧x≠x. Such a Y must be empty.

Call a set b with exactly 2 elements a 2-set if one element of b is empty and b∉b.

If there are distinct elements, then there is a set which a 2-set or a set whose only element is empty. Poof: Suppose a is empty and a≠b. Then there is a Y such that x∈Y⟺x≠Y∧(x=a∨x=b). Y is not empty and Y∉Y. If b∈Y then Y is a 2-set. If b∉Y then a is the only element of Y.

Let F(x) be the formula ∀t(t∉x∨x∉t). We observe that if F(s) for all s∈x, then F(x).

Suppose that b is not empty. Then there is a set Y whose only element is b.
Proof: There is a Y such that x∈Y⟺x≠Y∧x=b. Y is not empty because if it were then b∈Y. Therefore Y is a set whose only element is b.

Suppose there are distinct elements. Then there is a non-empty set Y such that F(Y).

Proof: If there is a set b whose only element is empty, then F(b). Suppose there is no such set. There is a W such that "x∈W⟺x≠W∧(x is not a 2-set)". W is not empty because there is a 2-set and a set whose only element is a 2-set. Suppose c is empty. There is a set Y such that x∈Y⟺x≠Y∧(x=c∨x=W). Y is a 2-set and F(Y).

There is an A such that x∈A⟺x≠A∧F(x). Then F(A).

There is only one element.
Proof: Suppose there are distinct elements. Then there is a non-empty element b such that F(b). But then there is a Y whose only element is b. One of b and Y must be in A. There is a B such that x∈B⟺x≠B∧(∃t(t∈x)∧((x∈A∨x=A))). A≠B since B only has non-empty elements and so A∈B. But F(B) and therefore B∈A. But this contradicts F(A).

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