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LSpice
LSpice
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  1. ∀a∃X((a ∉ a) → (a ∈ X))$\forall a\exists X((a \notin a) \implies (a \notin X))$
  2. ∀a∃X((a ∈ X) → (a ∉ a))$\forall a \exists X((a \in X) \implies (a \notin a))$.

The "upshot" is that you can deny (2) while retaining (1): you can say that there is some set that happens to have all noncircular elements to its name. However, if these are the only elements that set contains, then Russell's paradox arises, so we would have to say that if there is a set with all noncircular sets as elements, this set still is an element of itself on the side. This whole set does not have its elements just because they've been intensionally carved out as such, but either the whole set is "manually" self-inserted, or it satisfies a disjunctive property like, "All elements of X$X$ are either noncircular or are X$X$ itself."

Note, then, that there are set theories with universal sets, e.g. Quine's parafounded world, or Weber's paraconsistent ℵORD$\aleph_\text{ORD}$. At least in the consistency-friendly such cases, if there is a universal set, then the axiom scheme of separation has to fail for this set, the powerset axiom to boot: for a truly universal set, there is no difference between its elements and its subsets, or the concept of it having subsets is not well-thought, so there are no separable subsets or separate sets of subsets in play, here.

Yet another solution (from the separation-minus vantage, but comporting more strictly with the axiom of foundation) would be to invoke couniversal sets. The pure example would be $$\exists X \forall ab((a \neq X) \iff (a \in X))$$$$\exists X \forall ab((a \neq X) \iff (a \in X)).$$

The propriety of calling these "couniversal" is that they can be styled "sets of all other sets." In a world with a universal set simpliciter, then the deviant consequence obtains wherein the couniversal X$X$ contains its superior as an element, which also contains X$X$ in turn, so X$X$ ends up being cofounded. So to avoid this, one would have to go to a world where there's a couniversal set without a universal set above it. In this event, X$X$ is the set of all elements, not all sets (since X$X$ itself is not, in that world, an element of any set, not even itself).

  1. ∀a∃X((a ∉ a) → (a ∈ X))
  2. ∀a∃X((a ∈ X) → (a ∉ a))

The "upshot" is that you can deny (2) while retaining (1): you can say that there is some set that happens to have all noncircular elements to its name. However, if these are the only elements that set contains, then Russell's paradox arises, so we would have to say that if there is a set with all noncircular sets as elements, this set still is an element of itself on the side. This whole set does not have its elements just because they've been intensionally carved out as such, but either the whole set is "manually" self-inserted, or it satisfies a disjunctive property like, "All elements of X are either noncircular or are X itself."

Note, then, that there are set theories with universal sets, e.g. Quine's parafounded world, or Weber's paraconsistent ℵORD. At least in the consistency-friendly such cases, if there is a universal set, then the axiom scheme of separation has to fail for this set, the powerset axiom to boot: for a truly universal set, there is no difference between its elements and its subsets, or the concept of it having subsets is not well-thought, so there are no separable subsets or separate sets of subsets in play, here.

Yet another solution (from the separation-minus vantage, but comporting more strictly with the axiom of foundation) would be to invoke couniversal sets. The pure example would be $$\exists X \forall ab((a \neq X) \iff (a \in X))$$

The propriety of calling these "couniversal" is that they can be styled "sets of all other sets." In a world with a universal set simpliciter, then the deviant consequence obtains wherein the couniversal X contains its superior as an element, which also contains X in turn, so X ends up being cofounded. So to avoid this, one would have to go to a world where there's a couniversal set without a universal set above it. In this event, X is the set of all elements, not all sets (since X itself is not, in that world, an element of any set, not even itself).

  1. $\forall a\exists X((a \notin a) \implies (a \notin X))$
  2. $\forall a \exists X((a \in X) \implies (a \notin a))$.

The "upshot" is that you can deny (2) while retaining (1): you can say that there is some set that happens to have all noncircular elements to its name. However, if these are the only elements that set contains, then Russell's paradox arises, so we would have to say that if there is a set with all noncircular sets as elements, this set still is an element of itself on the side. This whole set does not have its elements just because they've been intensionally carved out as such, but either the whole set is "manually" self-inserted, or it satisfies a disjunctive property like, "All elements of $X$ are either noncircular or are $X$ itself."

Note, then, that there are set theories with universal sets, e.g. Quine's parafounded world, or Weber's paraconsistent $\aleph_\text{ORD}$. At least in the consistency-friendly such cases, if there is a universal set, then the axiom scheme of separation has to fail for this set, the powerset axiom to boot: for a truly universal set, there is no difference between its elements and its subsets, or the concept of it having subsets is not well-thought, so there are no separable subsets or separate sets of subsets in play, here.

Yet another solution (from the separation-minus vantage, but comporting more strictly with the axiom of foundation) would be to invoke couniversal sets. The pure example would be $$\exists X \forall ab((a \neq X) \iff (a \in X)).$$

The propriety of calling these "couniversal" is that they can be styled "sets of all other sets." In a world with a universal set simpliciter, then the deviant consequence obtains wherein the couniversal $X$ contains its superior as an element, which also contains $X$ in turn, so $X$ ends up being cofounded. So to avoid this, one would have to go to a world where there's a couniversal set without a universal set above it. In this event, $X$ is the set of all elements, not all sets (since $X$ itself is not, in that world, an element of any set, not even itself).

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Kristian Berry
Kristian Berry
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Russell's paradox depends on the local admissibility of universal exclusion, i.e. the "all and only" quantifier. Consider the difference between:

  1. ∀a∃X((a ∉ a) → (a ∈ X))
  2. ∀a∃X((a ∈ X) → (a ∉ a))

To get the Russell set, you have to merge (1) and (2) by composing the conditional arrow into the biconditional arrow, which is where the "and only" part of the universal exclusifier comes in.

The "upshot" is that you can deny (2) while retaining (1): you can say that there is some set that happens to have all noncircular elements to its name. However, if these are the only elements that set contains, then Russell's paradox arises, so we would have to say that if there is a set with all noncircular sets as elements, this set still is an element of itself on the side. This whole set does not have its elements just because they've been intensionally carved out as such, but either the whole set is "manually" self-inserted, or it satisfies a disjunctive property like, "All elements of X are either noncircular or are X itself."

Note, then, that there are set theories with universal sets, e.g. Quine's parafounded world, or Weber's paraconsistent ℵORD. At least in the consistency-friendly such cases, if there is a universal set, then the axiom scheme of separation has to fail for this set, the powerset axiom to boot: for a truly universal set, there is no difference between its elements and its subsets, or the concept of it having subsets is not well-thought, so there are no separable subsets or separate sets of subsets in play, here.

Given phenomena like the conflict between zero sharp and the constructible universe, or the Kunen inconsistency, it is possible to reformulate most or perhaps all axioms as dependent in the sense that they "run out" at some point in the cumulative hierarchy. For example, you can compromise on replacement, foundation, or usually general choice, to work around the Kunen inconsistency (Corazza's approach gives us Reinhardt cardinals below a certain iterate of the embedding function; antifoundation worlds have n.e.e.'s involving their parafounded fragments but can retain the Kunen wall for their well-founded fragments). These phenomena then support what we might call "deflection arguments," which contrast a universal set with relatively infinite sets and lend themselves to an anticlass principle: "If a set contains all of a certain kind of element, then it contains all elements of all kinds whatsoever." E.g., if there is a set of all ordinals, there is yet no set of all and only ordinals, but this set has to have all the cardinals in it, too, etc.

Yet another solution (from the separation-minus vantage, but comporting more strictly with the axiom of foundation) would be to invoke couniversal sets. The pure example would be $$\exists X \forall ab((a \neq X) \iff (a \in X))$$

The propriety of calling these "couniversal" is that they can be styled "sets of all other sets." In a world with a universal set simpliciter, then the deviant consequence obtains wherein the couniversal X contains its superior as an element, which also contains X in turn, so X ends up being cofounded. So to avoid this, one would have to go to a world where there's a couniversal set without a universal set above it. In this event, X is the set of all elements, not all sets (since X itself is not, in that world, an element of any set, not even itself).

So then a couniversally noncircular set would be "the set of all other noncircular sets," which indicates its own noncircularity.

All these solutions end up requiring a stronger distinction between extensional and intensional elementhood parameters. Cantor infamously mistook definability for orderability, and in fact per the natural language in use at the time, Cantor's system involved a partial collapse of the extension-intension distinction into the cardinal-ordinal one. We "know better," though, so we can talk about an elementhood parameter that is "manually extensional" rather than a matter of elements satisfying intensions. (I wish I could find it again, but I read an essay in which they explicitly brought up a difference between "is an extensional element of" and "is an intensional element of," which I saw as a way to expand on double-extension set theory, which itself involves a relatively novel attempt to deal with Russell's paradox.)