There are alternative set theories that allow for a universal set, e.g. NF(U), positive set theory and and topological set theory. There are also alternative set theories like ZFA that allow for the Axiom of Foundation to fail in a strong way, e.g. by satisfying Aczel's anti-foundation axiom. I am interested in alternative set theories that combine both of these properties.

In particular, I am looking for a set theory in which the following equation has a solution:

$S = B \times \mathcal{P}(S)^A$

Here, $A$ is a countable (possibly even finite) set, B is an infinite set, and $\mathcal{P}(S)$ denotes the power set of $S$. I believe that the simplified formula $S = B \times S^A$ can be solved in ZFA, but the addition of the $\mathcal{P}$ to the equation makes it unsolvable in ZFA due to size considerations. However, I suspect that an alternative set theory with a universal set might overcome this problem.

Note that the existence of a universal set is not a strict requirement for me. I just suspect that set theories that have it are more likely to be able to solve the above equation.

Can someone point to an alternative set theory that allows for a solution to the above equation?

  • $\begingroup$ I believe you can produce permutation models of NF to show that the existence of such an $S$ is consistent, but I doubt the existence of something like this would be a theorem of NF. Presumably you want it to be a theorem of the desired set theory? $\endgroup$ – Malice Vidrine Apr 24 '17 at 18:41

Check out the theory at the end of Vicious Circles by Barwise & Moss, which has the Universal Set U, though the “collection of all sets distinct from U will not be a class.” (p. 308). I crafted some of the axioms in “A Variant of Church’s Set Theory with a Universal Set in which the Singleton Function is a Set”† to avoid unnecessarily precluding Aczelian sets, but didn’t explicitly include them.

† Abridged in Logique et Analyse, Vol 59, No 233 (2016) pp. 81–131, doi:10.2143/LEA.233.0.3149532. The full version is available at the Centre National de Recherches de Logique.

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Define: $ |X|=\{Y| \exists f (f:Y \to X, f \text{ is injective})\}$

Then $\sf NFU$ proves:

$|V|=|V| \times \mathcal P(|V|)^{N}$

Given that $\times$ is understood as Cardinal mutlitplication, i.e. $S= B \times \mathcal P(S)^A$ is understood as $S, B, \mathcal P(S)^A$ are cardinals and $B \times P(S)^A$ is understood as the cardinality of the Cartesian product of a set whose cardinality is $B$ and a set whose cardinality is $P(S)^A$.

By the way $|V|=B \times \mathcal P(|V|)^A$ where $B$ is the cardinality of any nonempty set, and $A$ is nonempty. Is provable in $\sf NF$.

There is some ambiguity with $P(S)^A$ does it mean $P(S^A)$ or does it mean $(P(S))^A$? If the later then take $A=\emptyset$ and we have a simple solution that is $S=B \times \{\emptyset\}$; if the former then the solution is $S= B \times \{ \emptyset, \{ \emptyset \}\}$ both have a solution in $\sf ZFC$.

Appearantly you want $A$ to be nonempty countable set, and this would be easily solved in $\sf NF,NFU$, just let $S=|V|, B=|V|$.

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  • $\begingroup$ Is $|N|$ countable? I’d assume no, which would mean that this is an interesting almost-solution to the equation. $\endgroup$ – Matt F. Oct 17 '19 at 0:36
  • $\begingroup$ @MattF. in reality |N| can be countable if we assume NFU + negation of infinity, or it can be uncountable if we assume NFU+infinity. In reality it doesn't matter. I've changed |N| to N to enforce the countable condition in both cases. $\endgroup$ – Zuhair Al-Johar Oct 17 '19 at 5:13
  • $\begingroup$ Ok. What mapping establishes the equality? (It’s not obvious to me yet.) $\endgroup$ – Matt F. Oct 17 '19 at 8:13
  • $\begingroup$ For the case of NFU, its simply identity map on cardinality of $V$, i.e. on the equivalence class of all sets of the same size as $V$. For the rest, they are obvious. $\endgroup$ – Zuhair Al-Johar Oct 17 '19 at 16:58

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