# Anti-foundational set theory with a universal set

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?

• 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? – 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.

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 $$S=|V|, B=|V|$$.

• Is $|N|$ countable? I’d assume no, which would mean that this is an interesting almost-solution to the equation. – Matt F. Oct 17 '19 at 0:36
• @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. – Zuhair Al-Johar Oct 17 '19 at 5:13
• Ok. What mapping establishes the equality? (It’s not obvious to me yet.) – Matt F. Oct 17 '19 at 8:13
• 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. – Zuhair Al-Johar Oct 17 '19 at 16:58