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I was re-thinking Muller's criteria in Sets, Classes and Categories: page 14 for a theory that founds Mathematics. To him, it should be able to be a foundation of Category Theory. He lays down six points that a founding system must posses in order to qualify as such.

I see the following system easier to ponder, since it works with the common mainstream standard set theory line.

Start with $\sf ZC$, define set as an element of a class that is a limit to being element of, this is:

$$set(X) \iff \exists I: X \in I \land \forall Y \in I \, \exists Z \in I \, (Y \in Z)$$

Now add an axiom that a universe of all $sets$ exist, and call that universe $V$. Now, all Muller's criteria would be fulfilled by theory: $$ {\sf ZC + Universe + ZFC}^V$$

Where ${\sf ZFC}^V$ is $\sf ZFC$ axioms with all quantifiers relativized to $V$.

I don't personally see completeness by subset-hood (an axiom of Muller's $\sf ARC$ (page 17)) to be necessary condition to fulfill these six criteria, it is only necessay in Ackermannian setting to enact powersets, but here there is no need for it for that sake.

This theory doesn't suffer from the main two objections raised in that article, since it doesn't prove the existence of inaccessible sets, and so no Superabundancy; and the predicate $set$ is actually definable here, so there are no objects to be called sets that do not fulfill that predicate, so there is no Insufficiency, and moreover the concept of set is sharply demarcated from non-set classes by definition actually.

What I'm still not sure of is point (D) Categoristan. Hence my question:

Would this be enough to found Category theory in the way tackled in that article?

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  • $\begingroup$ So we have all of ZFC inside of $V$, and for ‘classes’ that aren’t sets we still have all of ZC to manipulate them? What is the additional set predicate doing that just specifying $V$ as a ‘class of sets’ and then requiring that $V\models ZFC$ wouldn’t do? Is there a reason we want ‘sets’ to be members of members? (your axiom seems to be saying that sets are classes which are members of classes such that all their members are already members of other members — what does this gain for us that just modeling all of ZFC wouldn’t give us?) $\endgroup$
    – Alec Rhea
    Commented Jun 12, 2022 at 3:58
  • $\begingroup$ @AlecRhea, No sets are something like deep classes, for a class X to be a set there must be a class $I$ such that there is an infinite membership chain raised above X such that $I$ has all members of that chain among its elements, that is $I$ is at infinite rank difference above $X$. While non-set classes cannot have that so there is not limit class over membership chains raised above them, we can say that the rank difference between any two non-set classes is finite. And this is what Muller wants. $\endgroup$
    – Zuhair Al-Johar
    Commented Jun 12, 2022 at 4:25
  • $\begingroup$ @AlecRhea, How we model ZFC? If you suggest to add to ZC the axiom that there exists a model of ZFC, then you'll increase the consistency strength of this theory, that is besides you'll be subject to the objection of insufficiency of Muller's and also the non-sharp discimination between sets and classes which he seems to demand. I think the system I suggested is equi-consistent with ZFC, I suppose, and that's an advantage. $\endgroup$
    – Zuhair Al-Johar
    Commented Jun 12, 2022 at 4:39
  • $\begingroup$ What does $ZFC^V$ mean? If it means that all axioms of ZFC relativized to V hold then V is a model of ZFC. (note that we basically said the same thing btw; if all members are already members of other members, then those members are also already members of other members, so on and so forth, so our descriptions of the axiom amount to the same thing) $\endgroup$
    – Alec Rhea
    Commented Jun 12, 2022 at 5:14
  • $\begingroup$ Take the axiom of pairing $$\phi=\forall x\forall y\exists z\forall a(a\in z\iff a=x\vee a=y).$$ I usually define its relativization to $V$ as $$\phi^V=\forall x\in V\forall y\in V\exists z\in V\forall a\in V(a\in z\iff a=x\vee a=y).$$ What does it mean in your notation? $\endgroup$
    – Alec Rhea
    Commented Jun 12, 2022 at 5:43

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