# Had this attempt to salvage naïve comprehension been studied before?

Is the following a possible way to overcome inconsistency with naive comprehension:

We add an $$\in_n$$ symbol for each natural $$n$$ to the signature of this theory, which is a first order theory with equality and all its axioms.

Then axiomatize:

Comprehension: $$\exists x \forall y \, (y \in_n x \leftrightarrow \phi^n(y))$$

where $$\phi^n(y)$$ has $$\in_{n-1}$$ as the maximal appearing membership symbol.

Extensionality: $$a \in_n x \land \forall z \, (z \in_n x \leftrightarrow z \in_m y) \to x=y$$

Membership: $$a \in_m x \land b \in_n x \to b \in_m x$$

had this been done before? References?

• Do we have $m\leq n$ or vice-verse for the indices in the last two axioms, or are they just universally quantified for all $m,n$? Commented Aug 6 at 9:18
• @AlecRhea, $n;m$ are metatheoretic, so we cannot quantify over them. The last two axioms are schemata having an axiom for each $n,m$, and there is no restriction to $m \leq n$. Commented Aug 6 at 11:37
• I presume @AlecRhea meant “metatheoretically universally quantified”, which is exactly what a schema is. Commented Aug 6 at 11:40
• @AlecRhea Yes, I think so. Commented Aug 6 at 12:32
• @AlecRhea, sorry for being late. Yes, it is as such, the base theory is first order logic with equality including the axioms of equality, and then you add the $\in_n$ membership symbols, and the axioms about them. There is no general $\in$ membership symbol. Commented Aug 6 at 12:35

Here is a model of your theory. Start with a countably infinite set of objects $$X=\bigsqcup_n X_n$$, partitioned into infinitely many infinite sections. We inductively define $$\in_n$$. Consider the various formulas $$\phi^n(y,z)$$ that use only at most $$\in_m$$ for $$m, with parameters $$z$$. With the various possible parameters $$z$$, these define certain subsets of $$X$$, but only countably many. For each such definable subset, which has not already arisen earlier, assign an object $$x$$ from $$X_n$$ to represent it, and define $$y\in_n x\iff\phi^n(y,z)$$ for this instance. That is, I pick the object to represent the set for each definable set, not for each formula. If the definable set had already occurred at an earlier stage $$m, then I use the same object $$x\in X_m$$ as previously, and define $$y\in_n x$$ for the same members $$y$$. Thus, we will satisfy comprehension.

One effect of using the earlier object when the set was already defined earlier is that whenever $$y\in_m x$$ then $$y\in_n x$$ for all $$n\geq m$$, since the same formula recurs at the higher level the way I described it, or if you insist that the higher membership symbols appear in the formula we can add a vacuous use of $$\in_{n-1}$$ to any formula, such as $$(y\in_{n-1} x\vee y\notin_{n-1} x)$$, and thereby pump up the syntactic level of the formula to any desired level $$n$$. So the level of membership doesn't matter—once you are in the set, you will be in at all higher levels as well, and all members of a given set start membership at the same level. This will give us the membership axiom.

And since I was careful always to reuse the old object whenever the definable set is the same, we will get extensionality.

Note that we do indeed get new definable sets at each stage. For example, if we take $$\phi(y)$$ as $$y\notin_{n} y$$, then this cannot be already represented at level $$n$$ by any object, and so it will define a new class at level $$n+1$$. And then we will have all the finite modifications of it, definable with parameters, making infinitely many new definable sets at that next level. So every object will get used as a set at its level.

This method of argument is rather similar to some of the solutions of Frege's Basic Law V for first-order definability, as discussed in my paper

See the prior art section there, where I discuss this method, used in various work by Parsons, Bell, and Burgess. In particular:

• Terence Parsons. “On the consistency of the first-order portion of Frege’s logical system”. Notre Dame J. Formal Logic 28.1 (1987), pp. 161–168. ISSN: 0029-4527. DOI:10. 1305/ndjfl/1093636853.

One can perform the construction over any given model already. That is, we can assume $$X$$ already has some other first-order structure, but then chop it into the sections and undertake this construction on top of it, allowing formulas $$\phi^n$$ in the full language. This is essentially how the Parsons method works.

• Do you think this theory can interpret each $n$-order arithmetic? So, it might be at the strength of Mac Lane set theory? Which is the same of $\sf TST + Infinity$. Commented Aug 6 at 17:12
• No, it is not strong enough (but I don't know anything about TST). If you start with $X$ as the standard model of arithmetic and do my construction on top, then this is rather low in the hyperarithmetic hierarchy, since you are just iterating a truth predicate $\omega$ many times. There is a model computable from $0^{(\omega^2)}$. Commented Aug 6 at 17:53
• If we just want a finite fragment of the theory instead of the whole thing, then we don't even need a truth predicate, and so this argument works in PA. So the theory is at most the strength of first-order arithmetic. I'm not convinced it even interprets PA though - we can make things that look sort of like $\mathbb{N}$, but then I don't know how to make induction work. Commented Aug 6 at 21:54
• @pastebee I was imagining starting with a model of arithmetic, and then defining the $\in_n$ relations as an expansion of this. I don't quite see how to define the $\in_n$ relations in PA, since we'd need to decide which formulas are equivalent, which is why I think a truth predicate is involved. Can you explain your idea a bit more? Commented Aug 7 at 1:51
• If we just start with a naked set, however, then truth would be trivial, and so we could do the whole construction in PA I think. Perhaps that is what you meant. Commented Aug 7 at 14:52