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Lets add a constant symbol $V$ to the signature of the language of set theory. So working in first order logic with equality, add the following axioms about $\in $ and $V$.

Extensionality: $\forall x \forall y (\forall z (z \in x \leftrightarrow z \in y) \to x=y)$

Set construction (reflection): if $\phi$ is a formula in which all and only symbols $y,x_1,..,x_n$ occur free, and non of them occur bound, in which the symbol $V$ doesn't occur, then: $$\forall x_1 \in V,...,\forall x_n \in V \\ \forall y (\phi \to y \in V) \to \exists x \in V \forall y (y \in x \leftrightarrow \phi)$$; is an axiom.

Set-hood: $\forall x \ (x \subsetneq V \leftrightarrow x \in V)$

Now this theory is formulated in the same langauge of Ackermann's, it share with it the first two axioms and the right to left implication of the third axiom. However the left to right implication of the third axiom is in some sense daunting! The idea here is that this theory doesn't have comprehension axioms about proper classes, clearly by the third axiom all classes the second axiom constructs are sets! If we just add the class comprehension schema of Ackermann, then we immediately get a contradiction, since the Russell class would be a set. Now this theory easily prove all axioms of Zermelo set theory, and I'd think that (if consistent) it might even be equi-interpretable with the full Ackermann's set theory itself. Its also to be noted that if instead of adding the symbol $V$ as a constant, we added it as a one place predicate symbol (and so every formula $x \in V$ would be turned to $V(x)$) [as it is the case in the original formulation of Ackermann's where he actually used the symbol $\mathcal M$ for that], then it appears that Infinity would not be provable, which proves that the way how $V$ is added as a primitive does matter as regards the consistency strength of extensions of fragments of Ackermann's set theory.

Question: What's the exact consistency strength of this theory?

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  • $\begingroup$ Is $x \subsetneq V$ necessary instead of the simpler $x \subseteq V$? $\endgroup$ – user76284 Mar 12 at 16:11
  • $\begingroup$ Might a parameter-free version of comprehension be sufficient, like it is for ZFC? $\endgroup$ – user76284 Mar 12 at 16:55
  • $\begingroup$ @user76284 if we allow $x \subseteq V$ then we'd have $V \in V$ therefore easily getting Russell's paradox by formula $x \in x_1 \land x \not \in x$ [just substitute$ x_1$ by $V$] and if you unleash parameters then you will encode $V$ in an indirect manner, so it would invoke Russell's paradox again. Yes parameter free would work but with the restriction of not containing $V$, of course. But I don't know its strength. $\endgroup$ – Zuhair Al-Johar Mar 12 at 17:37
  • $\begingroup$ Doesn’t that mean your formula contains $V$ somewhere, though? Do you have an example for the indirect encoding? $\endgroup$ – user76284 Mar 12 at 17:43
  • $\begingroup$ Wait, why did you edit the construction schema? $\endgroup$ – user76284 Mar 12 at 17:46
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The theory is inconsistent. Let ZG(x) be the formula ∀u∈x∀v∈u(v∈x)∧∀t(x∈t→∃s∈t(∀𝑦∈s(y∉t)))∧(∀t(∃s∈x(s∉t)→∃y∈x(y∉t∧∀u∈(x-t)(u∉y))))∧∀t∈x∃s∀v(v∈s↔(v=tνv=x))∧∀u∈x∃t∈x∀s(s∈t↔(s∈u∧∃r(r∈s)))∧∃t∀s(s∈t↔(s∈x∧∃r(r∈s))) (That is x is transitive; if x is in t, then t has an ∈-minimal element; if x is not contained in t, then there is an ∈-minimal element of x which is not in t; if t is in x then the pair {t,x} exists; if t is in x then t-{0} exists; and x-{0} exists.)

We note some simple properties of x for which ZG(x) holds: If x∈𝑉 then x⊆𝑉.(If x were not contained in V, then there is an ∈-minimal element m of x which is not in V. If m is not V, then by Set-hood m is in V. If m=V,then {V,x} exists and has no ∈-minimal element.) If x∈𝑉, then there is a W∈𝑉 such that t is in W iff t is contained in x.(By the above property x⊆𝑉, and so anything contained in x is properly contained in V since x is not in
x. Therefore by Set construction such a W must exist.) We will denote such a W by Px. If x∈𝑉 then ZG(Px).

Suppose that ZG(x) implies x∈𝑉. By Set construction, there is a z∈𝑉 which consists of all x for which ZG(x) holds. Let y be the union of z. Then y is in 𝑉 and ZG(y). Therefore ZG(Py).
Then Py is contained in y. But then y∈y which is impossible. Therefore there must be an x such that ZG(x) and x∉𝑉. If x is contained in 𝑉 then x-{0} is in 𝑉 and so x is in 𝑉. If x is not contained in 𝑉 then there is an ∈-minimal element m of x which is not in 𝑉. But then m-{0} is in 𝑉 and so m is in 𝑉.

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  • $\begingroup$ Forgive me if I'm missing something, but doesn't $\phi y = \bot$ prove the empty set $\varnothing$ exists, and then $\varnothing \in V$, and then $\phi y = \forall x (x \in y \leftrightarrow x = \varnothing)$ shows the unit set $\{\varnothing\}$ exists and is also in $V$? $\endgroup$ – user76284 Mar 12 at 15:41
  • $\begingroup$ @ZuhairAl-Johar I don't understand - user76284's comment seems correct to me. $\endgroup$ – Noah Schweber Mar 12 at 17:42
  • $\begingroup$ @NoahSchweber, you came late. Kirmayer is responding to my old presentation of the theory in which I missed stipulating the asserted class in set construction to be a set. the user 76284 comment in that old formulation fails because the empty class need not be a proper subclass of $V$. $\endgroup$ – Zuhair Al-Johar Mar 12 at 17:47
  • $\begingroup$ @ I think you are right! I think this proof can be simplified to require any object x to fulfill ZG if and only if x-{0} exists and whatever class t that x is not a subset of then x has a minimal element m that is not in t and for which m-{0} exists. This would enforce every ZG object to be a member of V. Clearly every von Neumann ordinal in V fulfills ZG, then the property "ZG(x) and x is a von Neumann ordinal" would enact Burali-Forti. Great! Thanks! $\endgroup$ – Zuhair Al-Johar Mar 13 at 13:43
  • $\begingroup$ I wonder if the theory presented at the below mentioned link would be subject to the same argument of inconsistency? math.stackexchange.com/questions/3251691/… $\endgroup$ – Zuhair Al-Johar Mar 13 at 13:49

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