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In posting about a reflection principle coupled with a limitation of size axiom over Ackermann set theory, the answer is that the theory is blown up to a Mahlo cardinal.

I'm here just wondering if this method can be iterated, and what is the maximal that it can reach to via this iteration process.

For example lets define a theory $\mathsf{K}^{+}(V_{\lambda})$ in the language of $FOL(=,\in, V_1, V_2,..,V_{\lambda})$ as long as $\lambda$ is some specific recursive ordinal having some specific ordinal notation, i.e. as long as $\lambda < \omega_1^{CK}$

Now the idea is that each theory $\mathsf{K}^{+}(V_{\lambda})$ has axioms of Extensionality, Class comprehension axiom schema for $V_{\alpha}$, a reflection axiom scheme for $V_{\alpha}$, and limitation of size axiom for $V_{\alpha}$, for each $\alpha < \lambda$, also we have the axiom schema:

if $\alpha < \beta$, then: $``\forall x (x \subset V_{\alpha} \to x \in V_{\beta})"$ is an axiom.

More specifically the formula of class comprehension for $V_{\alpha}$ is:

$$\forall x_1,..,x_n \subseteq V_{\alpha} \exists x \forall y (y \in x \leftrightarrow y \in V_{\alpha} \wedge \varphi(y,x_1,..,x_n))$$, where $\varphi(y,x_1,..,x_n)$ is a formula that do not use primitives $V_{\beta}$ when $\beta>\alpha$.

While the formula of reflection schema for $V_{\alpha}$ would be written as:

$$\forall x_1,..,x_n \in V_{\alpha} \\ [\exists y (\varphi(y,x_1,..,x_n)) \to \exists y \in V_{\alpha}(\varphi(y,x_1,..,x_n)) ]$$ where $\varphi(y,x_1,..,x_n)$ doesn't use any primitive symbol $V_{\beta}$ as long as $\beta \geq \alpha$.

Now what is the limit to the consistency strength of the $\mathsf{K}^{+}(V_{\lambda})$ theories?

Zuhair Al-Johar
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1 Answers1

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I claim $K^+(V^\lambda)$, for any $\lambda$ (I switched the notation so not to be confused with the von Neumann universe) is equiconsistent with the schema "$ORD$ is Mahlo" (Not to be confused with full stationarity, which could be called "$Ord$ is Mahlo" if you really wanted to distinguish them). First off each $V^\lambda$ is a Grothendieck universe, and so of the form $V_\kappa$ for inaccessible $\kappa$. Second of all, $V_\kappa\prec W$, where $W=\{x|x=x\}$. To see this, suppose $\exists x(\phi(x,x_0...x_n))$ where $\phi(x,x_0...x_n)$ is absolute.

Then $\exists x(\phi(x,x_0...x_n))$ if and only if $\exists x\in V^\lambda(\phi(x,x_0...x_n))$ if and only if $\exists x\in V^\lambda(\phi^{V^\lambda}(x,x_0...x_n))$ if and only if $V^\lambda\vDash\exists x(\phi(x,x_0...x_n))$. Note that at no point do we use reflection for a formula that uses $V^\lambda$. Therefore $K^+(V^\lambda)$ proves that there exists a reflection cardinal, and so the consistency strength of $K^+(V^\lambda)\ge$ the consistency strength of "$ORD$ is Mahlo."

Then, suppose $ORD$ is Mahlo. Then there exists a proper class of reflecting cardinals, and if we take $V^\lambda=V_\kappa$ for reflecting $\kappa$, we get $K^+(V^\lambda)$. Therefore the consistency strength of "$ORD$ is Mahlo"$\ge$ the consistency strength of $K^+(V^\lambda)$, and so the consistency of "$ORD$ is Mahlo"$=$ the consistency strength of $K^+(V^\lambda)$.

Master
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  • I just was under the impression that $K^+(V^2)$ already interprets "$Ord $ is Mahlo", I mean the second tier of this theory already proves the consistency of $Ord$ is Mahlo. The reason is because the proof in the linked answer already establishes $V^1$ as a a model of $\sf ZF + M$. Where $\sf M$ is the schema presented in the linked answer. So there exists $\kappa$ where $V^1=V_\kappa$ where $\kappa$ is a Mahlo. So I thought that we are already beyond Mahlo cardinals. – Zuhair Al-Johar Jul 05 '19 at 09:30
  • I believe the reason is that, while $V_\kappa$ might satisfy any individual axiom of "$ORD$ is Mahlo," you can't prove it satisfies all of them. – Master Jul 05 '19 at 15:49
  • Can you explicitly write $ORD$ is mahlo formally – Zuhair Al-Johar Jul 05 '19 at 16:18
  • It is a schema. If $C={\alpha|\phi(\alpha,p)}$ is club, there there is some regular $\kappa\in C$. It is not a single assertion. Here is a link to Cantors attic: http://cantorsattic.info/ORD_is_Mahlo – Master Jul 05 '19 at 16:20
  • I see what you mean. I myself made a mistake in my comment, I meant $V^2$ when I said $V^1$. So again I thought that Packomov's answer established that $V^2$ is a model of $ZFC+M$, this means that $V^2=V_\kappa$ where $\kappa$ is a Mahlo, so the set of all ordinals in $V^2$ is a Mahlo cardinal, and so it proves the consistency of $ORD$ is a Mahlo, so it is already stronger than $ORD$ is a Mahlo, so as I said we are already way beyond that. So there must be something wrong with your argument? – Zuhair Al-Johar Jul 05 '19 at 18:15
  • "The intended models are $V_\kappa$, where $\kappa$ is Mahlo." There a plenty of models of "$ORD$ is Mahlo" which are not of the form $V_\kappa$, where $\kappa$ is Mahlo be elementaryity (Take $V_\lambda\prec V_\kappa$, where $\kappa$ is the least Mahlo). But that is besides the point. Sure $V^2$ is a model of "$ORD$ is Mahlo," but $T\nvdash (V^2\vDash ORD,is,Mahlo)$. It does prove every individual instance of that schema. On the other hand, "$ORD$ is Mahlo" proves a proper class of $V_\kappa$, where $\kappa$ is reflecting, and each $V_\kappa$ satisfies the condition for $V^\lambda$. – Master Jul 05 '19 at 18:30
  • ($T$ is your theory) – Master Jul 05 '19 at 18:35
  • so according to that $Ord^2$ which is the class of all ordinals in $V^2$ is not provably a Mahlo in $T$, I just thought that the size limitation axioms coupled with class comprehension axioms would ensure that this must be the case, as they did it with the prior universe $V^1$ which has a similar situation with replacement, there they managed to prove that $V^1$ is inaccessible, and thus $T \vdash ZFC$, so I thought a similar thing would also apply for the case of $V^2$ such that $T \vdash Ord^2 is Mahlo$. Why that failed at this level? – Zuhair Al-Johar Jul 06 '19 at 06:18
  • They didn't prove $V^1$ is Mahlo, only that it is definably Mahlo. You can prove $V^1$ is inaccessible only be assuming limitation of size, which is much stronger then Replacement. – Master Jul 06 '19 at 17:18
  • but limitation of size is already an axiom of this theory. – Zuhair Al-Johar Jul 06 '19 at 19:32
  • I know. But the point is that there is a distinction between definable and true variants of properties. If you want $Ord$ to be actually Mahlo, you'll have to add it as a new axiom. – Master Jul 06 '19 at 19:41
  • I see your point. Thanks! – Zuhair Al-Johar Jul 06 '19 at 19:43