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Let take the first order set theory whose axioms are Extensionality, Separation and Universal reflection.

By $\operatorname {unv}(x)$, denoting "$x$ is a universe", we'll take it to mean that $x$ is a stage $V_\alpha$ of the cumulative hierarchy and such that $V_\alpha= \mathcal H_{V_\alpha}$, where $\mathcal H_x$ is the set of all sets hereditarily strictly subnumerous to $x$.

Universal reflection: $(\phi \to \exists \alpha: \operatorname {unv}(V_\alpha) \land \phi^{V_\alpha})$; if $\alpha$ not free in $\phi$.

Where $\phi^A$ is the result of bounding all quantifiers of $\phi$ by "$A$".

That is, every formula true of the whole set world is captured by some set that is a Grothendieck universe.

Is this set theory equivalent to $\sf TG$ set theory?

Clarification: by $y$ being hereditarily strictly subnumreous to $x$ it is meant that: $y$ is strictly subnumerous to $x$ and every element of the transitive closure of $y$ is also strictly subnumerous to $x$.

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  • $\begingroup$ What? Why? Of course not! TG implies there is the same as ZFC+"Proper class of inaccessible cardinals". Reflecting "There are unboundedly many inaccessible cardinals" will give you an inaccessible limit of inaccessible. This is equivalent to ZFC+"Ord is Mahlo". $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 22 at 15:48
  • $\begingroup$ @AsafKaragila, hmmm.., then this ought to be an answer? With some explication. $\endgroup$
    – Zuhair Al-Johar
    Commented Aug 22 at 16:18
  • $\begingroup$ @JoelDavidHamkins, $V_\alpha$ is defined in the language of this theory, then it is put in the axiom, so it is forced to be their by axiomatization. All of those axioms you mentioned would follow in this theory. $\endgroup$
    – Zuhair Al-Johar
    Commented Aug 22 at 20:51
  • $\begingroup$ @ZuhairAl-Johar You have an idiosyncratic meaning for $H_\kappa$, since when $\kappa$ is singular, this usually means the collection of sets whose transitive closure is less than $\kappa$, but this is evidently not what you are doing. Why not use the usual terminology and notation? Also, with just extensionality and separation, but not foundation, etc., I'm not entirely sure what it means to be "a stage of the cumulative hierarchy", since that theory simply does not prove that there is a robust theory of such stages, and so I don't know exactly how you intend to express the reflection axiom. $\endgroup$
    – Joel David Hamkins
    Commented Aug 23 at 5:56
  • $\begingroup$ @JoelDavidHamkins, by $x$ being hereditarily strictly subnumreous to $y$ I mean $x$ is strictly subnumerous to $y$ and every element of the transitive closure of $x$ is also strictly subnumerous to $y$. I'll add this for clarification. But for the $V_\alpha$ issue, this can be easily defined in the language of this theory, you can use for example level theory definitions, so $x = V_\alpha$ means $x$ is a level (see Button, Scott) and $\alpha$ is the set of all ordinals in $x$. Or you can do it using the usual definitions. to be continued... $\endgroup$
    – Zuhair Al-Johar
    Commented Aug 23 at 8:37

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The theory you suggest is significantly stronger than Tarski–Grothendieck. The latter theory is, essentially, $\sf ZFC$ augmented by "there is a proper class of inaccessible cardinals".

The theory you suggest implies that if $C$ is a closed and unbounded class of ordinals, then there is an inaccessible cardinal in $C$. In other words, "$\rm Ord$ is Mahlo" holds.

To see that, first note that you're going to reflect the fact that there is a proper class of inaccessible cardinals, and therefore there will be a $2$-inaccessible cardinal. So, we're already above $\sf TG$ as far as consistency strength.

Now, if $\varphi(x)$ defines a closed and unbounded class, $C$, there is some inaccessible $\kappa$ such that $V_\kappa$ thinks that $C\cap\kappa$ is a proper class. Since $C$ is closed, $\kappa\in C$. Therefore we proved that $\rm Ord$ is Mahlo holds.

In the other direction, if $\varphi(x)$ is a formula, then $\{\alpha\mid V_\alpha\models\varphi(x)\}$ is a closed and unbounded class. Since $\rm Ord$ is Mahlo, this class contains an inaccessible cardinal and therefore $\varphi$ reflects to a universe.

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    $\begingroup$ Note that the definition of unv in the question doesn't require $\alpha$ to be regular. In view of the question about TG, it's reasonable to suppose that regularity was intended, but as currently defined, unv is quite weak, and the proposed reflection schema seems to be provable in ZFC. $\endgroup$
    – Andreas Blass
    Commented Aug 22 at 19:16
  • $\begingroup$ @AndreasBlass, if $\operatorname {unv}(V_\alpha)$, then $\alpha$ must be regular, otherwise there would be a cofinal subset of $\alpha$ that is hereditarily strictly subnumerous to $V_\alpha$ and thus must be an element of $V_\alpha$ which cannot be. I think the point lies in the definition of "hereditarily". I've added a clarification about what I meant by it. $\endgroup$
    – Zuhair Al-Johar
    Commented Aug 23 at 9:06
  • $\begingroup$ @ZuhairAl-Johar: What exactly do you mean? $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 24 at 9:08
  • $\begingroup$ @ZuhairAl-Johar: Proof from what theory? $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 24 at 21:31

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