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Briefly, my question is about whether there exists conservative extensions of $T_0\colon=\mathbf{ZFC}$ other than $T_1\colon=\mathbf{NBG}$, specifically extensions of $\mathbf{NBG}$ obtained in a way similar to the extension of $\mathbf{ZFC}$ to $\mathbf{NBG}$, which will be clarified shortly. I believe it best to leave our variables unsorted, for clarity and to avoid cumbersome notation.

To clarify my remark regarding the means of extension, identifying languages with their set of formulas, $\mathbf{NBG}$ is an extension in the expansion of the usual language $\mathcal{L}_0=\mathrm{L}(\epsilon)$ of set theory (identifying languages with their set of formulas) to the language $\mathcal{L}_1=\mathrm{L}(\epsilon,\mu_0)$ and among the axioms of $\mathbf{NBG}$ is the sentence $\forall x(\varphi_0(x)\leftrightarrow\exists y(x\in y))$. Hence, I am referring to a similar approach by adding to $\mathcal{L}_0$ unary relation symbols $\mu_\alpha$ for each $\alpha<\beta$ and some ordinal $\beta$ while adding axioms similar to the previous one.

It goes without saying in moving from $T_0$ to $T_1$, for every $\mathcal{L}_0$-sentence $\sigma$ if its relativisation to $\mu_0(v)$ is $\sigma^{\mu_0(v)}$ ($v$ not occurring in $\sigma$ of course) then we want (and do have) \begin{equation} T_0\models\sigma \quad\iff\quad T_1\models\sigma^{\mu_0(v)} \end{equation}

To add some terminology for convenience, (also) refer to sets as $0$-classes, classes as $1$-classes and $\alpha$-classes for objects belonging to the domain of discourse of the theory $T_\alpha$, up to the ordinal $\alpha$ fo which this is possible or otherwise for every ordinal $\alpha$. Is this at least possible for all $\alpha<\omega$ or even $\alpha=\omega$? In case it is not clear, I would like that \begin{equation} T_\alpha\models\forall x(\mu_{\gamma}(x)\rightarrow \mu_{\delta}(x)),\quad\text{whenever }\gamma<\delta<\alpha. \end{equation}

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  • $\begingroup$ Both "$L_i$" and "$\Sigma_i$" notation here clashes with existing relevant notation, namely the constructible hierarchy and the Levy hierarchy respectively. $\endgroup$ – Noah Schweber Mar 20 '20 at 20:44
  • $\begingroup$ Note that you've written "consertative" in stead of "conservative" in the title. $\endgroup$ – Thibaut Demaerel Mar 20 '20 at 20:45
  • $\begingroup$ @NoahSchweber I shall edit it then $\endgroup$ – Jean-Pierre de Villiers Mar 20 '20 at 21:50
  • $\begingroup$ @NoahSchweber Do the new changes suffice? $\endgroup$ – Jean-Pierre de Villiers Mar 20 '20 at 21:58
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Yes, we can do this at least through the computable ordinals (so well beyond $\omega$). One way to do this is described here; basically, the idea is to have level $\alpha$ look like $L_\alpha(M)$ for some $M\models ZFC$ (analogously to how the conservativity of $NBG$ over $ZFC$ is to show that $L_1(M)$ is a model of $NBG$ whenever $M\models ZFC$.

(Note that here I'm using $L_\eta$ to denote the $\eta$th level of the constructible hierarchy, not as you've used it.)

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  • $\begingroup$ Somewhat related to this question. Why do category theorists want (uncountable) Grothendieck universes to exist if they could in stead work in a suitable conservative extension of ZFC? $\endgroup$ – Jean-Pierre de Villiers Mar 20 '20 at 22:03
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    $\begingroup$ @Jean-PierredeVilliers These conservative extensions don't buy you everything you'd want: iterating the definable powerset operator doesn't behave like iterating the powerset operator. E.g. $V_{\omega+1}$ has every real number in $V$ but we don't get every real number in $L$ until $L_{\omega_1}$. Grothendieck universes provide much stronger closure properties and if we try to add those closure properties to this construction we lose conservativity. Now of course in many (most?) situations universes are unnecessary, but - perhaps surprisingly - sometimes they are genuinely needed: (continued) $\endgroup$ – Noah Schweber Mar 20 '20 at 22:15
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    $\begingroup$ e.g. here we have a consistency result relative to a large cardinal assumption (two inaccessible cardinals) but don't currently know how to get rid of that consistency strength. $\endgroup$ – Noah Schweber Mar 20 '20 at 22:17
  • $\begingroup$ Does this mean that a serious category theorist should also be well-versed in some serious set theory as well? I'm only doing a course in basic category theory now so naturally these questions don't come up. I'm asking because I am very interested in specialising in set theory (which flavour I'm not yet sure, though I guess one doesn't have to completely box yourself in). $\endgroup$ – Jean-Pierre de Villiers Mar 20 '20 at 22:52
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    $\begingroup$ @Jean-PierredeVilliers I mean, everyone should be well-versed in some serious set theory :) More seriously, the relevant set theory can almost always be treated as a black box - the exceptions are if $(1)$ you want to optimize the "axiomatic overhead" (e.g. you actively prefer ZFC to ZFC + Universes) or $(2)$ you're working in some of the more recondite areas (e.g. my impression is that accessible categories are such a topic). My suggestion would be to first focus on the category theory, and then - based on what you find interesting - you can then decide how much you care about set theory. $\endgroup$ – Noah Schweber Mar 20 '20 at 22:53

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