In SGA 4, Grothendieck introduced a set-theoretic device called a Grothendieck universe. He and his collaborators worked in Bourbaki set theory, which is practically similar to $\mathsf{ZFC}$ but fundamentally different. To ease the work with universes, he introduced two addition axioms to the theory: they are mostly referred to as UA and UB.

While UA is widely known and doesn't specifically depend on Bourbaki set theory (it basically states that for any set $X$ there is a universe $\mathscr{U}$ containing it), UB apparently heavily relies on that version of set theory. In particular, it uses such concepts as relations and and the operator $\tau_x$, which $\mathsf{ZFC}$ lacks.

My question is the following one: is it possible to get a version of UB with respect to $\mathsf{ZFC}$ which would serve the same purposes for universes?

Here's the axiom UB in the language of Bourbaki set theory:

Let $R\{x\}$ be a relation and $\mathscr{U}$ a universe. If there is $y \in \mathscr{U}$ so that we have $R\{y\}$, then $\tau_xR\{x\} \in \mathscr{U}$.

P.S. I apologize if the question is not the best quality, I suspect it could even be a trivial one, but I personally don't understand Bourbaki set theory very well. Still, MO favors questions which a mathematical researcher potentially can ask and I can imagine a research who doesn't understand Bourbaki set theory but is interested in using universes à la Grothendieck in SGA.

  • $\begingroup$ What is $\tau_x$? It looks like a global choice function. $\endgroup$
    – Asaf Karagila
    Nov 18, 2018 at 14:25
  • $\begingroup$ Also what does it mean for $R\{x\}$ to be a relation here? $\endgroup$
    – Asaf Karagila
    Nov 18, 2018 at 14:27
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    $\begingroup$ @Asaf From what I know Bourbaki uses $\tau_x$ for the Hilbert operator, so yeah, a kind of global choice/Skolem function. UB would then seem to say that $\mathscr{U}$ is something like elementary in $V$. $\endgroup$ Nov 18, 2018 at 14:28
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    $\begingroup$ @MihaHabič UB looks considerably weaker than saying $\mathcal U$ is elementary in $V$. It just says th chosen witness $\tau_x\,R\{x\}$ is in $\mathcal U$ if some witness is in $\mathcal U$ --- not if there's just some witness in $V$. So if $\tau$ always chooses witnesses at the lowest possible rank, then UB is satisfied. $\endgroup$ Nov 18, 2018 at 16:17
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    $\begingroup$ @AsafKaragila Even if one required the universes to be elementary submodels of $V$, this would be considerably weaker than having sharps. It's more like having a Mahlo cardinal. Specifically, unless I"m overlooking something, if $\kappa$ is a (strongly) Mahlo cardinal, then there are cofinally (in fact stationarily) many inaccessible cardinals $\lambda<\kappa$ with $V_\lambda\prec V_\kappa$. $\endgroup$ Nov 18, 2018 at 16:21

1 Answer 1


If we work with some given universe $U$, we have to make sure that we do not leave it accidentally. The definition of a universe does this for most operations. But there is still a way to leave the universe, namely by the (global) axiom of choice, i.e., Hilbert's symbol $\tau$.

Consider a relation $R$ with a variable $x$. If there does not exist an object fulfilling this relation, then $\tau_x(R)$ is an arbitrary set of which we may say nothing (in particular not whether or not it is contained in $U$). If there exists an object fulfilling this relation, then $\tau_x(R)$ denotes such an object. Without any further axiom we cannot say whether or not $\tau_x(R)$ is contained in $U$. This is precisely the role played by the axiom scheme UB: It makes sure that if there exists an object fulfilling $R$ in $U$, then $\tau_x(R)$ lies in $U$. Using the fact that intersections of universes are again universes we can formulate UB as follows:

$\tau$ always chooses in the smallest possible universe.

Now, we do not have a global choice operator in ZFC, and as far as I understand it is not possible to leave a given universe with the usual axiom of choice in ZFC. Therefore, there is no need for an axiom similar to UB in ZFC.

  • $\begingroup$ Even for ZFC+GC, we can simply modify the universe axiom to say that every set is contained in a universe satisfying GC, right? $\endgroup$
    – user21820
    Nov 18, 2018 at 17:47
  • $\begingroup$ @user21820: I don't think this would yield the same, and moreover the choice operators defined like this need not coincide on intersections of universes. But in ZFC+GC we are given an operator $\tau$, and then UB makes sense in the same form a formulated by Bourbaki, and it is moreover necessary if we do not wish to leave a given universe when we use $\tau$. $\endgroup$ Nov 18, 2018 at 17:57
  • $\begingroup$ Yes my suggestion seems to be weaker than UA+UB over ZFC+GC, but why do we need the choice operator to coincide on intersection of universes? Is there a situation where it is insufficient that what we are dealing with is contained in a universe with a global choice operator? $\endgroup$
    – user21820
    Nov 18, 2018 at 18:12
  • $\begingroup$ With choice operators for each universe, change of universe may result in problems. For example, Remark 1.3.2 in SGA 4.I is no longer true. $\endgroup$ Nov 18, 2018 at 20:35
  • $\begingroup$ I see. I can't read that language so never mind I'll take your word for it. =) $\endgroup$
    – user21820
    Nov 19, 2018 at 6:12

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