Is {Ø,{Ø},{Ø,{Ø}}, ... } the only known universe?

Question

In the first pages of SGA4 I read

[...] Cependant le seul univers connu est l'ensemble des symboles du type {Ø,{Ø},{Ø,{Ø}}, ... } etc. (tous les éléments de cet univers sont des ensembles finis et cet univers est dénombrable). En particulier, on ne connaît pas d'univers qui contienne un élément de cardinal infini. [...]

(the sole known universe is like {Ø,{Ø},{Ø,{Ø}}, ... }, and we don't know any universe with a infinite cardinal).

Mais, c'est vrai? I wonder if during all these years somebody discovered a universe "bigger" than that exhibited by Grothendieck.

Answer 1

The universe that Grothendieck intends to suggest by his notation is known in set theory as HF, the class of hereditarily finite sets, the sets that are finite and have all elements finite and elements-of-elements, and so on (the transitive closure should be finite). The set HF is the same as $V_\omega$ in the Levy hiearchy, and can be built by starting with the emptyset and iteratively computing the power set, collecting everything together that is produced at any finite stage. This is the smallest nonempty transitive set that is closed under power set. It satisfies all the Grothendieck universe axioms, except that it doesn't have any infinite elements, since none appear at any finite stage of this consrtruction.

There is an interesting presentation of this universe by a simple relation on the natural numbers. Namely, define $n\ E\ m$ if the $n^{\rm th}$ bit in the binary expansion of $m$ is $1$. The structure $\langle\mathbb{N},E\rangle$ is isomorphic to $\langle HF,{\in}\rangle$ by the map $\pi(n)=\{\pi(m)\,|\,m\,E\,n\}$, which set-theorists will recognize as the Mostowski collapse of $E$.

Since HF doesn't have any infinite elements, it is a rather impoverished universe for many applications of that concept. And so we naturally seek larger universes. But the difficulty is that we cannot prove they exist. The difficulty is not one of "discovery," but rather just that we can prove that the hypothesis of the existence of a univese containing infinite sets is too strong for us to prove from our usual axioms. The reason is, as has been remarked in some of the other answers and comments, all other Grothendieck universes have the form $H_\kappa$, the hereditarily size less than $\kappa$ sets, for an inaccessible cardinal $\kappa$. So this is just like HF, which is $H_\omega$, but on a higher level, and in this sense, these higher universes are not so mysterious. They are intensely studied in set theory, a part of the research effort in large cardinals.

In this MO answer, I mention a number of weaker universe concepts that we can prove exist, and which I believe serve most of the uses of the universe concept in category theory, if one wanted to care more about such set theoretic issues.

Answer 2

Let me rephrase part of what Joel David Hamkins and Anon already said, but without mentioning inaccessible cardinals:

A Grothendieck universe strictly bigger than the one in the question would be a model of ZFC. (More precisely, it would become a model once we interpret the membership symbol of ZFC as actual membership.) So the existence of such a Grothendieck universe would imply the consistency of ZFC. G"oedel's second incompleteness theorem implies that ZFC cannot prove the consistency of ZFC. Therefore ZFC cannot prove the existence of a Grothendieck universe.

Answer 3

I assume you are referring to Grothendieck universes.

The existence of a bigger Grothendieck universe is equivalent to the existence of an inaccessible cardinal, which cannot be proved from ZFC, because it implies the consistency of ZFC.

There is a smaller example of a Grothendieck universe: the empty set. This is the only other Grothendieck universe that can be proven to exist in ZFC.

Answer 4

Just to make perfectly sure: Grothendieck is absolutely not questioning the existence of infinite sets in this quotation. (He had, and has, some eccentricities, but not in this direction!)

Remember that "universe" is a technical term for a certain type of set, essentially one which has maximally nice closure properties. The universe he is talking about corresponds to the cardinal $\aleph_0$, a countably infinite set whose elements may themselves be identified with the finite cardinals (as is a standard operating procedure since von Neumann, although those who don't think that much about infinite sets can and often do safely forget this point). He is not discussing universes as models of formal set theory or anything like that, so the idea that "internally" in this countably infinite universe, infinite sets do not exist, is not at all what he is getting at. Rather, since he has written down an example of an infinite set, we can conclude (from this passage alone, notwithstanding the rest of his work) that he believes in and is comfortable working with infinite sets.

Answer 5

There exist plenty of other universe. Recall that most of the times one proves the (relative) consistency of some axiom independent from ZF, one actually builds a model that satisfy that axiom. So for instance, the method of forcing invented by Cohen enables to list a infinite number of (elementary) different universes. Of course the universe you mentioned is much more tangible and intuitive then the one built with forcing, but if you accept AC, then they have the same dignity.