Talk:Grothendieck universe

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It's said here that every Grothendieck universe has the cardinality of a strongly inaccessible cardinal, and that, therefore, the existence of Grothendieck universes can't be proved within ZF (if ZF is consistent). However, both the empty set and the set of all hereditarily finite sets would be examples of Grothendieck universes, according to the formulation of this article, and, of course, can both be proven to exist in ZF. Indeed, the article seems to acknowledge at one point the possibility that a universe might be empty. Is this business about Grothendieck universes having strongly inaccessible cardinalities made under some additional assumption (such as, for example, that the universe contains an infinite set)? -Chinju 19:44, 18 April 2006 (UTC)

The definition explicitly says that U is a non-empty set. So you are wrong about the empty set satisfying the definition. As the definition stands, you are correct that Vω (the set of hereditarily finite sets) would satisfy the definition. I do not have access to the original definition of a Grothendieck universe, so I do not know whether that is a mistake (leaving out the axiom of infinity) or whether the error is in the statement about strong inaccessibles. But there are no other problems asside from Vω. Any U satisfying this definition is either Vω or Vκ for a strong inaccessible κ; and any such Vκ is a Grothendieck universe. JRSpriggs 07:07, 19 April 2006 (UTC)
Ah, whoops, I just looked at the four numbered properties and missed the opening bit about U necessarily being non-empty. (I was also kinda misled by the sentence starting "In particular, it follows from the last axiom that if U is non-empty..."). Thanks for clearing things up about Vω. -Chinju 18:53, 19 April 2006 (UTC)
My guess is that Vω is excluded by hand, in the same way that \aleph_0 is excluded from being an inaccessible cardinal (it satisfies the two defining features). If anyone has access to the proper literature and can confirm this nuance of the definition, please update the article thus. --expensivehat 22:47, 26 May 2006 (UTC)

See Bourbaki's article. Firstly, "L'ensemble vide est un univers noté U0". Secondly, he defines a strongly inaccessible cardinal as a regular strong limit cardinal, with no uncountability assumption. The article has been made consistent with modern usage. 141.211.62.20 20:36, 31 May 2006 (UTC)

[edit] Omissions

  1. It is not clear from the article why the cardinality of a Grothendieck universe must be strongly inaccessible. Perhaps it is obvious but there is no explicit statement about that.
  2. You have to prove (U) entails (C) for any cardinal, taking a strongly inaccessible cardinal is not enough.

Yecril 11:06, 30 March 2007 (UTC)

The article was unclear to the point of verging on error. Strong inaccessibility is true, but not obvious. (U) => (C), fortunately, is not too hard. You should try reading Bourbaki's article. It's really not that bad. 141.211.120.175 21:42, 6 June 2007 (UTC)

[edit] The other way round

The text says:

Since the existence of strongly inaccessible cardinals cannot be proved from the axioms of Zermelo-Fraenkel set theory, the existence of universes other than the empty set and Vω cannot be proved from Zermelo-Fraenkel set theory either.

It is exactly the other way round. By Gödel's 2nd incompleteness theorem, the consistence of Zermelo-Fraenkel set theory (and therefore the existence of a model) cannot be proved from Zermelo-Fraenkel set theory. Now, every strongly inaccessible cardinal would give rise to such a model. Therefore the existence of a strongly inaccessible cardinal cannot be proved.--91.23.239.166 (talk) 11:38, 22 December 2007 (UTC)

[edit] Explanation for "expert-subject" template: Replacement axiom

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Doesn't the definition of Grothendieck universe include some formulation of the Axiom schema of replacement too? Functor salad (talk) 12:47, 13 April 2008 (UTC)

Isn't that precisely what property (4) is? Benja (talk) 20:06, 5 May 2008 (UTC)
Oops, I see how it isn't quite; I missed the fact that the union of the xα is taken. But I still believe replacement is implied: Let I be an element of U, and let F be a function from I to U; then consider the family \{\{F(\alpha)\}\}_{\alpha \in I}, i.e., the family that assigns to each \alpha \in I the singleton {F(α)}. Then the union of these singletons is the image of F under I, right? Benja (talk) 20:16, 5 May 2008 (UTC)