Grothendieck universe

In mathematics, a Grothendieck universe is a set "U" with the following properties:

# If "x" is an element of "U" and if "y" is an element of "x", then "y" is also an element of "U". ("U" is a transitive set.)
# If "x" and "y" are both elements of "U", then {"x","y"} is an element of "U".
# If "x" is an element of "U", then "P(x)" is also an element of "U". ("P(x)" is the power set of "x".)
# If $\{x\_alpha\}\_\{alphain\; I\}$ is a family of elements of "U", and if "I" is an element of "U", then the union is an element of "U".

A Grothendieck universe is meant to provide a set in which all of mathematics can be performed. (In fact, it provides a model for set theory.) As an example, we will prove an easy proposition.

;Proposition 1.:If $x\; in\; U$ and $y\; subseteq\; x$, then $y\; in\; U$.;Proof.:$y\; in\; P(x)$ because $y\; subseteq\; x$. $P(x)\; in\; U$ because $x\; in\; U$, so $y\; in\; U$.

It is similarly easy to prove that any Grothendieck universe "U" contains:

* All singletons of each of its elements,
* All products of all families of elements of "U" indexed by an element of "U",
* All disjoint unions of all families of elements of "U" indexed by an element of "U",
* All intersections of all families of elements of "U" indexed by an element of "U",
* All functions between any two elements of "U", and
* All subsets of "U" whose cardinal is an element of "U".

In particular, it follows from the last axiom that if "U" is non-empty, it must contain all of its finite subsets and a subset of each finite cardinality. One can also prove immediately from the definitions that the intersection of any class of universes is a universe.

The idea of universes is due to Alexander Grothendieck, who used them as a way of avoiding proper classes in algebraic geometry.

Grothendieck universes and inaccessible cardinals

There are two simple examples of Grothendieck universes:
* The empty set, and
* The set of all hereditarily finite sets $V\_omega$.Other examples are more difficult to construct. Loosely speaking, this is because Grothendieck universes are equivalent to strongly inaccessible cardinals. More formally, the following two axioms are equivalent:

: (U) For all sets "x", there exists a Grothendieck universe "U" such that "x" $in$ "U".: (C) For all cardinals κ, there is a strongly inaccessible cardinal $lambda$ which is strictly larger than κ.

To prove this fact, we introduce the function c("U"). Define::$mathbf\{c\}(U)\; =\; sup\_\{x\; in\; U\}\; |x|$where by |"x"| we mean the cardinality of "x". Then for any universe "U", c("U") is strongly inaccessible: It is a strong limit cardinal because the power set of any element of "U" is an element of "U" and every element of "U" is a subset of "U". To see that it is regular, suppose that "c_λ" is a collection of cardinals indexed by "I", where the cardinality of "I" and of each "c_λ" is less than c("U"). Then, by the definition of c("U"), "I" and each "c_λ" can be replaced by an element of "U". The union of elements of "U" indexed by an element of "U" is an element of "U", so the sum of the "c_λ" has the cardinality of an element of "U", hence is less than c("U"). By invoking the axiom of foundation, that no set is contained in itself, it can be shown that c("U") equals |"U"|; see Bourbaki's article, which also has a counterexample when the axiom of foundation is not assumed.

Let κ be a strongly inaccessible cardinal. Say that a set "S" is strictly of type κ if for any sequence "s_n" $in$ ... $in$ "s₀" $in$ "S", |"s_n"| < κ. ("S" itself corresponds to the empty sequence.) Then the set "u(κ)" of all sets strictly of type κ is a Grothendieck universe of cardinality κ. The proof of this fact is long, so for details, we again refer to Bourbaki's article, listed in the references.

To show that the large cardinal axiom (C) implies the universe axiom (U), choose a set "x". Let "x₀" = "x", and for all "n", let "x_n+1" = "x_n" be the union of the elements of "x_n". Let "y" = "x_n". By (C), there is a strongly inaccessible cardinal κ such that |y| < κ. Let "u(κ)" be the universe of the previous paragraph. "x" is strictly of type κ, so "x" $in$"u(κ)". To show that the universe axiom (U) implies the large cardinal axiom (C), choose a cardinal κ. κ is a set, so it is an element of a Grothendieck universe "U". The cardinality of "U" is strongly inaccessible and strictly larger than that of κ.

In fact, any Grothendieck universe is of the form "u($kappa$)" for some $kappa$. This gives another form of the equivalence between Grothendieck universes and strongly inaccessible cardinals:

:For any Grothendieck universe "U", |"U"| is either zero, $aleph\_0$, or a strongly inaccessible cardinal. And if $kappa$ is zero, $aleph\_0$, or a strongly inaccessible cardinal, then there is a Grothendieck universe u($kappa$). Furthermore, u(|"U"|)="U", and |u($kappa$)|=$kappa$.

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 $V\_omega$ cannot be proved from Zermelo-Fraenkel set theory either.

See also

* Tarski-Grothendieck set theory
*Universe (mathematics)
*Von Neumann universe
*Constructible universe

References

cite conference
first = Nicolas
last = Bourbaki
authorlink = Nicolas Bourbaki
year = 1972
title = Univers
booktitle = Séminaire de Géométrie Algébrique du Bois Marie - 1963-64 - Théorie des topos et cohomologie étale des schémas - (SGA 4) - vol. 1 (Lecture notes in mathematics 269)
editor = Michael Artin, Alexandre Grothendieck, Jean-Louis Verdier, eds.
publisher = Springer-Verlag
location = Berlin; New York
language = French
pages = 185&ndash;217
url = http://modular.fas.harvard.edu/sga/sga/4-1/4-1t_185.html