Inset theory, aGrothendieck universe is a setU with the following properties:

1. Ifx is an element ofU and ify is an element ofx, theny is also an element ofU. (U is atransitive set.)
2. Ifx andy are both elements ofU, then${\displaystyle \{x,y\}}$ is an element ofU.
3. Ifx is an element ofU, thenP(x), thepower set ofx, is also an element ofU.
4. If${\displaystyle \{x_{\alpha }{\}}_{\alpha \in I}}$ is a family of elements ofU, and ifI is an element ofU, then the union$\bigcup _{\alpha \in I}{x}_{\alpha }$ is an element ofU.

A Grothendieck universe is meant to provide a set in which all of mathematics can be performed. (In fact,uncountable Grothendieck universes providemodels of set theory with the natural ∈-relation, natural powerset operation etc.). Elements of a Grothendieck universe are sometimes calledsmall sets. The idea of universes is due toAlexander Grothendieck, who used them as a way of avoidingproper classes inalgebraic geometry. Grothendieck’s original proposal was to add the followingaxiom of universes to the usual axioms of set theory: For every set${\displaystyle s}$, there exists a universe${\displaystyle U}$ that contains${\displaystyle s}$, i.e.,${\displaystyle s\in U}$.

The existence of a nontrivial Grothendieck universe goes beyond the usual axioms ofZermelo–Fraenkel set theory; in particular it would imply the existence ofstrongly inaccessible cardinals.Tarski–Grothendieck set theory is an axiomatic treatment of set theory, used in some automatic proof systems, in which every set belongs to a Grothendieck universe.The concept of a Grothendieck universe can also be defined in atopos.^[1]

As an example, we will prove an easy proposition.

Proposition. If${\displaystyle x\in U}$ and${\displaystyle y\subseteq x}$, then${\displaystyle y\in U}$.

Proof.${\displaystyle y\in P(x)}$ because${\displaystyle y\subseteq x}$.${\displaystyle P(x)\in U}$ because${\displaystyle x\in U}$, so${\displaystyle y\in U}$.

It is similarly easy to prove that any Grothendieck universeU contains:

• Allsingletons of each of its elements,
• Allproducts of all families of elements ofU indexed by an element ofU,
• Alldisjoint unions of all families of elements ofU indexed by an element ofU,
• All intersections of all nonempty families of elements ofU indexed by an element ofU,
• All functions between any two elements ofU, and
• All subsets ofU whosecardinality is an element ofU.

In particular, it follows from the last axiom that ifU 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.

There are two simple examples of Grothendieck universes:

• The empty set, and
• The set of allhereditarily finite sets${\displaystyle V_{\omega }}$.

Other examples are more difficult to construct. Loosely speaking, this is because Grothendieck universes are equivalent tostrongly inaccessible cardinals. More formally, the following two axioms are equivalent:

(U) For each setx, there exists a Grothendieck universeU such thatx ∈U.

(C) For each cardinalκ, there is a strongly inaccessible cardinalλ that is strictly larger thanκ.

To prove this fact, we introduce the functionc(U). Define:

${\displaystyle \mathbf{c}(U)=\underset{x\in U}{sup}|x|}$

where by |x| we mean the cardinality ofx. Then for any universeU,c(U) is either zero,${\displaystyle {\mathrm{\aleph }}_0}$, or strongly inaccessible. Assuming it is uncountable, it is a strong limit cardinal because the power set of any element ofU is an element ofU and every element ofU is a subset ofU. To see that it is regular, suppose thatc_λ is a collection of cardinals indexed byI, where the cardinality ofI and of eachc_λ is less thanc(U). Then, by the definition ofc(U),I and eachc_λ can be replaced by an element ofU. The union of elements ofU indexed by an element ofU is an element ofU, so the sum of thec_λ has the cardinality of an element ofU, hence is less thanc(U). By invoking the axiom of foundation, that no set is contained in itself, it can be shown thatc(U) equals |U|; when the axiom of foundation is not assumed, there are counterexamples (we may take for example U to be the set of all finite sets of finite sets etc. of the sets x_α where the index α is any real number, andx_α = {x_α} for eachα. ThenU has the cardinality of the continuum, but all of its members have finite cardinality and so${\displaystyle \mathbf{c}(U)={\mathrm{\aleph }}_0}$ ; see Bourbaki's article for more details).

Letκ be a strongly inaccessible cardinal. Say that a setS is strictly of typeκ if for any sequences_n ∈ ... ∈s₀ ∈S, |s_n| <κ. (S itself corresponds to the empty sequence.) Then the setu(κ) of all sets strictly of typeκ is a Grothendieck universe of cardinalityκ. 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 setx. Letx₀ =x, and for eachn, let${\displaystyle x_{n+1}=\bigcup x_n}$ be the union of the elements ofx_n. Lety =${\displaystyle \bigcup _n{x}_n}$. By (C), there is a strongly inaccessible cardinalκ such that|y| <κ. Letu(κ) be the universe of the previous paragraph.x is strictly of typeκ, sox ∈u(κ). To show that the universe axiom (U) implies the large cardinal axiom (C), choose a cardinalκ.κ is a set, so it is an element of a Grothendieck universeU. The cardinality ofU is strongly inaccessible and strictly larger than that ofκ.

In fact, any Grothendieck universe is of the formu(κ) for someκ. This gives another form of the equivalence between Grothendieck universes and strongly inaccessible cardinals:

For any Grothendieck universeU, |U| is either zero,${\displaystyle {\mathrm{\aleph }}_0}$, or a strongly inaccessible cardinal. And ifκ is zero,${\displaystyle {\mathrm{\aleph }}_0}$, or a strongly inaccessible cardinal, then there is a Grothendieck universe⁠${\displaystyle u(\kappa )}$⁠. Furthermore,u(|U|) =U, and|u(κ)| =κ.

Since the existence of strongly inaccessible cardinals cannot be proved from the axioms ofZermelo–Fraenkel set theory (ZFC), the existence of universes other than the empty set and${\displaystyle V_{\omega }}$ cannot be proved from ZFC either. However, strongly inaccessible cardinals are on the lower end of thelist of large cardinals; thus, most set theories that use large cardinals (such as "ZFC plus there is ameasurable cardinal", "ZFC plus there are infinitely manyWoodin cardinals") will prove that Grothendieck universes exist.

• Constructible universe
• Universe (mathematics)
• Von Neumann universe

• Bourbaki, Nicolas (1972)."Univers". InMichael Artin;Alexandre Grothendieck;Jean-Louis Verdier (eds.).Séminaire de Géométrie Algébrique du Bois Marie – 1963–64 – Théorie des topos et cohomologie étale des schémas – (SGA 4) – vol. 1 (Lecture Notes in Mathematics269) (in French). Berlin; New York:Springer-Verlag. pp. 185–217. Archived fromthe original on 2016-08-09. Retrieved2010-12-06.
• "universe".PlanetMath. 2013-03-22. RetrievedNovember 25, 2024.

• Set-theoretic universes
• Category theory
• Large cardinals

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