empty set, and The set of all hereditarily finite sets . 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 each set x, there exists a Grothendieck universe U such that x ∈ U. (C) For each cardinal κ, there is a strongly inaccessible cardinal λ that is strictly larger than κ. To prove this fact, we introduce the function c(U). Define: where by |x| we mean the cardinality of x. Then for any universe U, c(U) is either zero or strongly inaccessible. Assuming it is non-zero, 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 cλ is a collection of cardinals indexed by , where the cardinality of and of each cλ is less than c(U). Then, by the definition of c(U), and each cλ 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 cλ 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|; 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 xα where the index α is any real number, and xα = {xα} for each α. Then U has the cardinality of the continuum, but all of its members have finite cardinality and so ; see Bourbaki's article for more details). Let be a strongly inaccessible cardinal. Say that a set S is strictly of type if for any sequence sn ∈ ... ∈ s0 ∈ S, |sn| < κ. (S itself corresponds to the empty sequence.) Then the set of all sets strictly of type is a Grothendieck universe of cardinality . 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