limitation of size can be written: Implications of the axiom Von Neumann proved that the axiom of limitation of size implies the axiom of replacement, which can be expressed as: If F is a function and A is a set, then F(A) is a set. This is proved by contradiction. Let F be a function and A be a set. Assume that F(A) is a proper class. Then there is a function G that maps F(A) onto V. Since the composite function G ∘ F maps A onto V, the axiom of limitation of size implies that A is a proper class, which contradicts A being a set. Therefore, F(A) is a set. Since the axiom of replacement implies the axiom of separation, the axiom of limitation of size implies the axiom of separation. Von Neumann also proved that his axiom implies that V can be well-ordered. The proof starts by proving by contradiction that Ord, the class of all ordinals, is a proper class. Assume that Ord is a set. Since it is a transitive set that is strictly well-ordered by ∈, it is an ordinal. So Ord ∈ Ord, which contradicts Ord being strictly well-ordered by ∈. Therefore, Ord is a proper class. So von Neumann's axiom implies that there is a function F that maps Ord onto V. To define a well-ordering of V, let G be the subclass of F consisting of the ordered pairs (α, x) where α is the least β such that (β, x) ∈ F; that is, G = {(α, x) ∈ F: ∀β((β, x) ∈ F ⇒ α ≤ β)}. The function G is a one-to-one correspondence between a subset of Ord and V. Therefore, x < y if G−1(x) < G−1(y) defines a well-ordering of V. This well-ordering defines a global choice function: Let Inf(x) be the least element of a non-empty set x. Since Inf(x) ∈ x, this function chooses an element of x for every non-empty set x. Therefore, Inf(x) is a global choice function, so Von Neumann's axiom implies the axiom of global choice. In 1968, Azriel Lévy proved that von Neumann's axiom implies the axiom of union. First, he proved without using the axiom of union that every set of ordinals has an upper bound. Then he used a function that maps Ord onto V to prove that if A is a set, then ∪A is