Thus, König's theorem gives us an alternate proof of Cantor's theorem. (Historically of course Cantor's theorem was proved much earlier.) One way of stating the axiom of choice is "an arbitrary Cartesian product of non-empty sets is non-empty". Let "Bi" be a non-empty set for each "i" in "I". Let "Ai" = {} for each "i" in "I". Thus by König's theorem, we have: That is, the Cartesian product of the given non-empty sets "Bi" has a larger cardinality than the sum of empty sets. Thus it is non-empty, which is just what the axiom of choice states. Since the axiom of choice follows from König's theorem, we will use the axiom of choice freely and implicitly when discussing consequences of the theorem. König's theorem has also important consequences for cofinality of cardinal numbers. Choose a strictly increasing cf(κ)-sequence of ordinals approaching κ. Each of them is less than κ, so their sum, which is κ, is less than the product of cf(κ) copies of κ. According to Easton's theorem, the next consequence of König's theorem is the only nontrivial constraint on the continuum function for regular cardinals. Let formula_21. Suppose that, contrary to this corollary, formula_22. Then using the previous corollary, formula_23, a contradiction. Thus the supposition must be false, and this corollary must be true. Assuming Zermelo–Fraenkel set theory, including especially the axiom of choice, we can prove the theorem. Remember that we are given formula_24, and we want to show :formula_25  The axiom of choice implies that the condition "A" < "B" is equivalent to the condition that there is no function from "A" onto "B" and "B" is nonempty. So we are given that there is no function from "A""i" onto "B""i"≠{}, and we have to show that any function "f" from the disjoint union of the "A"s to the product of the "B"s is not surjective and that the product is nonempty. That the product is nonempty follows immediately from the axiom of choice and the fact that the factors are nonempty.