Well, the axiom of dependent choice is a bit stronger than that. And your contrast with the full axiom of choice makes it sound like dependent choice is just c

Well, the axiom of dependent choice is a bit stronger than that. And your contrast with the full axiom of choice makes it sound like dependent choice is just countable choice. For the onlookers: Axiom of choice -- Given any collection of disjoint sets, there's a set consisting of one element from each. Axiom of countable choice (this is weaker) -- Given any countable collection of disjoint sets, there's a set consisting of one element from each. Axiom of dependent choice (this is intermediate in strength) -- Suppose you have a set X and a binary relation R on X such that for any x in X, there's some y in X such that xRy holds. Then there's a sequence x_0, x_1, ..., consisting of elements in X, such that for any i, x_i R x_(i+1) holds. Countable choice is not enough to prove that aleph_0 is the smallest infinite cardinal; dependent choice is. Of course you could weaken dependent choice and still be able to prove it, but not all the way to countable choice. (Ignoring the AC/CH thing, that's been addressed already.)