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Just to elaborate for those not familiar with the terminology, the axion of dependent choice is a weaker version of the axiom of choice. It asserts that you ca

Just to elaborate for those not familiar with the terminology, the axion of dependent choice is a weaker version of the axiom of choice. It asserts that you can pick a countable(denumerable using the terminology in the post) sequence from any infinite set. The axiom of choice asserts (informally) that given any collection of sets, you can form another set which consists of one element from each set in the collection. The axiom of choice asserts that this holds true even when the number of sets is uncountable (non-denumerable). The axiom of choice and the continuum hypothesis are equivalent, meaning the assumption of either one implies the other. <<EDIT: This is wrong, see below>>

hypersoar 13y ago
This is a good reply, but I have to point out that the Continuum Hypothesis is certainly not equivalent to the Axiom of Choice. In fact, it's independent of the Zermelo Frankel axioms together with Choice. While broader mathematics community has pretty much now accepted Choice (it's too damn useful not to), no such consensus has arisen over the Continuum Hypothesis. However, unlike AC, I doubt many people outside set theory their have a strong opinion on CH.
- thejteam 13y ago
  Thanks, I temporarily got CH and well-ordering mixed up.
Sniffnoy 13y ago
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.)

hypersoar 13y ago

This is a good reply, but I have to point out that the Continuum Hypothesis is certainly not equivalent to the Axiom of Choice. In fact, it's independent of the Zermelo Frankel axioms together with Choice. While broader mathematics community has pretty much now accepted Choice (it's too damn useful not to), no such consensus has arisen over the Continuum Hypothesis. However, unlike AC, I doubt many people outside set theory their have a strong opinion on CH.

thejteam 13y ago
Thanks, I temporarily got CH and well-ordering mixed up.

Sniffnoy 13y ago

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.)