Editing Dedekind-infinite set (section)

==Relation to the axiom of choice==
Since every infinite well-ordered set is Dedekind-infinite, and since the AC is equivalent to the [[well-ordering theorem]] stating that every set can be well-ordered, clearly the general AC implies that every infinite set is Dedekind-infinite. However, the equivalence of the two definitions is much weaker than the full strength of AC.

In particular, there exists a model of '''ZF''' in which there exists an infinite set with no [[countable set|countably infinite]] subset. Hence, in this model, there exists an infinite, Dedekind-finite set. By the above, such a set cannot be well-ordered in this model.

If we assume the axiom of countable choice (i. e., AC<sub>ω</sub>), then it follows that every infinite set is Dedekind-infinite. However, the equivalence of these two definitions is in fact strictly weaker than even the CC. Explicitly, there exists a model of '''ZF''' in which every infinite set is Dedekind-infinite, yet the CC fails (assuming consistency of '''ZF''').