Is the axiom of choice really related to choice?

What do we lose out on if we only assume the axiom of countable choice?

• In other words, what interesting statements require the existence of choice functions for uncountably infinite collections of sets?

• Answer:

  As far as I know, countable choice is not strong enough to prove the ultrafilter lemma (http://en.wikipedia.org/wiki/Boolean_prime_ideal_theorem). This happens to be my favorite choice principle. It is equivalent to the following statements: http://en.wikipedia.org/wiki/G%C3%B6del's_completeness_theorem. The http://en.wikipedia.org/wiki/Compactness_theorem. http://en.wikipedia.org/wiki/Tychonoff's_theorem for compact Hausdorff spaces. The http://en.wikipedia.org/wiki/Banach%E2%80%93Alaoglu_theorem. And it implies the following statements: Every field has an algebraic closure (http://mathoverflow.net/questions/46566/is-the-statement-that-every-field-has-an-algebraic-closure-known-to-be-equivalent). Any two algebraic closures of a field are isomorphic (http://mathoverflow.net/questions/46566/is-the-statement-that-every-field-has-an-algebraic-closure-known-to-be-equivalent again). The http://en.wikipedia.org/wiki/Hahn%E2%80%93Banach_theorem. http://en.wikipedia.org/wiki/Arrow's_impossibility_theorem is false if the number of voters is infinite. Ultrafilters are wonderful. Terence Tao has written several blog posts (http://terrytao.wordpress.com/tag/ultrafilters/) about the relationship between ultrafilters and topics like nonstandard analysis, and I have also written a few blog posts (http://qchu.wordpress.com/tag/ultrafilters/) about the relationship between ultrafilters and logic, commutative algebra, topology, and Ramsey theory.