CC for finite sets and equivalent condition

Question

If we assume for all sequence of sets od cardinality exactly 2, there exist choice function. Can we prove Countable choice for finite sets

Answer 1

Not quite a complete answer to your question, but too long for a comment: http://projecteuclid.org/download/pdf_1/euclid.pjm/1102983625 shows that e.g. the axiom of choice for all familes of sets with exactly two elements, does not prove the axiom of choice for all families of sets with exactly 3 elements. (see page 235). This strongly suggests that the answer is "no" (the issue, of course, being that you ask if choice for all families of 2-element sets implies countable choice for finite sets, so it doesn't quite apply).

It's worth noting, though, that there are nontrivial implications here. For example, the axiom of choice for families of sets with either 2 or 3 elements does imply the full (not just countable) axiom of choice for families of sets with exactly 4 elements: given a family of disjoint sets $\{X_i: i\in I\}$ with $\vert X_i\vert=4$ for all $i\in I$, let $S_i$ be the set of all partitions of $X_i$ into two two-element sets. Note that $\vert S_i\vert=3$. Now we apply 3-choice to get a set $\{p_i: i\in I\}$, where each $p_i$ is a partition of $X_i$ into two two-element sets. Apply 2-choice to this family of partitions yields a family of sets $\{A_i: i\in I\}$ where $A_i$ is a two-element subset of $X_i$. Apply 2-choice once more, and we're done.

Answer 2

Let me give a different reference from the one Noah gave (although the mathematics is more or less the same).

Jech, "The Axiom of Choice", Chapter 7. The last part of the chapter deals with all sort of various implications and separations of choice from finite sequences, including a nontrivial condition which guarantees some sort of control on when one type of choice from finite sets implies another type of choice from finite sets.

To sum up, choice from any family of pairs, need not imply choice from families of larger sets (although in the case of sets of size $4$ it does, surprisingly enough). Additionally choice from any collection of finite sets will not imply countable choice, or even that there is no infinite Dedekind-finite set.