<< Am I missing something? According to mathworld, a free abelian group is an abelian group with no torsion, seemingly meaning that the only element with finite order is the identity (in this case the zero function). Clearly here, no function other than zero has finite order. Now if you ask for a set of generators, that could be a bit tricky. It's hard to list uncountably many things.
I'll post a proof soon (along the lines of Michael's sketch). But the definition of free abelian group above is valid only for *finite* abelian groups. Here's an example: Let P (S) denote the countable direct product (sum) of Z's, say indexed by 1,2,3,.... Then it's easy to see that the quotient group P/S has no torsion. But the the coset class [(1,2,4,8,...,2^n,...)] is divisible by arbitrarily high integers (since we may ignore any finite initial string). But no free abelian group can have an element divisible by arbitrarily high integers. (Def. Given any set X, the free abelian group on X is all expressions of the form {sum{k=1 to N} of x_k : N is a nonnegative integer and all x_k belong to X (with the obvious identifications made for the zero element).) --Dan