Dan asked last week:
Let P denote the direct product of countably many copies of the integers Z.
E.g., the set {f: Z+ --> Z | f is a function} under addition of functions.
Puzzle: Is P a free abelian group? Prove your answer.
And Gene just re-expressed interest in the subject. I started writing something about this yesterday, but realized I was on shaky ground. I thought I had a proof that P was not free abelian, but then realized I was making some sort of unwarranted assumptiong about continuous functions on P which depended on some notion of topology of P, which made me uncomfortable. Given a function f : P->Z, just because you know all the f(ei), where ei=(0,0,...,0,1,0,0,...), doesn't mean you know f(a1,a2,a3,a4,...). (The rest of yesterday's argument said that there were only countably many homomorphisms P->Z, since all but finitely many of the ei had to go to zero; but a free abelian group on uncountably many generators -- which P would have to be, it being uncountable itself -- should have uncountably many such homomorphisms. But as I said, this seems hollow to me now.) I second the motion for an answer. --Michael -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.