Assumption in Brown Representability Theorem

I'm reading Brown's theorem in Hatcher and Brown's original paper. I am confused by the assumptions concerning the generalized cohomology theory, which is defined axiomatically by the Eilenberg-Steenrod axioms except the dimension axiom. Brown replaced the dimension axiom by the condition that the nth cohomology group of a point is countable for nonzero n, and Brown said this condition couldn't be weakened. The wedge axiom used by Hatcher to define reduced cohomology theory, however, implies that the nth cohomology group of a point is trivial. This seems contradictory. Can somebody please help me?

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Reduced cohomology of a point is by definition trivial, whereas I guess Brown was defining non-reduced cohomology. –  David Roberts Jun 25 '13 at 3:56
Thanks! So Hatcher's version is more powerful than Brown's original one? –  Emmy Jun 25 '13 at 4:08
I wouldn't have thought so - one can recover ordinary cohomology from reduced cohomology by adding a disjoint basepoint to a space, so the two should be interconvertible to some extent. However, I don't see that assuming reducedness would give you the cohomology of a point to be countable, so you may be right. –  David Roberts Jun 25 '13 at 5:17
I haven't checked what versions are given in the sources that you mention. However, one important issue is as follows. If you assume that your original cohomology theory $H$ is defined only on finite CW complexes, then you need the coefficients to be countable. If you assume that $H$ is defined on all CW complexes, then countability is not required. –  Neil Strickland Jun 25 '13 at 8:26