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Well, now that you put it that way, I see the problem. I tried to eliminate that problem by writing the addends in the form of h(x,y)=f(x)g(y), but now I see that the sum of them does not quite conform to the way I wanted it to.mathwonk said:let me make my objection to your proof as simple as possible:

just because h = f+g, it does not follow that h(U) = f(U)+g(U), which is what you are apparently claiming in your argument. this is what hurkyl's counter example shows.

i.e. h(U) consists only of all points of form f(x)+g(x) for all x in U, while f(U)+g(U) consits of the much larger collection of points of form f(x)+g(y) for all x,y in U.

I.e. f(U) +g(U) is the image of the product set UxU under the map f+g, while h(U) is the image only of the diagonal.