I have a problem with this. The reason has to be added to is that, of course, is not normally a member of . And that is true because arithmetic operations cannot be extended to . So my question is, since, normally, fg(x)= (f(x))(g(x)), what does "fg" mean if either f(x) or g(x) is ?

I guess we could assume that for any but what is ? In order for the equation you give to be true, it must be some number other than or simply left undefined. But that should have been said.

Even then, I have a problem with . " " is not in the range of f! In order for that to make sense, you must have meant that f and g have ranges but then the equation you give is not correct. The first set on the right, consistes of numbers, x, such that , not .

I can only conclude that the " " terms are incorrect and your statement should have been .

That, with the definition of "product" I provided, is a true statement.

For any function f, . That is, it is the set of all values of x that give a value for f(x) contained in A.

So . And, of course, that is true if either and g(x) is non-zero or and f(x) is non-zero.

That's exactly what " " says.

The first term, img.top {vertical-align:15%;} and g(x)< 0. Such a product, by the definitions I gave, is The second term, , is all x such that f(x)= \infty" alt="f^{-1}( \{ \infty \} \cap g^{-1} ([ - \infty , 0))], is all x such that and g(x)< 0. Such a product, by the definitions I gave, is The second term, , is all x such that f(x)= \infty" /> and g(x)> 0[/tex]. The third term is the set of all x such that f(x)< 0 and . The final set, f ^{-1}(0, \infty ] \cap g^{-1} ( \{ - \infty \} ) [/tex] is the set of all x such that f(x)> 0 and .

The reason for that union of four sets was to separate from and to separate from [tex]g^{-1}((0, \infty]). That is to say "all xexceptx such that f(x)= 0 or g(x)= 0".