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If two functions are defined as set of points and there's a point that is defined only in one function but not in the other, e.g.
$A = \{(0,1)\}, B = \{(1,2)\}$.
In function arithmetic, what would be $A+B$ ?

Irregular User
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Ignas2526
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  • Usually in function arithmetic you can't add functions with different domains. One way to define this would be to define $(f+g) : A \cap B \to C,(f+g)(x)=f(x)+g(x)$. Thus you would restrict attention to the intersection of the two domains. Alternately you can choose some extension of $f$ and/or $g$ so that they share a common domain larger than $A \cap B$...but it is not entirely clear how this extension should be done. – Ian Jun 17 '16 at 02:01
  • An alternative perspective: If $f$ has domain $A$ and $g$ has domain $B$, and $A\neq B$, then to define $f+g$ on $A\cup B$, you would need to extend $f$ and $g$ (one method ix extending by $0$, i.e., $f$ remains the same on $A$, but is assigned the value $0$ on $(A\cup B)\setminus A$, and likewise for $g$), and then you could define $f+g$ in terms of their extensions, but not in terms of the original functions. Also, it's not clear how you would extend $f$ or $g$ without more context. (deleted and retyped comment b/c of a typo). – Nicholas Stull Jun 17 '16 at 02:16

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I don't fully understand the question. In general, we have notation $A \square B = \{ a\square b | a \in A, b \in B \}$. So here you have $A+B = \{ (1,3) \}$.

If $f,g$ are two functions in some space in which some operation $\square$ is defined, then $f \square g$ is only defined on $\operatorname{dom} A \cap \operatorname{dom} B$ which is empty here. In this context, the sum is $\emptyset$.

copper.hat
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