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Suppose we have two sets: $A=\{a_1,a_2,a_3\}$ and $B=\{b_1,b_2,b_3\}$.

Is there a way to define a function that simply adds/subtracts the elements of these two sets? For example, $$\mu(\cdot)=\sum_{i\in\mathbb{N}:\;a_i\in A}a_i+\sum_{j\in\mathbb{N}:\;b_j\in B}b_j.$$ I am not sure if I am allowed to say that $\mu(\cdot)$ is a function with domain $A\times B$.

Basically, my question is about rigorously defining a function that simply adds elements from two different sets. Is that possible? If yes, then how?

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2 Answers 2

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Hint: The cartesian product \begin{align*} A\times B&=\{(a_1,b_1),(a_1,b_2),(a_1,b_3),\\ &\qquad(a_2,b_1),(a_2,b_2),(a_2,b_3),\\ &\qquad(a_3,b_1),(a_3,b_2),(a_3,b_3)\} \end{align*} is not appropriate as domain for $\mu$, since a function $f:A\times B\to \mathbb{R}$ can only map elements of $A\times B$ i.e. pairs $(a_j,b_k)$ to $\mathbb{R}$. But we want to be able to sum up all elements from $A$ and $B$.

Here is an approach which might be useful:

  • We have to be careful to not mix up elements from $A$ and $B$ in case $A\cap B\neq \emptyset$. We consider instead $A\times\{0\}=\{(a_1,0),(a_2,0),(a_3,0)\}$ and $B\times\{1\}=\{(b_1,1),(b_2,1)(b_3,1)\}$ to overcome this problem.

  • We want to add all elements from $A$ and all elements from $B$. We take therefore the powerset $\mathcal{P}$ of $(A\times\{0\})\cup(B\times\{1\})$ as domain of $\mu$.

We define $\mu$ as follows.

\begin{align*} &\mu:\mathcal{P}\left((A\times\{0\})\cup(B\times\{1\})\right)\to\mathbb{R}\\ &\mu(Z)=\sum_{z\in Z}\pi_1(z) \end{align*}

where $z\in Z\subseteq(A\times\{0\})\cup(B\times\{1\})$ and $\pi_1(z)=x$ is the projection of $z=(x,y)$ to the first coordinate.

Taking $Z=(A\times\{0\})\cup(B\times\{1\})$ we obtain \begin{align*} \color{blue}{\mu(Z)}&=\mu((A\times\{0\})\cup(B\times\{1\}))\\ &=\sum_{z\in (A\times\{0\})\cup(B\times\{1\})}\pi_1(z)\\ &=\pi_1((a_1,0))+\pi_1((a_2,0))+\pi_1((a_3,0))\\ &\qquad+\pi_1((b_1,1))+\pi_1((b_2,1))+\pi_1((b_3,1))\\ &\,\,\color{blue}{=a_1+a_2+a_3+b_1+b_2+b_3}\\ \end{align*}

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You can define $\mu$ as a function with two variables (both finite subsets of a set of numbers) as $$\mu(A,B) :=\sum_{a\in A}a\, +\,\sum_{b\in B}b$$

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  • $\begingroup$ I understand that but my point is what if I want to specify the domain and codomain of that function? Could I write $\mu:A\times B\to\mathbb{R}$ given that $A,B\subset\mathbb{R}$? $\endgroup$
    – johnny09
    Commented Jan 20, 2019 at 0:57
  • $\begingroup$ It seems rather $P^{\text{fin}}(\Bbb R)\times P^{\text{fin}}(\Bbb R)$ where $P^{\text{fin}}(X)=\{A\subseteq X: A$ is finite $\}$. $\endgroup$
    – Berci
    Commented Jan 20, 2019 at 1:27

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