# $\sum$ over disjoint union of sets

In Discrete mathematics, rule of sum says that "If a first task can be performed in $m$ ways and another can be performed in $n$ ways and two task be independent, then whole work can accomplished in $m+n$ ways".

Now consider this statement: "If $\{A_n\}$ is a sequence of pairwise disjoint subsets of $\mathbb{N}$ and $A=\bigcup_{n=1}^\infty A_n$, then $$\sum_{k\in A} a_k = \sum_{n=1}^\infty\left(\sum_{k\in A_n}a_k\right)$$ where $a_k\ge 0,\forall k$ is hold.".

How i can proof this statement?

P.S.: Question material can be found on: "Principles of Real Analysis, Charalambos D. Aliprantis, 3rd Edition, p. 102, Exc.1".

• This holds in general only for $a_k\ge 0$. Mar 5, 2014 at 21:53
• In fact it's not really defined without order of summation for general $a_k\in\mathbb R$. Are you sure this is not a definition of $$\sum_{k\in A}a_k$$ as some sort of short notation for RHS where the $A_n$ are finite or something? Mar 5, 2014 at 21:55
• You are right! I edited my question. @Berci Mar 5, 2014 at 22:12
• $A_n$ is only a subset of $\mathbb{N}$. @AlfonsoFernandez Mar 5, 2014 at 22:13

This holds in general only for $a_k\ge 0$.
(For a counterexample consider e.g. $A_k:=\{2k,\ 2k+1\}$, and $a_n:=(-1)^n$.)
Now suppose that $a_k\ge 0$. Then $\sum_{k\in A}a_k\ =\ \sup\left\{\sum_{k\in A'}a_k\,\mid\, A'\right.\,$ is finite subset of $\,\left.A\right\}$.
Using this, for any $N\in\Bbb N$, we always have $$\sum_{k\in A}a_k \ \ge\ \sum_{n=0}^N\left(\sum_{k\in A_n}a_k\right)\,,$$ as the right hand side is a finite sum of supremum of finite sums (of positive numbers). By the same reasoning, we also have $$\sum_{k\in A}a_k\ \le\ \sum_{n=0}^\infty\left(\sum_{k\in A_n}a_k\right)\,.$$
• $\sum_{k\in A}a_k = \sup\left\{\sum_{k\in A'}a_k\mid A' \text{ is finite subset of } A \right\}$ is definition? Mar 5, 2014 at 22:16