Partition of the index set of the non-negative series

Question

Suppose that $a_j\geq 0$ for all $j\in \mathbb{N}$. Prove that $$\sum \limits_{j\in A}a_j+\sum \limits_{j\in B}a_j\leq \sum \limits_{j=1}^{\infty}a_j,$$ where $A$ and $B$ - partition of the natural numbers.

How to prove this inequality rigorously.

Answer 1

By definition ($a_j \geq 0$) we have $\sum\limits_{j \in A} a_j = \sup\limits_{\substack{C \subseteq A \\ |C| < + \infty}} \sum\limits_{j \in C} a_j$ and the same holds for $B$.

Consider now any pair of finite sets $(C,D)$ such that $C \subseteq A$ and $D \subseteq B$.

Then we have $$ \sum\limits_{j \in C} a_j + \sum\limits_{j \in D} a_j \leq \sum\limits_{j \in \mathbb{N}} a_j $$

Let's now consider $D$ fixed and take the sup of the LHS as $C$ varies over all finite subsets of $A$. We obtain $$ \sum\limits_{j \in A} a_j + \sum\limits_{j \in D} a_j \leq \sum\limits_{j \in \mathbb{N}} a_j $$ Let's now take the sup of the LHS as $D$ varies over all finite subsets of $B$, we get $$\sum\limits_{j \in A} a_j + \sum\limits_{j \in B} a_j \leq \sum\limits_{j \in \mathbb{N}} a_j $$