Summation of nonnegative numbers over countably infinite set Suppose that $T_1,T_2\subset \mathbb{N}$ and $a_i\geq 0$. I need to prove that $$\sum \limits_{i\in T_1\cap T_2}a_i+\sum \limits_{i\in T_1\cup T_2}a_i=\sum \limits_{i\in T_1}a_i+\sum \limits_{I\in T_2}a_i.$$

I do not know how to prove this fact at all because I even don't know what is the definition of $\sum \limits_{i\in K}a_i$ for $K\subset \mathbb{N}$.

I will highly appreciate if someone will explain my questions, please!

EDIT: Also I assume that $\sum \limits_{I\in T_1}a_i&lt;\infty,\sum \limits_{I\in T_2}a_i&lt;\infty$ and as I said above $a_i\geq 0$.
One definition for summations of non negative numbers over an arbitrary index set is $\sum_{n \in I} a_n = \sup_{J \subset I, J \text{ finite}} \sum_{n \in J} a_n$.

Note that it follows immediately that if $I \subset I'$ then
$\sum_{n \in I} a_n  \le \sum_{n \in I'} a_n$.

Now suppose $A \subset I$, then I claim that
$\sum_{n \in I} a_n = \sum_{n \in A} a_n + \sum_{n \in A^c} a_n$.

Suppose $J_1 \subset A, J_2 \subset A^c$ are finite, then since $J_1 \cup J_2$ is finite we have
$\sum_{n \in I} a_n \ge \sum_{n \in J_1 \cup J_2} a_n = \sum_{n \in J_1} a_n + \sum_{n \in J_2} a_n$, taking the $\sup$s on the right hand side gives
$\sum_{n \in I} a_n \ge \sum_{n \in A} a_n + \sum_{n \in A^c} a_n$.

Similarly, if $J \subset I$ is finite, then
$\sum_{n \in J} a_n = \sum_{n \in J\cap A} a_n + \sum_{n \in J \cap A^c} a_n \le \sum_{n \in A} a_n + \sum_{n \in A^c} a_n$.

Now, note that $T_1 \cup T_2 = T_1 \setminus T_2 \cup (T_1 \cap T_2) \cup T_2 \setminus T_1$, a disjoint union, hence
$\sum_{n \in T_1 \cup T_2} a_n+ \sum_{n \in T_1 \cap T_2} a_n=  \sum_{n \in T_1 \setminus T_2} a_n + 2 \sum_{n \in T_1 \cap T_2} a_n  + \sum_{n \in T_2 \setminus T_1} a_n$
and
$\sum_{n \in T_1 } a_n + \sum_{n \in T_2}a_n =  \sum_{n \in T_1 \setminus T_2} a_n + 2 \sum_{n \in T_1 \cap T_2} a_n + \sum_{n \in T_2 \setminus T_1} a_n$.

Note that this result holds whether or not the sum is finite.
For the notation part:
For example, let $K$ be the even numbers. Then $\sum\limits_{i\in K}a_i$ is just $a_2+a_4+a_6+....$

For the proof part:

Lemma: If $A$ and $B$ are disjoint and the terms are always positive and the sums converge: $\sum\limits_{i\in A\cup B}(a_i) = \sum\limits_{i\in A}a_i + \sum\limits_{i\in B}a_i$

Hence, since $T_2= T_1 \cap (T2-T1)$ with this to sets being disjoint:
$\sum\limits_{i\in T_1}a_i + \sum\limits_{i\in T_2}a_i = \sum\limits_{i\in T_1}a_i + \sum\limits_{i\in T_2-T_1}a_i + \sum\limits_{i\in T_2\cap T_1}a_i$

And again by the same lemma. combining the first and the second terms from the right side of the equality (remember that $T_1 \cup T_2 = T_1 \cup (T_2 - T_1)$):

$\sum\limits_{i\in T_1}a_i + \sum\limits_{i\in T_2}a_i = \sum\limits_{i\in T1\cup T_2}a_i + \sum\limits_{i\in T_2\cap T_1}a_i$.