# What is the definition of $\sum\limits_{0\leq i\leq m,\text{ }0\leq j\leq n}a_{ij}$

I understand the concept of double summations, at least intuitively, but I'm trying to understand it formally. So, to begin with, I have a question:

Is this double summation equality true by definition:$\sum\limits_{0\leq i\leq m}\sum\limits_{0\leq j\leq n}a_{ij}=\sum\limits_{0\leq i\leq m,\text{ }0\leq j\leq n}a_{ij}$ ?

If not, how can I prove it?

This is what I have till now:

1.- I understand that the symbol $\sum$ is defined for a sequence, in this case a finite sequence, of the form $\{a_1,a_2,...,a_n\}$. Then $\sum\limits_{0\leq i\leq n} a_i$ is defined recursively having $\sum\limits_{0\leq i\leq 0} a_i=a_0$ and $\sum\limits_{0\leq i\leq k} a_i=\sum\limits_{0\leq i\leq k-1} a_i+a_k$ for $k>0$.

2.- I understand that $\sum\limits_{0\leq i\leq m}\sum\limits_{0\leq j\leq n}a_{ij}=\sum\limits_{0\leq i\leq m}(\sum\limits_{0\leq j\leq n}a_{ij})$, meaning that $\sum\limits_{0\leq i\leq m}\sum\limits_{0\leq j\leq n}a_{ij}=\sum\limits_{0\leq i\leq m}b_i$ where $\{b_1,b_2,...,b_m\}$ having $b_i=\sum\limits_{0\leq j\leq n}a_{ij}$.

3.- I understand the meaning of the expression $\sum\limits_{0\leq i\leq m,\text{ }0\leq j\leq n}a_{ij}$. Here the problem is that I don't have a formal definition.

4.- From a general stand point if we use $f$ instead of $\sum$ and instead of a sequence we use an index set we can write this in the form $\mathop{f}_{i\in I}\mathop{f}_{j\in J}a_{ij}=\mathop f_{i\in I,\text{ } j\in J}a_{ij}$, and then again it's not very clear on how to make an interpretation for the right hand side. In my attempt I'd say that $f$ should be a function of the form $f:P(A)\longrightarrow A$, leting $A$ be a set that contains every $a_{ij}$, such that $\mathop{f}_{i\in I}\mathop{f}_{j\in J}a_{ij}=\mathop{f}_{i\in I}(\mathop{f}_{j\in J}a_{ij})$ and $\mathop f_{i\in I,\text{ } j\in J}a_{ij}=f\{a_{ij}\mid i\in I, j\in J\}$. But then I guess it's not always true that $\mathop{f}_{i\in I}\mathop{f}_{j\in J}a_{ij}=\mathop f_{i\in I,\text{ } j\in J}a_{ij}$ (intuitively). This makes me think that in the case of my double summation above I need to prove the statement instead of being true by definition. But then what is the definition of this expression on the right hand side of the equality?...

Thanks everyone for your help!

## 1 Answer

It is not true by definition, but is true by associativity and commutativity of addition.

In general, if you have some finite set $S$ and some function $f: S \to V$, where $V$ is some vector space, then $\sum_{s \in S} f(s)$ is defined as $0$ if $S = \emptyset$, and recursively as $\sum_{s \in S \cup \{\sigma\}} f(s) = \sum_{s \in S} f(s) + f(\sigma)$.

A little work is needed to show that the value is independent of order, this follows from associativity and commutativity of addition.

If $S_1,S_2 \subset S$ are disjoint, then a little work shows $\sum_{s \in S_1 \cup S_2} f(s) = \sum_{s \in S_1} f(s) + \sum_{s \in S_2} f(s)$.

Proving the equality of the iterated summations and the full summation is tantamount to showing that $\{ (i,j) | 0 \le i \le m, \ 0 \le j \le n \} = \cup_{i=0}^m \{ (i,j) | \ 0 \le j \le n \}$.

• But then what is the definition of $\sum\limits_{0 \leq i\leq m , \text{ }0 \leq j\leq n}a_{ij}$? – Daniela Diaz Oct 9 '13 at 4:47
• It depends on how formal you want to be, the above suffices for me. I have added a few more comments above that might help. – copper.hat Oct 9 '13 at 5:03