I'm trying to come up with a rigorous way to define sums using sigma notation. Obviously$^{1}$, the sum $\sum \limits_{i=1}^{n}a_{i}$ needs to be defined on a set $X$ on which we have previously defined an addition $+$, and it needs to be defined recursively for all $n\in\mathbb{N}$ as
\begin{align*}
\sum\limits_{i=1}^{0}a_{i}&=0,
\\
\sum\limits_{i=1}^{n+1}a_{i}&=\left(\sum\limits_{i=1}^{n}a_{i}\right)+a_{n+1}.
\end{align*}
What I'm struggling with are the details… What's the proper way to define the terms $a_{i}$? First, I was thinking about stating "Let $I\subseteq \mathbb{N}$ be a set and let $(a_{i})_{i\in I}$ be a sequence in $X$,'' but then kinda dismissed this idea because we want to define $\sum \limits_{i=1}^{n}a_{i}$ for all $n\in\mathbb{N}$, not just for a subset of $\mathbb{N}$. I'm also having concerns about dropping the requirement that $I\subseteq \mathbb{N}$ and using $(a_{i})_{i\in\mathbb{N}}$ instead. Then I considered to state "Let $\{\text{$a_{i}\in A\mid a_{1}\leq a_{i}\leq a_{n}$ for $n\in\mathbb{N}$}\}$'' but have similar concerns with this formulation. Also some text state that there are issues with infinite sums, guess it has to do with the Axiom of Choice, so I started questioning if we can actually define $\sum \limits_{i=1}^{n}a_{i}$ for all $n\in\mathbb{N}$ at all.
$^{1}$ I know, we should never use terms like "obviously" when trying to do rigorous math…
Best Answer
I think the cartesian product of sets is most convenient here. If you have $+: X \times X \to X$ defined (denoted $(a_1, a_2) \mapsto a_1 + a_2$), then we would define $\Sigma^n: X^n \to X$ to be $(x_1, \dots, x_n) \mapsto x_1 + (x_2 + (\cdots + (x_{n - 1} + x_n))$. Then if we have an $n$-tuple $(a_1, \dots, a_n) \in X^n$ we can define $$\sum_{i = 1}^n a_i := \Sigma^n(a_1, \dots, a_n).$$ The $n$ was arbitrary so this construction works for all $n \in \mathbb{N}.$ Notice that we needed very few assumptions on $+$ so this can be applied very generally.