In the book Concrete Mathematics (2nd) written by Ronald Graham, Donald Knuth and Oren Patashnik, they prove the next theorem.

Absolutely convergent sums over two or more indices can always be summed first with respect to any one of those indices.

I understood some parts of the proof, but I couldn't understand the crux idea beneath the proof. I quote their proof from the book page 61-62. (Because I have no background of real analysis, my questions might seem to be silly. I hope that you understand this.)

Formally, we shall prove that if $J$ and the elements of $\{K_j|j\in J\}$ are any sets of indices such that $$ \sum_{j\in J, k\in K_{j}}a_{j,k}$$ converges absolutely to A, then there exist complex number $A_{j}$ for each $j \in J$ such that $ \sum_{k\in K_j}a_{j,k}$ converges absolutely to $A_{j}$, and $\sum_{j\in J}A_j$ converges absolutely $A$.

Question 1. Why they prove this proposition other than showing two summation will be equal if the series absolutely converges? In this proof they do not show two summation will be same in the end. They only show that $ \sum_{k\in K_j}a_{j,k}$ converges absolutely to $A_{j}$, and $\sum_{j\in J}A_j$ converges absolutely $A$.

If suffices to prove this assertion all terms are nonnegative, because we can prove the general case by breaking everything into real and imaginary, positive and negative parts as before. Let's assume therefore that $a_{j,k}\ge 0$ for all pairs $(j,k)\in M$ where $M$ is the master index set $\{(j,k)|j\in J, k \in K_{j}\} $.

We are given that $\sum_{(j,k)\in M}a_{j,k}$ is finite , namely that $$\sum_{(j,k)\in F}a_{j,k}\le A$$

for all finite subsets $F\subseteq M$, and that $A$ is the least such upper bound. If $j$ is any element of $J$, each sum of the form $\sum_{k\in F_{j}}a_{j,k}$ where $F_{j}$ is a finite subset of $K_{j}$ is bounded above by A. Hence these finite sums have a least upper bound $A_{j}\ge 0$, and $\sum_{k\in K_{j}}a_{j,k}=A_{j}$ by definition.

We still need to prove that A is the least upper bound of $\sum_{j\in G}A_{j}$, for all finite subsets $G\subseteq J$.

Question 2. Why we need still to prove that A is the least upper bound?

Suppose that G is a finite subset of $J$ with $\sum_{j\in G}A_{j}=A' \gt A$. We can find finite subsets $F_{j}\subseteq K_{j}$ such that $\sum_{k\in F_{j}}a_{j,k} \gt (A/A')A_{j}$ for each $j\in G$ with $A_{j}\gt 0$ There is at least one such $j$.

Question 3. What's the meaning of $(A/A')A_{j}$? Why they use this amount in the proof? How we know that there is at least one such $j$?

But then $\sum_{j\in G, k \in F_{j}}a_{j,k}>(A/A')\sum_{j\in G}A_{j}=A$ contradicting the fact that we have $\sum_{(j,k)\in F}a_{j,k} \le A$ for all infinite subsets $F\subseteq M$. Hence $\sum_{j\in G}A_{j}\le A$, for all finite subsets $G\subseteq J$.

Finally, let $A'$ be any real number less than $A$. Our proof will be complete if we can find a finite set $G\subseteq J$ such that $\sum_{j\in G}A_{j}>A'$. We know that there's a finite set $F\subseteq M$ such that $\sum_{(j,k) \in F}a_{j,k}$. Then $\sum_{j\in G}A_{j}\ge \sum_{j\in G}\sum_{k\in F_{j}}a_{j,k}=\sum_{(j,k)\in F}a_{j,k}>A'$; QED

Question 4. Why we need to let $A'$ be any real number less than $A$ to complete the proof?


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