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This seems to be really basic and might be a case of begging the question here, but I can't seem to prove the that the right hand side implies the left hand side. Let $J$ be the set of all $j$ and $K(j)$ be the set of all $k$ that are valid for each $j\in J$. $K, J(k)$ are defined likewise.

So suppose I was working out a summation and I want to sum first on $j$ instead of $k$. Then by working out the set $K$ of valid $k$ from the LHS I arrive at the RHS. I also work out the valid $K(j)$ from the conditions of the LHS. So I can say that $J, K(j)\Rightarrow K,J(k)$.

Intuitively, this suggests the inverse is clearly true, but can it be proven rigorously?

$\sum_{j\in J}\sum_{k\in K(j)}f(j,k) = \sum_{k\in K}\sum_{j\in J(k)}f(j,k)$

Jia Cheng Sun
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  • I guess it is sufficient to show whatever is at the LHS will appear in the right summation, and whatever (j,k) that fails the left conditions will fail the right conditions, but this doesn't seem clear enough – Jia Cheng Sun Jan 25 '20 at 11:44

1 Answers1

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All you have to prove is $\left\{(j,k)\,|\, j\in J\text{ and }k\in K(j)\right\} = \left\{(j,k)\,|\, k\in K\text{ and }j\in J(k)\right\}$.

Nicolas FRANCOIS
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  • I realised that the nature of constructing the sets of K and J(k) proves the bijective relationship between the two sides. Because every valid (j,k) in J and K(j) has to belong to some K and J(k), and the opposite goes for invalid(j,k). So I was begging the question. – Jia Cheng Sun Jan 25 '20 at 13:54