Product of Sums/Corollary

Corollary to Product of Sums
Let $\ds \sum_{i \mathop \in X} a_{i j}$ be absolutely convergent sequences for all $j \in Y$.

Then:
 * $\ds \prod_{j \mathop \in Y} \paren {\sum_{i \mathop \in X} a_{i j} } = \sum_{f \mathop: Y \mathop \to X} \paren {\prod_{j \mathop \in Y} a_{\map f j j} }$

where $f$ runs over all mappings from $Y$ to $X$.

Proof
We will prove the case $X = Y = \N$ to avoid the notational inconvenience of enumerating the elements of $Y$ as $j_1, j_2, j_3 \dots$.

The general case where $X, Y$ are arbitrary sets has the same proof, but with more indices and notational distractions.

Consider that by the main theorem:
 * $\ds \prod_{j \mathop = 1, 2} \paren {\sum_{i \mathop \in \N} a_{i j} } = \sum_{x, y \mathop \in \N} a_{x_1}a_{y_2}$

and continuing in this vein:
 * $\ds \prod_{j \mathop = 1, 2, 3} \paren {\sum_{i \mathop \in \N} a_{i j} } = \paren {\sum_{x, y \mathop \in \N} a_{x_1} a_{y_2} } \paren {\sum_{z \mathop \in \N} a_{z_3} } = \sum_{x, y, z \mathop \in \N} a_{x_1} a_{y_2} a_{z_3}$

For an inductive proof of this concept for finite $n$, we assume that for some $n \in \N$:
 * $\ds \prod_{j \mathop = 1}^n \paren {\sum_{i \mathop \in \N} a_{i j} } = \sum_{u, v, \ldots, x, y \mathop \in \N} a_{u_1} a_{v_2} \dotsb a_{x_{\paren {n - 1} } } a_{y_n}$

Then:
 * $\ds \prod_{j \mathop = 1}^{n + 1} \paren {\sum_{i \mathop \in \N} a_{i j} } = \paren {\sum_{u, v, \dots, x, y \mathop \in \N} a_{u_1} a_{v_2} \dotsb a_{x_{\paren {n - 1} } } a_{y_n} } \paren {\sum_{z \mathop \in \N} a_{z_n} }$

which by Product of Sums is simply:
 * $\ds \sum_{u, v, \ldots, x, y, z \mathop \in \N} a_{u_1} a_{v_2} \ldots a_{x_{\paren {n - 1} } a_{y_n} a_{z_{\paren {n + 1} } } }$

completing the induction for finite $n$.