description/proof of that for \(2\) absolutely convergent series on \(1\)-dimensional Euclidean metric space, product of series is double series with terms as products of terms
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of convergence of series on metric space.
- The reader admits the proposition that for any convergent series on the \(1\)-dimensional Euclidean metric space and any real number, the series with the terms as the corresponding terms multiplied by the number converges with the convergence multiplied by the number.
- The reader admits the proposition that any real number is equal to or smaller than any another real number if it is equal to or smaller than the latter number plus any positive real number.
- The reader admits the proposition that for any absolutely convergent series on the \(1\)-dimensional Euclidean metric space and any partitions of the index set, the series of the series of the parts of the partitions converge to the same convergence.
- The reader admits the proposition that for any absolutely convergent finite-multiple series on the \(1\)-dimensional Euclidean metric space and any partitions of the domain, the series of the series of the parts of the partitions converge to the same convergence.
Target Context
- The reader will have a description and a proof of the proposition that for any \(2\) absolutely convergent series on the \(1\)-dimensional Euclidean metric space, the product of the series is the double series with the terms as the products of the terms.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(J_1\): \(\subseteq \mathbb{N}\)
\(J_2\): \(\subseteq \mathbb{N}\)
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\)
\(s_1\): \(: J_1 \to \mathbb{R}\), such that \(\sum_{j_1 \in J_1} \vert s_1 (j_1) \vert = r'_1 \in \mathbb{R}\)
\(s_2\): \(: J_2 \to \mathbb{R}\), such that \(\sum_{j_2 \in J_2} \vert s_2 (j_2) \vert = r'_2 \in \mathbb{R}\)
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Statements:
\((\sum_{j_1 \in J_1} s_1 (j_1)) (\sum_{j_2 \in J_2} s_2 (j_2)) = \sum_{j_1 \in J_1} \sum_{j_2 \in J_2} s_1 (j_1) s_2 (j_2)\)
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2: Proof
Whole Strategy: Step 1: see that \(\sum_{j_1 \in J_1} \sum_{j_2 \in J_2} \vert s_1 (j_1) \vert \vert s_2 (j_2) \vert \le r'_1 r'_2\); Step 2: see that \(r'_1 r'_2 \le \sum_{j_1 \in J_1} \sum_{j_2 \in J_2} \vert s_1 (j_1) \vert \vert s_2 (j_2) \vert\); Step 3: conclude that \(r'_1 r'_2 = \sum_{j_1 \in J_1} \sum_{j_2 \in J_2} \vert s_1 (j_1) \vert \vert s_2 (j_2) \vert\); Step 4: take the partition of \(J_1\), \(\{J_{1, +}, J_{1, -}\}\), and the partition of \(J_2\), \(\{J_{2, +}, J_{2, -}\}\), and see that \(\sum_{j_1 \in J_1} s_1 (j_1) = \sum_{j_1 \in J_{1, +}} s_1 (j_1) - \sum_{j_1 \in J_{1, -}} \vert s_1 (j_1) \vert\) and \(\sum_{j_2 \in J_2} s_2 (j_2) = \sum_{j_2 \in J_{2, +}} s_2 (j_2) - \sum_{j_2 \in J_{2, -}} \vert s_2 (j_2) \vert\); Step 5: see that \((\sum_{j_1 \in J_1} s_1 (j_1)) (\sum_{j_2 \in J_2} s_2 (j_2)) = \sum_{j_1 \in J_1} \sum_{j_2 \in J_2} s_1 (j_1) s_2 (j_2)\).
Step 0:
When \(r'_1 = 0\), each \(s_1 (j_1) = 0\), so, each \(s_1 (j_1) s_2 (j_2) = 0\), so, \((\sum_{j_1 \in J_1} s_1 (j_1)) (\sum_{j_2 \in J_2} s_2 (j_2)) = 0 = \sum_{j_1 \in J_1} \sum_{j_2 \in J_2} s_1 (j_1) s_2 (j_2)\).
When \(r'_2 = 0\), \((\sum_{j_1 \in J_1} s_1 (j_1)) (\sum_{j_2 \in J_2} s_2 (j_2)) = 0 = \sum_{j_1 \in J_1} \sum_{j_2 \in J_2} s_1 (j_1) s_2 (j_2)\), likewise.
Let us suppose that \(0 \lt r'_1, r'_2\), hereafter.
Step 1:
Let \({J_1}^` \subseteq J_1\) be any leading finite subset.
\(\sum_{j_1 \in {J_1}^`} \sum_{j_2 \in J_2} \vert s_1 (j_1) \vert \vert s_2 (j_2) \vert = \sum_{j_1 \in {J_1}^`} \vert s_1 (j_1) \vert \sum_{j_2 \in J_2} \vert s_2 (j_2) \vert\), by the proposition that for any convergent series on the \(1\)-dimensional Euclidean metric space and any real number, the series with the terms as the corresponding terms multiplied by the number converges with the convergence multiplied by the number, \(= (\sum_{j_1 \in {J_1}^`} \vert s_1 (j_1) \vert) \sum_{j_2 \in J_2} \vert s_2 (j_2) \vert\), because it is a finite addition, \(\le r'_1 r'_2\).
So, \(\sum_{j_1 \in J_1} \sum_{j_2 \in J_2} \vert s_1 (j_1) \vert \vert s_2 (j_2) \vert \le r'_1 r'_2\).
Step 2:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon \lt 3 r'_1 r'_2\).
When \(\vert J_1 \vert = n_1\), \(u_1 := r'_1 - \sum_{j_1 \in \{{J_1}_1, ..., {J_1}_{n_1}\}} \vert s_1 (j_1) \vert = 0\).
Otherwise, there is an \(N_1 \in \mathbb{N}\) such that for each \(n_1 \in \mathbb{N}\) such that \(N_1 \lt n_1\), \(u_{1, n_1} := r'_1 - \sum_{j_1 \in \{{J_1}_1, ..., {J_1}_{n_1}\}} \vert s_1 (j_1) \vert \lt \epsilon / (3 r'_2)\).
So, anyway, \(u_{1, n_1} \le \epsilon / (3 r'_1)\).
Likewise, \(u_{2, n_2} := r'_2 - \sum_{j_2 \in \{{J_2}_1, ..., {J_2}_{n_2}\}} \vert s_2 (j_2) \vert \le \epsilon / (3 r'_1)\).
\(r'_1 r'_2 = (\sum_{j_1 \in \{{J_1}_1, ..., {J_1}_{n_1}\}} \vert s_1 (j_1) \vert + u_{1, n_1}) (\sum_{j_2 \in \{{J_2}_1, ..., {J_2}_{n_2}\}} \vert s_2 (j_2) \vert + u_{2, n_2}) = (\sum_{j_1 \in \{{J_1}_1, ..., {J_1}_{n_1}\}} \vert s_1 (j_1) \vert) (\sum_{j_2 \in \{{J_2}_1, ..., {J_2}_{n_2}\}} \vert s_2 (j_2) \vert) + (\sum_{j_1 \in \{{J_1}_1, ..., {J_1}_{n_1}\}} \vert s_1 (j_1) \vert) u_{2, n_2} + u_{1, n_1} (\sum_{j_2 \in \{{J_2}_1, ..., {J_2}_{n_2}\}} \vert s_2 (j_2) \vert) + u_{1, n_1} u_{2, n_2} \le (\sum_{j_1 \in \{{J_1}_1, ..., {J_1}_{n_1}\}} \vert s_1 (j_1) \vert) (\sum_{j_2 \in \{{J_2}_1, ..., {J_2}_{n_2}\}} \vert s_2 (j_2) \vert) + (\sum_{j_1 \in \{{J_1}_1, ..., {J_1}_{n_1}\}} \vert s_1 (j_1) \vert) \epsilon / (3 r'_1) + \epsilon / (3 r'_2) (\sum_{j_2 \in \{{J_2}_1, ..., {J_2}_{n_2}\}} \vert s_2 (j_2) \vert) + u_{1, n_1} u_{2, n_2} \le (\sum_{j_1 \in \{{J_1}_1, ..., {J_1}_{n_1}\}} \vert s_1 (j_1) \vert) (\sum_{j_2 \in \{{J_2}_1, ..., {J_2}_{n_2}\}} \vert s_2 (j_2) \vert) + r'_1 \epsilon / (3 r'_1) + \epsilon / (3 r'_2) r'_2 + \epsilon / (3 r'_2) \epsilon / (3 r'_1) = (\sum_{j_1 \in \{{J_1}_1, ..., {J_1}_{n_1}\}} \vert s_1 (j_1) \vert) (\sum_{j_2 \in \{{J_2}_1, ..., {J_2}_{n_2}\}} \vert s_2 (j_2) \vert) + \epsilon / 3 + \epsilon / 3 + \epsilon / (3 r'_2) \epsilon / (3 r'_1)\).
But as \(\epsilon \lt 3 r'_1 r'_2\), \(\epsilon / (3 r'_2) \epsilon / (3 r'_1) \lt \epsilon / 3\).
So, \(\le (\sum_{j_1 \in \{{J_1}_1, ..., {J_1}_{n_1}\}} \vert s_1 (j_1) \vert) (\sum_{j_2 \in \{{J_2}_1, ..., {J_2}_{n_2}\}} \vert s_2 (j_2) \vert) + \epsilon / 3 + \epsilon / 3 + \epsilon / 3 = (\sum_{j_1 \in \{{J_1}_1, ..., {J_1}_{n_1}\}} \vert s_1 (j_1) \vert) (\sum_{j_2 \in \{{J_2}_1, ..., {J_2}_{n_2}\}} \vert s_2 (j_2) \vert) + \epsilon\).
\((\sum_{j_1 \in \{{J_1}_1, ..., {J_1}_{n_1}\}} \vert s_1 (j_1) \vert) (\sum_{j_2 \in \{{J_2}_1, ..., {J_2}_{n_2}\}} \vert s_2 (j_2) \vert) = \sum_{j_1 \in \{{J_1}_1, ..., {J_1}_{n_1}\}} (\vert s_1 (j_1) \vert \sum_{j_2 \in \{{J_2}_1, ..., {J_2}_{n_2}\}} \vert s_2 (j_2) \vert) = \sum_{j_1 \in \{{J_1}_1, ..., {J_1}_n\}} \sum_{j_2 \in \{{J_2}_1, ..., {J_2}_n\}} (\vert s_1 (j_1) \vert \vert s_2 (j_2) \vert)\), because they are some finite operations, \(\le \sum_{j_1 \in J_1} \sum_{j_2 \in J_2} (\vert s_1 (j_1) \vert \vert s_2 (j_2) \vert)\).
So, \(r'_1 r'_2 \le \sum_{j_1 J_1} \sum_{j_2 \in J_2} (\vert s_1 (j_1) \vert \vert s_2 (j_2) \vert) + \epsilon\) for each \(0 \lt \epsilon \lt 3 r'_1 r'_2\).
Then, \(r'_1 r'_2 \le \sum_{j_1 J_1} \sum_{j_2 \in J_2} (\vert s_1 (j_1) \vert \vert s_2 (j_2) \vert) + \epsilon\) for each \(3 r'_1 r'_2 \le \epsilon\) even more.
So, \(r'_1 r'_2 \le \sum_{j_1 J_1} \sum_{j_2 \in J_2} (\vert s_1 (j_1) \vert \vert s_2 (j_2) \vert) + \epsilon\) for each \(0 \lt \epsilon\).
So, \(r'_1 r'_2 \le \sum_{j_1 J_1} \sum_{j_2 \in J_2} (\vert s_1 (j_1) \vert \vert s_2 (j_2) \vert)\), by the proposition that any real number is equal to or smaller than any another real number if it is equal to or smaller than the latter number plus any positive real number.
Step 3:
So, \(r'_1 r'_2 = \sum_{j_1 \in J_1} \sum_{j_2 \in J_2} \vert s_1 (j_1) \vert \vert s_2 (j_2) \vert\).
Step 4:
Let us take the partition of \(J_1\), \(\{J_{1, +}, J_{1, -}\}\), such that for each \(j_1 \in J_{1, +}\), \(0 \le s_1 (j_1)\), and for each \(j_1 \in J_{1, -}\), \(s_1 (j_1) \lt 0\).
Let us take the partition of \(J_2\), \(\{J_{2, +}, J_{2, -}\}\), such that for each \(j_2 \in J_{2, +}\), \(0 \le s_2 (j_2)\), and for each \(j_2 \in J_{2, -}\), \(s_2 (j_2) \lt 0\).
\(\sum_{j_1 \in J_1} s_1 (j_1) = \sum_{j_1 \in J_{1, +}} s_1 (j_1) + \sum_{j_1 \in J_{1, -}} s_1 (j_1)\), by the proposition that for any absolutely convergent series on the \(1\)-dimensional Euclidean metric space and any partitions of the domain, the series of the series of the parts of the partitions converge to the same convergence, \(= \sum_{j_1 \in J_{1, +}} s_1 (j_1) + \sum_{j_1 \in J_{1, -}} - \vert s_1 (j_1) \vert = \sum_{j_1 \in J_{1, +}} s_1 (j_1) - \sum_{j_1 \in J_{1, -}} \vert s_1 (j_1) \vert\), by the proposition that for any convergent series on the \(1\)-dimensional Euclidean metric space and any real number, the series with the terms as the corresponding terms multiplied by the number converges with the convergence multiplied by the number.
\(\sum_{j_2 \in J_2} s_2 (j_2) = \sum_{j_2 \in J_{2, +}} s_2 (j_2) - \sum_{j_2 \in J_{2, -}} \vert s_2 (j_2) \vert\), likewise.
Step 5:
\((\sum_{j_1 \in J_1} s_1 (j_1)) (\sum_{j_2 \in J_2} s_2 (j_2)) = (\sum_{j_1 \in J_{1, +}} s_1 (j_1) - \sum_{j_1 \in J_{1, -}} \vert s_1 (j_1) \vert) (\sum_{j_2 \in J_{2, +}} s_2 (j_2) - \sum_{j_2 \in J_{2, -}} \vert s_2 (j_2) \vert) = (\sum_{j_1 \in J_{1, +}} s_1 (j_1)) (\sum_{j_2 \in J_{2, +}} s_2 (j_2)) - (\sum_{j_1 \in J_{1, +}} s_1 (j_1)) (\sum_{j_2 \in J_{2, -}} \vert s_2 (j_2) \vert) - (\sum_{j_1 \in J_{1, -}} \vert s_1 (j_1) \vert) (\sum_{j_2 \in J_{2, +}} s_2 (j_2)) + (\sum_{j_1 \in J_{1, -}} \vert s_1 (j_1) \vert) (\sum_{j_2 \in J_{2, -}} \vert s_2 (j_2) \vert)\).
\(= (\sum_{j_1 \in J_{1, +}} \sum_{j_2 \in J_{2, +}} s_1 (j_1) s_2 (j_2)) - (\sum_{j_1 \in J_{1, +}} \sum_{j_2 \in J_{2, -}} s_1 (j_1) \vert s_2 (j_2) \vert) - (\sum_{j_1 \in J_{1, -}} \sum_{j_2 \in J_{2, +}} \vert s_1 (j_1) \vert s_2 (j_2)) + (\sum_{j_1 \in J_{1, -}} \sum_{j_2 \in J_{2, -}} \vert s_1 (j_1) \vert \vert s_2 (j_2) \vert)\), by Step 3.
\(= (\sum_{j_1 \in J_{1, +}} \sum_{j_2 \in J_{2, +}} s_1 (j_1) s_2 (j_2)) + (\sum_{j_1 \in J_{1, +}} \sum_{j_2 \in J_{2, -}} - s_1 (j_1) \vert s_2 (j_2) \vert) + (\sum_{j_1 \in J_{1, -}} \sum_{j_2 \in J_{2, +}} - \vert s_1 (j_1) \vert s_2 (j_2)) + (\sum_{j_1 \in J_{1, -}} \sum_{j_2 \in J_{2, -}} \vert s_1 (j_1) \vert \vert s_2 (j_2) \vert)\), by the proposition that for any convergent series on the \(1\)-dimensional Euclidean metric space and any real number, the series with the terms as the corresponding terms multiplied by the number converges with the convergence multiplied by the number, \(= (\sum_{j_1 \in J_{1, +}} \sum_{j_2 \in J_{2, +}} s_1 (j_1) s_2 (j_2)) + (\sum_{j_1 \in J_{1, +}} \sum_{j_2 \in J_{2, -}} s_1 (j_1) s_2 (j_2)) + (\sum_{j_1 \in J_{1, -}} \sum_{j_2 \in J_{2, +}} s_1 (j_1) s_2 (j_2)) + (\sum_{j_1 \in J_{1, -}} \sum_{j_2 \in J_{2, -}} s_1 (j_1) s_2 (j_2))\).
\(= (\sum_{(j_1, j_2) \in J_{1, +} \times J_{2, +}} s_1 (j_1) s_2 (j_2)) + (\sum_{(j_1, j_2) \in J_{1, +} \times J_{2, -}} s_1 (j_1) s_2 (j_2)) + (\sum_{(j_1, j_2) \in J_{1, -} \times J_{2, +}} s_1 (j_1) s_2 (j_2)) + (\sum_{(j_1, j_2) \in J_{1, -} \times J_{2, -}} s_1 (j_1) s_2 (j_2))\), by the proposition that for any absolutely convergent finite-multiple series on the \(1\)-dimensional Euclidean metric space and any partitions of the domain, the series of the series of the parts of the partitions converge to the same convergence, because \(\{\{(j_1, j_2)\} \vert (j_1, j_2) \in J_{1, +} \times J_{2, +}\}\) is a partition of \(J_{1, +} \times J_{2, +}\) while \(\sum_{j_1 \in J_{1, +}} \sum_{j_2 \in J_{2, +}} \vert s_1 (j_1) s_2 (j_2) \vert \le \sum_{j_1 \in J_1} \sum_{j_2 \in J_2} \vert s_1 (j_1) s_2 (j_2) \vert = r'_1 r'_2\), and so on.
\(= \sum_{(j_1, j_2) \in J_1 \times J_2} s_1 (j_1) s_2 (j_2)\), by the proposition that for any absolutely convergent finite-multiple series on the \(1\)-dimensional Euclidean metric space and any partitions of the domain, the series of the series of the parts of the partitions converge to the same convergence, because \(\{J_{1, +} \times J_{2, +}, J_{1, +} \times J_{2, -}, J_{1, -} \times J_{2, +}, J_{1, -} \times J_{2, -}\}\) is a partition of \(J_1 \times J_2\).
\(= \sum_{j_1 \in J_1} \sum_{j_2 \in J_2} s_1 (j_1) s_2 (j_2)\), by the proposition that for any absolutely convergent finite-multiple series on the \(1\)-dimensional Euclidean metric space and any partitions of the domain, the series of the series of the parts of the partitions converge to the same convergence, because \(\{\{(j_1, j_2)\} \vert (j_1, j_2) \in J_1 \times J_2\}\) is a partition of \(J_1 \times J_2\).
So, \((\sum_{j_1 \in J_1} s_1 (j_1)) (\sum_{j_2 \in J_2} s_2 (j_2)) = \sum_{j_1 \in J_1} \sum_{j_2 \in J_2} s_1 (j_1) s_2 (j_2)\).
