description/proof of that for absolutely convergent series on \(1\)-dimensional Euclidean metric space and partitions of domain, series of series of parts converge to same convergence
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 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 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 for any \(2\) convergent series with any same domain on the \(1\)-dimensional Euclidean metric space, the series with the terms as the sums of the corresponding terms converges with the sum of the convergences.
Target Context
- The reader will have a description and a proof of 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.
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:
\(J\): \(\subseteq \mathbb{N}\)
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\)
\(s\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq \mathbb{R}\) and \(\sum_{j \in J} \vert s (j) \vert = r' \in \mathbb{R}\)
\(\{J_l \subseteq J \vert l \in L\}\): \(\in \{\text{ the partitions of } J\}\)
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Statements:
\(\sum_{l \in L} \sum_{j \in J_l} s (j) = \sum_{j \in J} s (j) \in \mathbb{R}\)
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\(J_l\) s and \(L\) are inevitably countable, and \(L\) can be ordered in any way.
2: Proof
Whole Strategy: Step 1: see that \(\sum_{j \in J_l} \vert s (j) \vert \le r'\); Step 2: see that \(\sum_{l \in L} \sum_{j \in J_l} \vert s (j) \vert \le r'\); Step 3: see that \(r' \le \sum_{l \in L} \sum_{j \in J_l} \vert s (j) \vert\); Step 4: conclude that \(r' = \sum_{l \in L} \sum_{j \in J_l} \vert s (j) \vert\); Step 5: see that \(\sum_{j \in J} s (j) = \sum_{j \in J} 1 / 2 (\vert s (j) \vert + s (j)) - \sum_{j \in J} 1 / 2 (\vert s (j) \vert - s (j)) = \sum_{l \in L} \sum_{j \in J_l} 1 / 2 (\vert s (j) \vert + s (j)) - \sum_{l \in L} \sum_{j \in J_l} 1 / 2 (\vert s (j) \vert - s (j)) = \sum_{l \in L} \sum_{j \in J_l} s (j)\).
Step 1:
Let \({J_l}^` \subseteq J_l\) be any leading finite subset.
There is a leading finite subset of \(J^` \subseteq J\) such that \({J_l}^` \subseteq J^`\).
So, \(\sum_{j \in {J_l}^`} \vert s (j) \vert \le \sum_{j \in J^`} \vert s (j) \vert \le r'\).
So, \(\sum_{j \in J_l} \vert s (j) \vert = r_l \le r'\).
Step 2:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
Let \(L^` \subseteq L\) be any leading finite subset.
For each \(l \in L^`\), when \(\vert J_l \vert = n_l \in \mathbb{R}\), let \(u_{l, n_l} := r_l - \sum_{j \in \{{J_l}_1, ..., {J_l}_{n_l}\}} \vert s (j) \vert = 0\); otherwise, there is an \(N_l \in \mathbb{N}\) such that for each \(n_l \in \mathbb{N}\) such that \(N_l \lt n_l\), \(\vert r_l - \sum_{j \in \{{J_l}_1, ..., {J_l}_{n_l}\}} \vert s (j) \vert \vert \lt \epsilon / \vert L^` \vert\), and let \(u_{l, n_l} := r_l - \sum_{j \in \{{J_l}_1, ..., {J_l}_{n_l}\}} \vert s (j) \vert\), so, anyway, \(\vert u_{l, n_l} \vert \le \epsilon / \vert L^` \vert\).
\(\sum_{l \in L^`} \sum_{j \in J_l} \vert s (j) \vert = \sum_{l \in L^`} r_l = \sum_{l \in L^`} (\sum_{j \in \{{J_l}_1, ..., {J_l}_{n_l}\}} \vert s (j) \vert + u_{l, n_l}) \le \sum_{l \in L^`} (\sum_{j \in \{{J_l}_1, ..., {J_l}_{n_l}\}} \vert s (j) \vert + \epsilon / \vert L^` \vert) = \sum_{l \in L^`} \sum_{j \in \{{J_l}_1, ..., {J_l}_{n_l}\}} \vert s (j) \vert + \sum_{l \in L^`} \epsilon / \vert L^` \vert = \sum_{l \in L^`} \sum_{j \in \{{J_l}_1, ..., {J_l}_{n_l}\}} \vert s (j) \vert + \epsilon\).
There is a leading finite subset, \(J^` \subseteq J\), such that \(\cup_{l \in L^`} \{{J_l}_1, ..., {J_l}_{n_l}\} \subseteq J^`\).
So, \(\sum_{l \in L^`} \sum_{j \in \{{J_l}_1, ..., {J_l}_{n_l}\}} \vert s (j) \vert \le \sum_{j \in J^`} \vert s (j) \vert \le r'\).
So, \(\sum_{l \in L^`} \sum_{j \in J_l} \vert s (j) \vert \le r' + \epsilon\).
So, \(\sum_{l \in L^`} \sum_{j \in J_l} \vert s (j) \vert \le r'\), 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.
So, \(\sum_{l \in L} \sum_{j \in J_l} \vert s (j) \vert \le r'\).
Step 3:
Let \(J^` \subseteq J\) be any leading finite subset.
\(J^`\) is contained in the union of some finite \(J_l\) s and each \(J^` \cap J_l \subseteq J_l\) is a finite subset, so, there is a leading finite subset, \(L^` \subseteq L\), and for each \(l \in L^`\), there is a leading finite subset, \({J_l}^` \subseteq J_l\), such that \(J^` \subseteq \cup_{l \in L^`} {J_l}^`\).
So, \(\sum_{j \in J^`} \vert s (j) \vert \le \sum_{l \in L^`} \sum_{j \in {J_l}^`} \vert s (j) \vert \le \sum_{l \in L} \sum_{j \in J_l} \vert s (j) \vert\).
So, \(r' = \sum_{j \in J} \vert s (j) \vert \le \sum_{l \in L} \sum_{j \in J_l} \vert s (j) \vert\).
Step 4:
So, \(r' = \sum_{l \in L} \sum_{j \in J_l} \vert s (j) \vert\).
Step 5:
Let us think of \(\sum_{j \in J} 1 / 2 (\vert s (j) \vert + s_j)\).
For each \(j \in J\), \(0 \le 1 / 2 (\vert s (j) \vert + s (j)) \le \vert s (j) \vert\).
So, \(\sum_{j \in J} 1 / 2 (\vert s (j) \vert + s (j)) = r_1 \in \mathbb{R}\).
\(\sum_{j \in J} 1 / 2 (\vert s (j) \vert + s (j)) = \sum_{l \in L} \sum_{j \in J_l} 1 / 2 (\vert s (j) \vert + s (j))\), by Step 4.
Let us think of \(\sum_{j \in J} 1 / 2 (\vert s (j) \vert - s (j))\).
For each \(j \in J\), \(0 \le 1 / 2 (\vert s (j) \vert - s (j)) \le \vert s (j) \vert\).
So, \(\sum_{j \in J} 1 / 2 (\vert s (j) \vert - s (j)) = r_2 \in \mathbb{R}\).
\(\sum_{j \in J} 1 / 2 (\vert s (j) \vert - s (j)) = \sum_{l \in L} \sum_{j \in J_l} 1 / 2 (\vert s (j) \vert - s (j))\), by Step 4.
\(r_1 - r_2 = \sum_{j \in J} 1 / 2 (\vert s (j) \vert + s (j)) - \sum_{j \in J} 1 / 2 (\vert s (j) \vert - s (j)) = \sum_{j \in J} (1 / 2 (\vert s (j) \vert + s (j)) - 1 / 2 (\vert s (j) \vert - s (j)))\), 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 and the proposition that for any \(2\) convergent series with any same domain on the \(1\)-dimensional Euclidean metric space, the series with the terms as the sums of the corresponding terms converges with the sum of the convergences, \(= \sum_{j \in J} s (j)\).
\(r_1 - r_2 = \sum_{l \in L} \sum_{j \in J_l} 1 / 2 (\vert s (j) \vert + s (j)) - \sum_{l \in L} \sum_{j \in J_l} 1 / 2 (\vert s (j) \vert - s (j)) = \sum_{l \in L} (\sum_{j \in J_l} 1 / 2 (\vert s (j) \vert + s (j)) - \sum_{j \in J_l} 1 / 2 (\vert s (j) \vert - s (j))) = \sum_{l \in L} (\sum_{j \in J_l} (1 / 2 (\vert s (j) \vert + s (j)) - 1 / 2 (\vert s (j) \vert - s (j)))\), 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 and the proposition that for any \(2\) convergent series with any same domain on the \(1\)-dimensional Euclidean metric space, the series with the terms as the sums of the corresponding terms converges with the sum of the convergences, \(= \sum_{l \in L} \sum_{j \in J_l} s (j)\).
So, \(\sum_{j \in J} s_j = \sum_{l \in L} \sum_{j \in J_l} s_j\).
