description/proof of that for absolutely convergent finite-multiple 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 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.
- The reader admits the proposition that for any absolutely convergent series on the \(1\)-dimensional Euclidean metric space, the series with orders changed converge to the same convergence.
Target Context
- The reader will have a description and a proof of 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.
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:
\(\{J_1 \subseteq \mathbb{N}, ..., J_k \subseteq \mathbb{N}\}\):
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\)
\(s\): \(: J_1 \times ... \times J_k \to \mathbb{R}\), such that \(\sum_{j_1 \in J_1} ... \sum_{j_k \in J_k} \vert s (j_1, ..., j_k) \vert = r' \in \mathbb{R}\)
\(\{J'_l \subseteq J_1 \times ... \times J_k \vert l \in L\}\): \(\in \{\text{ the partitions of } J_1 \times ... \times J_k\}\)
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Statements:
\(\sum_{l \in L} \sum_{j \in J'_l} s (j) = \sum_{j_1 \in J_1} ... \sum_{j_k \in J_k} s (j_1, ..., j_k) \in \mathbb{R}\)
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\(J'_l\) s and \(L\) are inevitably countable, and \(J'_l\) s and \(L\) can be ordered in any way.
2: Proof
Whole Strategy: Step 1: take any bijection, \(f: J \subseteq \mathbb{N} \to J_1 \times ... \times J_k\), and \(s \circ f: J \to \mathbb{R}\); Step 2: see that \(\{f^{-1} (\{j_1\} \times ... \times \{j_{k - 1}\} \times J_k) \vert (j_1, ... j_{k - 1}) \in J_1 \times ... \times J_{k - 1}\}\) is a partition of \(J\); Step 3: see that \(\sum_{j \in J} s \circ f (j) = \sum_{j_1 \in J_1} ... \sum_{j_k \in J_k} s (j_1, ..., j_k)\); Step 4: see that \(\{f^{-1} (J'_l) \vert l \in L\}\) is a partition of \(J\); Step 5: see that \(\sum_{j \in J} s \circ f (j) = \sum_{l \in L} \sum_{j \in f^{-1} (J'_l)} s \circ f (j) = \sum_{l \in L} \sum_{j \in J'_l} s (j)\); Step 6: conclude the proposition.
Step 1:
Let us take any bijection, \(f: J \subseteq \mathbb{N} \to J_1 \times ... \times J_k\), which is possible, because \(J_1 \times ... \times J_k\) is countable as a finite product of countable sets, as is well known: for example, \(0 \mapsto ({J_1}_1, ..., {J_k}_1)\), \(1 \mapsto ({J_1}_2, {J_2}_1, ..., {J_k}_1)\), \(2 \mapsto ({J_1}_1, {J_2}_2, {J_3}_1, ..., {J_k}_1)\), ... (1st, cover the combination that \(j_1 + ... + j_k = k\), 2nd, cover the combinations that \(j_1 + ... + j_k = k + 1\), 3rd, cover the combinations that \(j_1 + ... + j_k = k + 2\), ..., and so on) will do.
Let us think of \(s \circ f: J \to \mathbb{R}\).
Step 2:
\(\{\{j_1\} \times ... \times \{j_{k - 1}\} \times J_k \vert (j_1, ... j_{k - 1}) \in J_1 \times ... \times J_{k - 1}\}\) is a partition of \(J_1 \times ... \times J_k\).
So, \(\{f^{-1} (\{j_1\} \times ... \times \{j_{k - 1}\} \times J_k) \vert (j_1, ... j_{k - 1}) \in J_1 \times ... \times J_{k - 1}\}\) is a partition of \(J\), because \(f\) is a bijection.
Step 3:
\(\sum_{j \in J} s \circ f (j) = \sum_{(j_1, ... j_{k - 1}) \in J_1 \times ... \times J_{k - 1}} \sum_{j \in f^{-1} (\{j_1\} \times ... \times \{j_{k - 1}\} \times J_k)} s \circ f (j)\), 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: \(\sum_{j \in J} \vert s \circ f (j) \vert \le r'\), because for each leading finite subset, \(J^` \subseteq J\), there are some leading finite subsets, \({J_1}^` \subseteq J_1, ..., {J_k}^` \subseteq J_k\), such that \(J^` \subseteq {J_1}^` \times ... \times {J_k}^`\), and \(\sum_{j \in J^`} \vert s \circ f (j) \vert \le \sum_{j_1 \in {J_1}^`} ... \sum_{j_k \in {J_k}^`} \vert s (j_1, ... j_k) \vert \le r'\).
But \(\sum_{j \in f^{-1} (\{j_1\} \times ... \times \{j_{k - 1}\} \times J_k)} s \circ f (j) = \sum_{j_k \in J_k} s (j_1, ..., j_{k - 1}, j_k)\), because \(f \vert_{f^{-1} (\{j_1\} \times ... \times \{j_{k - 1}\} \times J_k)}: f^{-1} (\{j_1\} \times ... \times \{j_{k - 1}\} \times J_k) \to \{j_1\} \times ... \times \{j_{k - 1}\} \times J_k\) is a bijection and \(s \circ f (j) = s (j_1, ..., j_{k - 1}, j_k)\), and the proposition that for any absolutely convergent series on the \(1\)-dimensional Euclidean metric space, the series with orders changed converge to the same convergence applies: \(\sum_{j_k \in J_k} \vert s (j_1, ..., j_{k - 1}, j_k) \vert\) converges.
So, \(\sum_{j \in J} s \circ f (j) = \sum_{(j_1, ... j_{k - 1}) \in J_1 \times ... \times J_{k - 1}} \sum_{j_k \in J_k} s (j_1, ..., j_{k - 1}, j_k)\).
\(= \sum_{(j_1, ... j_{k - 2}) \in J_1 \times ... \times J_{k - 2}} \sum_{j_{k - 1} \in J_{k - 1}} \sum_{j_k \in J_k} s (j_1, ..., j_{k - 2}, j_{k - 1}, j_k)\), because \(\{\{j_1\} \times ... \times \{j_{k - 2}\} \times J_{k - 1} \vert (j_1, ..., j_{k - 2}) \in J_1 \times ... \times J_{k - 2}\}\) is a partition of \(J_1 \times ... \times J_{k - 1}\) and 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 applies: \(\sum_{(j_1, ... j_{k - 1}) \in J_1 \times ... \times J_{k - 1}} \vert \sum_{j_k \in J_k} s (j_1, ..., j_{k - 1}, j_k) \vert \le \sum_{(j_1, ... j_{k - 1}) \in J_1 \times ... \times J_{k - 1}} \sum_{j_k \in J_k} \vert s (j_1, ..., j_{k - 1}, j_k) \vert \le r'\), because for each leading finite subset, \(J^` \subseteq J_1 \times ... \times J_{k - 1}\), there are some leading finite subsets, \({J_1}^` \subseteq J_1, ..., {J_{k - 1}}^` \subseteq J_{k - 1}\), such that \(J^` \subseteq {J_1}^` \times ... \times {J_{k - 1}}^`\), and \(\sum_{j \in J^`} \sum_{j_k \in J_k} \vert s (j, j_k) \vert \le \sum_{j_1 \in {J_1}^`} ... \sum_{j_{k - 1} \in {J_{k - 1}}^`} \sum_{j_k \in J_k} \vert s (j, j_k) \vert \le r'\).
And so on, after all, \(= \sum_{j_1 \in J_1} ... \sum_{j_k \in J_k} s (j_1, ..., j_k)\).
Step 4:
\(\{f^{-1} (J'_l) \vert l \in L\}\) is a partition of \(J\), because \(f\) is a bijection.
Step 5:
\(\sum_{j \in J} s \circ f (j) = \sum_{l \in L} \sum_{j \in f^{-1} (J'_l)} s \circ f (j)\), 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.
\(= \sum_{l \in L} \sum_{j \in J'_l} s (j)\), because \(f \vert_{f^{-1} (J'_l)}: f^{-1} (J'_l) \to J'_l\) is a bijection and the proposition that for any absolutely convergent series on the \(1\)-dimensional Euclidean metric space, the series with orders changed converge to the same convergence applies: \(\sum_{j \in f^{-1} (J'_l)} \vert s \circ f (j) \vert \le \sum_{l \in L} \sum_{j \in f^{-1} (J'_l)} \vert s \circ f (j) \vert = \sum_{j \in J} \vert s \circ f (j) \vert \le r'\).
Step 6:
By Step 3 and Step 5, \(\sum_{j_1 \in J_1} ... \sum_{j_k \in J_k} s (j_1, ..., j_k) = \sum_{j \in J} s \circ f (j) = \sum_{l \in L} \sum_{j \in J'_l} s (j)\).
