description/proof of that for absolutely convergent series on \(1\)-dimensional Euclidean metric space, series with orders changed 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 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, the series with orders changed 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:
\(J\): \(\subseteq \mathbb{N}\)
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\)
\(s\): \(: J \to \mathbb{R}\), such that \(\sum_{j \in J} \vert s (j) \vert = r' \in \mathbb{R}\)
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Statements:
\(\exists r \in \mathbb{R} (\forall f: J \to J \in \{\text{ the bijections }\} (\sum_{j \in J} s \circ f (j) = r))\)
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2: Proof
Whole Strategy: Step 1: see that \(\sum_{j \in \{J_1, ..., J_n\}} \vert s \circ f (j) \vert \le r'\) and \(\sum_{j \in J} \vert s \circ f (j) = r''\); Step 2: see that \(\sum_{j \in \{J_1, ..., J_n\}} \vert s (j) \vert \le r''\) and \(r' \le r''\); Step 3: conclude that \(\sum_{j \in J} \vert s \circ f (j) \vert = r'\); Step 4: see that \(\sum_{j \in J} s \circ f (j) = \sum_{j \in J} 1 / 2 (\vert s \circ f (j) \vert + s \circ f (j)) - \sum_{j \in J} 1 / 2 (\vert s \circ f (j) \vert - s \circ f (j)) = \sum_{j \in J} 1 / 2 (\vert s (j) \vert + s (j)) - \sum_{j \in J} 1 / 2 (\vert s (j) \vert - s (j)) = \sum_{j \in J} s (j) = r\).
Step 1:
For each leading finite subset of \(J\), \(J^` = \{J_1, ..., J_n\}\), let us take \(J_{n'} := Max (\{f (J_1), ..., f (J_n)\})\), then, \(\{f (J_1), ..., f (J_n)\} \subseteq \{J_1, ..., J_{n'}\}\), so, \(\{s \circ f (J_1), ..., s \circ f (J_n)\} \subseteq \{s (J_1), ..., s (J_{n'})\}\), so, \(\sum_{j \in \{J_1, ..., J_n\}} \vert s \circ f (j) \vert \le \sum_{j \in \{J_1, ..., J_{n'}\}} \vert s (j) \vert \le r'\).
So, \(\sum_{j \in J} \vert s \circ f (j) \vert = r'' \le r'\).
Step 2:
As \(f\) is bijective, there is the inverse, \(f^{-1}: J \to J\).
For each leading finite subset of \(J\), \(J^` = \{J_1, ..., J_n\}\), let us take \(J_{n'} := Max (\{f^{-1} (J_1), ..., f^{-1} (J_n)\})\), then, \(\{J_1, ..., J_n\} \subseteq \{f (J_1), ..., f (J_{n'})\}\), because for each \(j \in \{J_1, ..., J_n\}\), \(f^{-1} (j) \le J_{n'}\), so, \(f (f^{-1} (j))\) appears in \(\{f (J_1), ..., f (J_{n'})\}\) as \(f (j')\), so, \(j = f (j') \in \{f (J_1), ..., f (J_{n'})\}\), so, \(\{s (J_1), ..., s (J_n)\} \subseteq \{s \circ f (J_1), ..., s \circ f (J_{n'})\}\), so, \(\sum_{j \in \{J_1, ..., J_n\}} \vert s (j) \vert \le \sum_{j \in \{J_1, ..., J_{n'}\}} \vert s \circ f (j) \vert \le r''\).
So, \(r' \le r''\).
Step 3:
So, \(\sum_{j \in J} \vert s \circ f (j) \vert = r'' = r'\).
Step 4:
Let us think of \(\sum_{j \in J} 1 / 2 (\vert s \circ f (j) \vert + s \circ f (j))\).
For each \(j \in J\), \(0 \le 1 / 2 (\vert s \circ f (j) \vert + s \circ f (j)) \le \vert s \circ f (j) \vert\).
So, \(\sum_{j \in J} 1 / 2 (\vert s \circ f (j) \vert + s \circ f (j)) = \sum_{j \in J} (1 / 2 (\vert s \vert + s)) \circ f (j) = \sum_{j \in J} (1 / 2 (\vert s \vert + s)) (j) = r_1 \in \mathbb{R}\), independent of \(f\), by Step 3.
Let us think of \(\sum_{j \in J} 1 / 2 (\vert s \circ f (j) \vert - s \circ f (j))\).
For each \(j \in J\), \(0 \le 1 / 2 (\vert s \circ f (j) \vert - s \circ f (j)) \le \vert s \circ f (j) \vert\).
So, \(\sum_{j \in J} 1 / 2 (\vert s \circ f (j) \vert - s \circ f (j)) = \sum_{j \in J} (1 / 2 (\vert s \vert - s)) \circ f (j) = \sum_{j \in J} (1 / 2 (\vert s \vert - s)) (j) = r_2 \in \mathbb{R}\), independent of \(f\), by Step 3.
\(\sum_{j \in J} (1 / 2 (\vert s \circ f (j) \vert + s \circ f (j)) - 1 / 2 (\vert s \circ f (j) \vert - s \circ f (j))) = r_1 - r_2 := r\), 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.
\(= \sum_{j \in J} s \circ f (j) = r\), independent of \(f\).
\(\sum_{j \in J} s (j) = r\) is the special case that \(f = id\).
