description/proof of that for \(2\) convergent series with same domain on \(1\)-dimensional Euclidean metric space, series with terms as sums of corresponding terms converges with sum of convergences
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.
Target Context
- The reader will have a description and a proof of 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.
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:
\(J\): \(\subseteq \mathbb{N}\)
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\)
\(s_1\): \(: J \to \mathbb{R}\), such that \(\sum_{j \in J} s_1 (j) = r_1 \in \mathbb{R}\)
\(s_2\): \(: J \to \mathbb{R}\), such that \(\sum_{j \in J} s_2 (j) = r_2 \in \mathbb{R}\)
\(s_1 + s_2\): \(: J \to \mathbb{R}, j \mapsto s_1 (j) + s_2 (j)\)
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Statements:
\(\sum_{j \in J} (s_1 + s_2) (j) = r_1 + r_2 \in \mathbb{R}\)
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2: Proof
Whole Strategy: Step 1: deal with the case that \(J\) is finite, and suppose otherwise, thereafter; Step 2: for each \(\epsilon\), take an \(N_1\) such that for each \(N_1 \lt n\), \(\vert r_1 - \sum_{j \in \{J_1, ..., J_n\}} s_1 (j) \vert \lt \epsilon / 2\) and an \(N_2\) such that for each \(N_2 \lt n\), \(\vert r_2 - \sum_{j \in \{J_1, ..., J_n\}\}} s_2 (j) \vert \lt \epsilon / 2\); Step 3: see that for each \(N_1, N_2 \lt n\), \(\vert r_1 + r_2 - \sum_{j \in \{J_1, ..., J_n\}} (s_1 + s_2) (j) \vert \lt \epsilon\).
Step 1:
When \(J\) is finite, it holds, because they are some finite operations.
Let us suppose otherwise, hereafter.
Step 2:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is an \(N_1 \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_1 \lt n\), \(\vert r_1 - \sum_{j \in \{J_1, ..., J_n\}} s_1 (j) \vert \lt \epsilon / 2\), by the definition of convergence.
There is an \(N_2 \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N_2 \lt n\), \(\vert r_2 - \sum_{j \in \{J_1, ..., J_n\}} s_2 (j) \vert \lt \epsilon / 2\), by the definition of convergence.
Step 3:
For each \(n \in \mathbb{N}\) such that \(N_1, N_2 \lt n\), \(\vert r_1 + r_2 - \sum_{j \in \{J_1, ..., J_n\}} (s_1 + s_2) (j) \vert = \vert r_1 - \sum_{j \in \{J_1, ..., J_n\}} s_1 (j) + r_2 - \sum_{j \in \{J_1, ..., J_n\}} s_2 (j) \vert \le \vert r_1 - \sum_{j \in \{J_1, ..., J_n\}} s_1 (j) \vert + \vert r_2 - \sum_{j \in \{J_1, ..., J_n\}} s_2 (j) \vert \lt \epsilon / 2 + \epsilon / 2 = \epsilon\).
That means that \(\sum_{j \in J} (s_1 + s_2) (j) = r_1 + r_2 \in \mathbb{R}\).
