description/proof of that for \(2\) convergent sequences with same index set on \(1\)-dimensional Euclidean metric space, if for each index, there is equal or larger index s.t. 1st sequence element with 2nd index is equal to or smaller than 2nd sequence element with 1st index, convergence of 1st sequence is equal to or smaller than convergence of 2nd sequence
Topics
About: metric space
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any \(2\) convergent sequences with any same index set on the \(1\)-dimensional Euclidean metric space, if for each index, there is an equal or larger index such that the 1st sequence element with the 2nd index is equal to or smaller than the 2nd sequence element with the 1st index, the convergence of the 1st sequence is equal to or smaller than the convergence of the 2nd sequence.
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:
\(J\): \(\subseteq \mathbb{N}\), such that \(J \neq \emptyset\)
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\)
\(s_1\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq \mathbb{R}\)
\(s_2\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq \mathbb{R}\)
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Statements:
\(\exists lim s_1 \land \exists lim s_2 \land \forall j \in J (\exists j' \in J \text{ such that } j \le j' (s_1 (j') \le s_2 (j)))\)
\(\implies\)
\(lim s_1 \le lim s_2\)
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2: Note
Also the proposition that for any \(2\) convergent sequences with any same index set on the \(1\)-dimensional Euclidean metric space, if for each index, there is an equal or larger index such that the 1st sequence element with the 1st index is equal to or smaller than the 2nd sequence element with the 2nd index, the convergence of the 1st sequence is equal to or smaller than the convergence of the 2nd sequence holds.
3: Proof
Whole Strategy: Step 1: deal with the case that \(J\) is finite, and suppose otherwise thereafter; Step 2: for each \(\epsilon\), take some \(N\) and \(N'\) such that for each \(N \lt n\), \(\vert lim s_2 - s_2 (J_n) \vert \lt \epsilon / 2\) and for each \(N' \lt n'\), \(\vert lim s_1 - s_1 (J_{n'}) \vert \lt \epsilon / 2\), and take any \(N, N' \lt n\) and any \(J_n \le j'\) such that \(s_1 (j') \le s_2 (J_n)\), and see that \(lim s_1 - \epsilon \lt lim s_2\).
Step 1:
Let us suppose that \(\vert J \vert \in \mathbb{N} \setminus \{0\}\).
\(lim s_2 = s_2 (J_{\vert J \vert})\).
There is a \(j' \in J\) such that \(J_{\vert J \vert} \le j'\) and \(s_1 (j') \le s_2 (J_{\vert J \vert})\), by the supposition, but \(J_{\vert J \vert} \le j'\) means that \(j' = J_{\vert J \vert}\), so, \(s_1 (J_{\vert J \vert}) \le s_2 (J_{\vert J \vert})\).
But \(lim s_1 = s_1 (J_{\vert J \vert})\).
So, \(lim s_1 = s_1 (J_{\vert J \vert}) \le s_2 (J_{\vert J \vert}) = lim s_2\).
Let us suppose otherwise, hereafter.
Step 2:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is an \(N \in \mathbb{N}\) such that for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), \(\vert lim s_2 - s_2 (J_n) \vert \lt \epsilon / 2\), so, \(s_2 (J_n) - \epsilon / 2 \lt lim s_2 \lt s_2 (J_n) + \epsilon / 2\).
There is an \(N' \in \mathbb{N}\) such that for each \(n' \in \mathbb{N} \setminus \{0\}\) such that \(N' \lt n'\), \(\vert lim s_1 - s_1 (J_{n'}) \vert \lt \epsilon / 2\), so, \(s_1 (J_{n'}) - \epsilon / 2 \lt lim s_1 \lt s_1 (J_{n'}) + \epsilon / 2\).
Let \(n \in \mathbb{N} \setminus \{0\}\) be any such that \(N, N' \lt n\).
As especially \(N \lt n\), \(s_2 (J_n) - \epsilon / 2 \lt lim s_2 \lt s_2 (J_n) + \epsilon / 2\).
There is a \(j' \in J\) such that \(J_n \le j'\) and \(s_1 (j') \le s_2 (J_n)\), by the supposition.
While \(j' = J_{n'}\), \(J_n \le j' = J_{n'}\) means that \(n \le n'\), so, \(N, N' \lt n \le n'\).
As especially, \(N' \lt n'\), \(s_1 (J_{n'}) - \epsilon / 2 \lt lim s_1 \lt s_1 (J_{n'}) + \epsilon / 2\).
So, \(lim s_1 - \epsilon \lt s_1 (J_{n'}) - \epsilon / 2 \le s_2 (J_n) - \epsilon / 2 \lt lim s_2\).
So, \(lim s_1 \le lim s_2 + \epsilon\).
So, \(lim s_1 \le lim s_2\), 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.
