description/proof of that for \(2\) convergent sequences with same domain on \(1\)-dimensional Euclidean metric space, sequence with elements as products of corresponding elements converges with product of convergences
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of convergence of sequence on metric space.
Target Context
- The reader will have a description and a proof of the proposition that for any \(2\) convergent sequences with any same domain on the \(1\)-dimensional Euclidean metric space, the sequence with the elements as the products of the corresponding elements converges with the product 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 \(lim s_1 = r_1 \in \mathbb{R}\)
\(s_2\): \(: J \to \mathbb{R}\), such that \(lim s_2 = r_2 \in \mathbb{R}\)
\(s_1 s_2\): \(: J \to \mathbb{R}, j \mapsto s_1 (j) s_2 (j)\)
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Statements:
\(lim s_1 s_2 = r_1 r_2 \in \mathbb{R}\)
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2: Proof
Whole Strategy: Step 1: deal with the case that \(J\) is finite, and suppose otherwise, thereafter; Step 2: take an \(0 \lt M_1\) such that \(\vert s_1 (n) \vert \le M_1\); Step 3: take \(u_n := r_2 - s_2 (n)\) and an \(N_2\) such that for each \(N_2 \lt n\), \(\vert u_n \vert \lt \epsilon / (2 M_1)\), and if \(r_2 \neq 0\), take an \(N_1\) such that for each \(N_1 \lt n\), \(\vert r_1 - s_1 (n) \vert \lt \epsilon / (2 \vert r_2 \vert)\); Step 4: see that for each \(N_1, N_2 \lt n\), \(\vert r_1 r_2 - s_1 s_2 (n) \vert \lt \epsilon\).
Step 1:
When \(\vert J \vert = n\), it holds, because \(lim s_1 s_2 = (s_1 s_2) (J_n) = s_1 (J_n) s_2 (J_n) = lim s_1 lim s_2 = r_1 r_2\).
Let us suppose otherwise, hereafter.
Step 2:
Let us see that there is an \(M_1 \in \mathbb{R}\) such that \(0 \lt M_1\) and for each \(j \in J\), \(\vert s_1 (j) \vert \le M_1\).
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is an \(N \in \mathbb{N} \setminus \{0\}\) such that for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), \(\vert r_1 - s_1 (J_n) \vert \lt \epsilon\).
So, for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), \(\vert s_1 (J_n) \vert = \vert s_1 (J_n) - r_1 + r_1 \vert \le \vert s_1 (J_n) - r_1 \vert + \vert r_1 \vert \lt \epsilon + \vert r_1 \vert\).
Let us take \(M_1 := Max (\{\vert s_1 (J_n) \vert \vert n \in \{1, ..., N\}\} \cup \{\epsilon + \vert r_1 \vert\})\).
Then, for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(n \le N\), \(\vert s_1 (J_n) \vert \le M_1\), and for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), \(\vert s_1 (J_n) \vert \lt \epsilon + \vert r_1 \vert \le M_1\).
So, for each \(n \in \mathbb{N} \setminus \{0\}\), \(\vert s_1 (J_n) \vert \le M_1\).
When \(M_1 = 0\), let us take any positive \(M_1\), which still satisfies \(\vert s_1 (J_n) \vert \le M_1\).
Step 3:
Let us take for each \(n \in \mathbb{N} \setminus \{0\}\), \(u_n := r_2 - s_2 (J_n)\).
There is an \(N_2 \in \mathbb{N} \setminus \{0\}\) such that for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N_2 \lt n\), \(\vert u_n \vert = \vert r_2 - s_2 (J_n) \vert \lt \epsilon / (2 M_1)\).
If \(r_2 = 0\), let us take \(N_1 = N_2\).
If \(r_2 \neq 0\), there is an \(N_1 \in \mathbb{N} \setminus \{0\}\) such that for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N_1 \lt n\), \(\vert r_1 - s_1 (n) \vert \lt \epsilon / (2 \vert r_2 \vert)\).
Step 4:
For each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N_1, N_2 \lt n\), \(\vert r_1 r_2 - s_1 s_2 (J_n) \vert = \vert r_1 r_2 - s_1 (J_n) s_2 (J_n) \vert = \vert r_1 r_2 - s_1 (J_n) (r_2 - u_n) \vert = \vert r_2 (r_1 - s_1 (J_n)) + s_1 (J_n) u_n \vert \le \vert r_2 (r_1 - s_1 (J_n)) \vert + \vert s_1 (J_n) u_n \vert = \vert r_2 \vert \vert r_1 - s_1 (J_n) \vert + \vert s_1 (J_n) \vert \vert u_n \vert \lt \vert r_2 \vert \vert r_1 - s_1 (J_n) \vert + M_1 \epsilon / (2 M_1) = \vert r_2 \vert \vert r_1 - s_1 (J_n) \vert + \epsilon / 2\).
If \(r_2 = 0\), \(= 0 \vert r_1 - s_1 (J_n) \vert + \epsilon / 2 = \epsilon / 2 \lt \epsilon\).
If \(r_2 \neq 0\), \(\lt \vert r_2 \vert \epsilon / (2 \vert r_2 \vert) + \epsilon / 2 = \epsilon / 2 + \epsilon / 2 = \epsilon\).
So, anyway, \(\vert r_1 r_2 - s_1 s_2 (J_n) \vert \lt \epsilon\).
That means that \(lim s_1 s_2 = r_1 r_2 \in \mathbb{R}\).
