description/proof of that for convergent sequence on metric space, subsequence converges to convergence of sequence
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.
- The reader knows a definition of subsequence of sequence.
Target Context
- The reader will have a description and a proof of the proposition that for any convergent sequence on any metric space, its any subsequence converges to the convergence of the 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 \(J \neq \emptyset\)
\(M\): \(\in \{\text{ the metric spaces }\}\)
\(s\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq M\)
\(J^`\): \(\subseteq \mathbb{N}\), such that \(J^` \neq \emptyset\)
\(s^`\): \(= s \circ f\), \(\in \{\text{ the subsequences of } s \text{ with } f: J^` \to J\}\)
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Statements:
\(\exists lim s\)
\(\implies\)
\(\exists lim s^` \land lim s^` = lim s\)
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2: Note
Compare with the proposition that for any sequence on any partially-ordered set and any subsequence, if the limit inferior of the sequence exists, the limit inferior of the subsequence does not necessarily exist, but if it exists, it is equal to or larger than the limit inferior of the sequence and the proposition that for any sequence on any partially-ordered set and any subsequence, if the limit superior of the sequence exists, the limit superior of the subsequence does not necessarily exist, but if it exists, it is equal to or smaller than the limit superior of the sequence.
3: Proof
Whole Strategy: Step 1: deal with the case that \(J\) is finite, and suppose otherwise thereafter; Step 2: take \(N\) such that for each \(N \lt n\), \(dist (lim s, s (J_n)) \lt \epsilon\), take \(N^`\) such that \(N \le f (N^`)\), and see that for each \(N^` \lt n\), \(dist (lim s, s^` (J^`_n)) \lt \epsilon\).
Step 1:
Let us suppose that \(\vert J \vert \in \mathbb{N} \setminus \{0\}\).
\(lim s = s (J_{\vert J \vert})\), by definition.
\(\vert J^` \vert \in \mathbb{N} \setminus \{0\}\), by Note for the definition of subsequence of sequence.
\(lim s^` = s^` (J^`_{\vert J^` \vert}) = s \circ f (J^`_{\vert J^` \vert}) = s (J_{\vert J \vert}) = lim s\).
Let us suppose otherwise, hereafter.
Step 2:
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is an \(N \in \mathbb{N} \setminus \{0\}\) such that for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), \(dist (lim s, s (J_n)) \lt \epsilon\).
There is an \(N^` \in \mathbb{N} \setminus \{0\}\) such that \(J_N \le f (J^`_{N^`})\).
For each \(n^` \in \mathbb{N} \setminus \{0\}\) such that \(N^` \lt n^`\), \(f (J^`_{N'}) \lt f (J^`_{n^`})\), so, \(J_N \le f (J^`_{N^`}) \lt f (J^`_{n^`})\), where \(f (J^`_{n^`}) = J_n\) such that \(N \lt n\), so, \(dist (lim s, s^` (J^`_{n^`})) = dist (lim s, s \circ f (J^`_{n^`})) = dist (lim s, s (J_n)) \lt \epsilon\).
So, \(lim s^` = lim s\).
