description/proof of that for metric space, iff each sequence on it has convergent subsequence, each sequence on it from natural numbers set has convergent subsequence
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 metric space, if and only if each sequence on it has a convergent subsequence, each sequence on it from the natural numbers set has a convergent subsequence.
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:
\(M\): \(\in \{\text{ the metric spaces }\}\)
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Statements:
\(\forall s: J \to M \in \{\text{ the sequences }\} (\exists s^`: J^` \to M \in \{\text{ the subsequences of } s\} (s^` \in \{\text{ the convergent sequences }\}))\)
\(\iff\)
\(\forall s: \mathbb{N} \to M \in \{\text{ the sequences }\} (\exists s^`: J^` \to M \in \{\text{ the subsequences of } s\} (s^` \in \{\text{ the convergent sequences }\}))\)
\(\iff\)
\(\forall s: \mathbb{N} \to M \in \{\text{ the sequences }\} (\exists s^`: \mathbb{N} \to M \in \{\text{ the subsequences of } s\} (s^` \in \{\text{ the convergent sequences }\}))\)
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2: Proof
Whole Strategy: Step 0: let \(\forall s: J \to M \in \{\text{ the sequences }\} (\exists s^`: J^` \to M \in \{\text{ the subsequences of } s\} (s^` \in \{\text{ the convergent sequences }\}))\) be called "statement A", let \(\forall s: \mathbb{N} \to M \in \{\text{ the sequences }\} (\exists s^`: J^` \to M \in \{\text{ the subsequences of } s\} (s^` \in \{\text{ the convergent sequences }\}))\) be called "statement B", and let \(\forall s: \mathbb{N} \to M \in \{\text{ the sequences }\} (\exists s^`: \mathbb{N} \to M \in \{\text{ the subsequences of } s\} (s^` \in \{\text{ the convergent sequences }\}))\) be callled "statement C"; Step 1: suppose the statement A; Step 2: see that the statement B holds; Step 3: suppose the statement B; Step 4: see that the statement C holds; Step 5: suppose that the statement C; Step 6: see that the statement A holds.
Step 0:
Let \(\forall s: J \to M \in \{\text{ the sequences }\} (\exists s^`: J^` \to M \in \{\text{ the subsequences of } s\} (s^` \in \{\text{ the convergent sequences }\}))\) be called "statement A".
Let \(\forall s: \mathbb{N} \to M \in \{\text{ the sequences }\} (\exists s^`: J^` \to M \in \{\text{ the subsequences of } s\} (s^` \in \{\text{ the convergent sequences }\}))\) be called "statement B".
Let \(\forall s: \mathbb{N} \to M \in \{\text{ the sequences }\} (\exists s^`: \mathbb{N} \to M \in \{\text{ the subsequences of } s\} (s^` \in \{\text{ the convergent sequences }\}))\) be callled "statement C".
Step 1:
Let us suppose the statement A.
Step 2:
Each \(s: \mathbb{N} \to M\) is an \(s: J \to M\), so, there is a convergent \(s^`: J^` \to M\), by the statement A, so, the statement B holds.
Step 3:
Let us suppose the statement B.
Step 4:
Let \(s: \mathbb{N} \to M\) be any.
By the statement B, there is a convergent \(s^`: J^` \to M\), which means that \(s^` = s \circ f\).
Note that \(J^`\) is inevitably an infinite set, by Note for the definition of subsequence of sequence.
There is the order-preserving bijection, \(g: J^` \to \mathbb{N}, {J^`}_n \mapsto n - 1\), which is indeed order-preserving, because for each \({J^`}_n \lt {J^`}_{n'}\), \(n \lt n'\), so, \(n - 1 \lt n' - 1\); that is indeed a bijection, because for each \({J^`}_n \neq {J^`}_{n'}\), \(n \neq n'\), so, \(n - 1 \neq n' - 1\), and for each \(n \in \mathbb{N}\), \({J^`}_{n + 1}\) is mapped to \(n\).
Note that also \(g^{-1}\) is order-preserving, because for each \(n \lt n'\), \(g^{-1} (n) = {J^`}_{n + 1} \lt {J^`}_{n' + 1} = g^{-1} (n')\).
Let \(\widetilde{s^`}: \mathbb{N} \to M = s^` \circ g^{-1} = s \circ f \circ g^{-1}\).
\(\widetilde{s^`}\) is a subsequence of \(s\), because for each \(n_1, n_2 \in \mathbb{N}\) such that \(n_1 \lt n_2\), \(f \circ g^{-1} (n_1) \lt f \circ g^{-1} (n_2)\), because \(g^{-1} (n_1) \lt g^{-1} (n_2)\), so, \(f \circ g^{-1} (n_1) \lt f \circ g^{-1} (n_2)\); for each \(n \in \mathbb{N}\), there is a \({J^`}_{n' + 1} \in J^`\) such that \(n \le f ({J^`}_{n' + 1})\), and \(n \le f ({J^`}_{n' + 1}) = f \circ g^{-1} (n')\).
\(\widetilde{s^`}\) converges to \(lim s^`\), because for each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is an \(N \in \mathbb{N}\) such that for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), \(dist (lim s^`, s^` ({J^`}_n)) \lt \epsilon\), then, for each \(N \lt n\), \(dist (lim s^`, s^` \circ g^{-1} (n)) \lt \epsilon\), because \({J^`}_{N + 1} = g^{-1} (N) \lt g^{-1} (n)\).
So, the statement C holds.
Step 5:
Let us suppose the statement C.
Step 6:
Let \(s: J \to M\) be any.
When \(J\) is finite, there is a convergent subsequence, because \(s\) itself is a convergent subsequence.
Let us suppose otherwise.
There is the order-preserving bijection, \(g: J \to \mathbb{N}, J_n \mapsto n - 1\), which is indeed order-preserving, because for each \(J_n \lt J_{n'}\), \(n \lt n'\), so, \(n - 1 \lt n' - 1\); that is indeed a bijection, because for each \(J_n \neq J_{n'}\), \(n \neq n'\), so, \(n - 1 \neq n' - 1\), and for each \(n \in \mathbb{N}\), \(J_{n + 1}\) is mapped to \(n\).
Note that also \(g^{-1}\) is order-preserving, because for each \(n \lt n'\), \(g^{-1} (n) = J_{n + 1} \lt J_{n' + 1} = g^{-1} (n')\).
By the statement C, there is a convergent subsequence of \(\widetilde{s}: \mathbb{N} \to M = s \circ g^{-1}\), \(\widetilde{s}^`: \mathbb{N} \to M\), which means that \(\widetilde{s}^` = \widetilde{s} \circ f = s \circ g^{-1} \circ f\).
\(\widetilde{s}^`\) is a subsequence of \(s\), because for each \(n_1, n_2 \in \mathbb{N}\) such that \(n_1 \lt n_2\), \(g^{-1} \circ f (n_1) \lt g^{-1} \circ f (n_2)\), because \(f (n_1) \lt f (n_2)\), so, \(g^{-1} \circ f (n_1) \lt g^{-1} \circ f (n_2)\), and for each \(J_n \in J\), \(J_n = g^{-1} (n - 1)\), and there is an \(n' \in \mathbb{N}\) such that \(n - 1 \le f (n')\), then, \(J_n = g^{-1} (n - 1) \le g^{-1} \circ f (n')\).
So, \(\widetilde{s}^`\) is a convergent subsequence of \(s\) with \(J^`\) specifically taken as \(\mathbb{N}\).
So, the statement A holds.
