description/proof of that for sequence on \(1\)-dimensional Euclidean metric space with canonical ordering, if limit superior exists, there is subsequence that converges to limit superior
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of limit superior of sequence on partially-ordered set.
- The reader knows a definition of subsequence of sequence.
- The reader knows a definition of convergence of sequence on metric space.
- The reader admits the proposition that for any linearly-ordered set and any subset, any element of the set is the infimum of the subset if and only if the element is equal to or smaller than each element of the subset and for each element of the set larger than the element, there is an element of the subset smaller.
- The reader admits the proposition that for any linearly-ordered set and any subset, any element of the set is the supremum of the subset if and only if the element is equal to or larger than each element of the subset and for each element of the set smaller than the element, there is an element of the subset larger.
- The reader admits the proposition that for any partially-ordered set, any subset, and any subset of the subset, if the infimum of the subset and the infimum of the subset of the subset exist, the infimum of the subset is equal to or smaller than the infimum of the subset of the subset, and if the supremum of the subset and the supremum of the subset of the subset exist, the supremum of the subset is equal to or larger than the supremum of the subset of the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any sequence on the \(1\)-dimensional Euclidean metric space with the canonical ordering, if the limit superior exists, there is a subsequence that converges to the limit superior.
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 \(J \neq \emptyset\)
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\) with the canonical ordering, \(\lt\)
\(s\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq \mathbb{R}\)
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Statements:
\(\exists lim sup s\)
\(\implies\)
\(s^` \in \{\text{ the subsequences of } s\} (lim s^` = lim sup s)\)
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2: Proof
Whole Strategy: Step 1: deal with the case that \(J\) is finite, and suppose otherwise thereafter; Step 2: choose \(l_1, l_2, ... \in \mathbb{N} \setminus \{0\}\) inductively such that \(l_1 \lt l_2 \lt ...\) and \(lim sup s - (1 / 2)^n \lt s (J_{l_n}) \lt lim sup s + (1 / 2)^n\) for each \(n \in \mathbb{N} \setminus \{0\}\); Step 3: take \(J^` = \mathbb{N} \setminus \{0\}\) and \(f: J^` \to J, n \mapsto J_{l_n}\).
Step 1:
Let us suppose that \(\vert J \vert = n \in \mathbb{N} \setminus \{0\}\).
\(lim sup s = s (J_n)\) inevitably exist, and \(lim s = s (J_n)\) exists, and \(lim s = lim sup s\).
So, let \(s^` := s = s \circ f\) with \(f: J^` \to J = id\), then, \(lim s^` = lim s = lim sup s\).
Let us suppose otherwise, hereafter.
Step 2:
Let us choose \(l_1, l_2, ... \in \mathbb{N} \setminus \{0\}\) inductively as this.
Let \(n = 1\).
As \(lim sup s = Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\), there is an \(m \in \mathbb{N} \setminus \{0\}\) such that \(Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \lt lim sup s + (1 / 2)^n\), by the proposition that for any linearly-ordered set and any subset, any element of the set is the infimum of the subset if and only if the element is equal to or smaller than each element of the subset and for each element of the set larger than the element, there is an element of the subset smaller: \(\mathbb{R}\) is linearly-ordered.
\(lim sup s \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\).
There is an \(l_n \in \mathbb{N} \setminus \{0\}\) such that \(m \le l_n\) and \(Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) - (1 / 2)^n \lt s (J_{l_n})\), by the proposition that for any linearly-ordered set and any subset, any element of the set is the supremum of the subset if and only if the element is equal to or larger than each element of the subset and for each element of the set smaller than the element, there is an element of the subset larger.
\(s (J_{l_n}) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\).
So, \(lim sup s - (1 / 2)^n \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) - (1 / 2)^n \lt s (J_{l_n}) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \lt lim sup s + (1 / 2)^n\).
So, \(lim sup s - (1 / 2)^n \lt s (J_{l_n}) \lt lim sup s + (1 / 2)^n\).
Let us suppose that \(l_1, ..., l_{n' - 1}\) have been chosen such that \(l_1 \lt ... \lt l_{n' - 1}\) and \(lim sup s - (1 / 2)^n \lt s (J_{l_n}) \lt lim sup s + (1 / 2)^n\) for each \(n \in \{1, ..., n' - 1\}\).
There is an \(m \in \mathbb{N} \setminus \{0\}\) such that \(Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \lt lim sup s + (1 / 2)^{n'}\), as before, but \(m\) can be chosen such that \(l_{n' - 1} \lt m\), because if \(m \le l_{n' - 1}\), take any \(m' \in \mathbb{N} \setminus \{0\}\) such that \(l_{n' - 1} \lt m'\), then, \(Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m' \le n\}) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \lt lim sup s + (1 / 2)^{n'}\), by the proposition that for any partially-ordered set, any subset, and any subset of the subset, if the infimum of the subset and the infimum of the subset of the subset exist, the infimum of the subset is equal to or smaller than the infimum of the subset of the subset, and if the supremum of the subset and the supremum of the subset of the subset exist, the supremum of the subset is equal to or larger than the supremum of the subset of the subset: \(\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m' \le n\} \subseteq \{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}\).
\(lim sup s \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\).
There is an \(l_{n'} \in \mathbb{N} \setminus \{0\}\) such that \(m \le l_{n'}\) and \(Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) - (1 / 2)^{n'} \lt s (J_{l_{n'}})\), as before, but \(l_{n' - 1} \lt m \le l_{n'}\).
\(s (J_{l_{n'}}) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\).
So, \(lim sup s - (1 / 2)^{n'} \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) - (1 / 2)^{n'} \lt s (J_{l_{n'}}) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \lt lim sup s + (1 / 2)^{n'}\).
So, \(lim sup s - (1 / 2)^{n'} \lt s (J_{l_{n'}}) \lt lim sup s + (1 / 2)^{n'}\).
So, we have chosen \(l_1, l_2, ... \in \mathbb{N} \setminus \{0\}\) inductively such that \(l_1 \lt l_2 \lt ...\) and \(lim sup s - (1 / 2)^n \lt s (J_{l_n}) \lt lim sup s + (1 / 2)^n\) for each \(n \in \mathbb{N} \setminus \{0\}\).
Step 3:
Let us take \(J^` = \mathbb{N} \setminus \{0\}\).
Let us take \(f: J^` \to J, n \mapsto J_{l_n}\).
Then, \(s^` = s \circ f: J^` \to \mathbb{R}\) is a subsequence of \(s\), because \(\forall j^`_1, j^`_2 \in J^` \text{ such that } j^`_1 \lt j^`_2 (f (j^`_1) \lt f (j^`_2)) \land \forall j \in J (\exists j^` \in J^` (j \le f (j^`)))\): \(j = J_m\) and as \(l_1 \lt l_2 \lt ...\), \(m \le l_n\) for an \(n\), and \(j = J_m \le J_{l_n} = f (n)\).
\(lim s^` = lim sup s\), because for each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is an \(N \in \mathbb{N} \setminus \{0\}\) such that \((1 / 2)^N \lt \epsilon\), and for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), \((1 / 2)^n \lt (1 / 2)^N \lt \epsilon\), and \(lim sup s - \epsilon \lt lim sup s - (1 / 2)^n \lt s^` (n) = s \circ f (n) = s (J_{l_n}) \lt lim sup s + (1 / 2)^n \lt lim sup s + \epsilon\), so, \(\vert s^` (n) - lim sup s \vert \lt \epsilon\).
