description/proof of that for sequence on \(1\)-dimensional Euclidean metric space with canonical ordering, if limit inferior exists, there is subsequence that converges to limit inferior
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of limit inferior 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 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 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 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 inferior exists, there is a subsequence that converges to the limit inferior.
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:
\(\exists lim inf s\)
\(\implies\)
\(s^` \in \{\text{ the subsequences of } s\} (lim s^` = lim inf s)\)
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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 \(lim inf s - (1 / 2)^n \lt s (J_{l_n}) \lt lim inf s + (1 / 2)^n\) 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 \(\vert J \vert = n \in \mathbb{N} \setminus \{0\}\).
\(lim inf s = s (J_n)\) inevitably exist, and \(lim s = s (J_n)\) exists, and \(lim s = lim inf s\).
So, let \(s^` := s = s \circ f\) with \(f: J^` \to J = id\), then, \(lim s^` = lim s = lim inf s\).
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 \(lim inf s = Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\), there is an \(m \in \mathbb{N} \setminus \{0\}\) such that \(lim inf s - (1 / 2)^n \lt Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\), 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: \(\mathbb{R}\) is linearly-ordered.
\(Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le lim inf s\).
There is an \(l_n \in \mathbb{N} \setminus \{0\}\) such that \(m \le l_n\) and \(s (J_{l_n}) \lt Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) + (1 / 2)^n\), 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.
\(Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le s (J_{l_n})\).
So, \(lim inf s - (1 / 2)^n \lt Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le s (J_{l_n}) \lt Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) + (1 / 2)^n \le lim inf s + (1 / 2)^n\).
So, \(lim inf s - (1 / 2)^n \lt s (J_{l_n}) \lt lim inf s + (1 / 2)^n\).
Let us suppose that \(l_1, ..., l_{n' - 1}\) have been chosen such that \(l_1 \lt ... \lt l_{n' - 1}\) and \(lim inf s - (1 / 2)^n \lt s (J_{l_n}) \lt lim inf s + (1 / 2)^n\) for each \(n \in \{1, ..., n' - 1\}\).
There is an \(m \in \mathbb{N} \setminus \{0\}\) such that \(lim inf s - (1 / 2)^{n'} \lt Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\), 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, \(lim inf s - (1 / 2)^{n'} \lt Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m' \le n\})\), 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: \(\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m' \le n\} \subseteq \{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}\).
\(Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le lim inf s\).
There is an \(l_{n'} \in \mathbb{N} \setminus \{0\}\) such that \(m \le l_{n'}\) and \(s (J_{l_{n'}}) \lt Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) + (1 / 2)^{n'}\), as before, but \(l_{n' - 1} \lt m \le l_{n'}\).
\(Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le s (J_{l_{n'}})\).
So, \(lim inf s - (1 / 2)^{n'} \lt Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le s (J_{l_{n'}}) \lt Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) + (1 / 2)^{n'} \le lim inf s + (1 / 2)^{n'}\).
So, \(lim inf s - (1 / 2)^{n'} \lt s (J_{l_{n'}}) \lt lim inf s + (1 / 2)^{n'}\).
So, we have chosen \(l_1, l_2, ... \in \mathbb{N} \setminus \{0\}\) inductively such that \(l_1 \lt l_2 \lt ...\) and \(lim inf s - (1 / 2)^n \lt s (J_{l_n}) \lt lim inf s + (1 / 2)^n\) 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 \(j = J_m \le J_{l_n} = f (n)\).
\(lim s^` = lim inf s\), 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 \(lim inf s - \epsilon \lt lim inf s - (1 / 2)^n \lt s^` (n) = s \circ f (n) = s (J_{l_n}) \lt lim inf s + (1 / 2)^n \lt lim inf s + \epsilon\), so, \(\vert s^` (n) - lim inf s \vert \lt \epsilon\).
