description/proof of that for sequence on \(1\)-dimensional Euclidean metric space with canonical ordering, if limit inferior and limit superior exist and are equal, convergence is limit inferior and limit superior
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 limit inferior of sequence on partially-ordered set.
- The reader knows a definition of limit superior of sequence on partially-ordered set.
- 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.
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 and the limit superior of the sequence exist and are equal, the convergence of the sequence is the limit inferior and the limit superior 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 \(J \neq \emptyset\)
\(\mathbb{R}\): \(= \text{ the Euclidean metric space }\) 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:
\(\exists lim inf s \land \exists lim sup s \land lim inf s = lim sup s\)
\(\implies\)
\(\exists lim s \land lim s = lim inf s = lim sup s\)
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2: Proof
Whole Strategy: Step 1: deal with the case that \(J\) is finite, and suppose otherwise thereafter; Step 2: let \(r := lim inf s = lim sup s\), and see that for each \(\epsilon\), \(r - \epsilon \lt Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le s (J_n)\) for an \(m\) and each \(m \le n\) and \(s (J_n) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m' \le n\}) \lt r + \epsilon\) for an \(m'\) and each \(m' \le n\) and see that for \(m, m' \lt N\), for each \(N \lt n\), \(r - \epsilon \lt s (J_n) \lt r + \epsilon\).
Step 1:
Let us suppose that \(\vert J \vert = n \in \mathbb{N} - \setminus \{0\}\).
\(lim inf s = s (J_n)\) and \(lim sup s = s (J_n)\) inevitably exist, and \(lim s = s (J_n)\) exists, and \(lim s = lim inf s = lims sup s\).
Let us suppose otherwise.
Step 2:
Let \(r := lim inf s = lim sup s\).
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
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 \(m \in \{N\} \setminus \{0\}\) such that \(r - \epsilon \lt Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\), 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.
\(\le s (J_n)\) for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(m \le n\).
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 \(m' \in \{N\} \setminus \{0\}\) such that \(Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m' \le n\}) \lt r + \epsilon\), 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.
\(s (J_n) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m' \le n\}) \lt r + \epsilon\) for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(m' \le n\).
Let \(N \in \mathbb{N} \setminus \{0\}\) be any such that \(m, m' \lt N\).
For each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), \(m, m' \lt n\), so, \(r - \epsilon \lt s (J_n) \lt r + \epsilon\), which means that \(dist (r, s (J_n)) \lt \epsilon\).
So, \(lim s = r = lim inf s = lim sup s\).
