description/proof of that for sequence on \(1\)-dimensional Euclidean metric space with canonical ordering, if convergence exists, 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 superior of sequence on partially-ordered set.
- The reader knows a definition of limit inferior of sequence on partially-ordered set.
- The reader admits the proposition that for any value-bounded sequence on the real numbers set, the limit inferior of the sequence and the limit superior of the sequence exist.
- 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.
- The reader admits the proposition that any real number is equal to or smaller than any another real number if it is equal to or smaller than the latter number plus any positive real number.
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 convergence of the sequence exists, the convergence is the limit inferior and 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:
\(\exists lim s\)
\(\implies\)
\(\exists lim inf s \land \exists lim sup s \land lim s = lim inf s = lim sup s\)
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2: Proof
Whole Strategy: Step 1: deal with the case that \(J\) is finite, and suppose otherwise thereafter; Step 2: see that \(s\) is value-bounded and that \(lim inf s\) and \(lim sup s\) exist; Step 3: see that there is an \(N\) such that for each \(N \lt m\), \(lim s - \epsilon \le Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le lim s + \epsilon\), and see that \(lim s - \epsilon \le lim inf s \le lim s + \epsilon\); Step 4: see that there is an \(N\) such that for each \(N \lt m\), \(lim s - \epsilon \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le lim s + \epsilon\), and see that \(lim s - \epsilon \le lim sup s \le lim s + \epsilon\).
Step 1:
Let us suppose that \(\vert J \vert = n \in \mathbb{N} \setminus \{0\}\).
\(lim s = s (J_n)\) inevitably exists, and \(lim inf s = s (J_n)\) and \(lim sup s = s (J_n)\) exist, and \(lim s = lim inf s = lims sup s\).
Let us suppose otherwise.
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 \(N \lt n\), \(dist (lim s, s (J_n)) \lt \epsilon\), which means that \(lim s - \epsilon \lt s (J_n) \lt lim s + \epsilon\).
So, \(s\) is value-bounded, because for each \(n \in \mathbb{N} \setminus \{0\}\), \(Min (\{Min (\{s (J_1), ..., s (J_N)\}), lim s - \epsilon\}) \le s (J_n) \le Max (\{Max (\{s (J_1), ..., s (J_N)\}), lim s + \epsilon\})\).
\(lim inf s\) and \(lim sup s\) exist, by the proposition that for any value-bounded sequence on the real numbers set, the limit inferior of the sequence and the limit superior of the sequence exist.
Step 3:
For each \(m \in \mathbb{N} \setminus \{0\}\) such that \(N \lt m\), \(lim s - \epsilon \le Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le lim s + \epsilon\), because as \(N \lt m \le n\), \(lim s - \epsilon \le s (J_n)\), so, \(lim s - \epsilon \in Lb (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\), and \(Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le s (J_n) \le lim s + \epsilon\).
Then, \(lim s - \epsilon \le Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) \le lim s + \epsilon\), because as \(Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\) is non-decreasing with respect to \(m\), 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, the \(m \lt N\) elements do not contribute to the supremum.
That means that \(lim s - \epsilon \le lim inf s \le lim s + \epsilon\).
\(lim s \le lim inf s + \epsilon\), so, \(lim s \le lim inf s\), by the proposition that any real number is equal to or smaller than any another real number if it is equal to or smaller than the latter number plus any positive real number.
\(lim inf s \le lim s\), by the proposition that any real number is equal to or smaller than any another real number if it is equal to or smaller than the latter number plus any positive real number.
So, \(lim s \le lim inf s \le lim s\), which implies that \(lim s = lim inf s\).
Step 4:
For each \(m \in \mathbb{N} \setminus \{0\}\) such that \(N \lt m\), \(lim s - \epsilon \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le lim s + \epsilon\), because as \(N \lt m \le n\), \(s (J_n) \le lim s + \epsilon\), so, \(lim s + \epsilon \in Ub (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\), and \(lim s - \epsilon \le s (J_n) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\).
Then, \(lim s - \epsilon \le Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) \le lim s + \epsilon\), because as \(Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\) is non-increasing with respect to \(m\), 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, the \(m \lt N\) elements do not contribute to the infimum.
That means that \(lim s - \epsilon \le lim sup s \le lim s + \epsilon\).
\(lim s \le lim sup s + \epsilon\), so, \(lim s \le lim sup s\), by the proposition that any real number is equal to or smaller than any another real number if it is equal to or smaller than the latter number plus any positive real number.
\(lim sup s \le lim s\), by the proposition that any real number is equal to or smaller than any another real number if it is equal to or smaller than the latter number plus any positive real number.
So, \(lim s \le lim sup s \le lim s\), which implies that \(lim s = lim sup s\).
