description/proof of that for value-bounded sequence on real numbers set, limit inferior and limit superior exist
Topics
About: set
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 limit superior of sequence on partially-ordered set.
Target Context
- The reader will have a description and a proof of 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.
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 \(J \neq \emptyset\)
\(\mathbb{R}\): 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:
\(\exists r_1, r_2 \in \mathbb{R} (\forall j \in J (r_1 \lt s (j) \lt r_2))\)
\(\implies\)
\(\exists lim inf s \land \exists lim sup s\)
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2: Note
When we say "\(lim inf s\) does not exist" or "\(lim sup s\) does not exist", that means that it does not exist in \(\mathbb{R}\): someone may say that it exists as \(\infty\) or \(- \infty\), but not \(\infty\) nor \(- \infty\) is any element of \(\mathbb{R}\): that someone is really thinking in the extended real numbers linearly-ordered set, \(\mathbb{R} \cup \{- \infty, \infty\}\), not in \(\mathbb{R}\).
3: Proof
Whole Strategy: use the well-known fact that any nonempty upper bounded subset of the real numbers set has the supremum and that any nonempty lower bounded subset of the real numbers set has the infimum; Step 1: see that \(r_1 \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\) for each \(m \in \mathbb{N} \setminus \{0\}\); Step 2: see that \(Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le r_2\) for each \(m \in \mathbb{N} \setminus \{0\}\).
Step 1:
For each \(m \in \mathbb{N} \setminus \{0\}\), \(\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\} \lt r_2\), so, \(Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\) exists.
For each \(m \in \mathbb{N} \setminus \{0\}\), \(r_1 \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\), because otherwise, \(Sup (...) \lt r_1 \lt s (J_n)\) for an (in fact, each) \(n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\), a contradiction against that \(Sup (...)\) is the supremum.
So, \(r_1 - 1 \lt \{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}\), and \(Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\) exists.
Step 2:
For each \(m \in \mathbb{N} \setminus \{0\}\), \(r_1 \lt \{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}\), so, \(Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\) exists.
For each \(m \in \mathbb{N} \setminus \{0\}\), \(Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le r_2\), because otherwise, \(s (J_n) \lt r_2 \lt Inf (...)\) for an (in fact, each) \(n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\), a contradiction against that \(Inf (...)\) is the infimum.
So, \(\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\} \lt r_2 + 1\), and \(Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\) exists.
