description/proof of that for sequence on linearly-ordered set, limit inferior is equal to or smaller than supremum of range of sequence
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of linearly-ordered set.
- The reader knows a definition of limit inferior 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 \(2\) elements, if each element that is smaller than the 1st element is smaller than the 2nd element, the 1st element is equal to or smaller than the 2nd element.
Target Context
- The reader will have a description and a proof of the proposition that for any sequence on any linearly-ordered set, if the limit inferior and the supremum of the range of the sequence exist, the limit inferior is equal to or smaller than the supremum of the range 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 \(J \neq \emptyset\)
\(S\): \(\in \{\text{ the linearly-ordered sets }\}\), with any linear ordering, \(\lt\)
\(s\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq S\)
//
Statements:
\(\exists lim inf s \land \exists Sup (Ran (s))\)
\(\implies\)
\(lim inf s \le Sup (Ran (s))\)
//
2: Note
There is no so simple relation between the existence of \(lim inf s\) and the existence of \(Sup (Ran (s))\).
For example, let \(J = \mathbb{N}\) and \(S = \mathbb{Q}\) with the canonical linear ordering and \(s\) be a sequence that starts from \(2\) and decreasingly approaches \(\sqrt{2}\), then, \(lim inf s\) does not exist but \(Sup (Ran (s))\) exists as \(2\).
For example, let \(J = \mathbb{N}\) and \(S = \mathbb{Q}\) with the canonical linear ordering and \(s\) be such that the index-even subsequence is constantly \(1\) and the index-odd subsequence starts from \(1\) and increasingly approaches \(\sqrt{2}\), then, \(lim inf s\) exists as \(1\) but \(Sup (Ran (s))\) does not exist.
3: Proof
Whole Strategy: Step 1: deal with the case that \(J\) is finite and suppose otherwise thereafter; Step 2: see that \(lim inf s \le Sup (Ran (s))\).
Step 1:
Let us suppose that \(\vert J \vert \in \mathbb{N} \setminus \{0\}\).
\(lim inf s = s (J_{\vert J \vert})\).
\(s (J_{\vert J \vert}) \le Min (Ub (Ran (s))) = Sup (Ran (s))\).
So, \(lim inf s \le Sup (Ran (s))\).
Let us suppose otherwise, hereafter.
Step 2:
\(lim inf s = Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\).
Let \(s' \in S\) be any such that \(s' \lt lim inf s\).
If there is no such \(s'\), it is OK.
There is an \(Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\) such that \(s' \lt Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\), 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.
For any \(n \in \mathbb{N} \setminus \{0\}\) such that \(m \le n\), \(Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le s (J_n)\).
But \(s (J_n) \le Sup (Ran (s))\).
So, \(s' \lt Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \le s (J_n) \le Sup (Ran (s))\), so, \(s' \lt Sup (Ran (s))\).
So, \(lim inf s \le Sup (Ran (s))\), by the proposition that for any linearly-ordered set and any \(2\) elements, if each element that is smaller than the 1st element is smaller than the 2nd element, the 1st element is equal to or smaller than the 2nd element.
