description/proof of that for sequence on linearly-ordered set, limit superior is equal to or larger than infimum 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 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 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 linearly-ordered set and any \(2\) elements, if each element that is larger than the 2nd element is larger than the 1st 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 superior and the infimum of the range of the sequence exist, the limit superior is equal to or larger than the infimum 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 \(Ran (s) \subseteq S\)
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Statements:
\(\exists lim sup s \land \exists Inf (Ran (s))\)
\(\implies\)
\(Inf (Ran (s)) \le lim sup s\)
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2: Note
There is no so simple relation between the existence of \(lim sup s\) and the existence of \(Inf (Ran (s))\).
For example, let \(J = \mathbb{N}\) and \(S = \mathbb{Q}\) with the canonical linear ordering and \(s\) be a sequence that starts from \(1\) and increasingly approaches \(\sqrt{2}\), then, \(lim sup s\) does not exist but \(Inf (Ran (s))\) exists as \(1\).
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 \(2\) and the index-odd subsequence starts from \(2\) and decreasingly approaches \(\sqrt{2}\), then, \(lim sup s\) exists as \(2\) but \(Inf (Ran (s))\) 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 \(Inf (Ran (s)) \le lim sup s\).
Step 1:
Let us suppose that \(\vert J \vert \in \mathbb{N} \setminus \{0\}\).
\(lim sup s = s (J_{\vert J \vert})\).
\(Inf (Ran (s)) = Max (Lb (Ran (s))) \le s (J_{\vert J \vert})\).
So, \(Inf (Ran (s)) \le lim sup s\).
Let us suppose otherwise, hereafter.
Step 2:
\(lim sup s = Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\).
Let \(s' \in S\) be any such that \(lim sup s \lt s'\).
If there is no such \(s'\), it is OK.
There is a \(Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\) such that \(Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \lt s'\), 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.
For any \(n \in \mathbb{N} \setminus \{0\}\) such that \(m \le n\), \(s (J_n) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\).
But \(Inf (Ran (s)) \le s (J_n)\).
So, \(Inf (Ran (s)) \le s (J_n) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \lt s'\), so, \(Inf (Ran (s)) \lt s'\).
So, \(Inf (Ran (s)) \le lim sup s\), by the proposition that for any linearly-ordered set and any \(2\) elements, if each element that is larger than the 2nd element is larger than the 1st element, the 1st element is equal to or smaller than the 2nd element.
