description/proof of that for infinite sequence on partially-ordered set and element of set, if there is any large index whose value is equal to or larger than element, limit superior is equal to or larger than element
Topics
About: set
The table of contents of this article
Starting Context
- 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 infinite sequence on any partially-ordered set and any element of the set, if the limit superior exists and there is any large index whose value is equal to or larger than the element, the limit superior is equal to or larger than the element.
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 \(\vert J \vert = \infty\)
\(S\): \(\in \{\text{ the partially-ordered sets }\}\), with any partial ordering, \(\lt\)
\(s\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq S\)
\(s'\): \(\in S\)
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Statements:
\(\exists lim sup s \land \forall j \in J (\exists j' \in J \text{ such that } j \lt j' (s' \le s (j)))\)
\(\implies\)
\(s' \le lim sup s\)
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2: Note
If \(J\) is finite, it is just a matter of \(s' \le s (J_{\vert J \vert})\) if and only if \(s' \le lim sup s\), because \(lim sup s = s (J_{\vert J \vert})\).
3: Proof
Whole Strategy: Step 1: see that \(s' \le lim sup s\).
Step 1:
\(lim sup s = Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\).
For each \(m \in \mathbb{N} \setminus \{0\}\), there is an \(n \in \mathbb{N} \setminus \{0\}\) such that \(m \lt n\) and \(s' \le s (J_n)\), by the supposition.
So, for each \(m \in \mathbb{N} \setminus \{0\}\), \(s' \le s (J_n) \le Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\).
So, \(s' \in Lb (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\).
So, \(s' \le Max (Lb (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})) = Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) = lim sup s\).
