description/proof of that for sequence on real numbers set with canonical ordering, if limit superior of minus sequence exists, it is minus limit inferior of sequence, and if limit inferior of minus sequence exists, it is minus limit superior of sequence
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.
- The reader knows a definition of limit inferior of sequence on partially-ordered set.
- The reader admits the proposition that for the real numbers set with the canonical ordering and any subset, if the supremum of the minus subset exists, it is minus the infimum of the subset, and if the infimum of the minus subset exists, it is minus the supremum of the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any sequence on the real numbers set with the canonical ordering, if the limit superior of minus the sequence exists, it is minus the limit inferior of the sequence, and if the limit inferior of minus the sequence exists, it is minus the limit superior 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:
\(\mathbb{R}\): \(= \text{ the real numbers set }\) with the canonical ordering
\(J\): \(\subseteq \mathbb{N}\), such that \(J \neq \emptyset\)
\(s\): \(: J \to \mathbb{R}\)
\(- s\): \(: J \to \mathbb{R}, j \mapsto - s(j)\)
//
Statements:
(
\(\exists lim sup - s\)
\(\implies\)
\(\exists lim inf s \land lim sup - s = - lim inf s\)
)
\(\land\)
(
\(\exists lim inf - s\)
\(\implies\)
\(\exists lim sup s \land lim inf - s = - lim sup s\)
)
//
2: Note
"\(lim sup - s = - lim sup s\)" or "\(lim inf - s = - lim inf s\)" does not hold in general.
For example, for \(s = (1, -1, 1, -1, ...)\), \(lim sup - s = 1\) while \(lim sup s = 1\) and \(lim inf - s = - 1\) while \(lim inf s = - 1\).
3: Proof
Whole Strategy: apply the proposition that for the real numbers set with the canonical ordering and any subset, if the supremum of the minus subset exists, it is minus the infimum of the subset, and if the infimum of the minus subset exists, it is minus the supremum of the subset; Step 1: deal with the case that \(J\) is finite, and suppose otherwise thereafter; Step 2: suppose that \(lim sup - s\) exists; Step 3: see that \(lim sup - s = - lim inf s\); Step 4: suppose that \(lim inf - s\) exists; Step 5: see that \(lim inf - s = - lim sup s\).
Step 1:
Let us suppose that \(\vert J \vert = n \in \mathbb{N} \setminus \{0\}\).
\(lim sup - s\) inevitably exists as \(= - s (J_n)\).
\(- lim inf s\) inevitably exists as \(= - s (J_n)\).
So, \(lim sup - s = - lim inf s\).
\(lim inf - s\) inevitably exists as \(= - s (J_n)\).
\(- lim sup s\) inevitably exists as \(= - s (J_n)\).
So, \(lim inf - s = - lim sup s\).
Let us suppose otherwise hereafter.
Step 2:
Let us suppose that \(lim sup - s\) exists.
Step 3:
\(lim sup - s = Inf (\{Sup (\{- s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\).
\(= Inf (\{- Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\), by the proposition that for the real numbers set with the canonical ordering and any subset, if the supremum of the minus subset exists, it is minus the infimum of the subset, and if the infimum of the minus subset exists, it is minus the supremum of the subset, \(= - Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\), by the proposition that for the real numbers set with the canonical ordering and any subset, if the supremum of the minus subset exists, it is minus the infimum of the subset, and if the infimum of the minus subset exists, it is minus the supremum of the subset, \(= - lim inf s\).
Step 4:
Let us suppose that \(lim inf - s\) exists.
Step 5:
\(lim inf - s = Sup (\{Inf (\{- s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) = Sup (\{- Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\), by the proposition that for the real numbers set with the canonical ordering and any subset, if the supremum of the minus subset exists, it is minus the infimum of the subset, and if the infimum of the minus subset exists, it is minus the supremum of the subset, \(= - Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\), by the proposition that for the real numbers set with the canonical ordering and any subset, if the supremum of the minus subset exists, it is minus the infimum of the subset, and if the infimum of the minus subset exists, it is minus the supremum of the subset, \(= - lim sup s\).
