description/proof of that for sequence on partially-ordered ring, if limit superior exists, limit superior of (sequence plus element) exists and equals (limit superior of sequence) plus element, and if limit inferior exists, limit inferior of (sequence plus element) exists and equals (limit inferior of sequence) plus element
Topics
About: ring
The table of contents of this article
Starting Context
- The reader knows a definition of partially-ordered ring.
- 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 any partially-ordered ring, any subset, and any element, if the supremum of the subset exists, the supremum of (the subset plus the element) exists and equals (the supremum of the subset) plus the element, and if the infimum of the subset exists, the infimum of (the subset plus the element) exists and equals (the infimum of the subset) plus the element.
Target Context
- The reader will have a description and a proof of the proposition that for any sequence on any partially-ordered ring, if the limit superior of the sequence exists, the limit superior of (the sequence plus any element) exists and equals (the limit superior of the sequence) plus the element, and if the limit inferior of the sequence exists, the limit inferior of (the sequence plus any element) exists and equals (the limit inferior of the sequence) plus 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 \(J \neq \emptyset\)
\(R\): \(\in \{\text{ the partially-ordered rings }\}\) with any partial ordering, \(\lt\)
\(s\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s) = J\) and \(Ran (s) \subseteq S\)
\(r\): \(\in R\)
//
Statements:
(
\(\exists lim sup s\)
\(\implies\)
\(\exists lim sup (s + r) \land lim sup (s + r) = (lim sup s) + r\)
)
\(\land\)
(
\(\exists lim inf s\)
\(\implies\)
\(\exists lim inf (s + r) \land lim inf (s + r) = (lim inf s) + r\)
)
//
2: Note
For some \(2\) sequences, \(s, s': J \to R\), "\(lim sup (s + s') = (lim sup s) + (lim sup s')\)" or "\(lim inf (s + s') = (lim inf s) + (lim inf s')\)" does not hold in general, by the proposition that for some \(2\) sequences on a partially-ordered ring with a same domain, the limit superior of the sum of the sequences is not necessarily the sum of the limits superior of the sequences, and the limit inferior of the sum of the sequences is not necessarily the sum of the limits inferior of the sequences. So, it is crucial for this proposition that \(r\) is a constant element instead of a sequence.
3: Proof
Whole Strategy: apply the proposition that for any partially-ordered ring, any subset, and any element, if the supremum of the subset exists, the supremum of (the subset plus the element) exists and equals (the supremum of the subset) plus the element, and if the infimum of the subset exists, the infimum of (the subset plus the element) exists and equals (the infimum of the subset) plus the element; 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 \(lim sup (s + r) = (lim sup s) + r\); Step 4: suppose that \(lim inf s\) exists; Step 5: see that \(lim inf (s + r) = (lim inf s) + r\).
Step 1:
Let us suppose that \(\vert J \vert = n \in \mathbb{N} \setminus \{0\}\).
\(lim sup s\) inevitably exists and equals \(s (J_n)\).
\(lim sup (s + r)\) inevitably exists and equals \(s (J_n) + r\).
So, \(lim sup (s + r) = s (J_n) + r = (lim sup s) + r\).
\(lim inf s\) inevitably exists and equals \(s (J_n)\).
\(lim inf (s + r)\) inevitably exists and equals \(s (J_n) + r\).
So, \(lim inf (s + r) = s (J_n) + r = (lim inf s) + r\).
Let us suppose otherwise, hereafter.
Step 2:
Let us suppose that \(lim sup s\) exists.
Step 3:
\(lim sup s = Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\), and those supremums and the infimum exist.
\(Sup (\{s (J_n) + r \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\) exists and \(= Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) + r\), by the proposition that for any partially-ordered ring, any subset, and any element, if the supremum of the subset exists, the supremum of (the subset plus the element) exists and equals (the supremum of the subset) plus the element, and if the infimum of the subset exists, the infimum of (the subset plus the element) exists and equals (the infimum of the subset) plus the element.
\(Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) + r \vert m \in \mathbb{N} \setminus \{0\}\})\) exists and \(= Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) + r\), by the proposition that for any partially-ordered ring, any subset, and any element, if the supremum of the subset exists, the supremum of (the subset plus the element) exists and equals (the supremum of the subset) plus the element, and if the infimum of the subset exists, the infimum of (the subset plus the element) exists and equals (the infimum of the subset) plus the element.
So, \(lim sup (s + r) = Inf (\{Sup (\{s (J_n) + r \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) = Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) + r \vert m \in \mathbb{N} \setminus \{0\}\}) = Inf (\{Sup (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) + r = (lim inf s) + r\), which exists.
Step 4:
Let us suppose that \(lim inf s\) exists.
Step 5:
\(lim inf s = Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\})\), and those infimums and the supremum exist.
\(Inf (\{s (J_n) + r \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\})\) exists and \(= Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) + r\), by the proposition that for any partially-ordered ring, any subset, and any element, if the supremum of the subset exists, the supremum of (the subset plus the element) exists and equals (the supremum of the subset) plus the element, and if the infimum of the subset exists, the infimum of (the subset plus the element) exists and equals (the infimum of the subset) plus the element.
\(Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) + r \vert m \in \mathbb{N} \setminus \{0\}\})\) exists and \(= Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) + r\), by the proposition that for any partially-ordered ring, any subset, and any element, if the supremum of the subset exists, the supremum of (the subset plus the element) exists and equals (the supremum of the subset) plus the element, and if the infimum of the subset exists, the infimum of (the subset plus the element) exists and equals (the infimum of the subset) plus the element.
So, \(lim inf (s + r) = Sup (\{Inf (\{s (J_n) + r \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) = Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) + r \vert m \in \mathbb{N} \setminus \{0\}\}) = Sup (\{Inf (\{s (J_n) \vert n \in \mathbb{N} \setminus \{0\} \text{ such that } m \le n\}) \vert m \in \mathbb{N} \setminus \{0\}\}) + r = (lim inf s) + r\), which exists.
