description/proof of that for \(2\) sequences on partially-ordered ring with same domain, limit superior of sum of sequences is not necessarily sum of limits superior of sequences, and limit inferior of sum of sequences is not necessarily sum of limits inferior of sequences
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.
Target Context
- The reader will have a description and a proof of 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 infimum of the sum of the sequences is not necessarily the sum of the limits inferior of the sequences.
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 \(Ran (s) \subseteq R\)
\(s'\): \(\in \{\text{ the sequences }\}\), such that \(Dom (s') = J\) and \(Ran (s') \subseteq R\)
\(s + s'\): \(: J \to R, j \mapsto s (j) + s' (j)\)
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Statements:
Not necessarily "\(lim sup (s + s') = (lim sup s) + (lim sup s')\)"
\(\land\)
Not necessarily "\(lim inf (s + s') = (lim inf s) + (lim inf s')\)"
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2: Note
Compare with 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.
3: Proof
Whole Strategy: Step 1: see an example that \(lim sup (s + s') \neq (lim sup s) + (lim sup s')\); Step 2: see an example that \(lim inf (s + s') \neq (lim inf s) + (lim inf s')\).
Step 1:
Let us see an example that \(lim sup (s + s') \neq (lim sup s) + (lim sup s')\).
Let \(J = \mathbb{N}\), \(R = \mathbb{R}\), and \(s: j \mapsto 1 \text{ when } j \text{ is even }; \mapsto - 1 \text{ when } j \text{ is odd }\) and \(s': j \mapsto - 1 \text{ when } j \text{ is even }; \mapsto 1 \text{ when } j \text{ is odd }\).
\(lim sup s = 1\) and \(lim sup s' = 1\).
But as \(s + s' = 0\), \(lim sup (s + s') = 0\).
So, \(lim sup (s + s') = 0 \neq 2 = (lim sup s) + (lim sup s')\).
Step 2:
Let us see an example that \(lim inf (s + s') \neq (lim inf s) + (lim inf s')\).
Let \(J = \mathbb{N}\), \(R = \mathbb{R}\), and \(s: j \mapsto 1 \text{ when } j \text{ is even }; \mapsto - 1 \text{ when } j \text{ is odd }\) and \(s': j \mapsto - 1 \text{ when } j \text{ is even }; \mapsto 1 \text{ when } j \text{ is odd }\).
\(lim inf s = - 1\) and \(lim inf s' = - 1\).
But as \(s + s' = 0\), \(lim inf (s + s') = 0\).
So, \(lim inf (s + s') = 0 \neq - 2 = (lim inf s) + (lim inf s')\).
