description/proof of that for partially-ordered ring, finite number of subsets with same index set, and subset as sum of subsets with same index set, supremum of subset is equal to or smaller than sum of supremums of subsets and infimum of subset is equal to or larger than sum of infimums of subsets
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 supremum of subset of partially-ordered set.
- The reader knows a definition of infimum of subset of partially-ordered set.
Target Context
- The reader will have a description and a proof of the proposition that for any partially-ordered ring, any finite number of subsets with any same index set, and the subset as the sum of the subsets with the same index set, if the supremums of the subsets exist, the sum of the supremums of the subsets is an upper bound of the subset and if furthermore the supremum of the subset exists, the supremum of the subset is equal to or smaller than the sum of the supremums of the subsets and if the infimums of the subsets exist, the sum of the infimums of the subsets is a lower bound of the subset and if furthermore the infimum of the subset exists, the infimum of the subset is equal to or larger than the sum of the infimums of the subsets.
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:
\(R\): \(\in \{\text{ the partially-ordered rings }\}\) with any partial ordering, \(\lt\)
\(J\): \(\in \{\text{ the nonempty index sets }\}\)
\(J'\): \(\in \{\text{ the nonempty finite index sets }\}\)
\(\{S_{j'} = \{s_{j', j} \in R \vert j \in J\} \vert j' \in J'\}\):
\(S\): \(= \{\sum_{j' \in J'} s_{j', j} \vert j \in J\}\)
//
Statements:
(
(
\(\forall j' \in J' (\exists Sup (S_{j'}))\)
\(\implies\)
\(\sum_{j' \in J'} Sup (S_{j'}) \in Ub (S)\)
)
\(\land\)
(
\(\forall j' \in J' (\exists Sup (S_{j'})) \land \exists Sup (S)\)
\(\implies\)
\(Sup (S) \le \sum_{j' \in J'} Sup (S_{j'})\)
)
)
\(\land\)
(
(
\(\forall j' \in J' (\exists Inf (S_{j'}))\)
\(\implies\)
\(\sum_{j' \in J'} Inf (S_{j'}) \in Lb (S)\)
)
\(\land\)
(
\(\forall j' \in J' (\exists Inf (S_{j'})) \land \exists Inf (S)\)
\(\implies\)
\(\sum_{j' \in J'} Inf (S_{j'}) \le Inf (S)\)
)
)
//
2: Note
\(Sup (S_{j'})\) or \(Inf (S_{j'})\) may not exist.
The existences of the \(Sup (S_{j'})\) s or the existences of the \(Inf (S_{j'})\) s do not guarantee the existence of \(Sup (S)\) or the existence of \(Inf (S)\): for example, \(R = \mathbb{Q}\), \(S_1 = \{q \in \mathbb{Q} \vert q \lt \sqrt{2}\} \cup \{2, 0\}\), \(S_2 = \{q \in \mathbb{Q} \vert q \lt \sqrt{2}\} \cup \{0, 2\}\), and \(S = \{2 q \in \mathbb{Q} \vert q \lt \sqrt{2}\} \cup \{2 + 0, 0 + 2\}\), then, \(Sup (S_1) = 2\) and \(Sup (S_2) = 2\), but \(Sup (S)\) does not exist, because \(2 \lt 2 \sqrt{2}\) and \(Ub (S) = \{q \in \mathbb{Q} \vert 2 \sqrt{2} \lt q\}\), which does not have any minimum.
3: Proof
Whole Strategy: Step 1: suppose that \(Sup (S_{j'})\) s exist; Step 2: see that \(\sum_{j' \in J'} Sup (S_{j'}) \in Ub (S)\); Step 3: suppose that \(Sup (S_{j'})\) s and \(Sup (S)\) exist; Step 4: see that \(Sup (S) \le \sum_{j' \in J'} Sup (S_{j'})\); Step 5: suppose that \(Inf (S_{j'})\) s exist; Step 6: see that \(\sum_{j' \in J'} Inf (S_{j'}) \in Lb (S)\); Step 7: suppose that \(Inf (S_{j'})\) s and \(Inf (S)\) exist; Step 8: see that \(\sum_{j' \in J'} inf (S_{j'}) \le Inf (S)\).
Step 1:
Let us suppose that \(Sup (S_{j'})\) s exist.
Step 2:
For each \(j' \in J'\), for each \(j \in J\), \(s_{j', j} \le Sup (S_{j'})\), because \(Sup (S_{j'}) \in Ub (S_{j'})\).
\(\sum_{j' \in J'} s_{j', j} \le \sum_{j' \in J'} Sup (S_{j'})\), where the left hand side is each element of \(S\).
That means that \(\sum_{j' \in J'} Sup (S_{j'}) \in Ub (S)\).
Step 3:
Let us suppose that \(Sup (S_{j'})\) s and \(Sup (S)\) exist.
Step 4:
By Step 2, \(\sum_{j' \in J'} Sup (S_{j'}) \in Ub (S)\).
As \(Sup (S)\) is the minimum of \(Ub (S)\), \(Sup (S) \le \sum_{j' \in J'} Sup (S_{j'})\).
Step 5:
Let us suppose that \(Inf (S_{j'})\) s exist.
Step 6:
For each \(j' \in J'\), for each \(j \in J\), \(Inf (S_{j'}) \le s_{j', j}\), because \(Inf (S_{j'}) \in Lb (S_{j'})\).
\(\sum_{j' \in J'} Inf (S_{j'}) \le \sum_{j' \in J'} s_{j', j}\), where the right hand side is each element of \(S\).
That means that \(\sum_{j' \in J'} Inf (S_{j'}) \in Lb (S)\).
Step 7:
Let us suppose that \(Inf (S_{j'})\) s and \(Inf (S)\) exist.
Step 8:
By Step 7, \(\sum_{j' \in J'} Inf (S_{j'}) \in Lb (S)\).
As \(Inf (S)\) is the maximum of \(Lb (S)\), \(\sum_{j' \in J'} inf (S_{j'}) \le Inf (S)\).
