description/proof of that for partially-ordered set and \(2\) subsets with same index set which have infimums, if for each index, element of 1st subset is equal to or smaller than element of 2nd subset, infimum of 1st subset is equal to or smaller than infimum of 2nd subset
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of infimum of subset of partially-ordered set.
- The reader admits the proposition that for any partially-ordered set and any subset, if the minimum of the subset exists, the minimum is the infimum of the subset, and if the maximum of the subset exists, the maximum is the supremum of the subset.
- The reader admits the proposition that for any partially-ordered set, any subset, and any subset of the subset, if the infimum of the subset and the infimum of the subset of the subset exist, the infimum of the subset is equal to or smaller than the infimum of the subset of the subset, and if the supremum of the subset and the supremum of the subset of the subset exist, the supremum of the subset is equal to or larger than the supremum of the subset of the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any partially-ordered set and any \(2\) subsets with any same index set which have some infimums, if for each index, the element of the 1st subset is equal to or smaller than the element of the 2nd subset, the infimum of the 1st subset is equal to or smaller than the infimum of the 2nd subset.
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:
\(S\): \(\in \{\text{ the partially-ordered sets }\}\) with any partial ordering, \(\lt\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(S_1\): \(= \{s_{1, j} \in S \vert j \in J\}\)
\(S_2\): \(= \{s_{2, j} \in S \vert j \in J\}\)
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Statements:
\(\exists Inf (S_1) \land \exists Inf (S_2) \land \forall j \in J (s_{1, j} \le s_{2, j})\)
\(\implies\)
\(Inf (S_1) \le Inf (S_2)\)
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2: Proof
Whole Strategy: Step 1: see that \(Lb (S_1) \subseteq Lb (S_2)\); Step 2: see that \(Max (Lb (S_1)) \le Max (Lb (S_2))\).
Step 1:
\(Lb (S_1) \subseteq Lb (S_2)\), because for each \(s \in Lb (S_1)\), for each \(j \in J\), \(s \le s_{1, j}\), so, \(s \le s_{1, j} \le s_{2, j}\), so, \(s \in Lb (S_2)\).
Step 2:
\(Inf (S_1) = Max (Lb (S_1)) \le Max (Lb (S_2)) = Inf (S_2)\), by the proposition that for any partially-ordered set and any subset, if the minimum of the subset exists, the minimum is the infimum of the subset, and if the maximum of the subset exists, the maximum is the supremum of the subset and the proposition that for any partially-ordered set, any subset, and any subset of the subset, if the infimum of the subset and the infimum of the subset of the subset exist, the infimum of the subset is equal to or smaller than the infimum of the subset of the subset, and if the supremum of the subset and the supremum of the subset of the subset exist, the supremum of the subset is equal to or larger than the supremum of the subset of the subset.
