description/proof of that for partially-ordered set, subset, and subset of subset, infimum of subset is equal to or smaller than infimum of subset of subset, and supremum of subset is equal to or larger than supremum of subset of 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 knows a definition of supremum of subset of partially-ordered set.
- The reader admits the proposition that for any partially-ordered set, any subset, and any subset of the subset, the set of the lower bounds of the subset is contained in the set of the lower bounds of the subset of the subset, and the set of the upper bounds of the subset is contained in the set of the upper bounds 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, 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.
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:
\(S\): \(\in \{\text{ the partially-ordered sets }\}\)
\(S^`\): \(\subseteq S\)
\(S^{``}\): \(\subseteq S^`\)
//
Statements:
(
\(\exists Inf (S^`) \land \exists Inf (S^{``})\)
\(\implies\)
\(Inf (S^`) \le Inf (S^{``})\)
)
\(\land\)
(
\(\exists Sup (S^`) \land \exists Sup (S^{``})\)
\(\implies\)
\(Sup (S^{``}) \le Sup (S^`)\)
)
//
2: Note
\(Inf (S^`)\) or \(Inf (S^{``})\) may not exist independently: for example, \(S = \mathbb{Q}\) with the canonical ordering, \(S^` = ((- \sqrt{2}, \sqrt{2}) \cap \mathbb{Q}) \subseteq \mathbb{Q}\), and \(S^{``} = ((- 1, 1) \cap \mathbb{Q}) \subseteq \mathbb{Q}\), and \(Inf (S^`)\) does not exist while \(S^{``}\) exists; \(S = \mathbb{Q}\) with the canonical ordering, \(S^` = ((- 2, 2) \cap \mathbb{Q}) \subseteq \mathbb{Q}\), and \(S^{``} = ((- \sqrt{2}, \sqrt{2}) \cap \mathbb{Q}) \subseteq \mathbb{Q}\), and \(Inf (S^`)\) exists while \(S^{``}\) does not exists.
\(Sup (S^`)\) or \(Sup (S^{``})\) may not exist independently, likewise.
3: Proof
Whole Strategy: Step 1: suppose that \(Inf (S^`)\) and \(Inf (S^{``})\) exist; Step 2: see that \(Lb (S^`) \subseteq Lb (S^{``})\); Step 3: see that \(Max (Lb (S^`)) \le Max (Lb (S^{``}))\); Step 4: suppose that \(Sup (S^`)\) and \(Sup (S^{``})\) exist; Step 5: see that \(Ub (S^`) \subseteq Ub (S^{``})\); Step 6: see that \(Min (Ub (S^{``})) \le Min (Ub (S^`))\).
Step 1:
Let us suppose that \(Inf (S^`)\) and \(Inf (S^{``})\) exist.
Step 2:
\(Lb (S^`) \subseteq Lb (S^{``})\), by the proposition that for any partially-ordered set, any subset, and any subset of the subset, the set of the lower bounds of the subset is contained in the set of the lower bounds of the subset of the subset, and the set of the upper bounds of the subset is contained in the set of the upper bounds of the subset of the subset.
Step 3:
\(Inf (S^`) = Max (Lb (S^`))\) and \(Inf (S^{``}) = Max (Lb (S^{``}))\).
Let us see that \(Max (Lb (S^`)) \le Max (Lb (S^{``}))\).
For each \(p \in Lb (S^{``})\), \(p \le Max (Lb (S^{``}))\), but as \(Max (Lb (S^`)) \in Lb (S^`) \subseteq Lb (S^{``})\), \(Max (Lb (S^`))\) is one of such \(p\) s, so, \(Max (Lb (S^`)) \le Max (Lb (S^{``}))\).
Step 4:
Let us suppose that \(Sup (S^`)\) and \(Sup (S^{``})\) exist.
Step 5:
\(Ub (S^`) \subseteq Ub (S^{``})\), by the proposition that for any partially-ordered set, any subset, and any subset of the subset, the set of the lower bounds of the subset is contained in the set of the lower bounds of the subset of the subset, and the set of the upper bounds of the subset is contained in the set of the upper bounds of the subset of the subset.
Step 6:
\(Sup (S^`) = Min (Ub (S^`))\) and \(Sup (S^{``}) = Min (Ub (S^{``}))\).
Let us see that \(Min (Ub (S^{``})) \le Min (Ub (S^`))\).
For each \(p \in Ub (S^{``})\), \(Min (Ub (S^{``})) \le p\), but as \(Min (Ub (S^`)) \in Ub (S^`) \subseteq Ub (S^{``})\), \(Min (Ub (S^`))\) is one of such \(p\) s, so, \(Min (Ub (S^{``})) \le Min (Ub (S^`))\).
