description/proof of that for partially-ordered set and nonempty subset, if infimum and supremum of subset exist, infimum is equal to or smaller than supremum
Topics
About: set
The table of contents of this article
Starting Context
- 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 set and any nonempty subset, if the infimum and the supremum of the subset exist, the infimum is equal to or smaller than the supremum.
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:
\(S'\): \(\in \{\text{ the partially-ordered sets }\}\)
\(S\): \(\subseteq S'\), such that \(S \neq \emptyset\)
//
Statements:
\(\exists Inf (S) \land \exists Sup (S)\)
\(\implies\)
\(Inf (S) \le Sup (S)\)
//
2: Note
When \(S = \emptyset\) (although we do not particularly expect to have to deal with that case), if \(Inf (S)\) and \(Sup (S)\) exist, \(Sup (S) \le Inf (S)\), because \(Lb (S) = Ub (S) = S'\) (refer to Note for the definition of set of lower bounds of subset of partially-ordered set and Note for the definition of set of upper bounds of subset of partially-ordered set), and \(Sup (S) = Min (S') \le Max (S') = Inf (S)\).
3: Proof
Whole Strategy: Step 1: take any \(s \in S\), and see that \(Inf (S) \le s \le Sup (S)\).
Step 1:
Let \(s \in S\) be any, which exists, because \(S \neq \emptyset\).
\(Inf (S) \le s\), because \(Inf (S) \in Lb (S)\).
\(s \le Sup (S)\), because \(Sup (S) \in Ub (S)\).
So, \(Inf (S) \le Sup (S)\).
