description/proof of that for partially-ordered set and subset, if minimum of subset exists, minimum is infimum, and if maximum of subset exists, maximum is supremum
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of subset of partially-ordered set with induced partial ordering.
- The reader knows a definition of minimum of partially-ordered set.
- The reader knows a definition of infimum of subset of partially-ordered set.
- The reader knows a definition of maximum of partially-ordered set.
- The reader knows a definition of supremum 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 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.
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, \(\lt'\)
\(S\): \(\subseteq S'\), with the induced partial ordering, \(\lt\)
//
Statements:
(
\(\exists Min (S)\)
\(\implies\)
\(Min (S) = Inf (S)\)
)
\(\land\)
(
\(\exists Max (S)\)
\(\implies\)
\(Max (S) = Sup (S)\)
)
//
2: Proof
Whole Strategy: Step 1: suppose that \(Min (S)\) exists; Step 2: see that \(Min (S) = Inf (S)\); Step 3: suppose that \(Max (S)\) exists; Step 4: see that \(Max (S) = Sup (S)\).
Step 1:
Let us suppose that \(Min (S)\) exists.
Step 2:
For each \(s \in S\), \(Min (S) \le s\) (which means that \(Min (S) = s \text{ or } Min (S) \lt s\)).
That implies that \(Min (S) \le' s\) for each \(s \in S\) (which means that \(Min (S) = s \text{ or } Min (S) \lt' s\)).
So, \(Min (S) \in Lb (S)\).
For each \(s' \in Lb (S)\), \(s' \le' Min (S)\), because \(Min (S) \in S\), which means that \(Min (S) = Max (Lb (S)) = Inf (S)\).
Step 3:
Let us suppose that \(Max (S)\) exists.
Step 4:
For each \(s \in S\), \(s \le Max (S)\) (which means that \(s = Max (S)\text{ or } s \lt Max (S)\)).
That implies that \(s \le' Max (S)\) for each \(s \in S\) (which means that \(s = Max (S) \text{ or } s \lt' Max (S)\)).
So, \(Max (S) \in Ub (S)\).
For each \(s' \in Ub (S)\), \(Max (S) \le' s'\), because \(Max (S) \in S\), which means that \(Max (S) = Min (Ub (S)) = Sup (S)\).
