description/proof of that for partially-ordered set, subset, and subset of subset, set of lower bounds of subset is contained in set of lower bounds of subset of subset, and set of upper bounds of subset is contained in set of upper bounds of subset of subset
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of set of lower bounds of subset of partially-ordered set.
- The reader knows a definition of set of upper bounds 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, 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.
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:
\(Lb (S^`) \subseteq Lb (S^{``})\)
\(\land\)
\(Ub (S^`) \subseteq Ub (S^{``})\)
//
2: Proof
Whole Strategy: Step 1: see that \(Lb (S^`) \subseteq Lb (S^{``})\); Step 2: see that \(Ub (S^`) \subseteq Ub (S^{``})\).
Step 1:
Let us see that \(Lb (S^`) \subseteq Lb (S^{``})\).
Let \(s \in Lb (S^`)\) be any.
For each \(p \in S^{``}\), \(p \in S^`\), so, \(s \le p\), because \(s \in Lb (S^`)\), which means that \(s \in Lb (S^{``})\).
So, \(Lb (S^`) \subseteq Lb (S^{``})\).
Step 2:
Let us see that \(Ub (S^`) \subseteq Ub (S^{``})\).
Let \(s \in Ub (S^`)\) be any.
For each \(p \in S^{``}\), \(p \in S^`\), so, \(p \le s\), because \(s \in Ub (S^`)\), which means that \(s \in Ub (S^{``})\).
So, \(Ub (S^`) \subseteq Ub (S^{``})\).
