A description/proof of that maximal element of set w.r.t. inverse of ordering is minimal element of set w.r.t. original ordering
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of set.
- The reader knows a definition of ordering.
- The reader knows a definition of inverse of ordering.
- The reader knows a definition of minimal of set with respect to ordering.
- The reader knows a definition of maximal of set with respect to ordering.
Target Context
- The reader will have a description and a proof of the proposition that any maximal element of any set with respect to the inverse of any ordering is a minimal element of the set with respect to the original ordering.
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: Description
For any set, \(S\), and any ordering, \(R \subseteq S \times S\), any maximal element of \(S\), \(m \in S\), with respect to \(R^{-1}\) is a minimal element of \(S\) with respect to \(R\).
2: Proof
There is no \(s \in S\) such that \(\langle m, s \rangle \in R^{-1}\). Then, there is no \(s \in S\) such that \(\langle s, m \rangle \in R\). So, \(m\) is a minimal element of \(S\) with respect to \(R\).
