(related to Proposition: Existence and Uniqueness of Greatest Elements in Subsets of Natural Numbers)
We will prove the proposition for the poset $(\mathbb N,\le).$ An analogous proof is possible for the strictly-ordered set $(\mathbb N, < ),$ in which we have to replace the order relation "$\ge$" by the order relation "$>$" and the word maximum by the word maximal element.
- Let \(M\) be a non-empty and finite subset of natural numbers \(M\subseteq\mathbb N.\)
- Since $M$ is non-empty, let \(n_0\in M\) be its element.
- We will construct an maximum of $M$ as follows:
- If for all $m\in M$ we have $m\ge n_0$ we set \(m_0:=n_0\), since it is the maximum we were looking for.
- Otherwise there exists an $n_1\in M$ with \(n_1 > n_0.\)
- We ask wether for all $m\in M$ we have $m\ge n_1.$
- If so, then we set \(m_0:=n_1\), otherwise we find an $n_2\in M$ with \(n_2 > n_1.\) and repeat the same procedure again.
- Since $M$ is finite, our searching process will be repeated at most \(n_0 - m_0+1\) times until we finally find the maximum of \(M\).
- By construction $m_0$ is unique.
- Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013