Proof

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.
∎

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References

Bibliography

Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013

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Proof

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