# 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

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