# 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 "$\le$" by the order relation "$<$" and the word minimum by the word minimal element.

• Let $$M$$ be a non-empty subset of natural numbers $$M\subseteq\mathbb N.$$
• Since $M$ is non-empty, let $$n_0\in M$$ be its element.
• We will construct a minimum as follows:
• If there is no smaller element $$m \le n_0$$ then we set $$m_0:=n_0$$, since it is the minimum we were looking for.
• Otherwise there exists an $n_1\in M$ with $$n_1 < n_0.$$
• We ask again if there is no smaller element $$m \le n_1$$.
• 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 search again.
• Our searching process will be repeated at most $$n_0 - m_0+1$$ times until we finally find the smallest element of $$M$$.
• This shows that $M$ contains a minimum, which is by construction unique.
• We have shown that $(\mathbb N,\le)$ is a well-ordered set.

Github: ### References

#### Bibliography

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