Proposition: principle of good ordination

The principle of good management that if (\mathbb{N}_0, \leq) is a well-ordered set (that is to say, any subset of \mathbb{N}_0 non empty) has minimal

Demonstration: principle of good ordination

We are going to prove by reduction to absurdity

Suppose that there is a subset A\subset\mathbb{N}_0, A\not=\emptyset, which has no minimum

Then we define the set S=\{n\in\mathbb{N}_0|n\leq a, \forall a\in A\}

We must realize that if n\in A\cap S, we would have to n=\text{min}(S) and we've assumed that S has no minimum, therefore A\cap S=\emptyset

Now let's test it for the whole set using the induction axiom:

We found that 0\in S
0+a=a then 0\leq a, \forall a\in A

\text{If }n\in S, as A\cap S=\emptyset then n < a, \forall a\in A, (by defining order in \mathbb{N}_0) in each case there will be a natural number n_a \geq 1 such that n+n_a=a, which is done whenever s(n)=n+1\leq n+n_a=a, and therefore s(n)\in S

In compliance with these conditions, the axiom of completeness assures us that S=\mathbb{N}_0

Which is absurd, for it A\cap S=\emptyset then A=\emptyset, which is clearly a contradiction of the original assumption A\not=\emptyset