Well-Ordering Principle "proof"

Question

Theorem. Well-Ordering Principle.

Every non-empty subset of natural numbers has a least element.

I have seen some proofs for the theorem, but is very "complex"proof really needed here?

My attempt of proof:

Let $$ D \subset \mathbb{N}= \left\{ 1, \ 2, \ 3, \ \dots \right\} $$ be an arbitrary non-empty subset of natural numbers. Therefore it has at least one element $$  n \in D.$$

Consider the finite set $$ \left\{1, \ \dots, \ n \right\}.$$ We check which of those natural numbers are elements of D. Then we choose the smallest one of those. There we have the least element.

Answer 1

aschepler's comment points out the problem exactly. First you construct the set $$D\cap \{ 1,2,\ldots n\}$$ and then you consider the smallest element of this set...

Except that without the well-ordering principle, you cant be sure that it has a smallest element!

It might be instructive to consider what happens in your argument if you try to apply it to prove that $\Bbb Z$ is well-ordered.

Answer 2

This is a proof by induction. The theorem states that every non-empty subset of $\mathbb{N}$ has a least element.

Let $A\subseteq \mathbb{N}$ be a set with no least element. We want to prove that $A=\varnothing $, that is $\forall n$ $\in \mathbb{N}$, $n\notin A$. For induction on $n$ we have that:

$0\notin A$, in fact if $0 \in A$ we would have that: $0=$min$\mathbb{N}=$min$A$, but $A$ has no least element.

Suppose now that $\begin{Bmatrix} 0,1,...,n \end{Bmatrix}\cap A=\varnothing $. What we want to prove is that: $n+1\notin A$. If $n+1\in A$, infact, $n+1\neq$ min$A$ , since $A$ has no minimum for hypotesis. So $\exists m\in A$, $m<n+1$, which is a contradiction with the inductive hypothesis. So $n+1 \notin A$.

Hence $A=\varnothing$, as we wanted.