The set of natural numbers are well-ordered

The Mathematician — Tue, 16 May 2023 13:00:44 +0000

Prove that the set of natural numbers is well-ordered

Before we go to the actual proof itself, note that the natural numbers are well-ordered under “

\leq

“.

We define the set

M

, which is an arbitrarily non-empty subset of

\mathbb{N}

. Further, we define the set

\begin{align*} L := \{l \in \mathbb{N} \ | \ l \leq m, \ \forall m \in \mathbb{N}\}. \end{align*}

We do know that

0 \in L

, so that is the minimal element of

L

. Now take

l \in L

such that

l + 1 \not \in L

, otherwise we will get the case that

L = \mathbb{N}

(note the induction). Then there exists an

m \in M

such that

m

l + 1

. This implies that

\begin{align*} m \geq l \quad \text{and} \quad m < l + 1 \iff l \leq m < l + 1. \end{align*}

It means that

m

can't be

l + 1

, but greater or equal to

l

. Since we work with positive integers, there is no integer between

l

and

l + 1

, so

m = l

. This implies that

m

must be the minimal element of

M

So the set of natural numbers is well-ordered.