Well-Ordering Principle

From ProofWiki
Jump to: navigation, search

Theorem

Every non-empty subset of $\N$ has a smallest (or first) element.

This is called the well-ordering principle.


The well-ordering principle also holds for $\N_{\ne 0}$.


Proof 1

Consider the natural numbers $\N$ defined as the naturally ordered semigroup $\struct {S, \circ, \preceq}$.

From its definition, $\struct {S, \circ, \preceq}$ is well-ordered by $\preceq$.

The result follows.


As $\N_{\ne 0} = \N \setminus \set 0$, by Set Difference is Subset $\N_{\ne 0} \subseteq \N$.

As $\N$ is well-ordered, by definition, every subset of $\N$ has a smallest element.

$\blacksquare$


Proof 2

Let $S$ be a non-empty subset of the set of natural numbers $\N$.

We take as axiomatic that $\N$ is itself a subset of the set of real numbers $\R$.

Thus $S \subseteq \R$.

By definition:

$\forall n \in \N: n \ge 0$

and so:

$\forall n \in S: n \ge 0$

Hence $0$ is a lower bound of $S$.

This establishes the fact that $S$ is bounded below.


By the Continuum Property, we have that $S$ admits an infimum.

Hence let $b = \inf S$.

Because $b$ is the infimum of $S$, it follows that $b + 1$ is not a lower bound of $S$.

So, for some $n \in S$:

$n < b + 1$

Aiming for a contradiction, suppose $n$ is not the smallest element of $S$.

Then there exists $m \in S$ such that:

$b \le m < n < b + 1$

from which it follows that:

$0 < n - m < 1$

But there exist no natural numbers $k$ such that $0 < k < 1$.

Hence $n = b$ is the smallest element of $S$.

$\blacksquare$


Also known as

This is otherwise known as:

  • the well-ordering property (of $\N$)
  • the least-integer principle
  • the principle of the least element.


Note that some authors cite this as the well-ordering theorem.

However, this allows it to be confused even more easily with the Well-Ordering Theorem, which states that any set can have an ordering under which that set is a well-ordered set.


Also see


Some authors extend the scope of this theorem to include:


Sources