Well-Ordering Principle

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Theorem

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

That is, the relational structure $\struct {\N, \le}$ on the set of natural numbers $\N$ under the usual ordering $\le$ forms a well-ordered set.

This is called the well-ordering principle.


Corollary

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


Let $\le_1$ denote the restriction of the usual ordering $\le$ to the set $\N_{\ne 0}$ of natural numbers without zero.

The relational structure $\struct {\N_{\ne 0}, \le_1}$ forms a well-ordered set.


Proof using Naturally Ordered Semigroup

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 using Von Neumann Construction

From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\omega$ is a minimally inductive class under the successor mapping.

From Successor Mapping on Natural Numbers is Progressing, this successor mapping is a progressing mapping.

The result is a direct application of Minimally Inductive Class under Progressing Mapping is Well-Ordered under Subset Relation.

$\blacksquare$


Proof by Restriction of Real Numbers

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 Zermelo"s 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

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