Natural Numbers with Extension fulfil Naturally Ordered Semigroup Axioms 1, 3 and 4
Theorem
Construction
Let $\N$ denote the set of natural numbers.
Let $\beta$ be an object such that $\beta \notin \N$
Let $M = \N \cup \set \beta$.
Let us extend the operation of natural number addition from $\N$ to $M$ by defining:
\(\ds 0 + \beta\) | \(=\) | \(\ds \beta + 0 = \beta\) | ||||||||||||
\(\ds \beta + \beta\) | \(=\) | \(\ds \beta\) | ||||||||||||
\(\ds n + \beta\) | \(=\) | \(\ds \beta + n = n\) |
There exists a unique total ordering $\le$ on $M$ such that:
- the restriction of $\le$ to $\N$ is the given total ordering $\le$ on $\N$
- $0 < \beta < 1$
such that the algebraic structure:
- $\struct {M, +, \le}$
is an ordered semigroup which fulfils the axioms:
- Naturally Ordered Semigroup Axiom $\text {NO} 1$: Well-Ordered
- Naturally Ordered Semigroup Axiom $\text {NO} 3$: Existence of Product
- Naturally Ordered Semigroup Axiom $\text {NO} 4$: Existence of Distinct Elements
but:
- does not fulfil Naturally Ordered Semigroup Axiom $\text {NO} 2$: Cancellability
- $\struct {M, +}$ is not isomorphic to $\struct {\N, +}$.
Proof
Recall the axioms:
A naturally ordered semigroup is a (totally) ordered commutative semigroup $\struct {S, \circ, \preceq}$ satisfying:
\((\text {NO} 1)\) | $:$ | $S$ is well-ordered by $\preceq$ | \(\ds \forall T \subseteq S:\) | \(\ds T = \O \lor \exists m \in T: \forall n \in T: m \preceq n \) | |||||
\((\text {NO} 2)\) | $:$ | $\circ$ is cancellable in $S$ | \(\ds \forall m, n, p \in S:\) | \(\ds m \circ p = n \circ p \implies m = n \) | |||||
\(\ds p \circ m = p \circ n \implies m = n \) | |||||||||
\((\text {NO} 3)\) | $:$ | Existence of product | \(\ds \forall m, n \in S:\) | \(\ds m \preceq n \implies \exists p \in S: m \circ p = n \) | |||||
\((\text {NO} 4)\) | $:$ | $S$ has at least two distinct elements | \(\ds \exists m, n \in S:\) | \(\ds m \ne n \) |
Some lemmata:
Lemma 1
The algebraic structure:
- $\struct {M, +}$
is a commutative monoid such that $0$ is the identity.
Lemma 2
There exists a unique total ordering $\preccurlyeq$ on $M$ such that:
- the restriction of $\preccurlyeq$ to $\N$ is the given total ordering $\le$ on $\N$
- $0 \prec \beta \prec 1$
This total ordering we will rename $\le$ to overload the notation for $\N$.
Lemma 3
The algebraic structure:
- $\struct {M, +}$
is not isomorphic to $\struct {\N, +}$.
Naturally Ordered Semigroup Axiom $\text {NO} 1$: Well-Ordered
Let $S \subseteq M$ such that $\beta \notin S$.
Then:
- $S \subseteq \N$
By the Well-Ordering Principle, $\struct {\N, \le}$ is a well-ordered set.
Hence $S$ is well-ordered.
Hence by definition, $S$ has a smallest element
Let $T \subseteq M$ such that $\beta \in T$.
- Case $1$
- $0 \in T$
Then as $0 < \beta$ we have that:
- $\forall x \in T: 0 \le x$
and so $T$ has a smallest element, that is, $0$.
- Case $2$
- $0 \notin T$
From Lemma $2$ we have that $\le$ is a total ordering such that:
- $\forall x \in T: \beta \le x$
and so $T$ has a smallest element, that is, $\beta$.
Hence it has been shown that every subset of $M$ has a smallest element.
That is, $\struct {M, \le}$ is a well-ordered set.
Hence Naturally Ordered Semigroup Axiom $\text {NO} 1$: Well-Ordered holds.
$\Box$
Naturally Ordered Semigroup Axiom $\text {NO} 3$: Existence of Product
By the construction of the natural numbers, $\struct {\N, +, \le}$ is a naturally ordered semigroup.
Hence Naturally Ordered Semigroup Axiom $\text {NO} 3$: Existence of Product holds for $\N$:
- $\forall m, n \in \N: m \le n \implies \exists p \in \N: m + p = n$
Hence:
- $\forall m, n \in M \setminus \set \beta: m \le n \implies \exists p \in \N: m + p = n$
Now consider $\beta \in M$.
- Case 1
Let $m \in M: m \le \beta$.
Then either:
- $m = \beta$
in which case:
- $\exists \beta \in M: m + \beta = \beta$
or:
- $m = 0$
in which case also:
- $\exists \beta \in M: m + \beta = \beta$
- Case 2
Let $n \in M: \beta \le n$.
Then:
- $\exists n \in \N: n + \beta = n$
and it is seen that Naturally Ordered Semigroup Axiom $\text {NO} 3$: Existence of Product holds.
$\Box$
Naturally Ordered Semigroup Axiom $\text {NO} 4$: Existence of Distinct Elements
We have that:
- $0 \in M$
and:
- $1 \in M$
and trivially Naturally Ordered Semigroup Axiom $\text {NO} 4$: Existence of Distinct Elements holds.
$\Box$
Naturally Ordered Semigroup Axiom $\text {NO} 2$: Cancellability
We have that:
- $\forall n \in M \setminus \set 0: n + 0 = n + \beta = n$
but it is not the case that $0 = \beta$.
That is, $\struct {M, +, \le}$ does not fulfil Naturally Ordered Semigroup Axiom $\text {NO} 2$: Cancellability.
$\Box$
Hence the result.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.2 \ \text{(c)}$