Natural Numbers with Extension fulfil Naturally Ordered Semigroup Axioms 1, 3 and 4/Lemma 2

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Lemma for Natural Numbers with Extension fulfil Naturally Ordered Semigroup Axioms 1, 3 and 4

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 $\preccurlyeq$ 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 $\N$.


Proof

We define $\preccurlyeq$ on $M$ as follows:

$a \preccurlyeq b \iff \begin {cases} a, b \in \N: a \le b \\ a = 0, b = \beta \\ a = \beta, b \in \N_{>0} \\ a = \beta, b = \beta \end {cases}$


Existence

The restriction of $\preccurlyeq$ to $\N$ is seen to be the total ordering $\le$ on $\N$.


We have that:

$0 \preccurlyeq \beta$ and $0 \ne \beta$

from which:

$0 \prec \beta$

and:

$\beta \preccurlyeq 1$ and $\beta \ne 1$

from which:

$\beta \prec 1$

and:

$\forall a \in \N_{>0}: \beta \preccurlyeq a$


Finally:

$\beta = \beta$

and so:

$\beta \preccurlyeq \beta$


This demonstrates that $\preccurlyeq$ is total.

Hence existence of the total ordering in question has been proven.

$\Box$


Uniqueness

Let $\preccurlyeq'$ be another total ordering on $M$ such that:

the restriction of $\preccurlyeq'$ to $\N$ is the given total ordering $\le$ on $\N$
$0 \prec' \beta \prec' 1$

We have by definition of $\le$ on $\N$ that:

$\forall a, b \in \N: a \le b \implies a \preccurlyeq' b$

From $0 \prec' \beta \prec' 1$ we have:

$0 \preccurlyeq' \beta$
$\beta \preccurlyeq' 1$

Because $\beta \preccurlyeq' 1$ and an ordering is transitive:

$\forall a \in \N: 1 \le a \implies \beta \preccurlyeq' a$

Then as an ordering is reflexive:

$\beta \preccurlyeq' \beta$


Hence $\preccurlyeq'$ is defined as:

$a \preccurlyeq' b \iff \begin {cases} a, b \in \N: a \le b \\ a = 0, b = \beta \\ a = \beta, b \in \N_{>0} \\ a = \beta, b = \beta \end {cases}$

which is seen to be exactly the same as $\preccurlyeq$.

Hence the result.

$\blacksquare$


Sources

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