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

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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:

\(\ds 0 + \beta\) \(=\) \(\ds \beta + 0 = \beta\)
\(\ds \beta + \beta\) \(=\) \(\ds \beta\)
\(\ds n + \beta\) \(=\) \(\ds \beta + n = n\)


The algebraic structure:

$\struct {M, +}$

is not isomorphic to $\struct {\N, +}$.


Proof

Aiming for a contradiction, suppose there exists a (semigroup) isomorphism $\phi$ from $\struct {M, +}$ to $\struct {\N, +}$.

By definition of isomorphism:

$\phi$ is a homomorphism
$\phi$ is a bijection.


As $\phi$ is a fortiori a surjection, $\phi$ is also an epimorphism.

Hence from Epimorphism Preserves Identity:

$\map \phi 0 = 0$


Let $a \in M$ such that $a \ne \beta$ and $a \ne 0$.

Then:

\(\ds \map \phi a\) \(=\) \(\ds \map \phi {a + \beta}\) Definition of $\beta$
\(\ds \) \(=\) \(\ds \map \phi a + \map \phi \beta\) Definition of Semigroup Homomorphism
\(\ds \leadsto \ \ \) \(\ds \map \phi \beta\) \(=\) \(\ds 0\) as $0$ is the identity of $\N$

But then:

$\map \phi \beta = \map \phi 0$

and so $\phi$ is not injective.

Hence, by definition, $\phi$ is not a bijection.

This contradicts our assertion that $\phi$ is an isomorphism.

Hence there can be no such semigroup isomorphism between $\struct {M, +}$ and $\struct {\N, +}$.

$\blacksquare$


Sources

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