Inductive Construction of Natural Numbers fulfils Peano's Axiom of Injectivity

Theorem

Let $P$ denote the set of natural numbers by definition as an inductive set.

Then $P$ fulfils:

Peano's Axiom $\text P 3$: $s$ is injective

where $s$ denotes the successor mapping.

Proof

Let $m$ and $n$ be natural numbers such that $n^+ = m^+$.

By construction:

$n \in n^+$

and:

$m \in m^+$

Thus as $n^+ = m^+$ we have:

$n \in m^+$

and:

$m \in n^+$

This gives us:

$n \in m \lor n = m$

and:

$m \in n \lor m = n$

Aiming for a contradiction, suppose that $n \ne m$.

Then from $n \in m \lor n = m$ we have:

$n \in m$

and from $m \in n \lor m = n$ we have:

$m \in n$

In summary, if $n \ne m$ we have

$n \in m$ and $m \in n$

But from Natural Numbers cannot be Elements of Each Other, this is not possible.

Hence by Proof by Contradiction:

$n^+ = m^+ \implies n = m$

and the result follows by definition of injection.

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $3$: The Natural Numbers: $\S 3$ Derivation of the Peano postulates and other results