Inductive Construction of Natural Numbers fulfils Peano's Axioms
Then $P$ fulfils Peano's axioms.
By definition of inductive set:
- $\O \in P$
Thus Peano's Axiom $\text P 1$: $0 \in P$ holds.
Let $x$ be a natural number.
Thus if $x \in P$ it follows that $x^+ \in P$.
Thus Peano's Axiom $\text P 2$: $n \in P \implies \map s n \in P$ holds.
For all $n$, $n$ is an element of $n^+$.
But $0$ is identified with the empty set $\O$.
By definition, $\O$ has no elements.
Therefore it is not possible for $\O$ to equal $n^+$ for any $n$.
Thus Peano's Axiom $\text P 4$: $0 \notin \Img s$ holds.
- 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