Natural number

The set of natural numbers ($\mathbb{N}$) consist of positive integers and 0. The set of non-negative integers is given by ${\mathbb{Z}}_{>0}$.

The inductive principle (that forms the basis of mathematical induction) states that if $S\subseteq \mathbb{N}$ and all of the following properties are true, then $S=\mathbb{N}$:

• $S$ contains 1, and
• If $S$ contains a natural number $n$, it also contains $n+1$, then:

Construction

We can rigorously construct the natural numbers with the idea of the successor set.

• Zero is the empty set $\varnothing$.
• One is the set $0^{+}$.
• Two is the set $1^{+}$.
• Any finite non-negative integer can be defined inductively.

More explicitly:

• $0=\varnothing$
• $1=0^{+}=\{0\}=\{\varnothing \}$
• $2=1^{+}=\{0,1\}=\{\varnothing ,\{\varnothing \}\}$
• $3=2^{+}=\{0,1,2\}=\{\varnothing ,\{\varnothing \},\{\varnothing ,\{\varnothing \}\}\}$
• and so on

We can assume that there’s a set containing the empty set and all of its subsequent successors. Let $S$ be a set with this property. Let us define a collection $\mathcal{A}$ of subsets of $S$ as:

$$\mathcal{A}=\{A\subseteq S\mid \varnothing \in A\text{and}{n}^{+}\in A\text{if}n\in A\}$$

i.e., any subset $A$ of $S$ such that the empty set is in $A$ and for any $n\in A$, it includes its successor set. A lemma tells us that for a set $\mathcal{B}$, if $\mathcal{B}\subseteq \mathcal{A}$, then the intersection of all subsets $B\in \mathcal{B}$ is in $\mathcal{A}$ (as defined above) $({\cap }_{B\in \mathcal{B}}B)\in \mathcal{A}$.

Then, we can define the set of non-negative integers ${\mathbb{Z}}_{\ge 0}=\mathbb{N}$ as the set ${\cap }_{A\in \mathcal{A}}A$.

We define the set of natural numbers as the set $Z_{\ge 0}\backslash \{0\}$.