NT Insights Vol 1.0 Axiomatizing Natural Numbers

One can not prove theorems by assuming nothing. But what is more fascinating is that a large number of concepts can be explained by a fixed number of elementary notions, known as axioms.
Axioms are assumed to be self-evident truths. One such set of axioms about Natural Numbers was proposed by Peano:

Peano’s Axioms

Let $\mathbb{N}$ denote a set. Suppose we are not permitted to see its contents. Instead, we are given the following information about $\mathbb{N}$:

1. $0 \in \mathbb{N}$
2. A mapping $s: \mathbb{N} \to \mathbb{N}$, different from indentity map, such that $\mathbb{N}$ is invariant under $s$, i.e., $a \in \mathbb{N} \implies s(a) \in \mathbb{N}$
3. Pre-image of $0$ does not exist under $s$,
   i.e., there does not exist an $a \in \mathbb{N}$ such that $s(a) = 0$
4. The mapping $s$ is injective, i.e., $s(a)=s(b)\implies a=b$
5. If $S \subseteq \mathbb{N}$ with $0 \in S$ and $a \in S \implies s(a) \in S$, then $S=\mathbb{N}$

With these axioms alone, we can describe/generate the entire set $N$, without actually seeing its elements. Let us see HOW.

Let us start by analyzing what are we provided with. We are sure that there exists an element denoted by $0$ in $\mathbb{N}$. So $\mathbb{N}$ is non-empty. Additionally there also exists a map $s$ from $\mathbb{N}$ to itself, which is different from identity map.
This map is defined in such a way that when fed with an element of $\mathbb{N}$, it gives us another element from $\mathbb{N}$ (because $s()$ is not identity). Notice here, that had the map been identity map, it would be useless to us, since it would return the same element. So $s$ must be a non-identity map, so that we can get new information about $\mathbb{N}$.
When we input $0$ into $s()$, we get $s(0)\neq 0$. Formally we can say that we knew $0\in \mathbb{N}$ and now $s(0)$ is the new element other than $0$ which is known to be inside $\mathbb{N}$. Using recursive compositions of $s()$, we can get successive elements from $s()$. Further, the map $s()$ is injective. This injective behaviour $s()$ leads us to a very important theorem about $\mathbb{N}$.

Theorem 1.0 $s()$ is injective $\implies$ $\mathbb{N}$ is infinite.
Proof (by contradiction): Suppose set $\mathbb{N}$ is finite with cardinality $m$ and let the last ($m^{th}$ element) be denoted by $a_m$. Notice two facts:

1. All elements of $\mathbb{N}$ (except $0$) $\in range\ s$, because all of them were produced by recursive compositions of $s()$, starting with $0$.
2. $s: \mathbb{N} \to \mathbb{N}$ is injective map.

From above two points, we see that while constructing $\mathbb{N}$, when it comes the turn to map $a_m$ under function $s()$, all previous elements (including $a_m$ itself, except $0$) $\in range\ s \implies$ all previous elements except $0$ have a pre-image in $\mathbb{N}$. Since $s:\mathbb{N} \to \mathbb{N}$ exists for all $\mathbb{N}$ and is injective, we have only 2 options to map $a_m$:

1. $a_m$ is mapped to $0$.
2. $a_m$ is mapped to a new element $\notin range\ s$ yet.

Option 1 is not possible, since $0$ does not have a pre-image under $s()$.
Since the map $s()$ exists and is defined for every element $\in$ $\mathbb{N}$, Option 2 is the only possible way to map $a_m$ under $s()$.
So, $s(a_m) \in range\ s \implies s(a_m) \in \mathbb{N}$ ($\because\ range\ s \subseteq \mathbb{N}$).
Thus, $s(a_m)$ is the new/unseen element in $\mathbb{N}$, next to $a_m$, which contradicts our assumption that $a_m$ is the last element in $\mathbb{N}$. $\space\blacksquare$

Since, the map $s()$ gives successive elements as outputs via recursive compositions, we can call $s()$ as the successor function. Notice that this successor function $s()$ induces an ordering onto the elements of $\mathbb{N}$. The elements of the set may actually be unordered, but it is the “extraction policy” (recursive compositions, with $0$ as starting input), that makes them ordered. Also note that we still don’t know the explicit definition of $s()$, neither is it required. Thus, ordering is an after-effect of extraction process, not inherent to $\mathbb{N}$. Mathematically speaking, the existence of $0$ as referance point of extraction and the extraction process $s()$ induces ordering on the set $\mathbb{N}$.

Remark: Since $0$ is the only natural number that does not have a pre-image (i.e., $0 \notin range\ s$), we can conclude that every non-zero natural number is a successor of some other natural number. We can prove this obvious fact by applying inductive ($5^{th}$) axiom of Peano.

Proposition 1.1 If $x\in \mathbb{N}$ and $x\neq0$, then there exists $y\in \mathbb{N}$ such that $x=s(y)$.
Proof: Notice that we have to prove the statement for all $x\in \mathbb{N}\setminus\{0\}$. So let us define set $T=\{ x\in \mathbb{N}: x=s(y)$ for some $y\in \mathbb{N}\}$.
Now we need to prove that $T=\mathbb{N}\setminus\{0\}$. But inductive axiom is not suitable to prove such equality. Thus, we instead define $S=T\cup\{0\}$ and prove that $S=\mathbb{N}$ using the inductive axiom.
$0\in S$, by definition.
Assume, $x\in S$, where $x\in T$, i.e., $\exists\ y\in\mathbb{N}$ such that $x=s(y)$.
Then, $s(x)=s(s(y))$.
Now, $y\in\mathbb{N}\implies s(y)\in\mathbb{N}$.
Thus, $\exists\ s(y)\in\mathbb{N}$ such that $s(x)=s(s(y))$
$\implies s(x)\in S$.
Hence, by inductive axiom of Peano, $S=N$. $\space\blacksquare$