Show that $\mathbb{N}=\{ 0\} \sqcup S(n)$ and also that that union is disjoint?

Question

Show that $\mathbb{N}=\{ 0\} \sqcup S(n)$ and also that that union is disjoint?

I've been assigned this exercise in my lectures of elements of mathematics 2. Three axioms have been given for a Peano system:

1. $S:N\to N$ is a defined injection in the set $N$;

2. $0\in N \setminus S(N) $;

3. Principle of finite induction (version in second order logic). If for all subset $X\subseteq N$, the property below is valid, then $X=N.$

$$0\in X \text{ and }\forall n \in N, n\in N \implies S(n) \in X$$

And the exercise is the following:

Show that $\mathbb{N}=\{ 0\} \sqcup S(\mathbb{N})$ (That is, $X$ consists precisely of zero and the immediate successors. And also, that that union is disjoint.)

I've made the following:

1. Let $X:=\{ 0\} \cup S(\mathbb{N})$

   1.1 $0 \in X$

   2.1 $S(n):=\{x:x\in N \setminus\{0\}\}$

I guess that with this, I show that $0 \in X$ and $S(\mathbb{N})$ are disjoint. I guess that I need to add something else. I'm not really sure what is it.

Answer 1

You need to translate formulas with sets to formulas without sets.

$A=B$ is translated to $\forall x(x\in A \leftrightarrow x\in B)$.

$x\in A\cup B$ to $x\in A \lor x\in B$.

$x\in \{0\}$ to $x=0$.

$S(\mathbb{N})$ is the image of $\mathbb{N}$ under $S$. $x\in S(\mathbb{N})$ to $\exists m (x=S(m))$.

Then $\mathbb{N}=\{ 0\} \cup S(\mathbb{N})$ to $\forall n (\top \leftrightarrow n=0 \lor \exists m (n=S(m)))$. This is equivalent to $\forall n (n=0 \lor \exists m (n=S(m)))$. Prove this.